Keywords: graphes arêtes-coloriés, algorithmes, chemin hamiltonien, arbres proprement arêtescoloriées, arbres faiblement arêtes-coloriées, nonapproximabilité, coloration Edge-colored graphs, algorithms, Hamiltonian paths, proper tree, weak tree, nonapproximability, coloring

Dans la présente thèse nous étudions l'extraction d'arbres dans des graphes arêtes-coloriés. Nous nous concentrons sur la recherche d'arbres couvrants proprement arête-coloriés et faiblement arête-coloriés, notée PST et WST. -Nous montrons que les versions d'optimisation de ces problèmes sont NP-Complete dans le cas général des graphes arêtes-coloriés, et nous proposons des algorithmes pour trouver ces arbres dans le cas des graphes arêtes-coloriés sans cycles proprement arêtes-coloriés. Nous donnons également quelques limites de nonapproximabilité. -Nous proposons des conditions suffisantes pour l'existence de la PST dans des graphes arêtes-coloriés (pas forcément propre), en fonction de différents paramètres de graphes, tels que : nombre total de couleurs, la connectivité et le nombre d'arêtes incidentes de différentes couleurs pour un sommet. -Nous nous intéressons aux chemins hamiltoniens proprement arêtes-coloriés dans le cas des multigraphes arêtes-coloriés. Ils présentent de l'intérêt pour notre étude, car ce sont également des arbres couvrants proprement arêtes-coloriés. Nous établissons des conditions suffisantes pour qu'un multigraphe contienne un chemin hamiltonien proprement arêtescoloriés, en fonction de plusieurs paramètres tels que le nombre d'arêtes, le degré d'arêtes, etc. -Puisque l'une des conditions suffisantes pour l'existence des arbres couvrants proprement arêtes-coloriés est la connectivité, nous prouvons plusieurs bornes supérieures pour le plus petit nombre de couleurs nécessaires pour la k-connectivité-propre. Nous énonons plusieurs conjectures pour les graphes généraux et bipartis, et on arrive á les prouver pour k = 1.

Introduction

Graph Theory is an important research area in mathematics and computer science, studied for over two hundred years. Graphs are one of the prime objects of study in Discrete Mathematics, these mathematical structures model pairwise relations between objects. Thus, a graph is a set of nodes (or "vertices") and a set of edges, connecting pairs of these nodes. Graphs are among the most universal structures for modeling various relations and processes in scientific domains such as physics, chemistry, biology, economics, social systems, as well as in different areas of Computer Science such as databases, networks, etc. Furthermore, many problems of practical interest can be represented by graphs. Consider the example of a city map presented in Figure 1. Imagine a pedestrian interested in finding the fastest way to go from point A to point B. How does he proceed ? He looks at the map and chooses the streets he should take in order to reach the desired destination. The same situation applies for a person who drives a car, however, the latter should take into account that some streets can be taken in a uni-directional way only. Clearly, in this example the city map can be represented as a graph, the intersections of the streets being the nodes and the streets being the oriented edges between these nodes. Each oriented edge represents a one-way street. This kind of graphs are called directed graphs. And the problem of finding the best itinerary for reaching the point B starting from the point A can be modeled as a problem of finding the shortest path connecting the nodes A and B in the directed graph.

In general, graph theory is divided in two branches : directed graphs and undirected (simple) graphs. Undirected graphs were studied extensively, then the directed graphs came into focus due to their theoretical and practical importance. Note, that undirected graphs can be seen as a subclass of directed graphs, with each undirected edge represented by two oppositely-directed edges. Nevertheless, even these two classes of graphs are not powerful enough to model all existing real-life constraints. For this reason new concepts in graph definitions are required. Going back to our above example, if one wants to differentiate highways from national roads, and main from secondary roads, a good idea would be to use colors. When the edges of graphs are colored, Figure 1.1 -The itinerary for a pedestrian and for a car we talk about edgecolored graphs, which generalize the directed graphs. Such generalization allows the formulation of a wide range of challenging problems, in addition to those existing in the case of usual graphs. Namely, suppose that we would also like to find an itinerary that uses highways exclusively. This problem corresponds to finding a monochromatic path of a given color between two given nodes. Thus, one can observe that problems related to edgecolored graphs often consist in finding (extracting) classical structures (subgraphs) such as paths, cycles and trees, with, in addition, certain constraints or properties imposed on colors. Trivial properties of the trees that can be found in edge-colored graphs, are monochromaticity and heterochromaticity. The former consists in having all the edges colored in same color, the latter requires distinct colored edges. Heterochromatic structures are also called rainbow, multicolored, colorf ul and polychromatic by different authors. Results on subgraphs and trees with these properties can be found in [START_REF] Rödl | Rainbow subgraphs in properly edge-colored graphs[END_REF][START_REF] Brualdi | Multicolored trees in complete graphs[END_REF][START_REF] Gyairfais | Edge colorings of complete graphs without tricolored triangles[END_REF][START_REF] Suzuki | A necessary and sufficient condition for the existence of a heterochromatic spanning tree in a graph[END_REF][START_REF] Akbari | Multicolored trees in complete graphs[END_REF][START_REF] Chen | Heterochromatic tree partition numbers for complete bipartite graphs[END_REF][START_REF] Fujita | Rainbow generalizations of ramsey theory : A survey[END_REF], as well as a recent survey on heterochromatic (rainbow colored) graphs in [START_REF] Kano | Monochromatic and heterochromatic subgraphs in edge-colored graphs-a survey[END_REF]. These results are of interest for us, since rainbow colored graphs are also proper -a property introduced in the sequel.

The constraint that we focus on in this thesis, is that of properness. The notion of properness consists in having no two adjacent edges of the same color in the structure we want to find. In the literature, algorithms are proposed to check the existence of properly edge-colored subgraphs of a given type in an edge-colored graph. Corresponding results are presented in [START_REF] Benkouar | On the complexity of some hamiltonian and eulerian problems in edge-colored complete graphs[END_REF]. Subgraphs with this property are also studied in [START_REF] Abouelaoualim | Paths and trails in edge-colored graphs[END_REF][START_REF] Saad | Finding a longest alternating cycle in a 2-edge-coloured complete graph is in RP[END_REF][START_REF] Benkouar | Alternating cycles in 2-edge colored complete graphs[END_REF][START_REF] Manoussakis | Minimal colorings for properly colored subgraphs[END_REF][START_REF] Manoussakis | Cycles of given color patterns[END_REF][START_REF] Gutin | Properly coloured cycles and paths : Results and open problems[END_REF][START_REF] Alon | Properly colored hamilton cycles in edge-colored complete graphs[END_REF][START_REF] Bang-Jensen | Alternating cycles and paths in edge-coloured multigraphs : a survey[END_REF]. In particular, Y. Manoussakis et al. [START_REF] Manoussakis | Alternating paths in edge-colored complete graphs[END_REF] investigated the existence, counting and enumeration of properly colored pairwise internally disjoint paths with common endpoints. They also studied properly colored subgraphs including Hamiltonian paths and cycles, cycles with a given lower bound on their In the present thesis we investigate the extraction of trees from edge-colored graphs. A tree is a connected acyclic subgraph. In graph theory there exist many important problems related to trees with different levels of difficulty. The difficulty depends on the properties of the tree we are trying to find. The properties can be related to the desired degree of each vertex, which is the number of edges incident to a vertex. For example, for a simple graph, it is difficult to find a bounded degree spanning tree 1 (see [START_REF] Fürer | Approximating the minimum degree spanning tree to within one from the optimal degree[END_REF][START_REF] Singh | Approximating minimum bounded degree spanning trees to within one of optimal[END_REF] for more details). It is also difficult to find a spanning tree maximizing the number of vertices of degree one (which are also called leafs). Since we are interested in edgecolored graphs, we focus on finding trees with properties based on coloring. Namely, we deal with proper and weak proper spanning trees (see figure 1.2), denoted PST and WST. A proper tree is a proper subgraph, whose uncolored version is acyclic. A proper spanning tree is a spanning proper subgraph, whose uncolored version is connected and acyclic. A weak proper tree is rooted, the root-to-leaf paths are proper and it's uncolored version is acyclic. Let us see an example of application of properly colored structures in edge-colored graphs. Consider some points that send and receive information. Different frequencies are used for receiving and for sending. Suppose now that certain points cannot send information directly to other points since there exists some natural barrier. Let us represent these points as graph vertices and the possibility to send and receive data between any two points as graph edges. Let us define a bijective function for the edges coloring : one color per frequency. Each vertex uses different frequencies for receiving and for sending operations, thus the corresponding edges have different colors. Various problems emerge in this context :

(1) Suppose we would like to test if one point can send information to another one. This problem corresponds to finding a proper colored path between two vertices in the edgecolored graph.

(2) Suppose we would also like to know if two given vertices can communicate. To discover that, it is sufficient to find a proper colored cycle or a closed trail.

(3) If we would like to determine how many vertices can communicate in this network simultaneously, then we have to find a proper colored subgraph.

If we want to broadcast the information to a maximum number of vertices from a given vertex, we have to find a maximum weak-proper tree for a given vertex as a root, since each point can receive on one frequency and send on all the others.

(4) Another possible problem is to find a maximum number of vertices that will always receive the broadcasted information and any of these vertices can serve as broadcasting point. In this case the subgraph to find corresponds to a maximum proper tree.

In addition, application areas of finding properly-colored Hamiltonian paths and cycles in edge-colored graphs include genetics [START_REF] Dorninger | Hamiltonian circuits determining the order of chromosomes[END_REF][START_REF] Pevzner | An algorithmic approach[END_REF], social sciences [START_REF] Chou | Paths through fixed vertices in edge-colored graphs[END_REF] and VLSI optimization [START_REF] Szeider | Finding paths in graphs avoiding forbidden transitions[END_REF].

After studying PST and WST in edge-colored graphs, we consider a similar problem in the case of edge-colored multigraphs. If we allowed more than one edges (but yet a finite number) between the same pair of vertices in a graph, the resulting structure is called a multigraph. Such edges are called parallel edges. If more than one edge is present between two vertices, then the edges can be of different colors or of the same color. In our case, we assume that parallel edges are of different colors. Clearly, multigraphs offer new kind of restrictions and possibilities. It is also clear that real life problems can be modeled more faithfully using such structures. The class of edge-colored multigraphs generalizes that of directed graphs. Several other generalizations of directed graphs exist, namely : edge-colored digraphs, hypertournaments and star hypergraphs. However, it is the class of edge-colored multigraphs that has been given the main attention in graph theory. This because many concepts and results on directed graphs can be extended to edge-colored multigraphs and also because edge-colored multigraphs have several important applications. Significant results were obtained concerning these graphs [START_REF] Bang-Jensen | Alternating cycles and trails in 2-edge-coloured complete multigraphs[END_REF][START_REF] Bang-Jensen | Alternating cycles and paths in edge-coloured multigraphs : a survey[END_REF]. In this work we considered proper and weak proper spanning trees in edge-colored multigraphs. This thesis is organized as follows. In chapter two, we give basic terminology and notations, as well as some notions of complexity, that are needed for the understanding of the results exposed in the next chapters.

In chapter three, we present an overview of the current state of the art of proper spanning trees (PST) and weak proper spanning trees (WST) problems in edge-colored graphs.

In chapter four, we consider maximum proper edge-colored trees and maximum weak proper trees in edge-colored graphs. We consider these problems from the graph theoretic as well as from the algorithmic viewpoints. We prove their optimization versions to be NP-hard in general and provide algorithms for finding such trees in graphs without properly edge-colored cycles. We also derive some nonapproximability bounds.

In chapter five, we study the existence of PST in edge-colored graphs under certain assumptions on the graph, both structural and related to the coloring. We consider sufficient conditions that guarantee the existence of proper spanning trees in graphs with edges colored (not necessarily properly) with any number of colors. The conditions we consider are the combinations of various parameters of colored graphs such as : total number of colors, number of vertices, connectivity and number of differently colored edges incident to a vertex. These particular conditions are of interest since there exist no good characterizations for the existence of such colored trees in general edge-colored graphs.

In the chapter six we consider proper Hamiltonian paths in edge-colored multigraphs. A proper Hamiltonian path is a path containing all the vertices of the multigraph such that no two adjacent edges have the same color. Since proper Hamiltonian paths are also proper spanning trees, they are of interest for our study. We establish sufficient conditions for a multigraph to have a proper Hamiltonian path, depending on several parameters such as the number of edges, the rainbow degree, etc.

In chapter seven, we focus on a problem related to color connectivity. Since one of the sufficient conditions for proper spanning trees to exist is the connectivity, we prove several upper bounds for the smallest number of colors needed for k-proper-connectivity. We state several conjectures for general and bipartite graphs, and prove them for k = 1. Then, some algorithmic results are proposed.

Finally, in chapter eight, we present the conclusion of our work and some open problems for future research. In this chapter we introduce the main notions, necessary to the understanding of this thesis. We begin by the terminology and recall the common notions of the graph theory in general and of edge-colored graphs in particular. Then we introduce the notions of complexity and approximation.

Notation and terminology

We use the notation and terminology given by Jean-Claude Fournier in [START_REF] Fournier | Basic Concepts[END_REF]. However, the reader is warned that there may arise some differences.

(Undirected) graph

An (undirected) graph G = (V, E) is defined by two finite sets : a nonempty set V of elements called vertices, a set E (which may be empty) of elements called edges. Each edge e has two associated vertices, x and y, (not necessarily distinct), called its endpoints. The graphs considered in this thesis are finite. The vertex set V and the edge set E of a graph G can be denoted by V (G) and E(G), respectively. The number of vertices of a graph, is usually denoted by n. The number of edges of a graph, is denoted by m. An edge e is simply denoted by xy (or yx) instead of the customary mathematical notation {x, y} when x = y.

Adjacent, incident, neighbors, loop, parallel edges

The endpoints x and y of an edge e are adjacent. An edge e is said to be incident to each of its endpoints. If the both the endpoints on an edge are same vertex then this edge is a loop. All adjacent vertices of a vertex x are the neighbors of this vertex. Edges e and e ′ is said to be parallel or multiple edges if they have the same endpoints. A simple graph is a graph without loops or multiple edges.

Independent set

An independent set or stable set is a set of vertices in a graph, no two of which are adjacent.

Underlying simple graph

We call underlying simple graph, of a non-simple graph G, the graph defined as follows : it has the same set of vertices as G and two vertices are joined by an edge if and only if they are different and joined by at least one edge in G.

Isomorphic We say that a graph G=(V, E) is isomorphic to an graph H=(S, F) if there exists a bijection : φ : V → S such that (u, v) ∈ E if and only if (φ(u), φ(v)) ∈ F .

Complete graph

A simple graph G where each two vertices are adjacent is called a complete graph. A complete graph is determined by the number n of its vertices and it is generally denoted by K n . The edge set of a complete graph is equal to n 2 = n(n-1)

2

.

Subgraph

A subgraph H = (S, F ) of a graph G = (V, E) is the graph where S ⊆ V and F ⊆ E and where each edge e ∈ F has both its endpoints in S.

Vertexinduced subgraph

A vertex-induced subgraph H is induced by a set of vertices S ⊆ V , if it is a graph of the form H = (S, F ), where F is the set of the edges of E both of whose endpoints are in S.

Edgeinduced subgraph

An edge-induced subgraph is induced by a subset of the edges F ⊆ E of a graph G = (V, E), if it is a graph of the form H = (S, F ), where S is the set of vertices that are their endpoints.

Spanning subgraph

A subgraph H = (S, F ) of G = (V, E) is called a spanning subgraph if S = V . It can be a spanning subgraph of set of edges F too.

Gx

Let G = (V, E) then G -S, where S ⊂ V , is a subgraph of G induced by V \ S. In other words, it is a subgraph of G obtained by removing the vertex set S from G and the incident edges. Let G = (V, E) then G -F , where F ⊂ E, is the subgraph of G induced by E \ F . In other words, it is the induced subgraph of G obtained by removing the edge set F from G. In this thesis we use Gx instead of G -{x} for x ∈ S and Ge instead of Ge for e ∈ F .

Walk

A walk of a graph G = (V, E) is a sequence of the form :

(x 0 , e 1 , x 1 , . . . , e k , x k )

where k ≥ 0, x i are vertices of G, such that there exists an edge e i+1 ∈ E for each x i x i+1 , and x 0 and x k are called the endpoints of the walk. We say that x 0 and x k are linked by the walk.

Length

The length of the walk in this case is k. A walk may have zero length, it consist of one vertex in this case. In the case when G is a simple graph, we define a walk by the sequence (x 0 , x 1 , • • • , x k ) of its vertices. A subwalk of a walk is a walk defined as a subsequence between two vertices of the initial walk.

Trail

A trail is a walk where all the edges are distinct. In other words, when the walk does not go twice through the same edge.

Path

A path is a walk where its vertices are all distinct.

Cycle

A cycle is a closed path, ie x 0 = x k . A cycle cannot have zero length. A loop is a cycle of length 1. In the case of simple graphs, a cycle may be defined by the sequence (x 0 , x 1 , • • • , x 0 ) of its vertices. A cycle is called even or odd, depending on whether its length is even or odd.

Degree

The degree of a vertex x in a graph G is the number of edges in G incident to x. The degree is denoted by d(x).

Isolated vertex

A vertex of degree zero is called an isolated vertex.

Minimum degree δ

The minimum degree of a graph G is the smallest degree of its vertices. The minimum degree is denoted by δ or δ(G). It is clear that any graph G, has at least one vertex with its degree equal to the minimum degree.

Maximum degree △

Similarly the maximum degree of G is the largest degree of its vertices.

The maximum degree of G is denoted by △ or △(G). Any graph G has at least one vertex with its degree equal to the maximum degree.

Regular graph k-regular graph

A graph G is called regular if the degrees of its vertices are all equal. In this case, the graph is called a k-regular graph, if the common vertex degree is k.

Connected disconnected graph

If every pair of vertices of a graph G are linked by a path in G, then we say that G is connected. If it is not the case, then we say that G is a disconnected graph.

Connected component

The maximal connected subgraphs of G are called its components or connected components. A disconnected graph has at least two connected components. Similary, a connected graph has only one connected component. Observe that the connected components of a graph are subgraphs with no common vertices and no common edges. It defines the decomposition of the graph into connected components. Observe that if a graph has a spanning connected subgraph, then it is itself connected.

Bipartite graph

A bipartite graph G, is any graph whose vertex set can be partitioned into two disjoint subsets, such that each edge has an endpoint in each subset. A bipartite graph is denoted by G = (V, S, E), where V and S are the two subsets of vertices, and E is the set of edges. Is evident that V ∪ S is the set of vertices of G.

Complete bipartite graph

A complete bipartite graph is the bipartite graph G = (V, S, E) where the set of its edges is E = {xy|x ∈ V, y ∈ S}, i.e. for each pair of a vertices one each from V and S is an edge in E(G). We denote it by K p,q , where p is the cardinality of V and q the cardinality of S.

Tree

A tree is a connected acyclic graph. Here, acyclic means a graph without a cycle. It follows that a tree is a simple graph. A path is a special type of tree.

Leaf

Any vertex in a tree of degree 1 is called a leaf.

Root, rooted tree

When we distinguish a special vertex in a tree and call it a root, we get a rooted tree.

Forest

A forest is an acyclic graph. The connected components of a forest are trees. Forests generalize trees.

Bridge

If a graph G has an edge e such that Ge has one more connected component than G, then this edge e is called a bridge of G. If x and y are endpoints of a bridge e then in Ge, x and y are no longer linked by a path. In fact the edge e separates the vertices x and y. If G is connected then after removing e, Ge is disconnected.

Spanning tree

A spanning tree of a graph G is a spanning subgraph of G which is a tree

Cut vertex A cut vertex of a graph G is a vertex x such that Gx has at least one more connected component than G. This idea leads to a classic and useful decomposition of graphs.

Block

A block of a graph G is a maximal connected subgraph without a cut vertex of G.

Block decomposition

The set of blocks of G constitutes the block decomposition of G.

Connectivity, κ(G)

The smallest number of vertices whose removal makes G disconnected or reduces G to one vertex is called connectivity of G. κ(G) denotes the connectivity of G.

k - connected A graph G is k -connected if κ(G) ≥ k.

Edge connectivity

The edge connectivity k ′ (G) of a graph G, with more than one vertex, is the smallest number of edges whose removal disconnects the graph. κ ′ (G) denotes the edge connectivity of G.

(Edge) cut Let B be a set of edges whose removal disconnects G, then the set of edges B is what we call an (edge)

cut of G. k -edge - connected A graph G is k -edge -connected if κ ′ (G) ≥ k.

M atching

A matching of a graph G is a set M of edges such that no two edges share a common endpoint.

Matched or saturated vertex

A vertex of the graph is said to be matched (saturated) by a matching M (or Msaturated), if it is an endpoint of an edge of M . Otherwise it is said to be unmatched (unsaturated) by M (or Munsaturated).

Perfect matching

If every vertex of G is matched, then matching M is said to be perf ect.

Maximal matching

A matching is said to be maximal if it is impossible to add an edge to it.

Maximum matching

A matching M is said to be maximum if it has the greatest possible number of edges. It is a matching with the greatest number of edges among all the possible maximal matchings.

Compatible (or alternating) path relative to a matching

For a given graph G = (V, E) and a matching M of G, we call a path P , compatible (or alternating) relative to M , or an Mcompatible (Malternating), if it is a path alternating between edges in M and E \ M .

Augmenting path for M

An M -alternating path is said to be an augmenting path for M, or an M-augmenting path, if its endvertices are unmatched.

Factor

A factor of a graph G is a spanning subgraph, i.e., a subgraph that has the same vertex set as G.

k -f actor A k -regular spanning subgraph is called a k -f actor.
Observe that 1f actor is a graph H on V vertices and the edge set corespond to a perfect matching.

kfactorization

A k-factorization partitions the edges of the graph into disjoint k-factors.

Euler trail, tour, Eulerian graph

A Euler trail of a graph G is a trail containing all the edges of G. Is the trail that goes exactly once through each edge of G. A Euler tour of G is a closed Euler trail. We call a graph Eulerian graph if it has an Euler tour.

Hamilton cycle, Hamiltonian graph

A Hamilton cycle of a graph G is a cycle going through all the vertices of the graph. A graph G is called Hamiltonian graph if it has a Hamilton cycle.

Hamilton path

A Hamilton path of a graph G is a path going through all the vertices of a graph.

Complement of a graph

Let G = (V, E), then we call a complement of a graph and denote them Ḡ the graph on V vertices and the edge set [V ] 2 \ E. In other terms let G = (V, E) and let F be the set of edges in K V , then the complement of G is Ḡ = (V, F \ E).

Edge coloring

An edge coloring of a graph G = (V, E) is a map c : E → S with c(e) = c(f ) of any adjacent edges e and f . Let k be the minimum number of colors used for colorings G, called kedgecolorings of G.

Then k is called the cromatic index or edge cromatic number, and its denoted by χ ′ (G).

2-edgecolored

A graph is 2-edge colored if each edge is colored either red or blue.

Proper or properly edgecolored

An edge-colored graph is properly edge-colored (or proper) if any two adjacent edges are colored differently.

Proper path A path in an edge-colored graph is said to be properly edge-colored (or proper ), if every two adjacent edges differ in color.

k-proper connected

An edge-colored graph G is k-proper connected if any two vertices are connected by k internally pairwise vertex-disjoint proper paths.

k-proper connected, pc k (G)
We define the k-proper connection number of a k-connected graph G, denoted by pc k (G), as the smallest number of colors that are needed in order to make G k-proper connected.

X c {1, 2, • • • , c}We denote by X c {1, 2, • • • , c} a set of given c ≥ 2 colors.

G c

By G c is denoted an edge-colored graph, so that each edge is colored with some color i ∈ X c and no two parallel edges joining the same pair of vertices have the same color.

V (G c ), E(G c ) The vertex and edge-sets of G c are denoted by V (G c ) and E(G c ), respectively. When no confusion arises, is used

V, E instead of V (G c ), E(G c ) respectively.

T he order of G c

T he order of G c is the number n of its vertices.

E i (G c ) For a given color i, E i (G c ) denotes the set of edges of G c on color i. When no confusion arises, E i is used instead of E i (G c ). K c n A edge-colored complete graphs, is denoted by K c n . N i (x) By N i (x)
We denote the neighbors of a vertex x in some color i ∈ X c , ie the vertices adjacent to x with edges colored in color i ∈ X c .

N (x)

The neighborhood of a vertex x, is denoted by N (x). The neighborhood of a vertex x is defined as the union

N (x) = ∪ i N i (x). N H (x)
The neighborhood of a vertex x in the subgraph H in some color i ∈ X c , is denoted by N i H (x), ie the vertices in H adjacent to x with edges colored in color i ∈ X c .

N i H (x)
The neighbors of a vertex x in the subgraph H, is denoted by N (x). The neighborhood of a vertex x is defined as the union

N (x) = ∪ i N i (x). i -degree, color degree, d i (x)
The colored i-degree of x, denoted by d i (x) equals |N i (x)|, i.e., the cardinality of N i (x).

xy, c(xy)

The edge between the vertices x and y is denoted by xy, and its color by c(xy).

Rainbow degree, rd(x)

The rainbow degree of x, denoted by rd(x) equals the maximum number of distinct colors present on edges with endvertex x.

rd H (x)

The rainbow degree of x in the subgraph H, denoted by rd H (x), is equals to the maximum number of distinct colors presented on edges with endvertices x and the vertices in H.

Rainbow degree of a graph, rd(G c )
The rainbow degree of a graph G c , denoted by rd(G c ) equals the minimum rd(x), ∀x ∈ V (G c ).

H rs

Let G c be an edge-colored graph. The rainbow spanning subgraph of G c , denoted by H rs , is the subgraph that spann all vertices of G c and all edges of H are colored by distinct colors. The E(H) ⊆ E(G c ) and

X H ⊆ X G c .

Proper cycle

A proper cycle in G c is a proper subgraph whose underlying non-colored graph is a cycle.

Acyclic edge-colored graph

An edge-colored graph is said to be acyclic if it does not contain proper cycles. Observe that in this case the underlying non-colored graph can contain cycles, but any properly colored subgraph is acyclic.

PT, PST, MPT

A proper edge-colored tree or simply a proper tree, denoted by P T , is a proper edge-colored subgraph whose underlying non-colored version is a tree. If the P T spans all the vertices of a G c then this is a proper edge-colored spanning tree or simply a proper spanning tree, denoted by P ST . For maximum proper edge-colored tree we use maximum proper tree as terminology, and denote it M P T .

WT, WST, MWT

A weak edge-colored tree or simply a weak tree, denoted W T , is an edgecolored subgraph whose underlying non-colred version is a rooted tree, with the property that all root-to-leaf paths are properly edge colored.

If the W T spans all vertices of a G c then this is a weak edge-colored spanning tree or simply a weak spanning tree denoted W ST . For maximum weak edge-colored tree we use maximum weak tree as terminology, and denote it M W T .

Heterochromatic, rainbow subgraph

A heterochromatic subgraph of a edge-colored graph G c is a subgraph whose edges are colored with distinct colors, i.e. the color set

|χ c | is equal to |V (G c )| -1.

k-path-cycle subgraph

A k-path-cycle subgraph of a graph G c is a subgraph of G consisting a disjoint collection of k paths and some cycles.

Notions and definition of Complexity

In this section we introduce the notion of complexity and some of the complexity classes, necessary for the understanding of the results of this thesis. The analysis of complexity of a problem allows to classify it along with other already identified problems, and constitutes one of the steps to its resolution. Thus, all computational problems are separated into complexity classes. We focus on two main classes : P and NP, these acronyms mean polynomial class and nondeterministic polynomial class, respectively. The non-determinism represents our incapacity (which may be temporary) to find directly, that is without the help of a certificate, the right answer, in polynomial time. After briefly presenting the concept of computational complexity, we introduce these two classes, P and NP, as well as the notion of optimization problem 1 .

The notion of Complexity

The concept of time complexity corresponds to the time required to run a program. The running time depends on several factors, such as the size of the data to be processed, the data encoding, the machine that executes the program and others. It is clear that processing a graph of 10 vertices and 1000 will not take the same time. So, the size of the data to be processed is an important factor that directly influences the running time. In fact we can decide if the running time is satisfying or not in terms of the data that the program processes, by using the notion of a complexity function.

When processing the data, a program executes some tasks. These tasks take some processing time. In order to measure the running time, we use the number of "elementary" operations performed by the program. The set of tasks taken together is called an algorithm. A complete formalization of the notion of an algorithm requires a model capable of describing an automatic process. Alan Turing in 1937, described the "Turing" machine or, as he called it, an "a(utomatic)-machine". The Turing machine represents a computing machine, reduced to its simplest expression. Each task can be done in some "elementary" operations on the "Turing" machine. The model of this machine is used by the complexity theory in order to formalize the concept of the algorithm. However, theoretical results on the complexity are independent of the model of the machine. The efficiency of the algorithm that solves a given problem on the data of size n is measured by a function, representing the number of elementary operations it performs in order to process the data of this size. This function is called the run-time complexity 2 . This function is obtained by calculating the upper bound for the number of operations, necessary to execute the algorithm on the data of any possible size. In this case we talk about the worst case complexity. We are only interested in the order of magnitude of this function. Therefore, the complexity is a good measure for algorithm's efficiency and is widely employed for categorizing the problems depending on their theoretical difficulty.

Class P

After this brief introduction, we go to the heart of complexity theory, relevant to this thesis, that is the polynomiality criterion.

As shown in Table 2.1, the size of the data to be processed is an important factor to take into consideration for the performance of the algorithm. When the input size increases the difference between a polynomial function and an exponential growth becomes manifest. The example shows the running time required for data of size n = 10, 20, . . . , 70, for different complexity functions. The running times remain reasonable for polynomial functions n and n 2 . In the case of the exponential function 2 n the time grows for n = 70 to 374 centuries, which is clearly not practical. It is well known that asymptotically an exponential function (with a positive base) has a faster growth than any power function.

This polynomial growth criterion of the complexity function was therefore introduced, in particular by J. Edmonds in 1965. This criterion is pertinent, because of its asymptotic nature and because of stability of this criterion relative to the composition of algorithms. Since the composition of polynomial functions is itself a polynomial function.

The classic notation of Landau is used to express the complexity of an algorithm. An algorithm is of complexity O(f (n)) when the number of elementary operations is upper bounded by 2. There exists a complexity in space, that measure the amount of space necessary for processing the data of a given size. However, usually we are not interested in this complexity, therefore, when saying "complexity", the run-time complexity is meant. f (n) multiplied by a constant, under the condition that the integer n is greater than a certain value. If the function f is polynomial, the algorithm is called polynomial. The problems that can be solved using these algorithms are also called polynomial. The set of all these problems define the complexity class denoted P . A polynomial function behaves as its highest degree term, and taking into account the constant involved in O, f (n) is replaced by an expression of the form O(n k ). The case k = 1 is the best possible, since the algorithm needs at least the time to read the data. This is the case of linear algorithms.

Class NP

A problem is recognized as class P, when it can be solved algorithmically in polynomial time. The class P contains a lot of problems, such as finding shortest paths or a spanning tree in a simple graph. But there exist problems that cannot be solved polynomially.

Take the Hamiltonian cycle problem for instance. It consists in finding a cycle H in a simple graph G, that contains all the vertices of G, ie V (H) = V (G). Now for a given graph G and a given cycle H, is it H hamiltonian, ie V (H) = V (G) ? Such a problem with a "yes" or "no" answer is called a decision problem. One way to solve it is to find all the possible cycles in G, take the subset with cycles of vertex set equal to V (G), and verify if H is one of these cycles. If, for a given H, V (H) = V (G) then H is hamiltonian. Verifying the "yes" answer is easy.

The set of all decision problems where the "yes"-answers can be verified in polynomial time form the NP class. The NP class contains problems for which it is possible to check a "yes"answer polynomially without being able to find polynomially this answer. Observe that the "no"-answers are not required to be easy (or polynomially) to check.

An algorithm that is used to check a "yes"-answer is called verifier or checking-algorithm. Therefore, another way to define the NP class, is to define the class of all the problems for which there exist checking-algorithms that executes in polynomial time.

The input data for a problem is called an instance of a problem. In our example these are the graph G and the cycle H. Formally, the class NP is defined as follows.

Definition 2.2.1. Let π be a decision problem. Now π is in N P if there exists a polynomial algorithm A and a polynomial p such that for any instance x of A the answer is "yes" if and only if there is a datum y such that |y| ≤ p(|x|) and algorithm A applied to x associated with y gives the answer "yes" (|x| designates the size of x, likewise for |y|). The algorithm A is the checking-algorithm and y is a certificate, for instance x of π. The certificate is necessarily of polynomial size compared to the size of the instance (the converse is not important). In this case such a certificate is said to be succinct.

Since a polynomial algorithm can be a verifier for the problems of the class P, and the certificate can be empty, we can conclude that P ⊆ N P . To this point, it has not been proved that the inclusion is strict.

NP-complete problems

Since P ⊆ N P , it is important to know which problems can be solved in polynomial time. Since has not been unproved that P = N P , it is natural to look for problems that cannot be solved in polynomial time under the condition that P = N P . Problems, that can not be solved in polynomial time unless P = N P , are called N Pcomplete problems. In order to prove that a problem is N Pcomplete, one has to show that the problem is at least as hard as another problem from this class. To do this, the concept of reduction is used. To check if a problem is N Pcomplete, the reduction has to be polynomial.

Definition 2.2.2. Polynomial reduction.

A problem π 1 can be polynomially reduced to a problem π 2 if there is a polynomial algorithm, called a reduction-algorithm from π 1 to π 2 , which calculates for each instance x 1 of π 1 , an instance x 2 of π 2 such that the answer for x 1 is "yes" if and only if the answer for x 2 is "yes" (x 1 and x 2 have the same answer). This is a very important and strong relation, since if there exists a polynomial algorithm solving the problem π 2 , then a polynomial time algorithm can be designed that solves π 1 . So if we denote the polynomial reduction by ≤ p , then : Definition 2.2.3. NP-complete problem. A decision problem π 1 is NP-complete if π 1 is in NP, and ∀π ∈ N P , π ≤ p π 1 .

Figure 2.1 -Classes of complexity

A problem to which all the others can be polynomially reduced is, in a way, a "universal" problem. For the class NP such a problem is called NP-complete. The first NP-complete problem, found at the beginning of the 1970s, is a problem of logic, called a satisfiability problem denoted SAT.

Class coNP

For a problem of class N P , the "no"-answer may not have an obvious succinct certificate. For example, concerning the problem of existence of a Hamiltonian cycle in a graph : if it is easy to see how to "certify" the answer "yes" (by giving a Hamiltonian cycle), it is not obvious how to certify the answer "no". The Class coN P is defined as the class of decision problems for which the "complement" problem, that is the problem set in order to invert the answers "yes" and "no", is in N P .

It is not established if N P = coN P . Consider the Hamiltonian problem, that is in NP : "Given a graph G, is it Hamiltonian ? 3 ". In order to verify the "yes" answer we provide a Hamiltonian cycle. The complementary problem is in coNP : "Given a graph G is it non-Hamiltonian ?". In order to verify the "no" answer we provide a Hamiltonian cycle. A possible configuration of the previous classes is shown in Figure 2.1.

Best, worst and average case complexity

The deterministic sorting algorithm quicksort, solves the problem of sorting a list of integers given as the input. If the input is already sorted, the algorithm takes time O(n logn) to sort. If the input is sorted in the reverse order, the algorithm takes O(n 2 ) time to sort. If all possible permutations of the input list are likely, the average time for sorting is O(n logn). The quicksort algorithm has the average performance of O(n logn). The best, worst and average case complexity refers to three different ways of measuring the time complexity. It is clear that some inputs of size n may be faster to solve than the others. For this reason, the following complexities are defined :

-Best-case complexity is the complexity of solving the problem for the best input of size n.

-Worst-case complexity is the complexity of solving the problem for the worst input of size n.

a Hamiltonian graph is a graph, having a Hamiltonian cycle

-Average-case complexity is the complexity of solving the problem on an average.

Approximability

To this point, we have seen what a decision problem is, now we introduce the optimization problems. An optimization problem is the problem of finding the best solution among a collection of possible solutions. But how the best solution is defined ? For example, let G be a simple graph where x, y are two vertices of G. Consider the following decision problem : "Does G have a Hamiltonian path that starts and ends at x and y respectively ?". An optimization problem in this case would be : "For a given graph G and its vertices x, y, what is the longest path between these two vertices ?". Another example of an optimization problem is : "Given G and two vertices x and y, what is the shortest path between these two vertices ?". Suppose the solution is 1 edge, which means that x and y are adjacent. Then the corresponding decision problem would be : "Given graph G and two vertices x and y, is there a path from x to y of length 2 or less ?" As one can notice, for each optimization problem there exists a corresponding decision problem. And the best solution, called optimal solution OPT, consists in satisfying the goal function, i.e. minimum or maximum. Definition 2.2.4. An optimization problem denoted P is a quadruple (I, f, m, g) where -I is a set of instances ; -if n ∈ I, then f (n) is the set of feasible solutions ; -if, for given n, s ∈ f (n) is a feasible solution, then m(n, s) is the measure of s ; -g is the goal function, either min or max.

Using this definition we can say that the goal is to find an optimal solution for some instance, and we denote it m(n, s) = g{m(n, s ′ )|s ′ ∈ f (n)}. If an optimization problem π has an NP-complete decision version, then π is NP-hard. But if the corresponding decision problem is in NP, then π is also in NP-optimization problems class. An NP-optimization problem (NPO) is an optimization problem with additional conditions :

-the size of every feasible solution s ∈ f (n) is polynomially bounded in the size of the given instance n, -the languages { n | n ∈ I } and { (n, s) | s ∈ f (n) } can be recognized in polynomial time, and -m is polynomial-time computable.

When an optimization problem is NP-hard, no polynomial time algorithm exists that finds the best solution, unless P = N P . Since finding the best solution is not easy, one may be interested in a solution close to the optimal. In practice, a solution that is nearly optimal may be good enough and may be much easier to find. Such a solution can be found by using approximation algorithms. When the solution given by the approximation algorithm differs from the optimum only by a constant, such an algorithm is called an absolute approximation algorithm. We use the definition below, given by Korte [START_REF] Korte | Approximation algorithms[END_REF]. Definition 2.2.5. An absolute approximation algorithm for an optimization problem P is a polynomial-time algorithm A for P for which there exists a constant k such that

|A(I) -OP T (I)| ≤ k,
for all instances I of P .

Unfortunately, very few classical NP-hard optimization problems have an absolute approximation algorithm to solve them. In most cases one has to be satisfied with relative performance guarantees. Here we have to restrict ourselves to the problems with nonnegative weights. Definition 2.2.6. Let P be an optimization problem with nonnegative weights and k ≥ 1. A k-factor approximation algorithm for P is a polynomial-time algorithm A for P, such that

1 k OP T (I) ≤ A(I) ≤ kOP T (I)
for all instances I of P. We also say that A has performance ratio k.

The first inequality applies to maximization problems, the second one to minimization problems. Observe that for instances I where OP T (I) = 0 we require an exact solution. The 1-factor approximation algorithms are precisely the exact polynomial-time algorithms.

The set of NPO problems, that have an approximation algorithm and a positive real number r such that k ≤ r, form the APX class.

Non-approximability

When it is NP-hard to approximate some specific optimization problems with a given approximation ratio, we talk about non-approximabilty. A basic tool for establishing such results is the so-called "gap-preserving reduction".

Let P 1 and P 2 be two optimization problems. Depending on whether P 1 and P 2 are minimization or maximization problems, we have slightly different definitions. Definition 2.2.7. Let P 1 and P 2 be the maximization problems. A gap-preserving reduction from P 1 to P 2 is a polynomial time algorithm which, given an instance x ∈ P 1 , produces an instance y ∈ P 2 such that :

-if OP T (x) ≥ h(x), then OP T (y) ≥ h ′ (y), and -if OP T (x) < g(|x|)h(x), then OP T (y) < g ′ (|y|)h ′ (y) for some functions h(x), g(|x|), h ′ (y), g ′ (|y|) with g(|x|), g ′ (|y|) ≤ 1.

Likewise, if P 1 is a minimization problem and P 2 is the maximization problem, we have : Definition 2.2.8. Let P 1 be a minimization problem and P 2 be a maximization problem. A gap-preserving reduction from P 1 to P 2 is a polynomial time algorithm which, given an instance x ∈ P 1 , produces an instance y ∈ P 2 such that :

-if OP T (x) ≤ h(x), then OP T (y) ≥ h ′ (y), and -if OP T (x) > g(|x|)h(x), then OP T (y) < g ′ (|y|)h ′ (y) for some functions h(x), g(|x|), h ′ (y), g ′ (|y|) with g(|x|) ≥ 1 and g ′ (|y|) ≤ 1.

If P 1 and P 2 are minimization problems, we have : Definition 2.2.9. Let P 1 and P 2 be minimization problems. A gap-preserving reduction from P 1 to P 2 is a polynomial time algorithm which, given an instance x ∈ P 1 , produces an instance y ∈ P 2 , such that :

-if OP T (x) ≤ h(x), then OP T (y) ≤ h ′ (y), and -if OP T (x) > g(|x|)h(x), then OP T (y) > g ′ (|y|)h ′ (y) for some functions h(x), g(|x|), h ′ (y), g ′ (|y|) with g(|x|), g ′ (|y|) ≥ 1.

Finally, if P 1 is a maximization problem and P 2 is a minimization problem, we have : Definition 2.2.10. Let P 1 be a maximization problem and P 2 be a minimization problem. A gap-preserving reduction from P 1 to P 2 is a polynomial time algorithm which, given an instance x ∈ P 1 , produces an instance y ∈ P 2 such that :

-if OP T (x) ≥ h(x), then OP T (y) ≤ h ′ (y), and -if OP T (x) < g(|x|)h(x), then OP T (y) > g ′ (|y|)h ′ (y) for some functions h(x), g(|x|), h ′ (y), g ′ (|y|) with g(|x|) ≤ 1 and g ′ (|y|) ≥ 1.

As one can see, by using P 1 we can found some bounds for approximation ratio of P 2 .

Conclusion

In this chapter we introduced different concepts, used extensively in the next chapters. These notions and definitions are commonly used in graph theory and by a majority of researchers in edge-colored graphs and multigraphs. Such diversity of tools and notions, used to study the edge-colored graphs, shows the richness and the complexity of the questions that are addressed by this field. In this chapter we present an overview of the prominent results in edgecolored graphs obtained during the last 20 years. Edgecolored graphs being a relatively new area of research in Graph Theory, the questions about the existence of the most simple structures such us paths, cycles, trees, connectivity, edge connectivity, matching as well as others problems, require more sophisticated techniques, compared to the traditional graphs. In this chapter we present the important results connected to our topic.

Proper edge-colored Hamiltonian cycles and paths

It is obvious that from a proper edgecolored Hamiltonian cycle we can find a proper edgecolored Hamiltonian path, and this path is also a proper spanning tree. So the results on properly edge-colored Hamiltonian cycles and paths are directly connected to our problem.

The problem of finding a properly edge-colored Hamiltonian cycle and path is NP-Complete for general graphs. It is NP-Complete for the properly edge-colored graphs as well. It is enough to color every edge with a different color and this graph will be properly edge-colored, however it will be Hamiltonian only if the initial graph was Hamiltonian. So the results that have been obtained to this day deal with special conditions for the edge-colored graphs that have properly edge-colored Hamiltonian cycles and paths. A properly edge-colored cycle and path is also called an alternating cycle and path. The most natural way is to see what happens in edgecolored complete graphs. Bollóbas and Erdos [START_REF] Bollobás | Alternating Hamiltonian cycles[END_REF] studied the problem of finding a properly edge-colored Hamiltonian path in K c n , -a complete graph colored with c colors. Let △(G c ) denote the maximum degree in same color in G c . The following conjecture is given in [START_REF] Bollobás | Alternating Hamiltonian cycles[END_REF].

Conjecture 3.1.1. [18] Every K c n with △(K c n ) ≤ ⌊ n 2 ⌋ -1 contains an alternating Hamiltonian cycle.
Clearly, such a graph is not properly edge-colored Hamiltonian if there is an odd number of vertices and c = 2.

Bollóbas and Erdos proved that if △(K c n ) < n 69 , then it contains a properly edge-colored Hamiltonian cycle. This condition was improved by Chen and Daykin [START_REF] Chen | Graphs with Hamiltonian cycles having adjacent lines different colors[END_REF] to

△(K c n ) ≤ n
and later, by Shearer [START_REF] Shearer | A property of the colored complete graph[END_REF] to △(K c n ) ≤ n 7 and again by Alon and Gutin in [START_REF] Alon | Properly colored hamilton cycles in edge-colored complete graphs[END_REF] stating that

△(K c n ) ≤ (1 -1 √ 2 -ǫ)n
, which is the best improvement until today. It is shown that, for ǫ > 0 and n > n 0 (ǫ), any complete graph K on n vertices, whose edges are colored so that no vertex is incident to more than (1 -1 √ 2ǫ)n edges of the same color, contains a properly edge-colored Hamilton cycle. Moreover, for every k between 3 and n any such K contains a properly edge-colored cycle of length k. Theorem 3.1.2. [START_REF] Alon | Properly colored hamilton cycles in edge-colored complete graphs[END_REF] For every ǫ > 0 there exists an n 0 = n 0 (ǫ) so that, for every n > n 0 , every K c n satisfying

△(K c n ) ≤ (1 -1 √ 2 -ǫ)n [= (0.2928... -ǫ)n]
contains alternating cycles of all lengths between 3 and n.

In the case where n = 2m + 1 they show that every edge of K c n is contained in a properly edge-colored cycle of each desired length between 4 and n (but not necessarily of length 3). Theorem 3.1.3. [START_REF] Alon | Properly colored hamilton cycles in edge-colored complete graphs[END_REF] For every ǫ > 0 there exists an m 0 = m 0 (ǫ) so thats for every m > m 0 , every K c m,m satisfying

△(K c m,m ) ≤ (1 -1 √ 2 -ǫ)m [= (0.2928... -ǫ)m]
contains an alternating cycle of every even length between 4 and 2m.

Another interesting result was obtained by Barr [START_REF] Barr | Properly coloured Hamiltonian paths in edge-coloured complete graphs without monochromatic triangles[END_REF].

Theorem 3.1.4. Every K c n without monochromatic triangles contains an alternating Hamiltonian path.

For a set F of connected graphs, a spanning subgraph F of a graph is called an F -factor, if every component of F is isomorphic to a member of F . So one properly edge-colored cycle factor with one properly edge-colored cycle is a properly edgecolored Hamiltonian cycle. Thus finding the sufficient conditions for an edgecolored graph to have a minimum properly edge-colored cycles factor can help in our quest (of finding PST and WST). M. Bánkfalvi and Z. Bánkfalvi [START_REF] Bánkfalvi | Alternating Hamiltonian circuit in two-coloured complete graphs[END_REF] considered a similar problem for K 2 2p .

Theorem 3.1.5. [START_REF] Bánkfalvi | Alternating Hamiltonian circuit in two-coloured complete graphs[END_REF] Let K 2 2p be a two edge-colored complete graph with vertex-set

V (K 2 2p ) = {x 1 , x 2 , ..., x 2p }. Assume that d r (x 1 ) ≤ d r (x 2 ) ≤ ... ≤ d r (x 2p ). The graph K 2
2p contains an alternating factor with a minimum number m of alternating cycles, if and only if there are m numbers k i , 2 ≤ k i ≤ p -2, such that for each i, i = 1, 2, ..., m, we have :

d r (x 1 ) + d r (x 2 ) + • • • + d r (x k i ) + d r (x 2p ) + d r (x 2p-1 ) + d r (x 2p-2 ) + • • • + d r (x 2p-k i +1 ) = k 2 i .
A.Benkouar, Y.Manoussakis, V.T. Paschos and R.Saad [START_REF] Benkouar | Hamiltonian problems in edge-colored complete graphs and Eulerian cycles in edge-colored graphs : some complexity results[END_REF] propose an algorithm that can find the minimum properly edge-colored factor in K 2 n in O(n 2.5 ) time. This algorithm is based on the algorithm of maximum matching in general graphs made by S.Even and O.Kariv [START_REF] Even | On O(N 2.5 ) algorithm for maximum matching in general graphs[END_REF]. The algorithm for finding the minimum properly edge-colored cycle factor works as follows. Take a maximum matching induced in one color graph K 2 n , and after in the second color. Now if neither of the matchings is perfect then, it is trivial that there are not properly edge-colored cycle factor. If they are perfects then the union of these two matchings give us the properly edge-colored Hamiltonian path. Theorem 3.1.6. [START_REF] Benkouar | Hamiltonian problems in edge-colored complete graphs and Eulerian cycles in edge-colored graphs : some complexity results[END_REF] There exists an O(n The first to study this problem were Grossman and Haggkvist [START_REF] Grossman | Alternating cycles in edge-partitioned graphs[END_REF]. Grossman and Haggkvist gave a characterization of two-edge-colored graphs, which have an alternating cycle. The case c ≥ 3 was proved recently by Yeo [START_REF] Yeo | A note on alternating cycles in edge-coloured graphs[END_REF] : Theorem 3.1.9. [START_REF] Grossman | Alternating cycles in edge-partitioned graphs[END_REF][START_REF] Yeo | A note on alternating cycles in edge-coloured graphs[END_REF] Let G be a c-edge-colored graph, c ≥ 2, such that every vertex of G is incident to at least two edges of different colors. Then either G has a cut vertex separating colors, or G has an alternating cycle.

There are three interesting corollaries of the Theorem 3.1.9 (for the case c = 2), all proved in [START_REF] Grossman | Alternating cycles in edge-partitioned graphs[END_REF].

Corollary 3.1.10. Let M be a perfect matching in a graph G. If no edge of M is a cut edge of G, then G has a cycle whose edges are taken alternately from M and G -M . Corollary 3.1.11. Let G be a 2-edge-colored Eulerian graph so that all monochromatic degrees are odd. Then G has an alternating cycle.

Corollary 3.1.12. Let G be a 2-edge-colored graph so that both red and blue subgraphs of G are regular and non-trivial. Then G has an alternating cycle.

A kpathcycle subgraph of an edge-colored graph G c is a subgraph of G c consisting a disjoint collection of k paths and some cycles. If the number of cycles is zero we call it a kpath subgraph. A cycle subgraph is a nonempty 0pathcycle. subgraph. Feng et al. [START_REF] Feng | Characterization of edge-colored complete graphs with properly colored hamilton paths[END_REF] proved the following : Working on the following conjecture proposed by Abouelaoualim et al. [START_REF] Abouelaoualim | Cycles and paths in edge-colored graphs with given degrees[END_REF], Gutin [START_REF] Gutin | Note on edge-colored graphs and digraphs without properly colored cycles[END_REF] proved that it is false. The conjecture is based on colordegree constraint for the existence of long properly edge-colored cycles, and in his theorem Gutin proves that this constraint is not relevant. Conjecture 3.1.14. [START_REF] Abouelaoualim | Cycles and paths in edge-colored graphs with given degrees[END_REF] Let G be a c-edge-colored undirected graph of order n with δ mon (G) = d ≥ 1. Then G has a properly colored cycle of length at least min{n, cd}. Moreover, if c > 2, then G has a properly colored cycle of length at least min{n, cd + 1}. Theorem 3.1.15. [START_REF] Gutin | Note on edge-colored graphs and digraphs without properly colored cycles[END_REF] For each d ≥ 1 there is an edge-colored graph G with δ mon (G) = d and with no properly colored cycle.

In the absence of characterization of K c n with a properly edge-colored Hamiltonian cycle, sufficient conditions are of interest. Manoussakis, Spyratos, Tuza and Voigt [START_REF] Manoussakis | Minimal colorings for properly colored subgraphs[END_REF] proved the following results. Theorem 3.1.17. [START_REF] Manoussakis | Minimal colorings for properly colored subgraphs[END_REF] If c ≥ 1 2 (n -3)(n -4) + 2, then there exists n 0 = n 0 (c) such that for all n ≥ n 0 , K c n has a properly edge-colored Hamiltonian cycle.

Abouelaoualim et al. [START_REF] Abouelaoualim | Paths and trails in edge-colored graphs[END_REF] studied the s-t properly edge-colored longest paths in c-edge-colored graph graphs. They propose a polynomial time procedure for finding the longest properly edgecolored st path in graphs with no properly edge-colored cycles.

Theorem 3.1.18. [START_REF] Abouelaoualim | Paths and trails in edge-colored graphs[END_REF] Assume that G c has no properly edge-colored cycles. Then we can always find in polynomial time a longest properly edge-colored st path or otherwise to conclude that such a path does not exist in G c .

Using same techniques they proved the following theorem that deal with a properly edgecolored path between two given vertices. Theorem 3.1.19. [START_REF] Abouelaoualim | Paths and trails in edge-colored graphs[END_REF] Let G c be a c-edge colored graph with no properly edge-colored closed trails and two vertices s, t ∈ V (G c ). Then, we can always find in polynomial time a longest properly edge-colored st trail in G c , provided that one exists.

Recently Abouelaoualim et al. [START_REF] Abouelaoualim | Cycles and paths in edge-colored graphs with given degrees[END_REF] gave some sufficient degree conditions for the existence of properly edge-colored cycles and paths in edge-colored graphs, multigraphs and random graphs. In particular, they prove that an edgecolored multigraph of order n of at least three colors and with minimum colored degree greater than or equal to ⌈ n+1 2 ⌉, has properly edge-colored cycles of all possible lengths, including Hamiltonian cycles. Longest properly edge-colored paths and Hamiltonian paths between given vertices are considered as well.

Theorem 3.1.20. [START_REF] Abouelaoualim | Cycles and paths in edge-colored graphs with given degrees[END_REF] Let G c be a 2-edge colored graph such that for every vertex x, d i (x) ≥ d ≥ 1, i ∈ {1, 2}. Then G c has a properly edge-colored path of length at least 2d.

In the case where c ≥ 3, the following corollary was proved. For edgecolored multigraphs they give similar theorems. For the following theorem they give a construction of a graph H 1 that is used in the theorem. In fact, the property of the graph is that any longest properly edge-colored cycle in H 1 has a length d. Theorem 3.1.22. [START_REF] Abouelaoualim | Cycles and paths in edge-colored graphs with given degrees[END_REF] Let G c be an c-edge-colored graph multigraph, c ≥ 2. Assume that for every vertex x, d i (x) ≥ d ≥ 1, i ∈ {1, 2, ..., c}. Then G c has either a properly-edge colored path of length at least min{n -1, 2d} or otherwise a properly-edge colored cycle of length d + 1, unless G c is isomorphic to H1, in which case it has a cycle of length d.

Using this theorem 3.1.22 they give some sufficient degree conditions for an edge-colored multigraph to have a properly edge colored Hamiltonian cycle. This result may be viewed as a counterpart to Dirac's well known result for general graphs [START_REF] Dirac | Some theorems on abstract graphs[END_REF] , insofar as the conditions involved deal only with degree conditions and with nothing else. Based on this theorem they obtained a series of corollaries for properly edge-colored Hamiltonian paths, namely : Corollary 3.1.24. [START_REF] Abouelaoualim | Cycles and paths in edge-colored graphs with given degrees[END_REF] Let G c be a c-edge colored multigraph, c ≥ 3. Assume that for each color i ∈ {1, 2, ..., c} and for each vertex

x of G c , d i (x) ≥ ⌈ n 2 ⌉.
Then G c has a properly edge-colored Hamiltonian path.

In next corollary they give interesting results for properly edge-colored Hamiltonian paths with fixed end-points.

Corollary 3.1.25. [3] Let x, y be two fixed vertices in G c , c ≥ 2. Assume that ∀v ∈ V (G c ), d i (v) ≥ ⌈ n+3
2 ⌉ for each color i ∈ {1, 2, ..., c}. Then G c has a properly edge-colored Hamiltonian path with endpoints x, y.

In next theorem, they go further by showing that under the conditions of Theorem 3.1.23, G c has cycles of many lengths. An edge-colored multigraph G c of order n is called pancyclic if it contains properly edge-colored cycles of all possible lengths 2, 3, 4, 5, . . . , n. Similarly, G c is evenpancyclic if it contains properly edge-colored cycles of all possible even lengths 2, 4, 6, 8, . . . , 2⌊ n 2 ⌋.

Theorem 3.1.26. [3] Let G c be a c-edge colored multigraph, c ≥ 2. Assume that ∀x ∈ V (G c ), d i (x) ≥ ⌈ n+1 2 ⌉ for each colori ∈ {1, 2, ..., c}. i)If c = 2, then G c is even-pancyclic. ii)If c ≥ 3, then G c is pancyclic.
Now let us see another interesting result on edgecolored multigraphs. Clearly, it is interesting to find a maximum 1-path-cycle subgraph or maximum cycle subgraph of a c-edge-colored graph multigraph. Based on the polynomial algorithm for finding a maximum weight perfect matching in a weighted graph in O(n 3 ) time ( [START_REF] Papadimitriou | Combinatorial optimization : algorithms and complexity[END_REF] Ch. 11), Bang-Jensen and Gutin [START_REF] Bang-Jensen | Alternating cycles and paths in edge-coloured multigraphs : a survey[END_REF] give the following theorem. In order to explain some of the following results, we introduce the notion of color-connectivity, proposed by Saad [START_REF] Saad | Sur quelques problémes de compléxite dans les graphes[END_REF]. Definition 3.1.28. A pair of vertices x, y of a c-edge-colored multigraph G is called colorconnected if there exist alternating (x, y)-paths P = xx ′ , ..., y ′ y and P ′ = xu ′ , ..., v ′ y, such that c(xx ′ ) = c(xu ′ ) and c(y ′ y ′ ) = c(v ′ y). We define a vertex x to be color-connected to itself. We say that G is color-connected if every pair of vertices of G is color-connected.

A very useful observation is that every Hamiltonian c-edge-colored graph multigraph is colorconnected. The next logical question is how to check if the given two vertices are or not colorconnected. Bang-Jensen and Gutin [START_REF] Bang-Jensen | Alternating cycles and trails in 2-edge-coloured complete multigraphs[END_REF] show that it is possible to check the color-connectivity between two given vertices in O(|E|) time.

Lemma 3.1.29. [START_REF] Bang-Jensen | Alternating cycles and trails in 2-edge-coloured complete multigraphs[END_REF] Let G = (V, E) be a connected 2-edge-colored multigraph and let x and y be distinct vertices of G. For each choice of i, j ∈ {1, 2} we can find an alternating path

P = x 1 x 2 ...x k with x 1 = x, x k = y, c(x 1 x 2 ) = i and c(x k-1 x k ) = j in time O(|E|) (if one exists).

Bipartite multigraphs

Let us see now some results on properly edge-colored bipartite multigraphs. Hilton [START_REF] Hilton | Alternating Hamiltonian circuits in edge-coloured bipartite graphs[END_REF] proved the following theorem : Theorem 3.2.1. [START_REF] Hilton | Alternating Hamiltonian circuits in edge-coloured bipartite graphs[END_REF] Let G be a 2-edge-colored regular bipartite graph such that each of the partite sets of G has m vertices and let G" be the blue subgraph of G. If the degree d(G) of G is at least m 2 + 1 and G" is regular of degree at least m 2 and at most d(G) -1, then G has a Hamiltonian alternating cycle.

X ⊂ U , Y ⊂ W , d 1 (X) + d 2 (Y ) > k 2 and d 2 (X) + d 1 (Y ) > k 2 .
The longest properly edge-colored cycle problem for 2edgecolored complete bipartite multigraphs is polynomially solvable. Theorem 3.2.4. [START_REF] Bang-Jensen | Alternating cycles and trails in 2-edge-coloured complete multigraphs[END_REF] The length of the longest alternating cycle in a color-connected 2-edgecolored complete bipartite multigraph G is equal to the number of vertices in maximum cycle subgraph of G. There is an algorithm for finding a longest alternating cycle in a color-connected 2-edge-colored complete bipartite multigraph on n vertices in time O(n 3 ).

Let r be a positive integer. The 2-edge-colored complete bipartite graph B(r) has partite sets {x 1 , ..., x 2 } and {y 1 , ..., y 2 }. The set of red edges of B(r

) is{x i y j : 1 ≤ i ≤ r, 1 ≤ j ≤ r} ∪ {x i y j : r + 1 ≤ i ≤ 2r, r + 1 ≤ j ≤ 2r}.
Theorem 3.2.5. [START_REF] Bang-Jensen | Alternating cycles and trails in 2-edge-coloured complete multigraphs[END_REF] A 2-edge-colored complete bipartite multigraph is vertex evenpancyclic iff it is Hamiltonian and not isomorphic to one of the graphs B(r) (r = 2, 3, ...). Corollary 3.2.6. [START_REF] Bang-Jensen | Alternating cycles and trails in 2-edge-coloured complete multigraphs[END_REF] A 2-edge-colored complete bipartite multigraph is evenpancyclic if and only if it is Hamiltonian and not isomorphic to one of the graphs B(r) (r = 2, 3, ...).

Heterochromatic subgraphs

In this section, we present some important results on heterochromatic subgraphs in edgecolored graphs. More details can be found in a survey by Kano and Li [START_REF] Kano | Monochromatic and heterochromatic subgraphs in edge-colored graphs-a survey[END_REF]. As mentioned before, heterochromatic subgraph is also called multicolored, rainbow colored, colorful or polychromatic subgraph. As one can easily understand from its name, a heterochromatic subgraph is a subgraph where each edge is of a distinct color, i.e. there is as many colors as edges in the subgraph. One of the problems on heterochromatic subgraphs is the Partition problem. There are two natural ways to make a partition, one is vertex partition and the second one is edge partition. We present some existing results on special types of heterochromatic subgraphs such that heterochromatic trees, cycles and paths. Clearly the heterochromatic spanning subgraphs are important as well car they are properly edge-colored and we can use classical algorithms i order to find properly edge-colored spanning trees for the initial graph.

Vertex Partitions

The vertex partition problem is the problem of partition by heterochromatic subgraphs, the initial c-edge-colored graph. In fact the goal is to find the minimum number of vertex-disjoint heterochromatic subgraphs. If the minimum is equal to one, then the heterochromatic trees, cycles and paths becomes spanning heterochromatic subgraphs.

If the c-edge-colored graph is monochromatic, i.e. all it's edges are colored in same color, then the minimum heterochromatic tree partition problem is equivalent to finding a maximum matching. Li and Zhang [START_REF] Li | On the minimum monochromatic or multicolored subgraph partition problems[END_REF] studied the complexity of the heterochromatic tree, cycle and path partition problems.

Theorem 3.3.1. [START_REF] Li | On the minimum monochromatic or multicolored subgraph partition problems[END_REF] The minimum heterochromatic tree (cycle) partition problem is NP-Complete, and there does not exist a constant factor approximation algorithm for it. Actually, the minimum heterochromatic tree partition problem is NP-Complete for bipartite graphs.

Theorem 3.3.2. [START_REF] Li | On the minimum monochromatic or multicolored subgraph partition problems[END_REF] The minimum heterochromatic path partition problem is NP-Complete for 2-edge-colored graphs, and therefore, it is NP-Complete for general graphs.

Another result for complete bipartite graphs is due to Chen et al. [START_REF] Chen | Heterochromatic tree partition numbers for complete bipartite graphs[END_REF]. They gave an explicit formula for the heterochromatic tree partition number (hetetree r ) of an r-edge-colored complete bipartite graph K m,n .

Theorem 3.3.3. [START_REF] Chen | Heterochromatic tree partition numbers for complete bipartite graphs[END_REF] Let n, m and r be integers such that 2 ≤ m ≤ n, 1 ≤ r ≤ mn. Then the heterochromatic tree partition number of an r -edge-colored complete bipartite graph K m,n is

hetetree r (K m,n ) =                n if 1 ≤ r ≤ n; 1 if m(n -1) + 1 ≤ r ≤ mn; 2 if m = n and r = n 2 -2n + 2; n -⌊ n m ⌋ if m + 1 ≤ r ≤ m(n -1)
, and r ≡ 0, 1 (mod m); n -⌊ n m ⌋ -1 otherwise.

Jin and Li [START_REF] Jin | Partitioning complete graphs by heterochromatic trees[END_REF] give the heterochromatic tree partition number of an r-edge-colored complete graph.

Theorem 3.3.4. [49] Let 3 ≤ n, 2 ≤ r ≤ n 2 and t 2 + 2 ≤ r ≤ t+1 2 + 1. Then hetetree r (K n ) = ⌈ (n-t) 2 ⌉.

Edge Partitions

The next results concern a heterochromatic tree partition of the edge set of a properly edge colored complete graph. Constantine [START_REF] Constantine | Colorful isomorphic spanning trees in complete graphs[END_REF] gave a result on the existence of a proper edge coloring of complete graphs, such that their edges can be partitioned into heterochromatic spanning trees.

Theorem 3.3.5. [START_REF] Constantine | Colorful isomorphic spanning trees in complete graphs[END_REF] For n = 1, 3, K 2n can be properly edge-colored with 2n -1 colors in such a way that the edges can be partitioned into edge-disjoint heterochromatic isomorphic spanning trees.

In graph theory, a factor of a graph G is a spanning subgraph, i.e., a subgraph that has the same vertex set as G. A k-factor of a graph is a spanning k-regular subgraph, and a kf actorization partitions the edges of the graph into disjoint k-factors. A graph G is said to be kf actorable if it admits a k-factorization. In particular, a 1f actor is a perf ect matching, and a 1f actorization of a k-regular graph is an edge coloring with k colors.

The degree condition for a properly edge-colored graph to have a heterochromatic H-factor, was studied by Yuster [START_REF] Yuster | Rainbow H-factors[END_REF]. He found the following result. Theorem 3.3.6. [START_REF] Yuster | Rainbow H-factors[END_REF] Let H be a graph and n ≥ 2 be an integer such that |H| divides n. Then there exists an integer k = k(H) such that every properly edge-colored graph of order n and with minimum degree at least (1

-1 χ(H) )n + k has a heterochromatic H -f actor.
Suppose that the edges of the complete K 2n are colored with 2n -1 colors in such a way that the edges of any single color form a perfect matching. Such a coloring is called a coloring of f actorization. It is easy to see that every complete graph K 2n with a coloring of factorization contains a heterochromatic spanning tree : namely, the star K 1,2n-1 at any vertex.

A complete graph on 2n vertices K 2n can be partitioned into n spanning trees. So Brualdi and Hollingsworth [START_REF] Brualdi | Multicolored trees in complete graphs[END_REF] formulated the following conjecture. Kaneko, Kano and Suzuki [START_REF] Kaneko | Three edge disjoint multicolored spanning trees in complete graphs[END_REF] extended the theorem 3.3.8 to properly edge-colored complete graphs and gave the following theorem and conjecture. Theorem 3.3.9. [START_REF] Kaneko | Three edge disjoint multicolored spanning trees in complete graphs[END_REF] Every properly edge-colored complete graph K n (n ≥ 6) has three edgedisjoint heterochromatic spanning trees.

Conjecture 3.3.10. [1] Every properly edge-colored complete graph K n (n ≥ 6) has ⌊ n
2 ⌋ edgedisjoint heterochromatic spanning trees. In particular, if n is even then the edges of K n can be partitioned into n 2 heterochromatic spanning trees.

Akbari and Alipourn [START_REF] Akbari | Multicolored trees in complete graphs[END_REF] generalized the theorem 3.3.8 for a graph colored by r colors with colorsdistribution (a 1 , a 2 , • • • , a r ) there a i is the number of edges colored in color i , 1 ≤ i ≤ r, as follows.

Theorem 3.3.11. [5] If the r -edge-colored complete graph K n has a color distribution (a 1 , • • • , a r ) with 1 ≤ a 1 ≤ • • • ≤ a r ≤ (n + 3)/2 and r ≥ n -1, then K n has a heterochromatic spanning tree.
Theorem 3.3.12. [START_REF] Akbari | Multicolored trees in complete graphs[END_REF] Suppose that the r-edge-colored complete graph K n has a color distribution

(a 1 , • • • , a r ) with 2 ≤ a 1 ≤ • • • ≤ a r ≤ (n + 1)/2.
If T is a non-star heterochromatic spanning tree of K n , then K n -T has a heterochromatic spanning tree, where T is regarded as an edge subset.

Theorem 3.3.13. [START_REF] Akbari | Multicolored trees in complete graphs[END_REF] If the complete graph K n is r -edge-colored so that its color distribution

(a 1 , • • • , a r ) satisfies 1 ≤ a 1 ≤ • • • ≤ a r ≤ (n)/2 and r ≥ n -1, then K n has two edge-disjoint heterochromatic spanning trees. Theorem 3.3.14. [5] If the complete graph K n , n ≥ 3, is r-edge-colored and r ≥ n-2 2 + 2, then K n has a heterochromatic spanning tree. Theorem 3.3.15. [5] If the complete graph K n , n ≥ 6, is r-edge-colored and r ≥ n-2 2 + 3, then K n has two edge-disjoint heterochromatic spanning trees.
Brualdi and Hollingsworth [START_REF] Brualdi | Multicolored forests in complete bipartite graphs[END_REF] obtained a necessary and sufficient condition using color distribution for an r-edge-colored complete bipartite graph to have a heterochromatic spanning tree.

Theorem 3.3.16. [22] Every (2n -1)-edge-colored complete bipartite graph K n,n with color distribution (a 1 , • • • , a 2n-1 ) such that 1 ≤ a 1 ≤ • • • ≤ a 2n-1 has a heterochromatic spanning tree if and only if for every integer k with k ≤ 2n -1, it follows k i=1 (a i ) > k 2 4 .

Heterochromatic Hamiltonian Cycle

A k -bounded edge-colored graph is a graph there the edges are colored using no color more than k times. The problem to find a large bound k = k(n) such that every kbounded edgecolored complete graph K n contains a heterochromatic Hamiltonian cycle was mentioned by Erdos, Nesetril and Rodl [START_REF] Erdős | On some problems related to partitions of edges of a graph[END_REF]. They addressed it as an Erdos-Stein problem and showed that k can be any constant. Hahn and Thomassen [START_REF] Hahn | Path and cycle sub-Ramsey numbers and an edge-colouring conjecture[END_REF] obtained that k could grow as fast as n 1 3 to guarantee that a kbounded edge-colored K n contains a heterochromatic Hamiltonian cycle. They conjectured that the growth rate of k could in fact be linear. Theorem 3.3.17. [START_REF] Hahn | Path and cycle sub-Ramsey numbers and an edge-colouring conjecture[END_REF] There exists a constant number c such that if n ≥ ck 3 then every kbounded edge-colored complete graph K n has a heterochromatic Hamiltonian cycle.

Frieze and Reed [START_REF] Frieze | Polychromatic Hamilton cycles[END_REF] made further progress. Theorem 3.3.18. [START_REF] Frieze | Polychromatic Hamilton cycles[END_REF] There exists a constant number c such that if n is sufficiently large and k ≤ n c ln(n) , then every k-bounded edge-colored complete graph K n contains a heterochromatic Hamiltonian cycle.

In [START_REF] Albert | Multicoloured Hamilton cycles[END_REF], Albert, Frieze and Reed proved the conjecture of [START_REF] Hahn | Path and cycle sub-Ramsey numbers and an edge-colouring conjecture[END_REF].

Theorem 3.3.19. [6] Let c < 1 32 .
If n is sufficiently large and k ≤ ⌈cn⌉, then every k-bounded edge-colored complete graph K n contains a heterochromatic Hamiltonian cycle.

Albert, Frieze and Reed also obtained a similar result in the case of directed graphs. Hahn and Thomassen [START_REF] Hahn | Path and cycle sub-Ramsey numbers and an edge-colouring conjecture[END_REF] considered the heterochromatic Hamiltonian path in infinite complete graphs K w . Since we are not dealing with infinite graphs in this thesis, we are not giving further details about these results.

Corollary 3.3.21. [START_REF] Hahn | Path and cycle sub-Ramsey numbers and an edge-colouring conjecture[END_REF] If K w is edge-colored so that no color is used more than k times for a fixed k, then it contains a heterochromatic Hamiltonian path.

Heterochromatic Spanning Tree

In the previous section we gave some results on the heterochromatic spanning trees. We continue with different results. An edge coloring of a graph G is called an edge coloring with complete bipartite decomposition if each color class form a complete bipartite subgraph of G. In unpublished work, Caen conjectured that if a complete graph K n is edge-colored with complete bipartite decomposition using n -1 colors, then K n has a heterochromatic spanning tree. Alon, Brualdi and Shader [START_REF] Alon | Multicolored forests in bipartite decompositions of graphs[END_REF] proved the following stronger result. Theorem 3.3.22. [START_REF] Alon | Multicolored forests in bipartite decompositions of graphs[END_REF] Every complete graph K n having an edge coloring with complete bipartite decomposition contains a heterochromatic spanning tree.

Suzuki [START_REF] Suzuki | A necessary and sufficient condition for the existence of a heterochromatic spanning tree in a graph[END_REF] studied necessary and sufficient conditions for the existence of a heterochromatic spanning tree in an edge-colored graph.

Theorem 3.3.23. [START_REF] Suzuki | A necessary and sufficient condition for the existence of a heterochromatic spanning tree in a graph[END_REF] An edge-colored connected graph G of order n has a heterochromatic spanning tree, if and only if, for any k colors (1 ≤ k ≤ n -2), the removal of all the edges colored with these k colors from G results in a graph having at most k + 1 components.

Using this Theorem 3.3.23 Suzuki proved the following. Theorem 3.3.24. [START_REF] Suzuki | A necessary and sufficient condition for the existence of a heterochromatic spanning tree in a graph[END_REF] The ( n 2 )-bounded edge-colored complete graph K n has a heterochromatic spanning tree.

Jin and Li [START_REF] Jin | Spanning trees with many colors in edge-colored graphs[END_REF] generalized Theorem 3.3.23 to the following theorem. Theorem 3.3.25. [START_REF] Jin | Spanning trees with many colors in edge-colored graphs[END_REF] Let G c be an edge-colored connected graph and 1 ≤ k ≤ |G| -1 be an integer. Then G has a spanning tree with at least k colors if and only if for any color subset S ⊆ c(e) : e ∈ E(G), it holds that

w(G -E S ) ≤ n -k + |S|,
where E S is the set of edges with colors in S, and w(G -E S ) denotes the number of components in G -E S .

Akbari and Alipour [START_REF] Akbari | Multicolored trees in complete graphs[END_REF] gave another necessary and sufficient condition for the existence of a heterochromatic spanning tree in an edge-colored connected graph.

Theorem 3.3.26. [START_REF] Akbari | Multicolored trees in complete graphs[END_REF] An edge-colored connected graph G has a heterochromatic spanning tree if and only if for every partition of V (G) into t parts, where 1 ≤ t ≤ |V (G)|, there exist at least t -1 edges with distinct colors that join different partition sets. [START_REF] Broersma | Spanning trees with many or few colors in edge-colored graphs[END_REF] obtained some results on the complexity problem of finding a spanning tree with as many different colors as possible, and of finding one with as few different colors as possible .

Broersma and Li

Theorem 3.3.27. [START_REF] Broersma | Spanning trees with many or few colors in edge-colored graphs[END_REF] Finding a spanning tree with as many different colors as possible is equivalent to finding a common independent set of maximum cardinality in two matroids, in particular, there is a polynomial time algorithm. Theorem 3.3.28. [START_REF] Broersma | Spanning trees with many or few colors in edge-colored graphs[END_REF] Finding a spanning tree with as few different colors as possible is NPhard.

Later, Chang and Leu [START_REF] Chang | The minimum labeling spanning trees[END_REF] showed that finding a spanning tree with as few different colors as possible is NP-Complete even for the complete graphs by using set cover problem, which implies that there is no constant factor approximation algorithm unless P = NP.

Proper trees and weak proper trees in edge colored graphs

Abouelaoualim, Manoussakis et al. [START_REF] Abouelaoualim | Exploration des graphs aretes-colorees : topologie, algorithmes, complixite et (non)-approximabilite[END_REF] defined for the first time a proper spanning tree (PST) and the PST problem. They gave some results on that using the m-tree-cycle-proper-edge-colored subgraph. Using this definition they prove the following lemma :

Definition 3.3.29. Let G c be a c-edge-colored graph, a m-tree-cycle-proper-edge-colored subgraph of C c
Lemma 3.3.30. Let K c
n be an c-edge-colored complete graph, T a proper-colored tree and C a proper colored cycle, such that T and C a vertex disjoint and

V (k c n ) = V (T ) ∪ V (C). If at least one of two following conditions holds, then K c n has a PST. (i) It exist x ∈ V (C) and t, t ′ ∈ V (T ) such that tt ′ ∈ E(T ) is an edge of T and (c(tx) = c(tt ′ ),
where t is a leaf of T) or (c(t

′ x) = c(tt ′ ), t ′ is a leaf of T). (ii) It exist x ∈ V (C) and t ∈ V (T ) such that c(tx) = c(tt ′ ), ∀t ′ ∈ N T (t). (iii) It exist t, t ′ ∈ V (T ) such that tt ′ ∈ E(T ) is an edge of T and c(tC) = c(tt ′ ) = c(t ′ C).
They also gave an algorithm that uses the fact of existence of a tree and of a cycle that spans the graph, and proved the following lemma. Lemma 3.3.31. Let K c n be an c-edge-colored complete graph, T a proper-edge-colored tree and C a proper edge colored cycle, such that C and T are vertex disjoint and

V (k c n ) = V (T ) ∪ V (C). Then K c
n has a PST, and its can be found in

O(|V (T )||V (C)|) time.
The following theorem is a generalization of the lemma.

Theorem 3.3.32. Let K c n be a c-edge-colored complete graph, T a proper-edge-colored tree and

C 1 , C 2 • • • , C k proper edge colored cycles , such that C 1 , C 2 • • • , C k and T are pairwise vertex disjoint and V (K c n ) = V (T ) ∪ V (C 1 ) ∪ V (C 2 ) • • • ∪ V (C k ). Then K c n has a PST and it can be found in O((k -1) 2 s 2 ) where s = max{V (T ), V (C i ) i : 1, • • • , k}.
Theorem 3.3.33. Let p be an even integer, and G c a c-edge-colored graph, where c

= n p . If the monochromatic subgraph G i in color ∀i ∈ {1, 2, • • • , c} is p 2 -edge -connected, then G c has a PST.

Conclusion

As we have seen in this chapter, there is a lot of results concerning Hamiltonian paths, Hamiltonian cycles, heterochromatic subgraphs and trees. The authors give necessary and sufficient conditions for the existence of proper Hamiltonian paths, cycles, heterochromatic spanning trees in an edge-colored connected graph. The proper spanning tree and the weak spanning tree problems have first been defined by A. Abouelaoualim, Y. Manoussakis et al. [START_REF] Abouelaoualim | Exploration des graphs aretes-colorees : topologie, algorithmes, complixite et (non)-approximabilite[END_REF]. They provided some results by using the m-tree-cycle-proper-edge-colored subgraphs. However, both problems are relatively new and have not been much investigated. Therefore, in this thesis our aim is to contribute to the study of these problems, and we begin with the maximum proper and weak proper trees in edge-colored graphs, in the next chapter.

Chapitre 4

Algorithmics aspects of PST and WST problems In this chapter we introduce the optimization version of the two problems PST and WST1 . Also in this chapter are studied the maximum proper colored tree (MPT) and the maximum weak-proper tree (MWT) problems, in a general edge-colored graph.

NP-completeness and non-approximability results

Definition 4.1.1. Let G c be a c -edge -colored graph. T is a proper spanning tree of G c if and only if V (T ) = V (G c ) and any two adjacent edges in E(T ) have different colors. Definition 4.1.2. Let G c be a c -edge -colored graph. T is a weak proper spanning tree of G c if and only if T is rooted, V (T ) = V (G c
) and all root-to-leaf paths are proper colored.

Problem :

maximum proper colored tree problem (MPT) Input :

Edge-colored graph and a constant k ∈ N.

Question :

Is there a proper tree of order k ?

Problem : maximum weak-proper tree problem (MWT) Input :

Edge-colored graph and a constant k ∈ N.

Question :

Is there a weak-proper tree of order k ?

Consider initially a 2-edge colored graph G c on n vertices. It is evident that a PST in this case corresponds to Hamiltonian path. But the maximum proper tree (MPT) corresponds to the Longest path in the graph. The PST and MPT problems obviously belong to NP. As you can see for 2-edge colored graph both problems are NP-Complete. Now consider a c-edge colored graph G c on n vertices. If the number of colors c is constant, we proved that the MPT problem is NP-Complete.

The next result shows that the proper tree problem remains NP-hard even for graphs with c = Ω(n 2 ) colors. The proof is based in the example in Démonstration. Let G c be an instance of the mpt problem with n nodes and c colors. Construct a complete graph K c ′ n with n nodes and color each of its edges with a different color. Add new edges between some fixed vertex of K c ′ n and all vertices of G c and give them all the same color different from those used on G c and K c ′ n . Clearly, this new graph has c = Ω(n 2 ) colors and contains a proper tree on n + t nodes if and only if G c has one on t nodes.

Before giving some nonapproximability results on MPT Problem we will present the proof of the NP-Completness of the mwt problem.

Theorem 4.1.4. Given a vertex r in G c , c ≥ 2, the wst problem rooted at r is NP-complete.
Démonstration. The wst problem obviously belongs to NP. To show that wst is NP-hard we construct a reduction from the 3-SAT problem as follows. Consider a boolean expression B in conjunctive normal form (CNF) with variables x 1 , . . . , x s and clauses c 1 , . . . , c t . In addition, suppose that B constains exactly 3 literals per clause (actually, we may also consider clauses of arbitrary size). We show how to construct a 2-edge-colored graph G c such that, B is true if and only if G contains a wst with root r.

The vertex set V (G c ) consists of 2s + t + 1 nodes and is defined as :

V = {r, a 11 , a 12 , a 21 , a 22 , . . . , a s1 , a s2 , c ′ 1 , . . . , c ′ t }.
The vertex r is the root, vertices a i1 , a i2 for i = 1, . . . , s are associated respectively with variables x i of B, and all vertices c ′ j (for j = 1, . . . , t) are associated with the clauses c j of B. The edge set E(G c ) is constructed in the following way. All edges between the root r and the vertices a i1 , a i2 , for each i ∈ {1, . . . , s}, are added with color blue. Each pair a i1 , a i2 , for each i ∈ {1, . . . , s}, is connected by a red edge. For each occurrence of x i in the positive form in the clause c j we add a blue edge a i2 c ′ j . Analogously, for each ocurrence of x i in the negative form in the clause c j we add a red edge a i2 c ′ j . See the example of Figure 4.2 (a). Therefore given a truth assignment for B, we obtain a Weak Spanning Tree T in G c as follows. For each variable x i which is true, we select edges ra i1 , a i1 a i2 and all blue incident to a i2 . Similarly, for each variable x i which is false, we select edges ra i2 and all red edges incident to a i2 .

Conversely, if T is a Weak Spanning Tree, an assigment for all variables of B is obtained as follows. If the last edge on the path from the root r to some c ′ j , namely a i2 c ′ j is colored blue, then the corresponding variable x i is set to true, otherwise it is set to false.

In the rest of this section, we present nonapproximability results for the mwt and mpt problems. Recall that in the mwt, the objective is to maximize the number of nodes covered by a tree T with root r. Initially, consider the following auxiliary result relating the maximum number of covered vertices Opt(G) and the maximum number of satisfied clauses Opt(B). Démonstration. Consider the construction described in the proof of the previous theorem, correlating a 3-CNF-formula and a corresponding graph. As in that proof, for every assignment of values to the variables of the formula, we have a weak tree rooted at r, covering all the a 11 , a 12 , . . . , a s1 , a s2 and all c ′ j associated with the set of satisfied clauses of B. Conversely, given any weak tree rooted at r, we obtain an assignment of values to the variables in B, such that we satisfy at least as many clauses as the number of paths from r to c ′ j . It follows that Opt(G) = Opt(B) + 2s + 1. Now lets look to the following nonapproximabilty results for mwt problem. To prove the nonapproximability bounds we are using a classical negative result for the Max-3-Sat problem. As proved in Hastad [START_REF] Håstad | Some optimal inapproximability results[END_REF], the Max-3-Sat cannot be approximated within 7/8 + ǫ, unless P=NP. Theorem 4.1.6. The mwt problem is nonapproximable within 63/64 + ǫ, for ǫ > 0, unless P = N P . Démonstration. Again, consider a boolean expression B with s variables and t clauses. In addition, suppose that B constains exactly 3 literals per clause. Then, by the gap reduction technique we prove that :

1) if Opt(B) ≥ t then Opt(G) ≥ f (s, t), where f (s, t) = 2s + t + 1 and, 2) if Opt(B) < (7/8 + ǫ)t then Opt(G) < (63/64 + ǫ)f (s, t), for ǫ > 0.
The first condition follows directly since Opt(G) = Opt(B) + 2s + 1 by Lemma 4.1.5. Thus, consider Opt(B) < (7/8 + ǫ)t. From this inequality, and Lemma 4.1.5, it follows that Opt(G) < (7/8 + ǫ)t

+ 2s + 1 = (7/8 + ǫ)t + f (s, t) -t. Therefore Opt(G) < (ǫ -1/8)t + f (s, t).
Now, from the definition of the 3-SAT problem it follows that 3 ≤ s ≤ 3t. Therefore,

f (s, t) ≤ 2s + t + s/3 = 7s/3 + t ≤ 8t. Thus, for 0 < ǫ < 1/8, it follows that Opt(G) < (ǫ -1/8)f (s, t)/8 + f (s, t) = (ǫ/8 -1/64 + 1)f (s, t). Finally, Opt(G) < (63/64 + ǫ ′ )f (s, t) where ǫ ′ = ǫ/8.
Now, we deal with the mst problem. Démonstration. We construct a tree associated with any formula in 3-CNF as follows. We have three vertices y i , a i1 , a i2 , for each variable x i , 1 ≤ i ≤ s, in the formula. We also have vertices c ′ 1 , . . . , c ′ t corresponding to each of the t clauses in the formula. We have a red edge between a i1 and c ′ i if variable x j occurs in positive form in the clause C j . In case the variable x i occurs in the negative form in clause C j , we put a red edge between a i2 and c ′ j . We have blue edges between y i and a i1 and y i and a i2 , for each i ∈ {1, . . . , s}. The vertex pairs y i and y i+1 are connected each by different colors other than red and blue for each i ∈ {1, . . . , s -1}.

It can be easily deduced by looking at Figure 4.2 (b), that for any i ∈ {1, . . . , s}, at most two of y i , a i1 , a i2 can be covered by a proper tree. The vertices corresponding to the satisfiied clauses can always be covered. Thus, clearly Opt(G) ≥ Opt(B) + 2s.

Now suppose that we are able to also cover a vertex corresponding to an unsatisfied clause. This implies that the vertex, say c ′ j is connected to some a i1 or a i2 , only one of which is present in the tree. Also, every satisfied clause is covered, and this is independent of the assignment made to the variable x i . This means we can satisfy an extra clause contradicting the optimality of the assignment. Thus Opt(G) ≤ Opt(B) + 2s. Theorem 4.1.8. The mst problem is non approximable within 55/56 + ǫ, for ǫ > 0, unless P = N P . Démonstration. Consider a boolean expression B with s variables, t clauses and exactly 3 literals per clause. We want to show that :

1) if Opt(B) ≥ t, then Opt(G) ≥ f (s, t), where f (s, t) = 2s + t and, 2) if Opt(B) < (7/8 + ǫ)t, then Opt(G) < (53/54 + ǫ)f (s, t), for ǫ > 0.
Case (1) follows immediately from Lemma 4.1.7. For case (2), we again apply the gap reduction technique using the Max-3-Sat problem. Clearly, Opt(G) = Opt(B) + 2s < (7/8 + ǫ)t + 2s = (7/8 + ǫ)t + (f (s, t)t) = (ǫ -1/8)t + f (s, t). Now, from the definition of the 3-SAT problem it follows that 3 ≤ s ≤ 3t. Therefore, f (s, t) ≤

2s + t ≤ 6t + t = 7t. Thus, for 0 < ǫ < 1/8, it follows that Opt(G) ≤ (ǫ -1/8)f (s, t)/7 + f (s, t) = (ǫ/7 -1/56)f (s, t) + f (s, t). Finally, Opt(G) < (55/56 + ǫ ′ )f (s, t) where ǫ ′ = ǫ/7.
It is known that it exist a polynomial algorithm to find Hamiltonian path in 2 edge-colored complete graphs [START_REF] Benkouar | Hamiltonian problems in edge-colored complete graphs and Eulerian cycles in edge-colored graphs : some complexity results[END_REF]. Looking for PST Problem for complete edge-colored graphs with at least 3 colors, we have the following result. Observe, that all the new vertices, being monochromatic, are necessarily leaves in any pst. Additionally, no two of them may be adjacent to the same vertex, since all edges incident to the entire set of new vertices are of the same color. Thus, it is necessary for the original graph G c to have a pst. It is also a sufficient condition, since a pst of G c does not use any edges of the new color, and hence the set of new vertices can be attached as leaves to such a tree to get a pst of H c+1 .

Since the pst problem is NP-complete for arbitrary graphs colored with two or more colors, it is also NP-complete for complete graphs colored with three or more colors.

Conclusion

In this section we has proved that the problem of determining whether an edge-colored graph has a PST or a WST is NP-Complete for general edge-colored graphs. We proved that the proper tree problem remains NP-hard even for graphs with c = Ω(n 2 ) colors. We proved that the proper tree problem remains NP-complete for complete edge-colored graphs with at least 3 colors. It is evident that in the case of properly edge-colored graphs both problems are polinomialy solvable, since we can use Prim's or Kruskal's algorithm to solve this problems. In the rest of this thesis we study same problems in the case of special families of edge-colored graphs.

Maximum colored trees in acyclic edge-colored graphs

As mentioned and proved in an earlier section, the problem of determining whether an edgecolored graph has a pst is NP-Complete. This is also the case with the wst problem. In this section, we prove that these problems can be solved efficiently when we restrict our attention to the class of colored acyclic graphs. When there is no spanning tree, our algorithms can be directly adapted to find a tree of maximum cardinality.

While some ideas are common, the proofs and conditions for the pst and wst problems on acyclic edge-colored graphs, differ significantly. Consequently, we have divided the presentation into two seperate subsections, for the sake of clarity.

pst

We begin with a very simple theorem which shows, when a given acyclic graph has a pst. The algorithm which results from this proof has complexity O(cn 2.5 ). Subsequently, we present an alternative, more complicated, proof but with a much better algorithmic complexity of O(n 2.5 ). Since that proof is more involved, we divide it into two parts. We prove the result first for acyclic complete graphs and then generalize it to all acyclic graphs. In addition to the superior running time, in case the graph does not contain a pst this latter algorithm can be modified to produce a mpt. Finally, it is possible that the ideas used in the latter results can be adapted to develop approximation algorithms for the mpt problem on general graphs. Démonstration. We use, here, M i to denote a maximum matching in color i and T i to denote the edge set of color i in a pst. Any subgraph induced by c i=1 M i is clearly a proper one. Thus if

| c i=1 M i | ≥ n, then the subgraph contains a proper cycle, contradicting the acyclicity of G c . Now suppose | c i=1 M i | < (n -1). Clearly, if G c has a pst then for some color i, |M i | < |T i |, since | c i=1 M i | < (n -1) = | c i=1 T i |.
This is a contradiction of the maximality of M i .

We now provide alternative results specifying conditions under which an acyclic graph has a pst. We also develop an efficient algorithm to construct one if it exists. If the given graph does not have an pst, then our algorithm can be adapted in a straightforward manner to produce a proper spanning forest (psf) with the smallest possible number of components (trees). This adaptation is just a greedy procedure, which finds a tree of largest size, and then repeats the procedure on the subgraph induced by the residual vertices. We first prove all these results for acyclic edge-colored complete graphs and then extend them for general acyclic edge-colored graphs.

Let G 1 , . . . , G p , p ≥ 1 be the components of G c \ {u}. By Theorem 3.1.9 if G c is acyclic, for some u, the edges between u and G i are monochromatic, for all i ∈ {1, . . . , p}. We call such a vertex u, a yeo-vertex. We refer to a yeo-vertex which uses different colors for the edges to different components as a rainbow-yeo-vertex. We call any other yeo-vertex non-rainbow-yeo, if the need arises to distinguish them.

We first prove an elementary result on colored acyclic graphs, which we use in Section 4.3.

Lemma 4.2.2. For any colored acyclic graph G c , c ≤ n -1.

Démonstration. Assume, G c is an acyclic colored graph with c ≥ n. Consider a subgraph induced by a set of n edges of distinct colors. Since, this subgraph has at most n vertices, the presence of n edges implies a cycle. It is a properly colored cycle as the set of selected edges are all of distinct colors. This contradicts the assumption that G c is acyclic.

Based on Theorem 3.1.9, we state below an easy lemma. Démonstration. This is because there are at least two components in the graph obtained by deleting this vertex which must be connected to each other in a potential pst by a path of length two through this vertex. These edges are necessarily of the same color and hence the resulting spanning tree is not a proper one.

In Theorem 4.2.5 we characterize acyclic edge-colored complete graphs having proper spanning trees. With a view to proving that result, we first give a structural characterization of acyclic edge colored complete graphs. Proposition 4.2.4. An acyclic edge-colored complete graph consists of a unique sequence of induced cliques (called blocks) B 1 , . . . , B k such that : i) each B i is a maximal monochromatic clique in the subgraph induced by k j=i B j , ii) all edges between each B i and k j=i+1 B j are monochromatic in the same color as the edges of B i .

Démonstration. For a complete graph, the induced subgraph obtained by deleting any vertex has exactly one component, because it is again a complete graph with one fewer vertex. Thus for any such graph which is acyclic, a yeo-vertex is necessarily a rainbow-yeo-vertex. In other words it is a monochromatic vertex (all edges incident to it are of the same color). Thus if there is more than one yeo-vertex in the given graph, then they are monochromatic in the same color, since they also have a direct edge between themselves (as the graph is complete).

Thus, the entire set of yeo-vertices of an acyclic edge-colored complete graph induce a monochromatic clique, and additionally, they are all connected to every other vertex in the graph by edges of this same color.

Note that if we remove this entire clique of vertices, the resultant smaller graph is also an acyclic edge-colored complete graph on fewer vertices. Thus, it follows, that there exists a similar clique (of a color distinct from the one just removed) in the remainder of the graph. We delete this clique and place the constituent vertices in a second group. We repeat this procedure of collecting vertices in this type of groups and obtaining a smaller graph until all the vertices have been removed. The partial order we use in finding a pst, is precisely the order determined by these groupings.

We now state a theorem, which given an acyclic edge-colored complete graph, determines whether or not it contains a pst. The decision is made on the basis of the cardinalities of the blocks B 1 , . . . , B k in the ordering described in Proposition 4.2.4. To facilitate the description of our results we let the color of the ith block be denoted c i . For any fixed i, 1 ≤ i ≤ k, we let t i denote the total number of vertices in the set of blocks from

B i to B k , i.e., t i = |V (B i )| + . . . + |V (B k )|. We define S i = {B j |i ≤ j ≤ k; c j = c i }. Define S i ′ = x ∈ S i . Now
t c i i ≤ t i 2 + 2. Else t c i i ≤ t i 2 .
Démonstration. First assume that the given graph has a pst, and let T represent any such possible pst. Notice that in the partial ordering of the vertices according to Proposition 4.2.4, the group of vertices in block B i are necessarily leaves in the subtree induced by the vertices of k j=i B j . Moreover, each of these vertices are attached to distinct internal vertices of the subtree (since all edges incident to these vertices in the subgraph induced by k j=i B j are of the same color). A necessary condition is, thus, that there is a proper tree T i spanning k j=i B j for each i ∈ {1, . . . , k}.

Thus, in order to add the vertices of B i+1 as leaves to the subtree T i , to construct T i+1 , there must be at least |B i+1 | vertices not having any edge of color c i incident to them in T i . However, all vertices, in blocks i, . . . , k whose color is c i+1 necessarily use an edge of this color in the tree. Additionally, each of them (except if they are in the last block B k ), are also appended to some other vertex in the tree T i , with color c i . The vertices of B i+1 must therefore be attached as leaves to |B i+1 | distinct vertices different from the vertices accounted for above. This proves the second condition of the theorem.

The first condition states that the last block B k must be of size exactly two. From the earlier arguments, these vertices must contain a proper tree spanning them. Since they induce a monochromatic clique, it is possible only if there are exactly two vertices.

Conversely, if the conditions are on the cardinalities are satisfied, we show that a pst exists. We construct a partial tree consisting of the edge between the two vertices of the last block B k . Subsequently, we consider in order the vertices of the blocks B k-1 , . . . , B 1 and at each stage, we pick a set of vertices of size |B i | in the partial tree T i-1 , which are free of the color c i and attach the vertices of B i as leaves to distinct vertices in this set, to augment the tree to T i . Now based on Theorem 4.2.5 we describe, an algorithm which computes a pst for any acyclic edge-colored complete graph, if one exists. If one does not exist then the algorithm can easily be adapted to produce a psf with the minimum possible number of components.

We now describe our algorithm to construct the pst.

The running time of the above algorithm is O(n 2 ). This is the cost using Breadth-First-Search (BFS), to compute the order in Step 1. The rest of the algorithm consists in finding for each new block of vertices, a certain type of matching saturating them, which enables them to be attached as leaves of the partial tree. If there is no pst, our algorithm finds a maximum size proper tree, and then repeats the procedure on the subgraph induced by the residual vertices, resulting in an ssf with the fewest possible number of trees.

We now show how Theorem 4.2.5 as well as the algorithm can be extended to find a pst in a general acyclic graph, if one exists. We define a canonical auxiliary tree associated with any acyclic edge-colored graph. It is similar to the linear order for acyclic edge-colored complete graphs. We assume that the graph has no non-rainbow-yeo-vertex. The root of the auxiliary tree is any yeo-vertex of the graph. The number of children of the root is the number of components resulting from the deletion of this vertex from the graph. The root of the auxiliary tree is connected to each subtree using an edge of the same color as the edge between the original Algorithm 1 pst for K c n -acyclic 1: compute the order described above 2: if last block B k has more than two vertices then return "No pst" 3: if last block B k has two vertices then connect the two vertices of B k to get an initial Proper Tree 4:

for i = k -1 to 1 do 5:
if condition 2 of Lemma 4.2.5 is true then 6: join the vertices of B i as leaves, to distinct vertices already incorporated in the tree which have not used an edge of color c i in the partial proper tree obtained in the previous iteration. First we need to define some associated auxiliary graphs, which we use to characterize acyclic edge-colored graphs that have a pst. Definition 4.2.6 (auxiliary tree). Given an acyclic edge-colored graph G c , we define an associated auxiliary tree T as follows. i) n(T ) = n(G c ).

ii) The root r(T ) is associated with a yeo-vertex v 0 of G c . iii) There is one subtree corresponding to each component G c \ {v 0 }. iv) The roots of these subtrees are computed recursively and are attached as children of v 0 . v) The color of the edge between two nodes of T is the same as the color of the edges between the corresponding component and yeo-vertex in G c . vi) Note that the presence of an edge between two nodes in T does not imply the presence of an edge between the corresponding vertices in G c . We now define a set of bipartite graphs one for each color in {1, . . . , c}. Definition 4.2.7. Associated with each color l ∈ {1, . . . , c} we have a bipartite graph as follows. i)If a node in the auxiliary tree (of Definition 4.2.6) corresponding to a vertex has a child of color l then it is placed in the left part of the bipartite graph corresponding to color l. ii) Any other vertex with an edge of color l incident to it in the graph G c is placed in the right part.

iii) The edges are all those in the original graph of the color l, crossing this partition. Theorem 4.2.8. An acyclic edge-colored graph has a pst iff for each color in {1, . . . , c} the corresponding bipartite graph has a matching saturating the vertices of the left partite set. Démonstration. By Lemma 4.2.3, if there exists a non-rainbow-yeo-vertex, then we immediately conclude that the given graph has no pst. It follows that we need only consider the case where all the yeo-vertices are of the rainbow type. Assume that v 0 is one such vertex. Let the components of G c \ {v 0 } be C 1 , . . . , C t . It is straightforward to verify that G c has a pst iff the subgraph of G c induced by the vertex set V (C i ) ∪ {v 0 } has a pst for each i ∈ {1, . . . , t}.

We are in effect able to use a rainbow-yeo-vertex to divide the problem into smaller and independent subproblems. This immediately suggests a recursive appproach to solving the problem. We use the recursion tree of Definition 4.2.6 and the corresponding bipartite graphs of Definition 4.2.7 to divide the problem.

In fact, if any C i does not have a pst then G c also does not have a pst. We conclude that G c has a pst iff each C i has a pst which can be extended to include the vertex v 0 . The connection of component C i to v 0 must be by an edge of color i. The feasibility of this is checked using cardinalities, like in the case of complete acyclic graphs. We do not have a simple linear structure here, unlike in that case, and hence check the condidions using matching in the auxiliary bipartite graph instead. The existence and computation of the matching can be done using any of the standard algorithms. The subtrees rooted at the children of v 0 are computed recursively in the same way.

Thus, using the recursion tree obtained, we construct the bipartite graphs described above and then solve the pst problem by transforming it to a series of matching problems.

Suppose the graph G c does not have a pst. This either means that there is no rainbow yeo-vertex, or the cardinality conditions fail.

In the former case, let us denote the components of G c \ v 0 , by C 1 , . . . , C t . We find the maximum proper trees for each C i . We also compute the maximum proper trees for each G i ∪ v 0 , which use touch vertex v 0 . From the latter set of trees, we pick the ones of largest cardinality corresponding to distinct colors of the edge incident to v 0 and take their union. A maximum proper tree is the largest among the first set of trees and the tree obtained by combining the second set as described above.

The second case is merely a special case of the previous one, wherein, when we compute the tree combining the second sequence, we consider all the components.

The next theorem follows easily from the preceeding analysis and results.

Theorem 4.2.9. Given an acyclic edge-colored graph G c , a pst, if any, can be found in G c in time O(n 2.5 ).

Démonstration. This cost is dominated by the time to compute matchings in graphs bipartite graphs, which we use in the algorithm for spt.

wst

We now show how to construct a wst, if one exists, in an acyclic edge-colored graph G c , with a given root vertex r. Like we did for the case of the pst, we define here as well, an associated auxiliary tree. Definition 4.2.10 (auxiliary tree). Given an acyclic edge-colored graph G c , and a specified vertex r, we define an associated auxiliary tree T ′ as follows.

i) The tree T is computed according to the Definition 4.2.6.

ii) The tree T ′ is then obtained by re-rooting T at the node R, corresponding to the vertex r in G c . Theorem 4.2.11. An acyclic edge-colored graph G c , has a wst iff in the auxiliary tree T ′ every path from the root R to any leaf is properly colored. Démonstration. The above auxiliary tree T ′ , obtained by re-rooting at the vertex corresponding to the specified root of the wst provides a direct way to compute a wst if one exists. A wst rooted at r is computed for each subgraph induced by {r} ∪ C i , i ∈ {1, . . . , t}. Here, C 1 , . . . , C t are the components of G c \{r}. These trees are then merely fused together at their only common vertex r, to get a wst. Such a tree is a weak one rooted at r, because in the auxiliary tree any path from R to any leaf is a properly colored one. Theorem 4.2.12. Given an acyclic edge-colored graph G c , a wst, if any, can be found in G c in time O(n 2.5 ).

Démonstration. It is almost identical to the proof of Theorem 4.2.9.

Proper spanning forests in acyclic graphs

In this section, we show how to find in polynomial time, a maximum proper spanning forest (mpsf) in G c , the maximality being in terms of the number of edges. If the graph has a pst, then naturally, one is produced by our algorithm for maximum forest. This section differs from the previous one, for graphs which do not contain a pst. In the previous section, the algorithm we describe produces a largest possible tree, whereas, here the algorithm produces a forest with, possibly, many trees, in such a way that the total number of edges is maximized. We conclude the section by showing how to decide, in polynomial time, whether G c contains a proper spanning forest (psf), satisfying given degree constraints on each vertex. We also show how to construct the forest if one exists in this latter case.

Basically, for the mpsf problem, the idea is to construct a new colored multigraph G c ′ , with color set χ c ′ = χ c ∪ {0} (where 0 is a new color) and an associated non-colored weighted graph G, which always contains a perfect matching. The multigraph G c ′ , has multiplicity at most two. Moreover, the color of two edges between the same pair of vertices always differ. If there are two edges between a pair of vertices, then one of them is always colored 0. We prove that a maximum weight perfect matching in G, whose construction is described below, is associated with a set of properly edge-colored closed trails in G c ′ , such that the number of edges with a color from χ c is G c ′ is constructed by taking the union of G c with a complete monochromatic graph (in color 0) on the same vertex set. Let K 0 n denote the resulting monochromatic complete subgraph of G c ′ with the edges of color 0. Thus, we have

V (G c ′ ) = V (G c ) = V (K 0 n ) and E(G c ′ ) = E(G c )∪E(K 0 n ).
Before we describe the construction of G, we first define gadgets G i associated with each vertex v i of G c ′ , as depicted in the sequel (see Figure 4.5). We use these gadgets in our construction of G. Formally :

-V (G i ) = ( γ∈χ c ′ {v i,γ , v ′ i,γ : N γ G c ′ (v i ) = ∅}) ∪ ( α,β∈χ c ′ {p i α,β , q i α,β : α < β, N α G c ′ (v i ) = ∅ and N β G c ′ (v i ) = ∅}) -E(G i ) = ( γ∈χ c ′ {v i,γ v ′ i,γ : N γ G c ′ (v i ) = ∅}) ∪ ( α,β∈χ c ′ {v ′ i,α p i α,β , p i α,β q i α,β , q i α,β v ′ i,β : α < β}) Now, the weighted non-colored graph G = (V ′ , E ′ ) with w : E ′ → {0, 1} is constructed as follows : -V ′ = v i ∈V (G c ′ ) V (G i ), and -E ′ = ( v i ∈V (G c ′ ) E(G i )) ∪ ( γ∈χ c ′ {v i,γ v j,γ : v i v j ∈ E γ (G c ′ )}) -w(v i,γ v j,γ ) = 1, for every v i,γ v j,γ ∈ E ′ with γ ∈ χ c .
The remaining edge weights of E ′ will be settled to 0, i.e., w(v i,0 v j,0 ) = 0 for every v i,0 v j,0 ∈ E ′ and w(xy) = 0, for every xy ∈ E(G i ).

After constructing G c ′ and G as above, we solve the maximum perfect matching problem over G (see [START_REF] Even | On O(N 2.5 ) algorithm for maximum matching in general graphs[END_REF]). Observe that graph G always contains a perfect matching. To see that, it suffices to exhibit a perfect matching (with cost 0) by only choosing edges of E(G i ). Thus, if M i denotes a perfect matching in G i (which is unique in this case), the subset M = v i ∈V (G c ′ ) M i obviously defines a perfect matching in G. Therefore, we can establish the following result : Démonstration. Basically, the idea is to prove that maximum proper spanning forests in G c are associated with maximum perfect matchings in G, and vice-versa. Initially, suppose we have a proper spanning forest T * in G c with the maximum number of edges. Let T * 1 , T * 2 , . . . , T * k (for k ≥ 1) be the subtrees of T * . Note that, for every subtree T * i of T * , the number of vertices of T * i with odd degree is even. As a consequence of that, the total number of vertices with odd degrees in T * is also even. Let P T * be this subset of vertices. Now, consider the multigraph G c ′ as above, obtained after the addition of edges with color 0 to G c . Further, consider H 0 (with all edges colored 0) the complete subgraph of G c ′ induced by the vertices of P T * . Let M 0 ⊆ E(H 0 ) be an arbitrary perfect matching of H 0 and

H c ′ = (V , E) with V = V (G c ′ ) and E = M 0 ∪ E(T *
), the associated subgraph. Note that subgraph H c ′ is not necessarily connected and all edges incident to v ∈ V have a different color. Further, notice that all vertices of H c ′ have an even degree and each connected component, say CT i (for 1 ≤ i ≤ k ′ and k ′ ≤ k), contains a properly edge-colored closed trail, i.e., each CT i defines an eulerian trail. Now, given E, we construct a perfect matching M * in G = (V ′ , E ′ ) as described in the sequel. Initially, we set M * = ∅ and add to M * all edges v i,0 v j,0 of E ′ with v i v j ∈ M 0 . Now, we increase M * by choosing all edges v i,γ v j,γ of E ′ with v i v j ∈ E γ (T * ) and γ ∈ χ c . The remaining edges in the gadgets G i (with cost 0) are now directly obtained. Note that c(M * ) = |E(T * )|. Finally, we prove that M * is a maximum perfect matching in G. Suppose, by contradiction, we have some new perfect matching M ′ with cost c(M ′ ) > c(M * ) and an associated proper edge-colored subgraph H ′ of G c with c(H ′ ) > |E(T * )|. In this case, there are two possibilities : a) If H ′ defines a new proper spanning forest, this contradicts the fact that T * is a maximum proper forest in G c ; b) If H ′ does not define a proper spanning forest, we would have some properly edge-colored cycle in H ′ , contradicting the fact that G c is acyclic.

Conversely, consider the weighted graph G = (V ′ , E ′ ) associated to G c as above, and M * a maximum perfect matching with cost c(M * ) in G. Let M = (E ′ \ E(G i )) ∩ M * be a subset of M * . Now, to obtain G c ′ from G it suffices to color edges v i,γ v j,γ of G with γ ∈ χ c ′ and contract all gadgets G i to vertex v i . Note that all edges of G c ′ associated to M define a subset of proper spanning closed trails in G c (denoted by CT 1 , . . . , CT k ′ ) with the maximum number of edges with colors in χ c . Hence, since G c is acyclic, all properly edge-colored cycles in CT i (for i ∈ {1, . . . , k ′ }) contain at least one edge-colored 0. After deleting all these edges, one directly obtains a proper spanning forest T * = k ℓ=1 T ℓ in G c with cost c(T * ) = c(M * ) and k ≥ k ′ . Finally, note that T * contains a maximum number of edges since no edges with unitary costs in G were eliminated in the process. Now, we show that the algorithm above has a polynomial time complexity in the order of G c . Initially, observe that each gadget G i of G (with c ≥ 2) contains at most c(c + 1) vertices. From Lemma 4.2.2, we know that c = Ø(n). Thus, the non-colored graph G contains Θ(n 3 ) vertices, in the worst-case. However, note that a maximum perfect matching in any graph G can be obtained in time O(|V (G)| 2.5 ) (see [START_REF] Even | On O(N 2.5 ) algorithm for maximum matching in general graphs[END_REF] for details). Therefore, the algorithm for the proper spanning forest has total complexity equal to O(n 7.5 ). Now, we conclude with the following result regarding proper spanning forests with given degrees.

≤ d(v i ) ≤ |{γ ∈ χ c : N γ G c (v i ) = ∅}|.
Then we can find in polynomial time, provided that one exists, a proper spanning forest in G c satisfying d.

Démonstration. Initially, given an acyclic edge-colored graph G c , we construct a non-colored graph G, as described in the sequel. For each v i ∈ V (G c ) with d(v i ) > 0 we define gadgets G i in the following manner :

• V (G i ) = ( γ∈χc {v i,γ , v ′ i,γ : N γ G c (v i ) = ∅}) ∪ {v i 1 , . . . , v i d(v i ) } • E(G i ) = ( γ∈χc {v i,γ v ′ i,γ : N γ G c (v i ) = ∅}) ∪ ( γ∈χc {v ′ i,γ v i j : j = 1, . . . , d(v i )} Above, if d(v i ) = 0 for some v i ∈ V (G c ), we just set V (G i ) = ( γ∈χc {v i,γ , v ′ i,γ : N γ G c (v i ) = ∅}) and E(G i ) = ( γ∈χc {v i,γ v ′ i,γ : N γ G c (v i ) = ∅}). Now, a non-colored graph G = (V ′ , E ′
) is constructed as follows :

• V ′ = v i ∈V (G c ) V (G i ),
and

• E ′ = ( v i ∈V (G c ) E(G i )) ∪ ( γ∈χc {v i,γ v j,γ : v i v j ∈ E γ (G c ))
We show that G c contains a proper spanning forest T satisfying the degree constraint d, if and only if, G = (V ′ , E ′ ) contains a perfect matching. Hence, our result follows since the perfect matching problem can be solved polynomial time.

Initially, consider M * , a perfect matching in G, if any. In this case, we can obtain a proper spanning forest H c ′ in G c (for c ′ ≤ c) in the following manner. Initially, let M = (E ′ \E(G i ))∩M * be a subset of M * . Now, we color all edges v i,γ v j,γ ∈ E ′ with color γ ∈ χ c and contract all gadgets G i to vertex v i . Finally, we construct H c ′ by choosing all edges of G c associated to M . Notice by the construction of G i , that we have exactly d(v i ) > 0 edges incident to v i in the resulting edgecolored subgraph H c ′ . Further, all edges incident to v i have a different color and v i is isolated if d(v i ) = 0. Therefore, since G c is an acyclic edge-colored graph, the subgraph H c ′ contains no properly colored cycles and defines a proper spanning forest in G c satisfying d.

Conversely, consider H c ′ a proper spanning forest in G c , and the graph G as above. Initially, set M * = Ø in G. Now, we obtain the associated matching M * in two steps :

(1) For every edge v i v j ∈ E γ (H c ′ ) with γ ∈ χ c we add edges v i,γ v j,γ to M * ;

(2) The remaining edges of M * present in the gadgets G i are now directly obtained.

It is easy to see that M * constructed as above defines a perfect matching in G.

Random graphs

Let c ≥ 3 be a fixed integer and let G(n, p) be the random graph on n vertices where each edge is present with independent probability p. We define G (c) (n, p) as the edge union of c independent copies of G(n, p c ) where for each j ∈ {1, . . . , c}, the edges of the j th copy are colored by j. n , where λ is a sufficiently large constant, then with probability tending to 1 as n tends to infinity, G (c) (n, p) contains a proper spanning tree.

Démonstration. Let G = G (c) (n, p) and let c 0 be a fixed color. We describe an algorithm to prospectively construct a proper spanning tree in our graph and prove that it works with probability close to 1. The algorithm proceeds in two stages.

(1) Construct a sequence of proper trees T 0 , T 1 , T 2 , ...T t . Here, T 0 is a star with an arbitrary root and with ν edges with colors pairwise distinct and different from c 0 . We define t and ν later. For each i ∈ {1, . . . , t}, T i is obtained from T i-1 by expanding a pendant vertex with edges of all the colours distinct from c 0 , and also distinct from the color of the edge incident to this vertex in T i-1 .

(2) From T t obtain a proper spanning tree by adding a pairing of the internal vertices of T t other than the root with the remaining vertices of G using edges of color c 0 .

This concludes the description of the algorithm. Note that T t has a total of t(c-2)+ν +1 vertices among which exactly t are internal vertices, distinct from the root. We want to match these t vertices with the remaining vertices of G, that is, we want

t = n -t(c -2) -ν -1 or, t = n -ν -1 c -1 ,
and this will be an integer for precisely one value of ν with 1 ≤ ν ≤ c -1. We fix this value for ν.

To prove the correctness of the first stage of the algorithm it suffices to prove that at each point in the construction we can find with probability 1o(1/n) an edge with any particular fixed colour linking a point v, say, already in the current tree and the external set which has size at least t. Now, when we look for such an edge of colour j, say, we have not looked previously at the edges of colour j linking v to the remaining vertices. Hence, the conditional distribution of these edges given the previous steps of the algorithm is the same as their unconditional distribution, namely, each is present with probability λ log n cn and they are independent. Therefore the probability that at least one is present is at least

1 -(1 - λ log n cn ) t ≥ 1 -exp(- λ log n c(c -1) ) = 1 -o(1/n) if λ ≥ c 2 .
We turn now to the second stage. Since we have not looked yet at the edges of colour c 0 , we just have to check that with high probability there exist in a random graph with edge probability p a pairing between to given disjoint sets with sizes t. (We have not looked before at the edges with color c.) This amounts to asking for a perfect matching in the random bipartite graph B(t, p). Now p = λ log n cn ≥ θ log n t with θ = 2λ 3c(c-1) . This graph is known to have, almost surely, a perfect matching for any fixed θ greater than 1/2 and thus for λ ≥ c(c -1)/2 (see Corollary 13, page 159 in [START_REF] Bollobás | Random graphs[END_REF]). Putting together this estimate with the bound already found, we infer that the theorem holds for λ ≥ c 2 . On the other direction, the theorem does not hold for λ ≤ c by connectivity considerations.

Conclusions

In this chapter we recapitulate the notion of various types of colored trees in edge-colored graphs. We obtain results reflecting the computational difficulties involved in their solution and provide efficient algorithms for the specific family of acyclic graphs. We give a mathematical characterization of complete graphs which contain a pst. We study the trends of random graphs with reference to the problem of pst.

We list here some possible future directions for research in this area.

(1) Algorithms to solve the pst and wst problems on other special classes of graphs like planar graphs or hypercubes.

(2) Approximation algorithms with performance guarantee of at least logarithm of the optimal solution.

(3) The minimum number of edges whose colors need to be changed in order to render the graph acyclic. This is closely related to the feedback arc set problem.

Chapitre 5

Sufficient conditions for PST In this chapter, we consider sufficient conditions that guarantee the existence of proper spanning trees in graphs whose edges are colored (not necessarily properly) with any number of colors. The conditions we impose are relationships between various parameters of colored graphs, such as total number of colors, number of vertices, connectivity and number of differently colored edges incident to a vertex. These special conditions are of interest because there are no good characterizations for the existence of such colored trees in general edge-colored graphs. We state some general results about rainbow spanning subgraphs using all colors in the edgecoloredgraph. After that, we give sufficient conditions on the total number of colors of the graph which guarantee the existence of proper spanning trees in G c . The number of colors depends on the connectivity. In next section, we obtain conditions on the number of colors and the rainbow degree of the graph for the existence of proper spanning trees. All the above results hold for graphs with a lower bound on the number of vertices, which are specified in the theorems'statements.

Some general results

Before explaining the main results of this section we present some general results for edgecolored connected graphs1 . We consider here a rainbow spanning subgraph using all colors in G c which gives us a tool to construct proper spanning trees. In the following lemmas we consider the largest component of such a rainbow spanning subgraph and study its order and connectivity. Démonstration. Consider a rainbow spanning subgraph H rs in G c using all the colors. We shall use H 0 to denote its largest component and H 1 to denote the subgraph induced on the rest of the vertices.

Let p ≥ 0 represent |V (H 1 )|. Thus, |V (H 0 )| = np. As the subgraph H rs is a rainbow using all the colors which are present in the graph, the number of edges in such a subgraph is equal to the number of colors in the original graph, whose lower bound is

(n -k -1) (n -k -2) 2 + k + 1.
Further, the number of edges in the subgraph must be at most

p (p -1) 2 + (n -p) (n -p -1) 2
, assuming all possible edges within the vertices of each part of the partition are present.

Combining these two facts, we get

p (p -1) 2 + (n -p) (n -p -1) 2 ≥ (n -k -1) (n -k -2) 2 + k + 1,
which reduces to the following inequality

2p 2 -2np + 2n + 2nk -k 2 -5k -4 ≥ 0
Solving this inequality on p, we get that p ∈ {0, . . . , k + 1} ∪ {nk -1, . . . , n} .

Moreover, since |V (H 1 )| = p, we have p ≤ k +1. Otherwise, let us suppose that p ≥ n-k -1, then the size of the largest component H 0 -and hence the size of each component in H rs -is at most k+1. In this case the total number of edges in the subgraph is at most

n k + 1 (k + 1) k 2 = nk 2 ,
which is strictly less than (nk -1) (nk -2) 2 + k + 1, a lower bound on the number of colors

for n ≥ k 2 + 3k + 4 2
. This proves the lemma.

Lemma 5.1.2. Let G c be a c-edge-colored connected graph such that |V (G c )| = n with n > k 2 + 3k + 4 2 and c ≥ (n -k -1) (n -k -2) 2 + k + 1.
Let us denote by H rs a rainbow spanning subgraph using all colors in G c and by H 0 its largest component, where

|H 0 | = n -p ≥ n -k -1.
Then the number of bridges in H 0 is less or equal to kp + 1.

Démonstration. Recall that the number of edges in H rs is exactly c, which is at least

(n -k -1)(n -k -2) 2 + k+1.
As in the previous lemma, H 1 denotes the subgraph induced on the vertices V (G c )-V (H 0 ), so |V (H 1 )| = p and the number of edges in H 1 is at most p(p -1) 2 . Suppose that the number of bridges in E (H 0 ) is kp + 2 or greater, thus the number of edges in H rs is less or equal to

(n -k -2)(n -k -3) 2 + k -p + 2 + p(p -1) 2
, corresponding to the case where the deletion of the kp + 2 bridges decomposes H 0 into a "big" component and kp + 2 singletons.

The following inequality holds :

(n -k -1)(n -k -2) 2 + k + 1 ≤ (n -k -2)(n -k -3) 2 + k -p + 2 + p(p -1) 2
This inequality reduces to

n -k -2 ≤ (p -1) (p -2) 2 ,
and since p ≤ k + 1, implies that

n -k -2 ≤ k(k -1) 2 which contradicts the hypothesis n > k 2 + 3k + 4 2 . Lemma 5.1.3. Let G c be a c-edge-colored connected graph such that |V (G c )| = n with n > k 2 + 3k + 4 2 and c ≥ (n -k -1) (n -k -2) 2 + k + 1.
Let us denote by H rs a rainbow spanning subgraph using all colors in G c and by H 0 its largest component. If the deletion of the bridges in H 0 disconnects H 0 into a component H 0 and m singletons, then H 0 is at least (km + 2)edge-connected.

Démonstration. Recall that the number of edges in H rs is exactly c, which is at least

(n -k -1)(n -k -2) 2 + k +1.
As in the previous results H 1 denotes the subgraph induced on the vertices V (G c )-V (H 0 ).

The number of edges in

H 1 is at most (k + 1)k 2 since |H 1 | ≤ k + 1 by Lemma 5.1.1.
We have m bridges in E (H 0 ) connecting m singletons to the "big" component H 0 , so that the number of edges in H 0 is at least

(n -k -1)(n -k -2) 2 + k + 1 - (k + 1)k 2 -m. Therefore at least (n -k -1)(n -k -2) 2 - k(k + 1)
2 edges remain after the removal of any km + 1 edges from that component, i.e., since

(n -k -1)(n -k -2) 2
precisely corresponds to the number of edges in a complete graph on nk -1 vertices, the former deletion results in a deficit of at most

(k + 1)k 2 edges. Since n > k 2 + 3k + 4 2 , we have k(k + 1) 2 = k 2 + 3k + 4 2 -(k + 2) < n -k -2
, the edge-connectivity of a complete graph on nk -1 vertices.

Minimum number of colors in terms of the connectivity

In this section we consider G c to be a c -edge-colored simple k-connected graph. Our main theorem of this section, Theorem 5.2.3, is stated at the end. It follows as a direct consequence of two propositions which consider different cases depending on the number of bridges that can appear in the largest component of a rainbow spanning subgraph using all colors in G c .

Proposition 5.2.1. Let G c be a c -edge-colored simple k-connected graph such that |V (G c )| = n with n ≥ k 2 + 3k + 4 2 and c ≥ (n -k -1) (n -k -2) 2 + k + 1.
Let H rs be a rainbow spanning subgraph using all colors in G c and H 0 its largest component, where |H 0 | = n-p and the number of bridges in H 0 is m ≤ kp. Then G c has a properly edge-colored spanning tree. Démonstration. By Lemma (5.1.1) H 0 has at least nk -1 vertices. In this case we set the parameter k to be the connectivity of the graph and, since the subgraph is rainbow, every spanning tree of its components is proper.

Let H 1 denote, again, the subgraph induced on the vertices

V (G c )-V (H 0 ). Since |V (H 1 )| = p and G c is k-connected, then |N H 0 (x)| ≥ k -p + 1 for all x ∈ H 1 .
Let B 0 be the set of vertices of H 0 that have neighbors both in H 0 and in H 1 and let B be the set of all vertices in B 0 is the following set :

x ∈ B 0 , such that E(x, H 0 ) = B(x, H 0 ) and c(E(x, H 0 )) = c(E(x, H 1 )) .
Let us now consider the bipartite graph

G := V (G c ) , E(G c ) -E(H 0 ) ∪ E(H 1 ) ∪ E B, H 1 .
Let us apply Hall's theorem (see [START_REF] Bondy | Graph theory[END_REF]) to the subgraph G. This subgraph contains a matching of size k crossing the cut (H 0 , H 1 ) if and only if |S| ≤ N G H 0 (S) for all subsets S of H 1 where N G H 0 (S) denotes the neighbors of S in H 0 joined by edges in E G , i.e., To see this, let us assume that such a matching does not exist. Then there is some subset S of H 1 such that N G H 0 (S) < |S| = s. Under this assumption, let us study the order of N G (S). This number is given by :

N G H 0 (S) = x ∈ H 0 such that xy ∈ E G for some y ∈ S .
|N G (S)| = |N H 0 (S)| + |N H 1 (S)| = N G H 0 (S) + N G-G H 0 (S) + |N H 1 (S)| < s + k -p + p -s = k, since N G-G H 0
(S) is bounded by the number of bridges in H 0 . This contradicts the fact that G is k-connected, since there is a subset S of G that can be disconnected from the rest of the graph by removing a number of vertices strictly smaller than k. Therefore we have proved that G contains a matching of H 0 into H 1 and therefore G contains a matching of size k crossing the cut (H 0 , H 1 ) where all the problems to attach the vertices in H 1 to H 0 have been avoided and the result holds.

Proposition 5.2.2. Let G c be a c -edge-colored simple k-connected graph such that |V (G c )| = n with n ≥ k 2 + 3k + 4 2 and c ≥ (n -k -1) (n -k -2) 2 + k + 1.
Let H rs be a rainbow spanning subgraph using all colors in G c and H 0 its largest component, where |H 0 | = n-p and the number of bridges in H 0 is m = kp + 1. Then G c has a properly edge-colored spanning tree.

Démonstration. We shall denote Y 0 = {y 1 , . . . , y m } the set of vertices in H 0 , such that for every i ∈ {1, . . . , m} there exists a unique x i ∈ N H 0 (y i ) and b i = x i y i is a bridge. The deletion of {b 1 , . . . , b m } divides H 0 in a total of m + 1 components where m of them are singletons.

Let us denote H

0 := (V (H 0 ) -Y 0 , E (H 0 ) -{b 1 , . . . , b m }). By Lemma (5.1.3) H 0 is at least (k -m + 2)-edge-connected.
Since the graph G c is k-connected, there is a matching of size k crossing the cut H 0 , Y 0 ∪ H 1 . Denote the edges of this matching by E M = {b 1 , . . . , b m , e m+1 , . . . , e k }. Let the endpoints of these edges in H 0 be respectively X = {x 1 , . . . , x k } and the corresponding ones in H 1 be Y 1 = {y m+1 , . . . , y k }. Let Y be Y 0 ∪ Y 1 and z be the unmatched vertex in H 1 , then

H 1 := (V (H 1 ) ∪ Y 0 , E (H 1 )) where V (H 1 ) ∪ Y 0 = Y ∪
{z}, and we shall distinguish two cases according to N (z).

Case 1 : N (z) X ∪ Y. Let z ′ ∈ N H 0 (z) -X.
In this case, we add zz ′ and b 1 , . . . , b m , e m+1 , . . . , e k to H rs deleting any edge in H rs incident at z ′ with color c(zz ′ ), and also any edge incident to e i with color c(e i ) for all i ∈ {m + 1, . . . , k}. Any such deletion on account of e i happens at exactly one of x i or y i , but not both. Observe that there is no edge in H 0 incident to b i for i ∈ {1, . . . , m} with color c(b i ) since b i was the unique edge in H rs colored in c(b i ). So this entire process leads to the deletion of at most km + 1 edges from H 0 , which thus remains connected.

All the vertices of H 1 have been connected to H 0 to get a connected spanning properly colored subgraph. Hence this subgraph, and also G c , has a proper spanning tree.

Case 2 : N(z) ⊆ X ∪ Y If there is a vertex x i 0 ∈ N X (z) such that c(z, x i 0 ) = c(b i 0 ) if i 0 ∈ {1, .
. . , m} (respectively, c(z, x i 0 ) = c(e i 0 ) if i 0 ∈ {m + 1, . . . , k}), then we simply add all the edges in E M and (z, x i 0 ), deleting any edge incident to x i of the same color for i ∈ {m + 1, . . . , k} (recall again that there is no edge in H 0 incident to b i 0 for i ∈ {1, . . . , m} with color c(b i ) since b i 0 was the unique edge in H rs colored in c(b i )). Similarly, if there is an edge of the form (z, y i 0 ) with c(z, y i 0 ) = c(e i 0 ) if i 0 ∈ {1, . . . , m} (respectively, c(z, x i 0 ) = c(b i 0 ) if i 0 ∈ {m + 1, . . . , k}). In either case, this connects all the vertices of H 1 to the rest of the graph, while preserving the connectedness of H 0 . This yields a spanning connected properly edge colored subgraph implying the existence of a proper spanning tree for this subgraph and G c . We now consider the scenario where for every edge (z, x i ) and (z, y j ) we have c(z, x i ) = c(e i ) and c(z, y j ) = c(e j ) for all i ∈ {1, . . . , k} such that (z, x i ) , (z, y j ) ∈ E (G c ). Since the number of edges in H rs is exactly c, which is at least

(n -k -1)(n -k -2) 2 + k + 1 and V H 0 ≤ (n -k -1)(n -k -2) 2
, there are at least k + 1 differently colored edges in H 1 . Thus one of these edges must be of the form (y i , y j ) in E H 1 where c (y i , y j ) / ∈ c (E (z, Y 1 )). We may assume, by relabeling the edges in E M , if necessary, along with their endpoints in X and Y 2 , that this edge is indeed (y m+1 , y m+2 ).

If there exists a vertex z ′ ∈ N (z) ∩ {x m+1 , x m+2 , y m+1 , y m+2 }, we connect z using e = zz ′ and add up (y m+1 , y m+2 ) and all edges in E M except for the one which is incident to e in order to connect all the points in Y 1 . We again must delete all the edges in H 0 which are incident to any e i colored with c(e i ) for all i ∈ {m + 1, . . . , k}. This preserves the connectedness of H 0 , the properness and the resultant subgraph is spanning and connected.

If N (z) ∩ {x m+1 , x m+2 , y m+1 , y m+2 } is the empty set, then N G (H 1 ) ⊆ {x 3 , . . . , x k }. Then we disconnect H 1 by deleting k -2 vertices from original graph G c , which contradicts the kconnectivity of G c . This bound is the best possible, in the sense that there exist edge-colored simple k-connected graphs colored with (n-k-1)(n-k-2) 2 + k colors and without a proper spanning tree. More precisely, consider a rainbow complete edge-colored simple graph on nk -1 vertices. Add k + 1 new vertices x 1 , . . . , x k+1 . Consider, also, k new colors, say c 1 , . . . , c k , not already used in the rainbow complete graph. Pick now k vertices y 1 , . . . , y k in the complete graph and then add all monochromatic edges on color c i , between each x i and all y j , 1 ≤ j ≤ k, i = 1, . . . , k. Finally add edges x k+1 x i each on color c i , for i = 1, . . . , k. The resulting graph, although colored with

(n-k-1)(n-k-2) 2
+ k colors, has no proper spanning tree as vertex x k+1 cannot be appended to each component of any proper spanning forest of the rest of the graph.

Minimum number of colors in terms of the rainbow degree

In this section we consider G c to be a connected graph with rainbow degree rd (G c ) = k. As in the previous section, the main result, Theorem 5.3.3, is presented as a corollary of two propositions which consider different cases depending on the order of the largest component of a rainbow spanning subgraph using all colors in G c . Proposition 5.3.1. Let G c be a c-edge-colored connected graph with rainbow degree rd

(G c ) = k such that |V (G c )| = n with n ≥ k 2 + 3k + 4 2 and c ≥ (n -k -1) (n -k -2) 2 + k + 1.
Let H rs be a rainbow spanning subgraph using all colors in G c such that the order of its largest component is nk -1, then G c has a properly edge-colored spanning tree.

Démonstration. Let H 0 the largest component of H rs and H 1 the subgraph induced on the rest of the vertices. Since

|V (H 0 )| = n -k -1, |V (H 1 )| = k + 1. Let X be the set {x 1 , x 2 , . . . , x k } and V (H 1 ) = {x} ∪ X.
We shall consider two different cases depending on rd H 1 (H 1 ).

Case 1 : rd H 1 (H 1 ) = k.
The graph G c is connected, then N H 0 (H 1 ) = ∅. Suppose without loss of generality that there exists N H 0 (x) = ∅. Since rd H 1 (x) = k, then xx i ∈ E (G c ) and c (xx i ) = c (xx j ) for all i, j ∈ {1, . . . , k} such that i = j. The subgraph H 0 remains connected after the deletion of the edge b since it is (k + 2)-edge-connected (see Lemma 5.1.3). We then have a connected, spanning and properly colored subgraph and we can extract a properly edge-colored spanning tree.

1.B. : If b ∈ E (H 1 ), i.e., b is of the form xx i for some i ∈ {1, 2, . . . , k}, say i = 1. We consider

N H 1 (x 2 ), the neighborhood of x 2 in H 1 . Since N H 1 (x 2 ) = V (H 1 ), necessarily x 1 ∈ N H 1 (x 2 ). We also know that c (x 1 x 2 ) = c (xx 2 ) since rd H 1 (x 2 ) = k, then we join x 1 to H 0 via the path yxx 2 x 1 .
Case 2 : rd H 1 (H 1 ) < k.

We shall consider two cases : 2.A. : There is a (k + 1)-matching M between H 0 and H 1 .

Let us denote E M = {e, e 1 , . . . , e k } the set of edges running between H 0 and H 1 . We shall use this matching in order to join every vertex in V (H 1 ) to H 0 and delete any edge b, b i ∈ E (H 0 ) adjacent to e and e i , respectively, such that c(e) = c(b) and c (e i ) = c (b i ). There are at most k + 1 such edges, therefore the deletion of all of them preserves the connectivity of H 0 since H 0 is (k + 2)-edge-connected.

2.B. :

There is no (k + 1)-matching M between H 0 and H 1 . Let l ∈ N such that kl is the maximum rainbow degree in H 1 for all vertices in V (H 1 ). Suppose again without loss of generality that kl = rd H 1 (x) and {x 1 , . . . , x k-l } = N H 1 (x).

Since rd (x) ≥ k, there also exist at least l vertices, y 1 , . . . , y l , in N H 0 (x) such that all edges xx i and xy j are pairwise differently colored. Let Y be {y 1 , . . . , y l }. Our aim is to connect the vertices of the set V (H 1 ) -(N H 1 (x) ∪ {x}) = {x k-l+1 , . . . , x k } to H 0 . For the sake of readability, let us rename this set as Z = {z 1 , z 2 , . . . , z l }. Since rd H 1 (z i ) ≤ k-l and rd(z i ) ≥ k for all i ∈ {1, . . . , l}, thus all these vertices must have at least l neighbors in H 0 and therefore we can always find a l-matching (Z, N H 0 (Z)). If for some i ∈ {1, . . . , l}, there exists y l+1 ∈ N H 0 (z i ) -Y , then there is a (l + 1)-matching (Z ∪ x, N H 0 (Z ∪ x)) and by deleting, if necessary, all edges adjacent to the extreme vertices of the matching in H 0 colored the same way, we get the desired result.

Otherwise Y = N H 0 (z i ) = N H 0 (x) for all i ∈ {1, . . . , l}, and all but one vertex in Z, say z 1 , are connected via the edges y i z i for all i ∈ {2, . . . , l} and y 1 x. We have to find a new way to connect z 1 to H 0 .

Since the number of colors in c (E (H 1 )) is greater or equal to k + 1, there is at least one extra color apart from the ones in E (x, X). We shall study different cases depending on where this new colored edge lies, i.e., the new colored edge is of the form : z i z j or z i x j . Then we can add the edge z 1 y i , delete the edge of the matching y i z i and use the path y j z j z i or y 1 xx j z i , respectively to connect back z i .

x i x j , for some i, j ∈ {1, . . . , k -l}. In this case, as |N H 1 (z 1 )| = kl, we simply change the roles that x and z 1 play and we will be back to the previous case.

(n -k -1) (n -k -2) 2 + k + 1.
Let H rs be a rainbow spanning subgraph using all colors in G c such that the order of its largest component is greater than nk -1, then G c has a properly edge-colored spanning tree.

Démonstration. Let H 0 the largest component of H rs and H 1 the subgraph induced on the rest of the vertices. Let p ∈ N represent |V (H 1 )|, so that |V (H 0 )| = np and let l ∈ N such that pl is the maximum rainbow degree in H 1 for all vertices in V (H 1 ). Let x ∈ V (H 1 ) be a vertex such that rd H 1 (x) = pl, then |rd H 0 (x)| ≥ kp + l. This number is greater or equal to kp + 1, which is the maximum number of bridges that can appear in E (H 0 ) as it was proved in Lemma (5.1.2).

We shall study two different cases depending on the value of l. Case 1 : l = 1.

Since rd H 1 (x) = p -1, x is adjacent to every point in V (H 1 ) and the edges of type xx i are pairwise differently colored. Additionally, |rd H 0 (x)| ≥ kp + 1 ≥ 1 since p ≤ k. Therefore, there exists y ∈ N H 0 (x) such that c(xy) / ∈ c (x, H 1 ). The subgraph H 0 remains connected after removing a possible edge in E (H 0 ) adjacent to xy and colored in c(xy), so we get the desired result.

Case 2 : l > 1.

Let us set N H 1 (x) = {x 1 , x 2 , . . . , x p-l }, since rd H 0 (x) ≥ kp + l and the number of bridges in H 0 is less or equal than kp -1, there are at least l -1 vertices, y 1 , y 2 , . . . , y l-1 , in N H 0 (x) such that all edges xy i are pairwise differently colored and such that there is no bridge in E (H 0 ) incident to them. Let Y be {y 1 , . . . , y l-1 } (|Y | = l-1). Consider Z = V (H 1 )-(N H 1 (x) ∪ {x}) = { z 1 , z 2 , . . . , z l-1 }. Our aim is again to connect the vertices of the set Z to H 0 . Since rd H 0 (z i ) are greater or equal to kp + l for all i ∈ {1, 2, . . . , l -1}, there is a (l -1)-matching between N H 0 ({x} ∪ Z) and H 0 so that none of the adjacent edges in E (H 0 ) is a bridge. Then only one vertex, say z 1 , is left to be joined. Since rd(z 1 ) ≥ k and rd H 1 (z 1 ) ≤ p-l, then rd H 0 (z 1 ) ≥ k-p+l which is greater or equal to l since k ≥ p. Thus there exists a vertex y ∈

V (H 0 ) -Y such that yz 1 ∈ E (G c ) and c(yz 1 ) / ∈ c (z 1 , V (H 1 ) ∪ Y )
and we obtain a l-matching between N H 0 ({x} ∪ Z) and H 0 . Hence the proof is finished.

Theorem 5.3.3. Let G c be a c-edge-colored connected graph with rainbow degree rd (G c ) = k and n ≥ k 2 + 3k + 4 2 . If c ≥ (n -k -1) (n -k -2) 2 + k + 1,
then G c has a properly edge-colored spanning tree.

Démonstration. Let us consider H rs , a rainbow spanning subgraph in G c using all colors and denote H 0 its largest component and H 1 the subgraph induced on the rest of the vertices. By Lemma (5.1.1), we know that |H 0 | ≥ nk -1, so the proof of the Theorem follows directly from Proposition 5.3.1 and Proposition 5.3.2.

Conclusion

In this chapter, we considered sufficient conditions that guarantee the existence of proper spanning trees in edge-colored graphs. The conditions that we imposed are relationships between various parameters of colored graphs, such as total number of colors, number of vertices, connectivity and number of differently colored edges incident to a vertex.

In the next chapter we will consider edge-colored multigraps. Since proper Hamiltonian paths are also proper spanning trees, we give sufficient conditions in relationships between total number of edges, number of colors and number of vertices, guarantee the existence of proper Hamiltonian paths in edge-colored multigraphs.

Chapitre 6

Colored trees in Multigraphs

Introduction

In this chapter we focus on edge-colored multigraphs1 . Since we try to find proper and weak proper trees, there is no interest to suppose that the parallel edges are colored with the same color. We give sufficient conditions involving various parameters, such as the number of edges, rainbow degree, etc., in order to guarantee the existence of properly edge-colored hamiltonian paths in edge-colored multigraphs. Results involving only degree conditions can be found in [START_REF] Abouelaoualim | Cycles and paths in edge-colored graphs with given degrees[END_REF].

In this chapter all multigraphs are assumed to be connected. We consider two families of multigraphs H k,k+3 and H k,k+2 without proper hamiltonian paths :

(1) H k,k+3 denotes a 2-edge-colored multigraph on 2k + 3 vertices, k ≥ 1, defined as follows : consider a complete red graph on k vertices joined with red edges to an independent set of k + 3 vertices. Finally, superpose it with a complete blue graph on 2k + 3 vertices.

(2) H k,k+2 denotes a c-edge-colored multigraph on 2k + 2 vertices, k ≥ 1 and c ≥ 3. Consider a rainbow complete graph on k vertices joined with edges of all possible colors to an independent set of k + 2 vertices.

Before presenting the main results of this chapter, we introduce some of our preliminary results for connected graphs, that can be of independent interest.

Preliminary results

Lemma 6.2.1. Let G be a simple non-colored graph on n ≥ 14 vertices. If m ≥ (n-3)(n-4)

2 + 4, then G has a matching M of size |M | ≥ ⌈ n-2 2 ⌉.
Démonstration. We can assume that the graph is connected, otherwise it is easy to see that the only possible case is when G has two components, the first one having two adjacent vertices and the second one having n -2 vertices and at least (n-3)(n-4) 2 + 3 edges. Then, the result immediately follows from the well known Theorem in [START_REF] Byer | Edge bounds in nonhamiltonian k-connected graphs[END_REF]. Therefore, let G be a graph of order n. Let M be a maximum matching in G and let U be the set of unmatched vertices. We shall prove that |U | ≤ 2 by contradiction.

Assume that |U | ≥ 3. In fact, it is enough to prove it for |U | = 3. Set U = {u 1 , u 2 , u 3 }. Since M is a maximum matching, neither we can add any new edge to M nor we can replace a set of edges in M , in order to obtain a new matching in G which is larger than M . Necessarily, U is an independent set, otherwise we could add a new edge to M .

Let us denote the complement of G by G. We will count the number of edges in E(G), i.e. those edges which cannot be present in E(G) because otherwise the matching M would not be maximum.

Since m = |E(G)| ≥ (n-3)(n-4) 2 + 4, then E(G) ≤ 3n -10. In the sequel we will prove this lemma for n odd, the case even being very similar.

Since the graph is connected, we shall prove the result depending on the size of a maximum matching M between U and G -U .

We distinguish three cases, depending on the cardinality of | M |.

Case 1 : M = 1. There is a unique vertex in N (U ) and therefore there are at least three vertices of degree 1 in V (G). This leads to a contradiction, since the number of edges in E(G) would be at most (n -3)(n -4) 2 + 3 (the edges which are present in a complete graph on n -3 vertices, and three more, which connect the vertices of degree one to the complete graph), which contradicts the hypothesis of the lemma.

Case 2 : M = 2. In this case, there are two vertices

v 1 , v 2 ∈ G -U , v 1 = v 2 , such that u 1 v 1 , u 2 v 2 , u 3 v 2 ∈ E(G) and N ({u 2 , u 3 }) ⊆ {v 1 , v 2 },
otherwise we would have a larger matching between the sets U and G -U . Therefore, all edges of the type u 2 v and u 3 v, for any v ∈ G -U -{v 1 , v 2 }, are in E(G). Also, {u 1 , u 2 , u 3 } is an independent set, there are at least three more edges in E(G). Since M is a perfect matching in G -U , the vertices v 1 and v 2 are the extremities of some edge in M . If v 1 v 2 ∈ M , then we can replace v 1 v 2 by u 1 v 1 and u 2 v 2 , contradicting the fact that M is maximum. Thus, there exist

w 1 , w 2 ∈ G -U -{v 1 , v 2 }, w 1 = w 2 , such that v 1 w 1 , v 2 w 2 ∈ M . Necessarily, w 1 w 2 ∈ E(G). Otherwise we can replace {v 1 w 1 , v 2 w 2 } by {u 1 v 1 , u 2 v 2 , w 1 w 2 } and
obtain a larger matching. The edge u 1 w 2 is also in E(G), otherwise we can replace v 2 w 2 in M by u 1 w 2 and u 2 v 2 in order to get a larger matching. So far we have at least 2(n -5) + 3 + 1 + 1 = 2n -5 edges in E(G).

There are n -7 2 edges, e 1 , e 2 , ..., e n-7

2

, in M -{v 1 w 1 , v 2 w 2 }. We shall study the conditions given by the possible connections between the edges e i , for i = 1, ..., n -7 2 , and the set {w 1 , w 2 }.

Necessarily,

|N e i ({w 1 , w 2 })| ≤ 2, ∀ i = 1, 2, ..., n -7 2 .
Otherwise we can find a matching of size two, say {f i1 , f i2 }, connecting e i and {w 1 , w 2 } and we can replace

{e i , v 1 w 1 , v 2 w 2 } by {u 1 v 1 , u 2 v 2 , f i1 , f i2 }.
We can assure then, that there are at least n -7 new edges in E(G). By adding these edges to the 2n -5 edges obtained before, we conclude that there are at least 3n -12 edges in E(G).

We shall study two different subcases.

(1) N ({v 1 }) ∩ {u 2 , u 3 , w 2 } = ∅. Then 3 new edges belong to E(G). By adding these new edges to those we have counted previously, we have 3n -12 + 3 = 3n -9 edges in E(G). This number is larger than 3n -10, the result follows.

(2) N (v 1 ) ∩ {u 2 , u 3 , w 2 } = ∅. Then u 1 w 1 ∈ E(G). In fact, if for example, u 2 v 1 ∈ E(G), then we can replace v 1 w 1 in M by {u 1 w 1 , u 2 v 1 }, so one more edge is in E(G). Now, if u 2 v 1 or u 3 v 1 are in E(G) then all edges of the type u 1 v, for any v ∈ G -U -{v 1 , v 2 }, are in E(G) since M = 2. Therefore both u 2 v 1 and u 3 v 1 are missing, or n -5 -2 = n -7 edges are missing (since we have already counted u 1 w 1 and u 1 w 2 ). In the first case we arrive to 3n -9, which is a contradiction. In the second case we have 3n-11+n-7 = 4n-18 > 3n-10, which is again a contradiction.

Case 3 : M = 3. In this case there are three distinct vertices

v 1 , v 2 , v 3 ∈ G -U , such that u i v i ∈ E(G), for i = 1, 2, 3.
Since there is a perfect matching in G -U , the vertices v 1 , v 2 and v 3 are the extremities of some edge in M . If v i v j ∈ M , then we can replace v i v j by u i v i and u j v j contradicting that M is maximum. Thus, there exist three distinct vertices

w 1 , w 2 , w 3 ∈ G -U -{v 1 , v 2 , v 3 }, such that v 1 w 1 , v 2 w 2 , v 3 w 3 ∈ M .
Necessarily, w i w j ∈ E(G) for all i, j = 1, 2, 3. Otherwise we can replace v i w i and v j w j by u i v i , u j v j and w i w j and obtain a larger matching. In the same way, u i w j ∈ E(G) for all i, j = 1, 2, 3, i = j, otherwise we can replace v j w j by {u j v j , u j v j }. Since U is an independent set, there are at least 3 + 3 + 6 = 12 edges in E(G). If u i w i ∈ E(G) then N (v i ) ∩ {u j , w j } = ∅ for j = i. Otherwise, if for example u j v i ∈ E(G), we can replace v i w i in M by u i w i and u j v i . Thus, there are 3 more edges in E(G). Now, there are n -9 2 edges, e 1 , e 2 , ..., e n-9

2

, in M -{v 1 w 1 , v 2 w 2 , v 3 w 3 }. We shall study the conditions given by the possible connections between the edges e i , for i = 1, ..., n -9 2 , and the sets

U = {u 1 , u 2 , u 3 } and W = {w 1 , w 2 , w 3 }.
Necessarily, |N e i (W )| ≤ 3, for all i = 1, 2, ..., n -9 2 . Otherwise, we can find a matching of size two, say {f ij , f ik }, j, k ∈ {1, 2, 3}, connecting e i and {w j , w k }, and we can replace

{e i , v j w j , v k w k } by {u j v j , u k v k , f ij , f ik }.
Also |N e i (U )| ≤ 3, for all i = 1, 2, ..., n -9 2 . Otherwise, we can find a matching of size two, say {g ij , g ik }, j, k ∈ {1, 2, 3}, connecting e i and {u j , u k }, and we can replace e i by { g ij , g ik }. We can assure then, that there are at least 2(3 n -9 2 ) = 3n -27 new edges in E(G).

By adding up these edges to the 15 edges considered before, we obtain 3n -12 edges in E(G).

Recall that we have denoted the extremities of the edge e i by p i and q i . Without loss of generality, let us assume that if |N e i (u j )| = 1, for all i = 1, 2, ..., n -9 2 and j = 1, 2, 3 , then N e i (u j ) = p i . Thus, either |N e i (U )| ≤ 1 or N q i (U ) = ∅, i = 1, 2, 3. Otherwise we can find a matching of size 2 between U and e i and again, we can define a larger matching. Now, for n ≥ 15, there are at least 3 edges in M -{v 1 w 1 , v 2 w 2 , v 3 w 3 }. If N e i (U ) = ∅, then 3 more edges are in E(G) and, by adding these new edges to the 3n -12 that we have already found, the result follows.

If |N e i (U )| = 1, then either there exists j = i such that N e i (U ) = N e j (U ) or N e i (U ) = N e j (U ) for all i = j. In the first case q i q j ∈ E(G) and either p 1 u 1 or p 1 q 2 is in E(G). We again have 3 new edges in E(G) and the proof is finished. In the latter case all edges connecting the edges in M -{v 1 w 1 , v 2 w 2 , v 3 w 3 } and U have a common vertex, say u 1 . Again, at least 3 more edges are in E(G).

Only one case is left : if N q i (U ) = ∅, then q i q j ∈ E(G) for all i = 1, 2, 3 and we also get 3 new edges in E(G). This completes the proof. Observe also that m i ≥ (n -3)(n -4) 2 + 4 for i ∈ {r, b}, since this threshold is tight when the multigraph is complete on one of the colors.

Let us see the case where n is odd. By Lemma 6.2.1, there exist two matchings M r and M b each of the size n -1 2 , so the result is straightforward.

Let us see now the case where n is even. Then, again by Lemma 6.2.1, there exist two matchings, M r and M b , each of the size at least n-2 2 . We shall prove the result by contradiction.

Let us consider monochromatic subgraphs in colors r and b respectively, and suppose that

|M r | = |M b | = n -2 2 . Let U = {u 1 , u 2 }
denote the independent set of unmatched vertices in M r . The vertices u 1 and u 2 are connected to the extremities of the edges in M r . We claim that there exist two distinct vertices

v 1 , v 2 in V (G) -{u 1 , u 2 }, such that u 1 v 1 , u 2 v 2 ∈ E r (G c ). Otherwise, if N r ({u 1 , u 2 }) = v 1 ,
this vertex is the extremity of some edge v 1 w 1 in M r and then we distinguish two cases. First, if N r (w 1 ) = v 1 , then we have three distinct vertices of degree one which leads us to a contradiction with the total number of edges. Second, there exist w 2 ∈ N r (w 1 ) -{v 1 } and v 2 , such that v 2 w 2 ∈ M r . Then we can replace {v 1 w 1 , v 2 w 2 } by {u 1 v 1 , w 1 w 2 }. After this permutation, we have a new matching of same size but u 1 is replaced by v 2 . So, as we claimed, there always exist two distinct vertices

v 1 , v 2 in V (G) -{u 1 , u 2 } such that u 1 v 1 , u 2 v 2 ∈ E r (G c ) (u 2
and v 2 are those vertices in the case we have just seen). Now, the edge v 1 v 2 is not in M r , otherwise we can replace it by {u 1 v 1 , u 2 v 2 } and get a larger matching in color r, contradicting that M r is maximum. So, there are two vertices w 1 and w

2 in V (G) -{u 1 , u 2 , v 1 , v 2 } such that v 1 w 1 , v 2 w 2 ∈ M r .
Observe that there can be at most two edges connecting the endpoints of any edge in M r to the set {u 1 , u 2 }, i.e. there are at least two missing edges for each edge in M r , in total 2( n-2

2 ). Similarly, for w 1 and w 2 , we have the same constraints and therefore, this also means two edges are missing for each edge in M r -{v 1 w 1 , v 2 w 2 }, which gives the total of 2( n-6

2 ). Otherwise, suppose that there is an edge v 3 w 3 with three edges between its endpoints to the vertices w 1 and w 2 . So we can replace v 1 w 1 and v 2 w 2 by say w 1 v 3 , w 2 w 3 , v 1 u 1 and v 2 u 2 to obtain a perfect matching. Finally, as u 1 and u 2 are independent, the edge u 1 u 2 is missing. If we sum up these numbers, we conclude that there are at least (n -2) + (n -6) + 1 = 2n -7 edges missing in color red. The same reasoning can be done with the matching M b to obtain 2n -7 blue missing edges. So, the total number of missing edges in colors red and blue is 4n -14. Since the complement of G c has the edge set of the size less than or equal to 3n -10, for n ≥ 6 we have a contradiction. Therefore, the result holds. Lemma 6.2.3. Let G c , c ≥ 2 be a connected c-edge-colored multigraph. Suppose that G c contains a proper path P = x 1 y 1 x 2 y 2 . . . x p y p , p ≥ 2, such that each edge x i y i is red. If G c does not contain a proper cycle C with vertex set {x 1 , y 1 , x 2 , y 2 , . . . , x p , y p }, then there are at least (c -1)(2p -2) missing edges in G c . Démonstration. Let P = x 1 y 1 x 2 y 2 . . . x p y p be a proper path, p ≥ 2, such that each edge x i y i is red. Let blue be the second color. The blue edge x 1 y p cannot be in G c , otherwise C = x 1 y p . . . x 1 is a proper cycle. Suppose that the blue edges x 1 x i are present in G c , for i = 2, . . . , p. Then, the blue edges y i-1 y p can not be in G c , otherwise we have the proper cycle C = x 1 x i . . . y p y i-1 . . . x 1 , that contradicts our hypothesis. Therefore, for each edge y i-1 x i in the path, either the blue edge x 1 x i or the blue edge y i-1 y p is missing. So there are 2p-2 2 missing blue edges. Now, suppose that the blue edges x 1 y i are present in G c , for i = 3, . . . , p -2. Then, the blue edges x i+1 y p cannot be in G c at the same time as x i y i+1 , y i-1 x i+2 or y i-1 y i+1 , x i x i+2 , otherwise we have the proper cycles C = x 1 y i x i y i+1 x i+1 y p . . . x i+2 y i-1 . . . x 1 or C = x 1 y i x i x i+2 . . . y p x i+1 y i+1 y i-1 . . . x 1 . The minimum in this case corresponds to one missing edge x i+1 y p for each edge y i-1 x i in the path, for i = 2, . . . , p -1. Therefore, there are 2p-6 2 missing blue edges.

For the moment we have 2p -3 missing blue edges. To obtain the last missing edge, suppose that the blue edge x 2 y p is present in G c . Then, it cannot be at same time with x 1 y 2 , y 1 x 3 or x 1 x 3 , y 1 y 2 , otherwise we obtain the proper cycles C = x 1 y 2 . . . x 2 y p . . . x 3 y 1 x 1 or C = x 1 x 3 . . . y p x 2 . . . y 2 y 1 x 1 . The minimum in this case corresponds to one missing edge x 2 y p . We remark that the blue edges x 2 y p , y 1 y 2 and y 1 x 3 were not counted before. The edges x 1 x 3 and x 1 y 2 were supposed to exist, otherwise in order to obtain the last missing edge we consider the symmetric case, i.e. using the blue edge y p-1 x 1 (if exists).

In total, there are 2p-2 2 + 2p-6 2 + 2 = (2p -2) missing blue edges in G c . As we have c -1 colors different from red, that gives us (c -1)(2p -2) missing edges, as desired. Note that this number of missing edges is the same as in the simplest case, this is, if all edges different from red x 1 x i and x 1 y i , for i = 2, . . . , p are not present in G c . Lemma 6.2.4. Let G c be a connected c-edge-colored multigraph, c ≥ 2. Let M be a matching of G c in one color, say red, of size |M | ≥ ⌈ n-2 2 ⌉. Let P = x 1 y 1 x 2 y 2 . . . x p y p , p ≥ 2, be a maximum proper path compatible with M . Then, the following cases are possible :

- Démonstration. Before starting the proof, we remark that the edges x 1 y 1 and x p y p are red.

Otherwise we can easily extend the path by adding an edge of the matching to P .

First, suppose that n is even and 2p < n. Since M is of the size n 2 , there are n-2p 2 red edges outside P . Let us denote these edges by e i for i = 1, . . . , n-2p

2 . Suppose that P is not a proper cycle. Let blue be the second color. By Lemma 6.2.3 there are (2p -2) missing blue edges. As the path is maximum, we cannot extend P having an edge e i neither at the beginning nor at the end of it, thus there are no blue edges between the vertices x 1 , y p and the edges e i . Therefore, there are 4 n-2p 2 blue missing edges. Finally, as we cannot add any edge e i in between the path, then there at most 2 blue edges between the edges e i and the edges y i x i+1 , i = 1, . . . , p -1. So, there are 2 n-2p 2 2p-2 2 missing blue edges different from red. By summarizing and simplifying all these numbers and having c -1 colors different from red, we obtain that there are (n -2 + pn -2p 2 )(c -1) missing edges in G c different from red. If we calculate the minimum value of this function, we obtain (2n -4)(c -1) for p = n-2 2 and case (1a) holds. Now, if P is also a proper cycle, then there cannot exist any edge, different from red between all vertices of P and the edges e i , and, therefore there are 2 n-2p 2 2p(c -1) = (2pn -4p 2 )(c -1) missing edges different from red. Again, minimizing the function we obtain the same result as above and the case (1b) holds.

Suppose now that n is odd, M = n-1 2 and 2p < n -1, or n is even, M = n-2 2 and 2p < n -2. In both cases, same arguments as before apply, just by replacing n with n -1 or n -2 respectively, in the number of missing edges. This, because we have n -1 matched vertices and one non-matched vertex for the first case and n -2 matched vertices and two non-matched vertices for the second case. + 5 edges, has a hamiltonian cycle. So, if we add a new vertex v to G, joined to all its vertices, we obtain that G + {v} has m ≥ (n-2)(n-3) 2 + 3 + n = (n-1)(n-2) 2 + 5 edges. Therefore, G + {v} has a hamiltonian cycle, i.e. G has a hamiltonian path and this implies that there exists a matching M in G of the size |M | = ⌊ n 2 ⌋. Definition 6.2.6. A path P is compatible with a matching M if its edges belong alternatively to M and not to M . Definition 6.2.7. A vertex is called monochromatic if all its incident edges are colored by the same color.

Main results

Theorem 6.3.1. Let G c be a c-edge-colored multigraph, c = 2. Suppose that for every vertex x in G c , rd(x) = 2 and n ≥ 14. If m ≥ (n -3)(n -4) + 3n -2, then G c has a proper hamiltonian path.

Démonstration. Let us suppose that G c has not a proper hamiltonian path. We will show that E(G c ) has more than 3n -10 edges, i.e., G c has less than (n -3)(n -4) + 3n -2 edges, a contradiction to the hypothesis of the theorem.

-n even : By Lemma 6.2.2,

G c has two matchings M r , M b , such that |M r | = n 2 and |M b | ≥ n-2
2 . Take the longest proper path P = x 1 y 1 x 2 y 2 . . . x p y p compatible with the matching M r and the longest proper path

P ′ = x ′ 1 y ′ 1 x ′ 2 y ′ 2 . . . x ′ p ′ y ′ p ′ compatible with the matching M b . Suppose 2p < n and 2p ′ < n -2. Since |M r | = n 2 , c(x 1 y 1 ) = c(x p y p ) = r.
Otherwise, we can easily extend the path by adding an edge of the matching to P . Suppose also that c(x ′ 1 y ′ 1 ) = c(x ′ p ′ y ′ p ′ ) = b. So, as P (resp. P ′ ) is properly colored, the edges x i y i (resp. x ′ i y ′ i ) are red (resp. blue) for i = 1, . . . , p (resp. p ′ ) and the edges y i x i+1 (resp. y ′ i x ′ i+1 ) are blue (resp. red) for i = 1, . . . , p -1 (resp. p ′ -1). We have the following possibilites : -Neither P nor P ′ are proper cycles.

-P ′ is a proper cycle but P is not.

-P is a proper cycle but P ′ is not.

-Both P and P ′ are proper cycles. First, observe that in any case, by Lemma 6.2.4, there are at least 2n -4 blue missing edges and 2n -8 red ones. We obtain 4n -12 > 3n -10 missing edges. A contradiction. Therefore we conclude that 2p = 2p ′ = n -2. Now suppose that P is not a proper cycle, then again by Lemma 6.2.4, there are at least 2n -4 blue missing edges. Now, let v 1 , w 1 be the vertices in G c -P ′ . It is clear that if there exists a blue edge joining v 1 and w 1 , we are under the same conditions of the red matching, and therefore in contradiction with the number of edges. Otherwise we count red missing edges if we cannot extend P ′ to a proper hamiltonian path. First, if we have 3 or more red edges between v 1 , w 1 and x ′ 1 ,y ′ p ′ , we have a hamiltonian path and we are done. So, there are at least 2 red missing edges. Second, if the path P ′ is not a proper cycle again by Lemma 6.2.3 there are 2p ′ -2 = n -4 red missing edges. Now we combine and sum up all these number and we arrive to 2n -4 + n -4 + 2 = 3n -6 > 3n -10 missing edges. Again a contradiction. Finally, if the path P ′ is a proper cycle we have that, if there are 3 or more red edges between v 1 , w 1 and the endpoints of the red edges of the cycle, we can add v 1 and w 1 to the path P ′ as following. Suppose without losing generality that v 1 is adjacent to y ′ i and w 1 to x ′ i+1 , both with red edges and the red edge y ′ i x ′ i+1 is present in the proper cycle. So we form the proper hamiltonian path starting with the red edge v 1 y ′ i then following the proper cycle in the direction of x ′ i until we arrive to x ′ i+1 and finally we take the red edge x ′ i+1 w 1 . Therefore if there are not these 3 or more red edges, they are missing 2 2p ′ -2 2 = n -4 red edges. If we sum up again we arrive to 2n -4 + n -4 + 2 = 3n -6 > 3n -10 missing edges. Again a contradiction. Suppose now, that P is a proper cycle. Let e be the red edge of the matching outside the path P . If there existe a blue edge between any endpoint of e to the cycle, we easily obtain a proper hamiltonian path, taking the edge e, any blue edge going to the cycle and then following it in the appropriate direction. Otherwise, as the graph is connected, all edges between the endpoints of e to the path are blue. Now, as rd(G c ) = 2, we have that there must exist a blue edge parallel to e and therefore we obtain the proper hamiltonian path just as before but starting with the blue edge parallel to e. So, we covered all the cases for n even.

-n odd : Again by Lemma 6.2.2,

G c has two matchings M r , M b , such that |M r | = |M b | = n-1
2 . We take the longest proper paths P and P ′ compatibles with the matching just as before. Suppose now that 2pn -1 and 2p ′ < n -1. As the even case, we have exactly the same four possibilities. Observe that in any case, by Lemma 6.2.4, there are at least 2n -6 blue missing edges and 2n -6 red ones. We obtain 4n -12 > 3n -10 missing edges. A contradiction. Therefore we have that 2p = 2p ′ = n -1. We will count blue missing edges if we cannot extend P to a proper hamiltonian path. Same argument holds for red edges since we are under the same conditions. Now, let v be vertex outside the path P . It is clear that is P is also a proper cycle we trivially obtain a proper hamiltonian path since the graph is connected and rd(v) = 2. So, as P is not a proper cycle we have by Lemma 6.2.3, that there are 2p -2 = n -3 blue missing edges. If we have blue edges between v and x 1 or y p , we have a hamiltonian path and we are done. So, there are 2 blue missing edges. Finally, if there exists a blue edge between x 1 and x i (i = 2, . . . , p), then it cannot exist the blue edge vy i-1 (i = 2, . . . , p), otherwise we form the proper hamiltonian path vy i-1 . . . x 1 x i . . . y p . Similar, if we have the blue edges y p and y i , with vx i+1 (i = 1, . . . , p -1). Note that since P is not also a proper cycle we cannot have at the same time the blue edges x 1 x i and y p y i-1 (i = 2, . . . , p). Therefore, there are 2p-2 2 = n-3 2 blue missing edges. Concluding, if we make the sum and multiply it by two, since the same number of red missing edges is obtained with P ′ we obtain that there are 2(n -3 + 2 + n-3

2 ) = 3n -5 missing edges. A contradiction. Since we covered all the cases, the theorem is proved. Theorem 6.3.1 is the best possible for n odd, n ≥ 14, consider a complete blue graph, say A, on n -3 vertices. Add 3 new vertices v 1 , v 2 , v 3 and join them to a vertex v in A with blue edges. Finally, superpose the obtained graph with a complete red graph on n vertices. Although the resulting 2-edge-colored multigraph has (n -3)(n -4) + 3n -3 edges, it has no proper hamiltonian path since one of the vertices v 1 , v 2 , v 3 cannot belong to such a path.

The two lemmas below will be useful in view of Theorem 6.3.4. Démonstration. Set C = x 1 y 1 x 2 y 2 . . . x s y s x 1 , where x i y i are the red edges of C, i = 1, 2, . . . , s.

Then d b {x i ,y i } (x)+d b {x i ,y i } (y) ≤ 2, otherwise if d b {x i ,y i } (x)+d b {x i ,y i } (y) ≥ 3, then the cycle x 1 y 1 x 2 y 2 . . . x i xyy i . . . x s y s x 1 is the desired one. It follows that i d b {x i ,y i } (x) + d b {x i ,y i } (y) ≤ 2 |C| 2 =
|C|, a contradiction to the hypothesis of the lemma. This completes the proof.

Lemma below is established in view of Theorem 6.3.4

Lemma 6.3.3. If m ≥ f 0 (n) = (n -1)(n -2) + n, then G c
has a proper hamiltonian cycle if n is even, and a proper cycle of length n -1 otherwise.

Démonstration. Let red and blue be the two colors of G c . The proof is by induction on n. The theorem is true for small values of n, say n = 2, 3, 4. Let us suppose that n ≥ 5 and that the theorem is true until n -1. We will prove it for n. By a Theorem of [START_REF] Abouelaoualim | Cycles and paths in edge-colored graphs with given degrees[END_REF], if for any vertex d r (x) ≥ n+1

(x) = 0, then m ≤ n(n -1) -(n -1) = (n -1) 2 < (n -1)(n - 2) + n, a contradiction. Similarly d r (x) + d b (x) ≥ 3, otherwise, if d r (x) + d b (x) ≤ 2, then m ≤ n(n -1) -2n + 4 = n 2 -3n + 4 < (n -1)(n -2) + n,
(n -1) -2(n -2) + 2 = n 2 -3n + 4
which again is a contradiction unless all inequalities become strict equalities. In particular there is a red and a blue edge between x and y. In that case find a proper path starting at, say x, containing all the vertices of C. Then join y to x by using one edge xy with the appropriate color. The obtained proper path is a hamiltonian one. Theorem 6.3.4 is the best possible for n ≥ 14. In fact, consider a rainbow complete 2-edgecolored multigraph on n -2 vertices for n odd. Add two new vertices x 1 and x 2 . Then add the red edge x 1 x 2 and all red edges between {x 1 , x 2 } and the complete graph. Although the resulting graph has n 2 -3n + 3 edges, it has no proper hamiltonian path, since at least one of the vertices x 1 or x 2 cannot be attached to any such path. Indeed, for n odd, the two extremal edges of any proper hamiltonian path must differ in colors. If n = 5, 7, Theorem 6.3.4 does not hold for the graphs H k,k+3 , k = 1, 2.

In the rest of this chapter we will deal with c-edge-colored multigraphs, for c ≥ 3. Now, we will present a result that allows us to consider just the case c = 3. Lemma 6.3.5. Let G c be a c-edge-colored connected multigraph on n vertices, c ≥ 3 and m ≥ cf (n) + 1 edges. There exists one color c j such that if we color its edges with another used color and we delete parallel edges with same color, then the resulting (c -1)-edge-colored multigraph is connected and has

m ′ ≥ (c -1)f (n) + 1 edges, such that if G c-1 has a proper hamiltonian path then G c has one too. Moreover, if rd(G c ) = k, 1 ≤ k ≤ c, then rd(G c-1 ) = k -1.
Démonstration. Let c i denote the color i, for i = 1, . . . , c, in G c , and denote by |c i | the number of edges with color i. Let c j be the color with less number of edges. Color the edges on color c j with another used color, say c l , and delete (if necessary) parallel edges with that color. Call this graph G c-1 . By this, we delete at most |c j | edges. It is clear that this graph is connected since we delete just parallel edges. Also, if is G c-1 has a proper hamiltonian path, then, this path is also proper hamiltonian in G c but maybe with some edges on color c j (in the case that they have been recolored with c l ). Observe also that if rd(G c ) = k then rd = (G c-1 ) = k -1 since only the color c j disappeared. We will show now that m ′ ≥ (c -1)f (n) + 1. We have two cases. First, if

|c j | > f (n), then clearly m ′ ≥ (c -1)f (n) + 1 edges since for all i, |c i | > f (n).
For the second case, we have that

|c j | ≤ f (n). Now, m = c i=1 |c i | ≥ cf (n) + 1 and therefore c i=1,i =j |c i | ≥ cf (n) -|c j | + 1 = (c -1)f (n) + f (n) -|c j | + 1.
This last expression is greater or equal than (c-1)f (n)+1 since f (n)-|c j | ≥ 0. Finally, we have that G c-1 has m ′ ≥ (c-1)f (n)+1 edges as desired. Démonstration. First of all by Lemma 6.3.5 we can assume that c = 3 and m ≥ 3(n-1)(n-2) 2 + 1. Since there exists one color, say red, such that the number of edges are at least (n-1)(n-2) 2 + 1 then by a theorem in [START_REF] Byer | Edge bounds in nonhamiltonian k-connected graphs[END_REF], there is a hamiltonian red path and therefore a perfect or almost perfect matching. Also, as m ≥ 3(n-1)(n-2) 2 + 1 we have that |E(G c )| is at most 3n -4. Now, we will split our proof in the even and odd case.

-n even. Let M r be the perfect matching in color red, i.e. |M r | = n 2 . Take the longest proper path P = x 1 y 1 x 2 y 2 . . . x p y p compatible with the matching

M r . Set 2p = |P |. Since |M r | = n 2 , c(x 1 y 1 ) = c(x p y p ) = r.
Otherwise, we can easily extend the path by adding an edge of the matching to P . So, as P is properly colored, the edges x i y i are red (i = 1, . . . , p) and the edges y i x i+1 are in colors different from red (i = 1, . . . , p-1). Now, by Lemma 6.2.4, if 2p < n (i.e. P not hamiltonian) then there are at least 2(n -2 + pn -2p 2 ) missing edges in G c different from red if P is not also a proper cycle and 2(2pn -4p 2 ) otherwise. We start with the first case.

-P is not a proper cycle. Since there are at least 2(n-2+pn-2p 2 ) missing edges different from red the folllowing inequality must be satisfied : 2(n -2 + pn -2p 2 ) ≤ 3n -4. So, if we solve it, we can observe that it is only satisfied for n = 4 and p = n-2 2 . Therefore, for all other values we have a contradiction with the number of edges of G c . The case n = 4 and p = n-2 2 is easy to check. So, P must be a proper hamiltonian path. -P is a proper cycle. As before, 2(2pn -4p 2 ) ≤ 3n -4 must be satisfied, and this only happens, again, if n = 4 and p = n-2 2 . Finally P must be a proper hamiltonian path. -n odd. We will prove the result by induction. Cases n = 3, 5 are easy to check. Now, let M r be the almost perfect matching in color red, i.e. |M r | = n-1 2 . Take the longest proper path P = x 1 y 1 x 2 y 2 . . . x p y p compatible with the matching M r . As in the even case, c(x 1 y 1 ) = c(x p y p ) = r. By Lemma 6.2.4, if 2p < n -1 then there are at least 2(n -3p + pn -2p 2 ) missing edges in G c different from red if P is not also a proper cycle and 2(2pn -2p -4p 2 ) otherwise. We start with the first case.

-P is not a proper cycle. Since there are at least 2(n -3p + pn -2p 2 ) missing edges different from red the folllowing inequality must be satisfied : 2(n-2+pn-2p 2 ) ≤ 3n-4.

If we solve it, we have that is true for n = 5, 7 and p = n-3 2 . We will show first the case where 2p = n -1, this is, we have just one vertex, say v, not in the proper path P . We have the following cases depending on the degree and neighbors of v.

-v has at least two neighbors, say x and y, in two different colors, say c 1 and c 2 respectively. If we cannot attach v to the path P we have that there are 4 missing edges in colors different from red between v and the vertices x 1 , y p . Also, we have at most 2 edges different from red between v and the edges y i , x i+1 of the path. Therefore there are 2 n-3 2 missing edges. Adding up all this, we conclude that the degree of v in colors different from red is at most n -3. So, if we contract the three vertices v, x, y to a new one, say v ′ , we will delete n -3 edges different from red and n -1 red edges from v. At most n-3 edges of color c 1 from x, n-3 of color c 2 from y, 3 parallel edges between x and y. And finally all the edges in color c 3 from x but from y, we just delete the edges of color c 3 that are not incident to a common neighbor of x with color c 3 . All this is at most n-3. Adding up all these numbers we arrive that, after the contraction, we deleted 5n -10 edges. Now, it is easy to the new 3-edge-colored multigraph on n -2 vertices, say G ′c , has at least (n-1)(n-2) 2 + 1 -(5n -10) ≥ (n-3)(n-4) 2 + 1, for n ≥ 5, therefore, by inductive hyphotesis, we have a proper hamiltonian path P ′ in G ′c . To build the proper hamiltonian path for G c , we have the following situations. If v ′ is an endpoint of the path P ′ , this is v ′ P ′′ , and the edge in the path incident to v ′ has color c 1 (c 2 ), this means that, because of the edges that we have deleted, this edge must be incident to y (x) in the graph G c . Therefore, we obtain the proper hamiltonian path in G c , xvyP ′′ (yvxP ′′ ) with c(xv) = c 1 , c(vy) = c 2 and c(yP ′′ ) = c 1 (with c(yv) = c 2 , c(vx) = c 1 and c(xP ′′ ) = c 2 ). If the edge in the path P ′ incident to v ′ has color c 3 , we have the proper hamiltonian path xvyP ′′ or yvxP ′′ in G c , since the endpoint was a common neighbor with color c 3 to x and y. Now, it the path P ′ is of the form P 1 v ′ P 2 and let v 1 , v 2 be the vertices in P 1 , P 2 respectively adjacent to v ′ we have the following situations. If c(v

1 v ′ ) = c 1 and c(v ′ v 2 ) = c 2 , or c(v 1 v ′ ) = c 1 and c(v ′ v 2 ) = c 3
we have, by the way we deleted the edges for the contraction, the proper hamiltonian path in G c , P 1 yvxP 2 with c(v

1 y) = c 1 , c(yv) = c 2 , c(vx) = c 1 and c(xv 2 ) = c 2 , or P 1 yvxP 2 with c(v 1 y) = c 1 , c(yv) = c 2 , c(vx) = c 1 and c(xv 2 ) = c 3 .
The other cases are just symmetric or analog.

-v is monochromatic in red. We have these situations. First, d(v) ≤ n -2, consider the G c -{v} and delete from a neighbor of v, say w, all edges in two colors in order to have w monochromatic not in red. Call this graph G ′c . Observe that we can always do this since, it is impossible to have 2 monochromatic vertices. So, by this, we delete at most n -2 + n -2 + n -2 = 3n -6 edges. It is easy to see that the graph G ′c on n -1 vertices has at least (n-1)(n-2)

2 + 1 -(3n -6) = (n-2)(n-3) 2
+ 1, then, by even case, we have a proper hamiltonian path P ′ in G ′c . Since w is monochromatic not in red, w is either in the beggining or in the end of P ′ and therefore it is trivial to add v to P ′ in order to find a proper hamiltonian path in G c . If this does not hold, we have that d(v) = n -1. Therefore the graph G c -{v} on n -1 vertices has at least

(n-1)(n-2) 2 + 1 -(n -1) ≥ (n-2)(n-3)
2 + 1, then, again, we have a proper hamiltonian path P ′ in G ′c . Now, if the path either starts or ends with a color different from red, we trivially add v to the path. If not, both of them finish with red. Now, if we can take any parallel edge of these without losing the property of being properly colored we have again that it is easy to add v to the path. Otherwise, we have, without losing generality, the degree in some color, say c 1 of the first vertex of the path, say w, is at most n -3. So, we are in the same case as the first one, since we take the graph G c -{v} and we delete from w the edges in that color c 1 and in another color in order to have w monochromatic not in red. By this, we delete n -1 + n -2 + n -3 = 3n -6 edges and finally, the result follows exactly as in the first situation.

-v has one neighbor, say w, in at least two colors. This case is analog to the last one just taking the graph G c -{v} and deleting from w edges in the appropriate two colors in order to have it monochromatic. By this we delete 3 + 2(n -2) = 2n + 1 edges and we have that

(n-1)(n-2) 2 + 1 -(2n + 1) ≥ (n-2)(n-3) 2 + 1 for n ≥ 7.
As n = 5 was proved the case is finished. To finish the case that P is not a proper cycle, we have just to check when n = 7 and p = n-3 2 . So, let P = x 1 y 1 x 2 y 2 be the path compatible with the matching, let uv be the edge of the matching not in the path and let w be the non-matched vertex. Since P is maximum with the compatibility property, we cannot add the edge uv to it and since P is not a proper cycle we can check that either the degree of x 1 is at most 10 or the degree of y 2 is at most 10. So, if we contract without losing generality x 1 , y 1 and one neighbor of x 1 or y 1 we will delete as most 25 edges. We can check that the new graph satisfy the hyphotesis of the theorem for n = 5 and the result will hold with the same argument for the contraction as before.

-P is a proper cycle. As before, 2(2pn -2p -4p2 ) ≤ 3n -4 must be satisfied, and this only happens, again, if n = 5, 7 and p = n-3 2 . The case where 2p = n -1 it is trivial, since as the graph is connected, the only vertex outside the path, must be connected to it, and therefore, as the path is also a cycle, we can obtain a proper hamiltonian path just taking this vertex connected to the path and the following the cycle in the appropriate direction. For the case n = 7 and p = n-3 2 , let P = x 1 y 1 x 2 y 2 be the path compatible with the matching (that is also a proper cycle), let uv be the edge of the matching not in the path and let w be the non-matched vertex. Now, as P is maximum, we cannot add the edge uv to it, so we have just red edges between the vertices u, v and the vertices in the path, meaning that the degree of u and v is at most 10. So, we will contract the vertices u, v and one neighbor of u or v, as we did before. By this, we will delete at most 25 edges. The result will hold with the same argument as last case. Theorem 6.3.6 is the best possible. Indeed, consider a rainbow complete graph on n -1 vertices with c colors and add a new isolated vertex x. The resulting graph, although it has c (n-1)(n-2) 2 edges, contains no proper hamiltonian path, as x cannot belong to such a path.

Notice that in the above theorem there is no condition guaranting the connectivity of the underling graph. Démonstration. Before starting the proof, we will show that if there is no almost perfect matching in some color, then there are at least 15 missing edges in that color. We will use this result through the proof. CLAIM : If there exist a maximum matching M i such that |M i | ≤ 3 then there are at least 15 missing edges. Proof of the CLAIM : So, suppose that there is a maximum matching M in some color of size less than or equal to n-3 2 = 3 and just consider the subgraph of G c in that color. We distinguish two cases.

-There exist 2 isolated vertices. Then, clearly, there are at least n -1 + n -2 = 2n -3 = 15 missing edges. -There are no 2 isolated vertices. We consider that M is maximum. Let q be the number of edges of M , then M = {e 1 , • • • , e q } with 1 ≤ q ≤ 3. So, there are n -2q vertices without incident edges between them. Let I be the set of those vertices. Since any arbitrary couple of vertices in I can not have more than two incident edges with e j 1 ≤ j ≤ q, then the number of missing edges between M and I is at least

2q( n-2q
2 ) n-2q-1 = q(n -2q). This number is justified by the number of combinations of 2 among |I| = n -2q and each vertex of I in this number is counted n -2q -1 times. Inside of I, the number of missing edges of color r and b is (n -2q)(n -2q -1). Since I is an independent set, the number of missing edges is (n-2q)(n-2q-1)

2

. Adding up both numbers, we have at least (n-1)(n-2q) 2 = 8(9-2q) 2 missing edges. For q = 1, 2, we have 8(9-2q) 2 ≥ 20 missing edges and for q = 3, just 12. Therefore, we will study the case q = 3. Observe the subgraph induced by the vertices of M . Let us assume first that there are two edges of M such that the underlying subgraph is a clique. Otherwise we have 3 missing edges to add to Ḡ. We distinguish now the following cases :

-One vertex of I has 2 neighbors in the clique.

That implies that there is no edge between the others 2 vertices of I and the clique, so the number of missing edges grows up to 4. In same time, since at least one vertex from 2 vertices of I are connected to the edge not in the clique then the second extremity is not connected to the clique and results in 4 more missing edges. The total number in this case is 8(9-2q) In same time if the 3 vertices in I are connected to the edge not in the clique then the second extremity is not connected to the clique and results in 4 more missing edges, otherwise if not, then 3 more edges are missing. That give us 8(9-2q) 2 +3+4 = 12+7 = 19 or 8(9-2q) 2 + 3 + 3 = 12 + 6 = 18 missing edges. So we can conclude that the minimum number of missing edges is when M has no clique of size 4, and therefore there are at least 8(9-2q) 2 + 3 = 15 missing edges if M = 3.

Since there are no more cases to observe we conclude that if there is no almost perfect matching, there are always 15 missing edges. This completes the proof of the CLAIM.

Through the proof, if M i,j is a matching in colors i and j, we will denote the path compatible with it as P i,j = x 1 y 1 , . . . , x p y p Also, suppose now that |M i,j | = n-1 2 and v is the unmatched vertex. Let v k w k be the edges in M i,j for k = 1, . . . , n-1

2 . If we have an edge vv k in color i (or j) from v to some vertex v k in the matching and the color of the edge v k w k is j (i) we will use the set M ′ i,j = M i,j ∪ {vv k }. Since m ≥ 72 we have that E( Ḡc ) ≤ 36, so we will count the number of missing edges if G c has no proper hamiltonian path or it if has no matchings in different sizes. If we arrive to 37 missing edges we will have a contradiction.

We have then, the following cases :

-There exist two colors, say i and j, such that ∄M ′ i,j = n+1 2 . Take the matching M i,j such that ∄M ′ i,j = n+1 2 . Let v be the unmatched vertex. Since it is not possible to create M ′ i,j there are no edges with color i from v to the edges with color j of the matching, and vice-versa. Another situation is when v has no incident edges at all with colors i and j. Otherwise observe that if it exist an edge in color j from v to an edge in color j (or both in color i) of the matching then, we can replace that edges of the matching by the edge incident to v and restart the operation. Observe that we can continue to find new vertices with bad connections until we return to the same vertices. So if we can not create M ′ i,j , is because it exists a set of vertices S, such that there are no edges in color i between S and the edges in color j of the matching, and vice-versa. Observe that for each edge in M i,j , if v has an edge in color j connecting v with an edge in color j also, of M i,j in one extremity say v ′ , then the second one, say v ′′ , can take his place. Let us suppose vv ′ is in color j. So at this step |S| = 2 and there are 2 edges going to same vertex v ′ . We have three possibilities : -There is one more edge from v going to v ′′ in color j in this case S = {v, v ′ , v ′′ } and suppose there are no more edges in color j and i from vertices in S to the rest of edges in M i,j . In this case there are 2(3(n -3)) = 36 edges in color i and j missing between S and V (G c ) \ S, and 3 more in color i inside S. That makes 39 in total, a contradiction with the number of edges in G c .

-There is one more edge from v or v ′′ going to another edge of the matching with extremities w and w ′ . In this case S = {v, v ′′ , w}. Then there are 2(3(n -5)) = 24 missing edges between S and V (G c ) \ (S ∪ {v ′ , w}), 6 missing edges inside S and 6 missing edges parallel to those that can exist between S and {v ′ , w}. That gives us the total of 36 missing edges. Since, we can suppose that the rest V (G c ) \ S are complete in color i and j, there is a cycle in two colors i and j of order n -3. Now, using edges in color k and j we can insert say v in this cycle, after that using color k we can add the rest of vertices w and v ′′ to some edge in color i or j not incident to the edge of color k used to insert v. Unless there is at least one more missing edge that will give us a contradiction. -There are no more edges at v and v ′′ . Then in this case there are 2(2(n -3)) = 24 edges in color i and j missing and 2 more edges in color i parallel to vv ′ and v ′ v ′′ plus 2 between them, that give us 28 missing edges. Therefore the maximum matching in color i has size at most n-3 2 . Suppose now that M j = n-1 2 . Take the longest proper path P j = x 1 y 1 , . . . , x p y p compatible with M j using just the edges in colors j and k. The worst case is when |P j | = n -1. Let x be the vertex outside the path. If this path is also a proper cycle then we can find a proper hamiltonian path since G c is connected. Otherwise, there are, by Lemma 6.2.3, 2p -2 = n -3 = 6 missing edges in color k, and since we can not add x to P j there are 2 more missing edges in color k. Now, if we have the edge xy 1 with color k we can not have the edge x 1 x 2 in color k, otherwise we would have a proper hamiltonian path. Therefore there is one more missing edges and with it, we arrive to 37, a contradiction. The case when M i ≤ n-3 2 and M j ≤ n-3 2 as it is the same when the set S has size 1, will be analyzed later. As we have just showed, it remains to study just the case |S| = 1. Suppose that colors i and j are colors red and blue (r and b respectively) and suppose that given a matching M r,b , we have that ∄M ′ r,b = n+1 2 . Let v be the vertex outside M r,b . It is clear that there are 16 missing edges in colors r and b from v. Now we have the following situations depending on the size of maximum matchings in colors r and b.

-|M r | = |M b | = n-1
2 . Let P r and P b the longest proper paths compatibles with matchings M r and M b . Clearly v is outside both paths.

Now, if |P r | < n -1 (analog if |P b | < n -1)
then by Lemma 6.2.4 there are at least 2(2n -6) missing edges in colors b and g. Therefore, adding this number with the 16 missing edges from v we have 24 + 16 = 40 missing edges. A contradiction. If |P r | = |P b | = n -1 we have two situations. If any of them is also a proper cycle, it is clear that G 3 has a proper hamiltonian path, since it is connected therefore there exists at least one edge from v to the cycle. Otherwise, if neither P r nor P b are proper cycles, by Lemma 6.2.3, there are at least 3(n -3) = 18 missing edges in colors r, b and g. Also there are no edges in color g from v to the extremities of the paths. This adds two more missing edges (because the extremities of both paths can be the same vertices). Adding up all these numbers with the 16 missing edges in colors r and b from v, we arrive to 36 missing edges. Finally, for finding the last missing edge to get a contradiction, take for example the path P r . If we have the green edge vy 1 then we cannot have the green edge x 1 x 2 otherwise we obtain the proper hamiltonian path vy 1 x 1 x 2 , . . . , y p . Also, if we have the green edge vx 2 then we cannot have the green edge y p y 1 otherwise we obtain the proper hamiltonian path vx 2 , . . . , y p y 1 x 1 . Now, since in Lemma 6.2.3, just one of the green edges x 1 x 2 , y p y 1 were counted, we have at least one more missing edge, giving us 37 missing edges in total. Again a contradiction.

-|M r | = n-1 2 and |M b | ≤ n-3
2 (the symmetric case is analog). For this case we will observe the size of the maximum green matching

M g . If |M g | ≤ n-3
2 we have that there area at least 15 green missing edges and the same for M r . Also since we have 8 red missing edges from v (the blue ones might be already counted) we arrive to 38 missing edges. A contradiction. Now, consider when |M g | = n-1 2 . Let P r and P g the longest proper paths compatibles with matchings M r and M g . Clearly v is outside P r . If |P r | < n -1 we have by Lemma 6.2.4 that there are at least 24 missing edges in colors b and g. Adding these edges with the 16 missing edges from v we arrive to 40 missing edges. A contradiction. Now if |P r | = n -1 we have, as before, that clearly P r is not also a proper cycle. We obtain by Lemma 6.2.3 that there are 12 missing edges in colors b and g. Moreover, there are two green missing edges between v and the extremities of the path P r . Now, if we have the green edge vy 1 then we cannot have the green edge x 1 x 2 otherwise we obtain the proper hamiltonian path vy 1 x 1 x 2 , . . . , y p . Also, if we have the green edge vx 2 then we cannot have the green edge y p y 1 otherwise we obtain the proper hamiltonian path vx 2 , . . . , y p y 1 x 1 . Now, since in Lemma 6.2.3, just one of the green edges x 1 x 2 , y p y 1 were counted, we have at least one more missing edge. As same argument applies for green edges vy 2 or vx 3 , and vy 3 or vx 4 , we have at least 3 more missing edges. If |P g | < n -1, we have, by Lemma 6.2.4, that there are at least 12 edges in color red. Finally, we have 8 more missing edges from v in color b (red ones might be already counted). Adding up all these numbers we arrive to 37 missing edges, therefore, a contradiction. The last case to observe is when |P g | = n -1. We have already counted from P r , 17 missing edges in colors b and g. Now, same arguments apply to P g counting red missing edges. That give us 6 missing edges if P g is not a proper cycle, 2 more missing edges between the non-matched vertex and the extremities of P g and 3 more between the non-matched vertex and the interior vertices of P g . So we have 17 + 11 missing edges from the paths and 8 more edges in blue color from v (again red ones might be already counted). This gives us 37 missing edges. For the last one, observe that, since |M b | ≤ n-3 2 we have that at least one blue edge is not parallel to the edges in the matching M r , otherwise we have that

|M b | = n-1
2 . Finally, we have 37 missing edges. A contradiction.

-|M r | ≤ n-3 2 and |M b | ≤ n-3 2 .
Since both matchings are of size lower or equal than n-3

2
we have that there are at least 15 red missing edges and 15 blue ones. Suppose now that there exists a matching in two colors r and b of size n-1 2 . Take the longest proper path P r,b compatible with this matching. If |P r,b | < n -1 we have by Lemma 6.2.4 that there are 12 green missing edges therefore 42, a contradiction. If |P r,b | = n -1, it can not be a proper cycle, so by Lemma 6.2.3 there are 6 green missing edges and 2 more between v and the extremities of the path, therefore 38. Again a contradiction. Now, it does not exist a matching in two colors r and b of size n-1 2 . As colors r and b have at least 15 missing edges there are at most 21 edges in color r and 21 edges in color b. Therefore, there are at least 30 edges in color g where the maximum number is 36, so the green subgraph is almost complete. Let S r be the set of unmatched vertices by the red matching and let S b be the set of unmatched vertices by the blue matching. Clearly, S r and S b are independent sets in colors r and b respectively. So, if there are not blue edges between S r and M r , and no edges between S b and M b we have 42 missing edges. A contradiction. Then, there must be at least 6 edges between any of them. Suppose that we have a blue edge between S r and M r . Since the green color is almost complete, it is easy to find a proper hamiltonian path using those 6 edges alternating with the matching edges and the green ones.

-For all colors r, b, g we have that

M ′ r,b = M ′ r,g = M ′ b,g = n+1 2
. Let e i , f j the adjacent edges that complete the sets M ′ i,j for i = j, i, j ∈ {r, b, g}. Take the longest proper paths P i,j , compatibles with these matchings, where the edges e i , f j are taken as a single one. Clearly, for all colors i, j, |P i,j | < n -1. Now, by Lemma 6.2.4 we have that if each path P i,j is not a proper cycle, then there 12 are missing edges in color k, where i = j = k. If at least one of them is a proper cycle, it is trivial to find a proper hamiltonian path. That gives us 36 missing edges. Suposse that we have two colors say i, j such that we can find a proper path compatible with M ′ i,j of size 5, without using the edges e i , f j . Denote the endpoints of the edge e i by x and z, and the second endpoint of the edge f j by w. Then we will find one more missing edge in color k by the following observation. If e i f j (vzw) has an edge in color k from x 1 to z and an edge in color k from v to w, we can find the proper hamiltonian path P = wvzx 1 y 1 . . . x p y p . Otherwise, there is one more missing edge and therefore a contradiction. Otherwise suppose that for all colors we can not find such a paths. Then suppose |P i,j | = 3 and there the double edge vzw and the edge e 3 outside the path. We will count again the missing edges in color k. As we have so far 24 missing edges in colors i and j, we need at least 13 in color k to arrive to a contradiction. We distinguish two situations : -P i,j is a cycle. Then there are no edges in color k between P i,j and e 3 . This give us 8 missing edges. Suppose the case that there are no edges in color k between the extremities of e 3 and v, and between the extremities of e 3 and w. In this case there are 4 missing edges more in color k. Observe that P i,j and vzw form a cycle, otherwise there is at least one more missing edge. If there is no cycle P i,j with vzw, then we have 13 missing edges in color k, a contradiction. If there is a cycle P i,j with vzw then we can use the edges in color k from the extremities of e 3 and z to create a hamiltonian path, unless they are missing. In this case we arrive to 14 missing edges and again a contradiction. -P i,j is not a cycle. In this case, by Lemma 6.2.4, there are 8 missing edges in color k in interior of P i,j and from e 3 to the extremities of P i,j . There are at most 2 edges in color k from e 3 to the edge y 1 x 2 . We have now the possible situations :

(1) Suppose that both extremities of e 3 are connected with edges in color k to y 1 . And suppose that there are 2 edges connecting e 3 to v. In this case there are 2 missing edges from e 3 to w. Since we can not add e 3 and vzw directly to P i,j there are 3 missing edges more from w to x 1 , y 2 and x 2 . That give us 13 missing edges in color k. A contradiction.

(2) Suppose that both extremities of e 3 are connected with edges in color k to y 1 . And suppose that there are 4 edges connecting e 3 to v, and w. In this case there are 3 missing edges from w to x 1 , y 2 and x 2 , We have the same situation at v. That give us 14 missing edges in color k. A contradiction.

(3) Suppose that both extremities of e 3 are connected with edges in color k to y 1 . And suppose that there are no edges in color k connecting e 3 to v, and w. In this case there are 4 missing edges more. Denote the endpoints of e 3 by x 3 and y 3 . Suppose that there is and edge from any of them in color k to z. Suppose without losing generality that we have the edge y 3 z. Also suppose that there are the edges vx 2 and y 2 w, in color k. Then we have the hamiltonian path P = x 1 y 1 x 3 y 3 zvx 2 y 2 w, otherwise there is at least one more missing edge. A contradiction.

(4) suppose e 3 are not connected to P i,j . Then there are 2 more missing edges that give us a total of 10 missing edges in color k. If we suppose that there is no edges from e 3 to vzw then we have contradiction since there are 4 more missing edges. Suppose there is edges from extremities of e 3 to v. Since there is no to w there are 2 missing edges more in color k. There are 2 more missing edges from w to x 1 and y 2 . Suppose now that there are edges from e 3 to v and w then there can not be edges from v and w to x 1 and y 2 , that give us 4 more missing edges in color k. By this a contradiction.

(5) Suppose that one extremity of e 3 , say x 3 , is connected to y 1 and x 2 with edges in color k. We restart to count the missing edges in color k. There 3 missing edges in P i,j , 4 more missing from y 3 to P i,j , and 2 more from x 3 to x 1 and y 2 . This give us in total 9 missing edges in color k. If we suppose that there are no edges from e 3 to vzw then we have contradiction, since 4 missing edges more are added. Otherwise, if there are edges only to v, then there are no edges wx 1 and wy 2 in color k, or if there are edges from e 3 to w and v, then there can not be edge from wx 1 , wy 2 , vx 1 and vy 2 . Therefore a contradiction.

Suppose now that |P i,j | = 1 for all i = j, i, j ∈ {r, b, g}. This can arrive when the matchings are parallel. Therefore there are no edges between the edges of the matchings in any color. Now, we can see that these edges are fully connected to the vertices v, z and w. Then we can take alternatively the edges in colors i, j and k to make a hamiltonian path. Since we covered all cases, the proof is completed. . By addition, we have at least 9(10 -2q) missing edges. Getting a contradiction about the number of edges, means to have the number of missing edges greater than 41 which the maximum number of edges of Ḡc . For q = 1, 2, we have 9(10 -2q) > 41.

Then we have to study the particular case of q = 3. We shall focus our attention on the induced subgraph of M r,b . Let assume there are two edges of M r,b such that the underlying subgraph is a clique. Otherwise we add at least 6 more missing edges and by this we arrive to contradiction.

We distinguish now the following cases :

-Case 1 : One vertex of I has 2 neighbors in the clique That implies there no edge of color r or b between the others 3 vertices of I and the clique, so the number of missing edges is 24. The number of missing edges of color r and b inside of I is 12. We know also that the minimum number of missing edges between I and an edge of M r,b (we consider the no used edge of the clique) is 8. Then at least the total number of missing edges is 44.

-Case 2 : No vertex of I has more than one neighbor in the clique Then, for any vertex of I, the number of missing edges with the clique is at least 6. Then between the clique and I, the number of missing edges is at least 24. By adding the missing edges inside of I and since any arbitrary couple of vertices in I can not have more than two incident edges with color r and b with the edge of M r,b not considered in the clique, we obtain at least 44 missing edges.

In both cases, the number of missing edges is greater than 41. In conclusion, we say that |M i,j | ≥ 4, ∀i, j ∈ {r, b, g}, j = i. Therefore the claim is proved. Now let prove by contradiction that G c has no proper hamiltonian path.

-Let us assume that there is a perfect monochromatic matching in some color We will prove that G c has a proper hamiltonian path. Without loss of generality, let set

|M r | = n 2 and |M b,g | ≥ n-2
2 and let us consider the longest compatible paths P r and P b,g with M r and M b,g . Let us set P r = x 1 y 1 x 2 y 2 • • • x p y p and P b,g = s 1 t 1 s 2 t 2 • • • s p ′ t p ′ such that c(x i y i ) = r , c(s j t j ) = b or g. Let us assume that there is no proper hamiltonian path in G c . We are going to count the missing edges into the two following cases :

Let us first suppose that |M r | = n 2 and |M b,g | = n 2 .

(1) No one of the paths P r or P b,g is a cycle Since P r is not a cycle, by lemma 6.2.4, there are at least 2(n -2 + pn -2p 2 ) missing edges in G c in color b and g. There are also at least n -2 + p ′ n -2p ′2 missing edges in color r because P b,g is a not cycle. By addition the missing edges are at least

f (p, p ′ ) = -4p 2 + 2np -2p ′2 + np ′ + 3n -6
Now we compare f (p, p ′ ) with 5n -9 witch is the maximum number of Ḡc . It suffices to have the minimum value of f (p, p ′ ) greater than 5n -9 for to conclude that there is a proper hamiltonian path. We also know that 1 ≤ p, p ′ ≤ n-2

2

By studying the optimum of the function f (p, p ′ ), we find the minimum value for (1, 1) and ( n-2 2 , n-2 2 ).

f (1, 1) = f ( n -2 2 , n -2 2 ) = 6n -12
Since 6n -12 ≥ 5n -9 for n = 10, then we say G c has a proper hamiltonian path in this case.

(2) At least one of the paths P r or P b,g is a cycle we have to solve the three following combinations : -P r is a cycle and P b,g is not a cycle -P r is a not a cycle and P b,g is a cycle -P r is a cycle and P b,g is a cycle We apply the lemma 6.2.4 to obtain, the functions g 1 (p, p ′ ), g 2 (p, p ′ ) and g 3 (p, p ′ ), 1 ≤ p, p ′ ≤ n-2 2 , for the different combinations.

g 1 (p, p ′ ) = 2(2pn -4p 2 ) + n + np ′ -2p ′2 -2 g 2 (p, p ′ ) = 2n + 2np -4p 2 -4 + 2p ′ n -4p ′2 g 3 (p, p ′ ) = 2(2pn -4p 2 ) + 2p ′ n -4p ′2 we have g 1 (p, p ′ ) -f (p, p ′ ) = 2(g 2 (p, p ′ ) -f (p, p ′ )) = 2np -2n -4p 2 + 4 ≥ 0 if p ∈ {1, • • • , n 2 -1}. we have also g 3 (p, p ′ ) -f (p, p ′ ) = (g 1 (p, p ′ ) -f (p, p ′ )) + (g 2 (p, p ′ ) - f (p, p ′ )) ≥ 0.
We conclude that in this case G c has a hamiltonian path.

-Now we deal with the case that there is no monochromatic perfect matching Let say that the color r is the dominating color in G c . It verifies the number of edges of the conditions of lemma 6.2.5. It means that the underlying subgraph on color r is not connected. It will be easy to verify this subgraph contains a big component of size n-1 and an isolated vertex z. Since G c is connected, it is trivial that there is an edge in some color on z, such that this edge and the component have by lemma 6. 2 . Then it implies that d g (z) = 0. Now we are going to distinguish to cases : first we consider that |M g | = n-2 2 and second

|M g | < n-2 2 Case 1 : |M g | = n-2
2 . Let us consider the longest compatible path P g = x 1 y 1 x 2 y 2 • • • x p y p using edges from M g and edge of color r. From |M r,b | = n 2 , the missing edges in color g are 2n -4 in which we add one parallel edge to M r,b . Since M b is not perfect, there are n -1 missing edges in color b. There are n -1 missing edges in color r, since d r (z) = 0. Let assume that the order of P g is smaller than n -2. It means that there is at least one edge of color g out of the path P g and two vertices among which z. The missing edges in color r from P g are at least 2n -8 since we cannot add the last edge. So the total number of these missing edges is 6n -13 which 47 for n = 10. Now let suppose that the order of P g is exactly n -2. So, the vertex z and an other vertex, say z ′ are out of the path P g . Clearly, there is no edge in color b between z and z ′ , otherwise |M b,g | = n 2 which is not possible. Now let us suppose that P g is a cycle Let assume that z ′ cannot be inserted in P g so as to obtain a new proper cycle containing all the vertices of G c except z. Otherwise we get a proper hamiltonian path because G c is connected and we easily link z and choose an appropriate direction. Since we cannot insert z ′ to P g , then z ′ is incident to some edges y i x i+1 (if any) of P g in only color r or only b , or z ′ is adjacent on 2 colors (parallel edges) to only one extremity of some edges y i x i+1 .

Let's denote the set of edges in color r that has monochromatic edges to z ′ by Z and the rest by (P b \ M b ) \ Z. There is no edge from z to Z otherwise we get proper hamiltonian path with z and z ′ as extremities and there is one edge from z to each edge of (P b \ M b ) \ Z to same extremity. So in total there are 4 edges missing for each edge in Z ( 2 since z ′ can not be inserted and 2 since there is no edges from z to this edges, otherwise we can brake the cycle) and 3 for each in (P b \ M b ) \ Z ( 2 since z ′ can not be inserted and 1 since there is no edges from z to second extremity of the edges, otherwise we can brake the cycle). The minimum correspond to at least 3 n-2 2 = 12 missing edges, where |Z| = 0. Let us now consider the two following subcases : -There is edge zx i in color b and edges z ′ x i-1 in color b and r

Then we can make a new path almost alternating in color g in one edge P = z ′ x i-1 y i-2 . . . x i z such that y i-1 is outside the path. If now we can insert the vertex y i-1 in place of an edges in color r from the path, where are

|(P b \ M b ) \ {y i-1 x i }| = n-2 2 -1 = n-4
2 edges in color r to consider, then we have a hamiltonian path. Since is not the case there are at least 2 edges missing in color r and b at y i-1 for each edge in (P b \ M b ) \ {y i-1 x i }, that give us n -4 edges more. So 2n -3 in color g plus n -1 in color r at z, plus 12 plus n -4, give us 4n + 4 that is a contradiction with number of edges in G c . -There is edge zx i in color b and edges z ′ x i-1 in color b or r Suppose that we cannot make a new path almost alternating in color k in one edge P = z ′ x i-1 y i-2 . . . x i z such that y i-1 is outside the path. So there are n-2 2 possibilities we miss one edge per each possibility. Then we have 2n -3 in color k plus n -1 in color r at z, plus 3 n-2 2 + n-2 2 give us 5n -8 missing edges, result contradiction. Let us suppose that P g is a not cycle. In this case by Lemma 6.2.3 there is (2p -2) = n -4 missing edges in color r, between the vertices of P g . Let consider the longest compatible path P ′ g using edges from M g and edges of color b . Suppose that P ′ g is not a cycle of order n -2, otherwise we return to previous case. By the same lemma, there are n -4 edges missing in color b between the vertices of P ′ g , otherwise we return to the previous case. There are also 2 missing edges between z and z ′ and the two endpoints of P ′ g because we can not have more than 2 edges in color b between them, otherwise we will get a proper hamiltonian path. Then we distinguish two cases : -The missing edges in color b with the extremities of P ′ g are only at z Then observe that z ′ has parallel edges in color r and b connected to both extremities that allow to make a proper cycle of order n -1. Hence it will be easy to connect z so as to get a proper hamiltonian path. Then there are 2 more missing edges at z ′ for not to have a such path.

-The missing edges in color b with the extremities of P ′ g are on z and z ′ Then there are two edges in color b going in one of the extremities of P ′ g . So it

is not allowed to have an edge of color r between the other extremity and the vertex z ′ Otherwise, we will have a proper hamiltonian path. In total, there are 2n -3 edges in color g , n -1 at z in color r, n -4 between vertices of P g in color r, n -4 in color b between vertices of P g and at least 3 between z, z ′ and the extremities of P g , plus the edge between z and z ′ in color b , and we arrive to 5n -8 missing edges that contradicts the number of edges in G c . Case 2 :

|M g | < n-2 2
Since the matchings M r and M b are not perfect, there are at least (n -1) = 9 edges missing in color r and (n -1) = 9 in color b . This correspond to 18 edges in Ḡc . For |M g | = n-4 2 there are several situations but only one doesn't give direct contradiction. It is the case that there is no clique in color g using 2 edges from M g . Following the same reasoning presented above there are only one situation that don't give a contradiction, is in the case then we miss 21 edges in color g, but this case suppose that all vertices has at least 3 incident edges in color g . Since in our case now it exist a vertex v without color g incident to it we conclude that there are 3 edges more missed in color g , that give us 24 missing edges in color g . In total, there are 42 missing edges that contradicts the number of edges in G c . Theorem 6.3.9. Let G c be a connected c-edge-colored multigraph on n vertices, n ≥ 9 and c ≥ 3. If m ≥ c(n-2)(n-3)

2

+ n, then G c has a proper hamiltonian path. Démonstration. By Lemma 6.3.5 it is enough to prove the theorem for c = 3.

The proof is by induction on n. For n = 9, 10 the theorem is true, by Theorems 6.3.7 and 6.3.8, respectively. We distinguish 2 cases, depending if the multigraph contains or not a monochromatic vertex.

(1) There exists a monochromatic vertex in G c . Let us denote this vertex by x. Clearly, there exists a vertex y ∈ N (x), such that y is not monochromatic, otherwise there is no enough edges in G c . Suppose that the edge xy is of color b. Replace the vertices x and y by a new vertex, say z, such that

N b (z) = N r (z) = ∅ and N g (z) = N g (y) (or N b (z) = N g (z) = ∅ and N r (z) = N g (y)
). This process is called the contraction.

Observe that, if the resulting multigraph on n -1 vertices is connected (as we will show later) it has a proper hamiltonian path, then, since s is monochromatic, it can only be the endpoint of the path. We can replace back s with x and y using the edge xy in color b and find the proper hamiltonian path for the initial multigraph.

We now count the edges that are deleted by contraction process. We check the maximum number of edges that are deleted by the contraction process. We delete all the edges incident to x, at most n -1, and the edges in color b and say g (or r) incident to y, at most 2(n -2). This gives us a total of 3n -5 edges. We shall choose x, y for the contraction process, such that we can guarantee that we delete at most f (n)f (n -1) = 3 (n-2)(n-3)

2

+ n -3 (n-3)(n-4)

2

n + 1 = 3n -8 edges needed for the induction hypothesis to hold.

Let us study the different cases, depending on the degree in color b at x.

-d b (x) = n -1. Consider the subgraph G c -{x}. By a Theorem of [3], if for every vertex z ∈ V (G c -{x}) , d i (z) ≥ n 2
, where i ∈ {r, b, g} , then G c -{x} has a proper hamiltonian cycle. We can conclude that in G c -{x} exists a vertex such that d i (z) < n 2 . Without loss of generality let us suppose that say d r (z) < n 2 . Then in the contraction process we delete at most n -1 edges in color b at x and n -2 in color b at z and n 2 at z in color r. In total we have (5n-6) 2 deleted edges that is less than 3n -8 for n > 10.

-d b (x) = n -2. Using the same arguments as in the previous case we can conclude that in G c -{x} there exists a vertex such that d i (w) < n 2 , w ∈ V (G c -{x}) and i ∈ {r, b, g}. If we suppose that there exists the edge xw in color b then we return to the previous case, otherwise suppose there is no edge xw in color b. Then there exists a vertex z ∈ N b (x), such that there is no edge zw in color say r. Consider the subgraph V (G c )-{x, w} vertices with the edge set in color r and b. By a Theorem of [START_REF] Abouelaoualim | Cycles and paths in edge-colored graphs with given degrees[END_REF], if for any vertex z ∈ G c -{x, w}, d i (z) ≥ n-1

2

, where i ∈ {r, b} , then G c -{x, w} has a proper hamiltonian cycle, alternating in color b. Let us see now the two possible situations :

-There is an alternating cycle in color b of order n -2. Since x is connected to all vertices of the cycle we can use edges in color g and b at w to break the cycle at one end point of some edge in color b and add x and w to the extremities of this edge.

Since G c has no hamiltonian path we conclude that there is no edges in color b and g at w. Then we can use any vertex from the cycle for the contraction process as vertex x, such that we delete at most n -2 edges in color b at x and at most 2(n -3) in color b and g at the vertex in the cycle. Therefore there are 3n -8 deleted edges and the contraction process can be done. -There is no alternating cycle in color b of order n -2. If such is the case, then there exists a vertex z ′ such that the degree in color say r : d r (z ′ ) < n-1

2

. If we suppose that there are the edges in color r, z ′ w then d r (z) < n+1 Using the same arguments as in the previous case, we can conclude that in G c -{x} exists a vertex such that d i (w) < n 2 , w ∈ V (G c -{x}) and i ∈ {r, b, g}. If we suppose that there exists the edge xw in color b then we return to the previous case. Otherwise suppose that there is no edge xw. Then there exists a vertex z ∈ N b (x), such that there is no edge zw in color say r. Then we delete n -3 edges in color b of x and n -2 in color b of z plus n -3 in color r of z, which gives us the total of 3n -8 that is enough for the contraction process.

(2) There is no monochromatic vertex in G c . First suppose that there exists a vertex x such that N (x) = 1. Take then x and y, where y ∈ N (x), for the contraction process. We delete at most 3 edges at x and at most 2n -4 in color say r and b at y. So in total 2n -1 edges will be deleted, which guarantees the induction hypothesis for n ≥ 8.

In what follows, we suppose that N (x) ≥ 2 for all x ∈ G c . We describe now a new contraction process, between 3 vertices. Let us take a vertex x, y and z, where y, z ∈ N (x). Suppose that the edge xy is in color b and the edge xz in color r. Replace x, y and z by a new vertex say s, such that N r (s) = N r (y), N b (s) = N b (z) and N g (s) = (N g (y) ∩ N g (z)). Suppose now that the contraction of 3 vertices has been done and the resulting multigraph is connected (we will show this later). So, as the multigraph on n -2 vertices has a proper hamiltonian path, then s can be replaced back with x, y and z using the initial edges as follows. Suppose that s is neither at the beggining nor at the end of the path and suppose that usv Let us count now the maximum number of edges that are deleted in the contraction process.

In general, we would delete at most 3(n -1) edges at x and (n -3) at y in color b, (n -3) at z in color r and (n -3) in color g at z and y, plus 3 edges between z and y. This gives a total of 6n -9 deleted edges at x, z and y.

First of all we shall find a triplet of vertices, x, z and y. We shall furthermore choose x, z and y so that we guarantee that we delete at most 3 (n-2)(n-3)

2

+ n -(3 (n-4)(n-5)

2

+ n -2) = 6n -19 edges, necessary for the induction hypothesis. So we need to find some particular vertices, x, z and y, so that we can guarantee that the number of edges that have been deleted is at most 6n-19. We should prove that there exist a vertex in G c , such that its total color degree is less than or equal to 6n -19 -(3n -6) = 3n -13 and use this vertex as x, or that exist x y and z such that the total number of deleted edges in the contraction process is less than or equal to 6n -19.

The proof is split in two cases, depending on the parity of n. In what follows we show how to find this triplet.

Let E b be the set of edges in color b and let's suppose that the color b is the dominating Let us suppose that M b = (n-2) 2 . In this case, between the vertices in the cycle and the vertices in M b \ (P ∩ M b ), there is no edges in color r and g. Suppose that there is at least 1 edge in M b \ (P ∩ M b ). Take one vertex of the edges as x. Observe that the maximum number of edges in color r and color g is when the vertices (M b \ (P ∩ M b )) ∪ (G c \ M b ) form a complete multigraph in color r and g. Then we delete at x at most n -1 edges in color b plus 2(5) edges in color r and g. So using this triplet we can do the contraction. 2 . Since there is a cycle of order n -2 we can use the edges in color r and b at vertices in (G c \ M b ) to add them to the cycle using the extremities of some edge y i-1 x i . Observe that one vertex is enough to add to the cycle, since G c is connected, we can then construct a hamiltonian path. Therefore, there are at least

(n-2) 2
missing edges in color r and g. Using this vertex as x we respect the induction hypothesis.

Let us suppose the case when P is not a proper cycle.

(a) Let us suppose that M b = n 2 . Since P is not hamiltonian there is at least one edge in M b \ (P ∩ M b ). Clearly, there are no edges in color r and g at the extremities of P to the edges in M b \ (P ∩ M b ). By Lemma 6.2.4 we can check that |P | ≥ n -4 for n ≥ 11. Let us suppose that |P | = n -4. Since we can not add one more edge from M b \ (P ∩ M b ) to P there are at most 4 missing edges in color r and g from x p+1 and y p+1 , to each edge y i-1 x i in P . There are (n-6) 2 edges y i-1 x i , this implies that there are 4 (n-6) 2 edges from x p+1 and y p+1 to the vertices in P . Suppose that x p+1 is the vertex with minimum degree in r and g. Then, there can be at most (n -6) + 4 = n -2 edges incident in color r and g at x p+1 . We can use this vertex as x for the contraction process for n ≥ 9.

Suppose now that |P | = n -2. In this case doing the same observations as below we conclude that there is a vertex with minimum degree in color r and g that has at most n edges. For |P | = n -4 the proof is analog. Then, suppose that |P | = n -2. There are two vertices, w and s in G c \ M b . We insert both vertices into the path at vertices y i-1 and x i if it is possible. We distinguish two situations :

-Suppose we add s to P . But we cannot add w to P ∪ {s}. Since, there are now (n-6) 2

edges y i-1 x i (since one edge was used to insert s) we can conclude that at w there are at most (n -6) edges in color r and g plus 6 possible edges to s and 2 of its neighbors in P . This implies that the degree in color r and g is n, therefore we can use this vertex as x for the contraction process if n ≥ 11.

-If we cannot add any vertex to P , then there are at most 2 (n-4) 2 edges in color r and g from each vertex to the internal vertices in P and at most 2 to the extremities of P . Observe that there can not be more than 2 edges from one vertex to the extremities of P without having a cycle of order n -1, that will contradict the hypothesis of the nonexistence of a proper hamiltonian path in G c . By this observations we have that at, for example, w there are at most n -2 edges in color r and g and so we can use this vertex as x for our contraction process.

-Let us suppose that n is odd, and M b ≥ n-1 2 . Clearly there are no edges in color r and g at the extremities of P to the edges in M b \ (P ∩ M b ). By Lemma 6.2.4 we can check that |P | ≥ n -7 for n ≥ 11. Then let us distinguish two cases : -P is also a proper cycle. In this case, as for the case of n even, we can observe that if |P | = n -7 we can use a vertex from P as x for our contraction process. If n -5 ≤ |P | ≤ n -3 there are at most 5 vertices in G c \ P , therefore there can be a vertex w in M b \ (P ∩ M b ) with maximum degree in color r and g at most 2(4). This vertex can be used for the contraction process if n ≥ 10. -P is not a cycle. Since P is not cycle we will analyze the degree in color r and g at x 1 and y p . There since there are no crossing edges x 1 x i and y i-1 y p we can suppose that there are 2p -2 missing edges in both colors at x i and y p . There is 2 more missing edges between x 1 and y p . We can say that there exists a vertex with minimum degree in color r and g between x 1 and y p . This vertex has the missing degree in color r and g at most (2p-2) 2 + 2. Then we distinguish two cases depending on size of

M b \ (P ∩ M b ) (a) (M b \ (P ∩ M b )) = ∅.
Then, since p ≤ (n-3) 2 , there can be 2 more edges in color r and g to the vertex in G c \ M b . Let suppose without loosing generality that x 1 is the vertex with the minimum degree. Then the total degree at x 1 is 2

(2p -1) + 2 -(2p-2) 2 -2 = (6p-2)

2

. For the maximum value of p ≤ (n-5) 2 , (6p-2) 2 is less than or equal to 2n -12 and it is for n ≥ 11. Suppose now that p = (n-3) 2 then we use the fact that where can be at most 4 edges to each edge y i-1 x i from x p+1 y p+1 we can affirm that there are at most n -5 edges in color r and g at the vertex with minimum degree between x p+1 and y p+1 to the vertices in P and 2(2) to the rest of vertices in G c -P . Then since n -1 ≤ 2n -12 for n ≥ 11 we can use this vertex for the contraction process.

(b) (M b \ (P ∩ M b )) = ∅. In this case p = (n-1)
2 . If we cannot add the last vertex to P , it is because its degree in color r and g is at most n -3, so we can use this vertex for the contraction process.

We now check the connectivity of the resulting multigraph after the contraction of 2 and of 3 vertices.

Observe that, if the resulting multigraph on n -1 vertices is connected (as we will show later) it has a proper hamiltonian path by Theorem 6.3.9, then, since s is monochromatic, it can only be the endpoint of the path. We can replace back s with x and y using the edge xy in color b and find the proper hamiltonian path for the initial multigraph.

-There is no vertex x in G c , such that N r,g,b (x) = 1.

Let us suppose that there are 3 vertices x, y and z, such that xy is in color, say b, and xz in color, say r. Then we will replace x, y and z by a new vertex s, such that N r (s) = N r (y),

N b (s) = N b (z) and N g (s) = N g (y) ∩ N g (z)
. Clearly, we will delete at most 3(n -1) edges incident to x, n -3 edges in b incident to y, n -3 edges in r incident to z, and n -3 in color g incident to y and z, plus 3 edges between y and z. In total we will delete at most 6n -9 edges. But only f

(n) -g(n -2) = 3(n-2)(n-3) 2 + 7 -3(n-4)(n-5)

2

n + 2 = 5n -12 edges can be deleted to make the reduction. Therefore, we need to find a vertex such that its total degree is less than or equal to 2n -6. Suppose now that the contraction of 3 vertices has been done and the resulting multigraph is connected (we will show this later). So, as the multigraph on n -2 vertices has a proper hamiltonian path, then s can be replaced back with x, y and z using the initial edges as follows. Suppose that s is neither at the beggining nor at the end of the path and suppose that usv for n odd. The proof is divided in two cases depending on the parity of n. We will distinguish two sub-cases depending if the longest proper path P = x 1 y 1 x 2 y 2 . . . x i y i . . . x p y p compatible with M b is also a proper cycle or not. Let us suppose that we take M b such that the resulting P is the longest possible. Suppose that G c has no proper hamiltonian path. And suppose that the vertex x p+1 is the vertex with minimum degree (between x p+1 and y p+1 ), where x p+1 and y p+1 extremities of an edge in M b \ (M b ∩ P ).

(1) n is even.

-P is a proper cycle. By Lemma 6.2.4 we can check that |P | ≥ n -6. Clearly that there can not be edges in color r and g between M b \ (M b ∩ P ) and P . Since there are at most 3 edges in M b \ (M b ∩ P ) the degree in color r and g of any vertex is at most 2(n -2p -1).

by Lemma 6.2.4 there are 6n -26 more missing edges in color r and g. Now, since there are no 2 independent edges to those in M b \ (M b ∩ P ) there are at least 2 missing edges in color b, otherwise we will miss more parallel edges. Also there are 4 edges in color b between the extremities of the edges in M b \ (M b ∩ P ) and the unmatched vertex, since otherwise we can construct 4 different matchings and therefore 8 more missing edges. In conclusion we have at least 6 missing edges in color b and 6n -22 in color r and g, that give us a total of 6n -16 missing edges. Now suppose that there are parallel edges x 1 y 1 and x p y p in color r and g, then we can make a proper cycle of order n -1, C = x 1 y 1 . . . x p y p y p+2 y p+1 x p+1 x p+2 x 1 , if we have the edges x p y p and x 1 y 1 in color r or g, y p y p+2 and x p+2 x 1 in color b, x p+1 x p+2 and y p+1 y p+2 in color r or g and x p+1 y p+1 in color b. Now as it is trivial to attach the unmatched vertex to the proper cycle there is at least one of those edges missing and therefore a contradiction.

If |P | = n -3, we have as before that d r,g P (x p+1 ) ≤ n -5. As always there are no parallel edges in color r and g to the edge x p+1 y p+1 . Now, if there are no edges between x p+1 y p+1 and the unmatched vertex z in color r and g, then d r,g,b (x p+1 ) ≤ 2n -6. Otherwise if there are edges in color r and g to the vertex z then there can not be the edges in color b parallel to them since we could replace x p+1 y p+1 by one of these and therefore, the previous case. By rainbow degree there is an edge between z and some edge x i y i in color b. Suppose that the edge is zx i , then there can not be the edge x p+1 y i in color b, otherwise a longest proper path exists. We can conclude then that the degree in all colors of x p+1 are less than or equal to n -5 + 2 + n -3 = 2n -6. Let us suppose that |P | = n-1. Then for the same reasons as before d r,g (z) ≤ n-3. If we suppose that there are edges in color r and g between z and y i , then there can not be the edges x i z in color b, otherwise a hamiltonian path can be found. Suppose there is no such edges then either x 1 or y p have degree in colors r and g at most n -3 and all parallel. If there are at least one of the edges zy 1 or zx p in color b, then we can replace z by either x 1 or y p . In both cases we arrive that d r,g,b (x 1 ) ≤ 2n -6 or d r,g,b (y p ) ≤ 2n -6, otherwise d r,g,b (z) ≤ 2n -6. Since all cases were covered, there always exists a vertex x such that d r,g,b (x) ≤ 2n -6.

We will check now the connectivity of the resulting multigraph after the contraction of 2 or 3 vertices.

We start with the case when v has only one neighbor, say y. Let us suppose that after the deletion of two colors at y a vertex w is disconnected. This means that the rainbow degree of w is at most 2, contradicting the hypothesis of the theorem.

We continue with the case of 3 vertices.

Let us suppose that after the contraction of v, y and z, with d r,g,b (v) ≤ 2n -6, c(vy) = r The Theorem 6.3.10 is the best possible for n ≥ 9. In fact, consider a rainbow complete multigraph, say A, on n -2 vertices. Add 2 new vertices v 1 , v 2 and then join them to a vertex v of A with all possible colors. The resulting c-edge-colored multigraph has c(n-2)(n-3) 2 + 2c edges and clearly has no proper hamiltonian path. If n = 6, 8, the graphs H k,k+2 , k = 2, 3, are exceptions for Theorem 6.3.10.

Summary and remarks

In this chapter, we studied the existence of proper hamiltonian paths in edge colored multigraphs with bounded number of edges and fixed rainbow degree.

Here, the notable fact is that the proofs were sometimes long and tedious, despite the lower bounds for the edges in the considered multigraphs were really high.

It should be also interesting to study similar conditions for other patterns such as cycles, trees, etc.

Chapitre 7

Proper connection of graphs Clearly, if a graph is k-proper connected, then it is also k-connected. Conversely, any kconnected graph has an edge coloring that makes it k-proper connected ; the number of colors is easily bounded by the edge chromatic number which is well known to be at most ∆(G) or ∆(G) + 1 by Vizing's Theorem [START_REF] Vizing | On an estimate of the chromatic class of a p-graph[END_REF] (where ∆(G), or simply ∆, is the maximum degree of a vertex in G over all its vertices). Thus pc k (G) ≤ ∆ + 1 for any k-connected graph G.

An edge-colored graph is connected if the underlying non-colored graph is connected. We denote the connection number of a graph by κ(G). Throughout this chapter, all edge-colored graphs are considered to be connected unless otherwise specified. Given a colored path P = v 1 v 2 . . . v s-1 v s between any two vertices v 1 , v s , we denote by start(P ) the color of the first edge in the path, i.e. c(v 1 , v 2 ), and by end(P ) the last color, i.e. c(v

s-1 , v s ). If P is just the edge v 1 v s then start(P ) = end(P ) = c(v 1 , v s ).
This chapter is organized as follows : In Section 7.1 we study pc k (G) for bipartite graphs. We state a conjecture, prove several small results and finally we prove the conjecture for k = 1, that is, for pc(G). In Section 7.2, we study pc(G) for general graphs and prove non-trivial bounds, improving Vizing's trivial bound of ∆ + 1. Then, motivated by both of these sections, we state a conjecture regarding pc k (G) for general graphs. In Section 7.3 we prove a bound concerning the minimum degree of G. Finally we present the conclusions of our work 1 and some open problems.

show that this is the best possible. Consider the bipartition of G = V ∪ W such that |V | = n and |W | = 3. Let V = {v 1 , ..., v n } and W = {w 1 , w 2 , w 3 }. For each vertex v i ∈ V , we consider a 3-tuple C i = (c 1 , c 2 , c 3 ) so that c(v i , w j ) = c j for 1 ≤ j ≤ 3. Therefore, each vertex v i ∈ V has c 3 different ways of coloring its incident edges using c colors. We then color the edges of G as follows. If c ≥ 4 then we color the edges of (c -1) 3 vertices of V with all the different triples of c -1 colors and, for the remaining vertices, we choose different triples but this time using the c th color. If c = 3, we just choose different triples of colors but starting with the c! colorings with all three colors different. By this coloring we have that for each pair of vertices v i , v j ∈ V we have that C i = C j for all 1 ≤ i = j ≤ n.

Before proving that this coloring is 2-proper, it is easy to see that G cannot be colored to make it 2-proper connected using fewer than c colors by following the same argument as in Proposition 7.1.2. That is, if we use fewer than c colors, there must exist at least two vertices v i , v j ∈ V such that C i = C j , a contradiction. Now consider two vertices v i , v j ∈ V and we would like show the existence of 2-proper paths between them. Since C i = C j , we know that at least one of the three colors is different. If two or three are different, then we have 2-proper paths of the form v i , w k , v j and v i , w l , v j such that c(v i , w k ) = c(v j , w k ) and c(v i , w l ) = c(v j , w l ). Suppose now that exactly one of the three colors is different, say c 1 without losing generality, then v i , w 1 , v j is a proper path. For the second path, there exists a vertex v k ∈ V such that, by construction of the coloring, c(v i , w 2 ) = c(v k , w 2 ), c(v j , w 3 ) = c(v k , w 3 ) and c(v k , w 2 ) = c(v k , w 3 ). Therefore v i , w 2 , v k , w 3 , v j is a proper path between v i and v j .

Next consider w i , w j ∈ W , it is clear that there exist two vertices v k , v l ∈ V such that C k and C l have both colors different to w i , w j . Therefore w i , v k , w j and w i , v l , w j are proper paths. Finally, we consider the case where v i ∈ V and w j ∈ W . The edge v i w j provides a trivial proper path. For the second path, simply choose other appropriate vertices v k ∈ V and w l ∈ W such that v i , w l , v k , w k results in a proper path. These vertices exist by the constructed coloring of G. As no cases are left, the theorem holds. Now we prove the conjecture for complete bipartite graphs. Now we will study the case k = 1, that is pc(G). By König's Bipartite Theorem [START_REF] König | Über Graphen und ihre Anwendung auf Determinantentheorie und Mengenlehre[END_REF] we have that the edge chromatic number is ∆ for bipartite graphs and therefore ∆ is a trivial upper bound for pc(G) for any bipartite graph G. Then, we obtain this trivial corollary. Démonstration. Let u, w be two neighbors of v in G. Since we have assumed there is a 2-coloring of G so that G is properly connected, there is a properly colored path P from u to w in G. Color the edge uv so that c(u, v) = start(P ) and color vw so that c(v, w) = end(P ). Since every vertex of G has a properly colored path to a vertex of P , every vertex has a properly colored path to v through either u or w, thereby completing the proof.

The following theorem is the main result of the section. It improves upon the upper bound of ∆ by König to the best possible whenever the graph is bipartite and 2-edge-connected. Theorem 7.1.7. Let G be a graph. If G is bipartite and 2-connected then pc(G) = 2 and there exists a 2-coloring of G that makes it proper connected with the following strong property. For any pair of vertices v, w there exists two paths P 1 , P 2 between them (not necessarily disjoint) such that start(P 1 ) = start(P 2 ) and end(P 1 ) = end(P 2 ).

An ear of a graph G is a maximal path whose internal vertices have degree 2 in G. An ear decomposition of G is a decomposition P 0 , P 1 , . . . , P k such that P 0 is a cycle and P i for i ≥ 1 is an ear of P 0 ∪ . . . ∪ P k . A graph is 2-connected if and only if it has an ear decomposition. An close ear in a graph G is a cycle C such that all vertices of C expect one have degree 2 in G. An closeear decomposition of G is a decomposition P 0 , P 1 , . . . , P k such that P 0 is a cycle and P i for i ≥ 1 is either an (open) ear or a closed ear in P 0 ∪ . . . ∪ P k . A graph is 2-edge-connected if and only if it has an closed-ear decomposition.

Given a 2-connected graph G, let G 1 be an instance of the graph G \ P where P is the set of internal vertices of the last ear of an ear decomposition of a G. Similarly, if the graph is 2-edge-connected, there is a (closed) ear decomposition in which an ear may attach to the previous structure at a single vertex. Therefore, using the same argument, one could easily show the result also holds for a 2-edge-connected graph G.

Démonstration. Suppose G is 2-connected and bipartite and consider a spanning minimally 2connected subgraph (meaning that the removal of any edge would leave G 1-connected). For the sake of simplicity, we call this subgraph G. This proof is by induction on the number of ears in an ear decomposition of G. The base case of this induction is when G is simply an even cycle and we alternate colors on the edges.

Let P be the last ear added where the ends u and v of P are in G 1 and all internal vertices of P are in G \ G 1 . Since G is minimally 2-edge-connected, we know that the length of P is at least 2. By induction on the number of ears, we obtain a 2-coloring of G 1 so that G 1 has the strong property. Color P with alternating colors.

Finally we show that this coloring of G is proper connected with the strong property. Every pair of vertices in C has the strong property since C is an alternating even cycle. Also, by induction, every pair of vertices in G 1 has the strong property. Let x ∈ G \ C and let y ∈ P . The pair xu has the strong property so there exists a path Q u from x to u so that xQ u uP y forms a proper path Q ′ u . Similarly the pair xv has the strong property so there exists a path

Q v from x to v so that xQ v vP y is a proper path Q ′ v . Since C is a proper cycle, Q ′ u and Q ′ v must
have different colors on the edges incident to y. Note also that, since G is bipartite, the parity of the length of Q ′ u is the same as the parity of the length of Q ′ v . Hence, Q ′ u and Q ′ v must also have different colors on the edges incident to x. This shows that x and y have the strong property, thereby completing the proof.

As a result of Theorem 7.1.7 we obtain the following corollaries.

Corollary 7.1.8. Let G be a graph. If G is 3-connected, then pc(G) = 2 and there exists a 2-edge-coloring of G that makes it proper connected with the following strong property. For any pair of vertices v, w there exist two paths P 1 , P 2 between them (not necessarily disjoint) such that start(P 1 ) = start(P 2 ) and end(P 1 ) = end(P 2 ).

Démonstration. By [START_REF] Paulraja | A characterization of Hamiltonian prisms[END_REF], any 3-connected graph has a spanning 2-connected bipartite subgraph. Then the result holds by Theorem 7.1.7

Within this coloring, there exist two paths P 1 and P 2 from u to v such that start(P 1 ) = start(P 2 ) and end(P 1 ) = end(P 2 ). If possible, properly color P so that c(u, u 1 ) / ∈ {start(P 1 ), start(P 2 )} and c(u p , v) / ∈ {end(P 1 ), end(P 2 )}. Note that this is always possible if either P has at least 2 internal vertices or {start(P 1 ), start(P 2 )} ∪ {end(P 1 ), end(P 2 )} = {1, 2, 3}. It will become clear that this is the easier case so will assume this is not the case, namely that P has only one internal vertex x and {start(P 1 ), start(P 2 )} ∪ {end(P 1 ), end(P 2 )} = {1, 2}.

Color the edge xu with color 3 and xv with color 2 (supposing that end(P 2 ) = 2). We will show that this coloring of G is proper connected with the strong property. For any pair of vertices in G 1 , there is a pair of proper paths connecting them with the strong property by induction. Since P ∪ P 1 forms a proper cycle, any pair of vertices in this cycle also have the desired paths. Let y ∈ G 1 \ P 1 and note that our goal is to find two proper paths from x to y with the strong property.

Since y and u are both in G 1 , there exist a pair of paths P u 1 and P u 2 starting at y and ending at u with the strong property. Similarly, there exist two paths P v 1 and P v 2 starting at y and ending at v with the strong property. Since these paths have the strong property, we know that Q 1 = xuP u i y (note that the implied orientation on P u i is reversed when traversing the path from u to y) is a proper path for some i ∈ {1, 2} (suppose i = 1) and similarly Q 2 = xvP v j y is a proper path for some j ∈ {1, 2} (suppose j = 1). These paths form the desired pair if end(Q 1 ) = end(Q 2 ) so suppose start(P v 1 ) = start(P u 1 ).

Next consider walk R 1 = xuP 1 vP v 2 y and the path R 2 = Q 2 . If R 1 is a path, then R 1 and R 2 are the desired pair of paths since end(P 1 ) = c(x, v) = end(P v 2 ), meaning that R 1 is a proper walk. Hence, suppose R 1 is not a path and let z be the vertex closest to y on P v 2 which is in P 1 ∩ P v 2 . Now if the path R ′ 1 = xuP 1 zP v 2 y is a proper path, then R ′ 1 and R 2 are the desired pair of paths so we may assume that end(uP 1 z) = start(zP v 2 y).

Finally we show that the paths S 1 = xvP 1 zP v 2 y and S 2 = Q 1 = xuP u 1 y are proper paths from x to y with the strong property. Certainly, as noted above, S 2 is a proper path. Also, S 1 is a proper path since P 1 is proper so end(vP 1 z) = end(uP 1 z) = start(zP v 2 y). Finally since end(zP v 2 y) = start(P v 2 ) = start(P v 1 ) = start(P u 1 ), we see that S 1 and S 2 have the strong property.

It is important to mention that there exist 2-connected graphs with pc(G) = 3 and therefore the bound obtained by Theorem 7.2.1 is reached. Now we give an example (see Fig. 7.2) of such a graph and prove why two colors are not enough. Proposition 7.2.2. Any graph G consisting of an even cycle with the addition of three ears creating disjoint odd cycles such that each uninterupted segment has at least 4 edges has pc(G) = 3.

The assumption that each uninterupted segment has length at least 4 is mostly for convenience. Note that the graph G (in Figure 7.2) does not satisfy this condition but it can still be shown that pc(G) = 3 by a similar argument.

Démonstration. By Theorem 7.2.1, we know that pc(G) ≤ 3 so it suffices to show that pc(G) = 2. Suppose we have a 2-coloring of G which is properly connected. Label the segments of G as in Figure 7.2. Note that we may assume there are no three edges in a row of the same color within an uninterupted segment since we could switch the color of the middle edge (making that subsegment alternating) without disturbing the proper connectivity.

We would first like to show the segments A, B and C are all alternating. If two of these segments are not alternating, suppose A and B, then any vertex in D cannot be properly connected to any vertex of C so this is clearly not the case. This means that at most one segment, suppose A, is non-alternating. Suppose the edges uv and vw have the same color for some u, v, w ∈ A (see Figure 7.3). There must exist a proper path from u to w so suppose there is such a path using the segments F CEBD. Since the following argument does not rely on the parity of this path, this assumption, as opposed to using any of D ′ , E ′ or F ′ , does not lose any generality.

Let x be a vertex in the interior of B. We already know there is a proper path from x to v using D. Since D ∪ D ′ forms an odd cycle, there can be no proper path from x to v through D ′ . Let y ∈ E ′ . In order for y to have a proper path to w, it must use the segments BD (as opposed to BD ′ ) and similarly to reach u, it must use CF (as opposed to CF ′ ). Since E ∪ E ′ forms an odd cycle, and yet y can reach both u and w, we know that the edges on either side of y must have the same color. This holds for all y ∈ E ′ , clearly a contradiction. Therefore we know that A, B and C are all alternating segments.

Next we would like to show that at least one of D or D ′ must be alternating (and similarly If we let c ∈ C, then each of u, w, x and z must have an alternating path to c. Suppose the edge of A incident to a has color 1. Then both edges incident to a in Q must have color 2. This means that both edges of Q which are incident to a must be the same color (and similarly both edges of Q incident to b must have the same color). Therefore, there are exactly 4 vertices in Q for which both edges of Q have the same color. Unless x = a (or possibly z = b, u = a or w = b), this means that Q is even, a contradiction. Suppose x = a so, in order for z = b to have a proper path to w, we must also have w = b, meaning that u = a and z so again Q is even for a contradiction. Hence, we know that at least one of D or D ′ must be alternating (and similarly for the other odd ears). Without loss of generality, suppose D, E and F are all alternating.

Our next goal is to show that Q = A ∪ B ∪ C ∪ D ∪ E ∪ F forms an alternating cycle (with the possible replacement of D with D ′ , E with E ′ or F with F ′ ). As we have shown, the only places where we can have a problem is at the intersections so let a and b be (as before) the end-vertices of D (the same argument may be applied for E or F ) and suppose a is between two edges of the same colors (suppose color 1) on Q. Let u, v, w be the neighbors of a with u ∈ A, v ∈ D ′ and w ∈ D so we have assumed the edges au and aw both have color 1 (see Figure 7.5 where the darker edges represent edges that must have color 1). In order for an alternating path to get from u to w, we must either use D ′ ∪ D or Q (with the possible replacements noted above). If the path uses D ′ , then D ∪ D ′ forms an alternating (and hence even) cycle, a contradiction. Hence, we may assume there is an alternating path from u to w through BECF A (recall again that E may be replaced with E ′ or F with F ′ in this argument). Let x ∈ E ′ . There is an alternating path from u to x and from w to x. Since E ∪ E ′ forms an odd cycle but x has an alternating path through B (to get to w) and through C and A (to get to u), we know that x must have two edges of the same color within E ′ . Since x was chosen arbitrarily, this is clearly a contradiction. This means that Q is an alternating (and hence even) cycle.

Now we simply consider one vertex in each of D ′ , E ′ and F ′ . Since these ears form odd cycles, there exists a vertex in each segment from which (and to which) an alternating path can only go one direction on Q. By the pigeon hole principle, at least two of them must go the same direction, meaning there is no alternating path between them. This completes the proof of Proposition 7.2.2.

If the diameter is small, then the proper connection number is also small. More formally, we get the following result. Démonstration. If G is 3-connected, Corollary 7.1.8 implies that pc(G) = 2 so we may assume κ(G) = 2. Let C = {c 1 , c 2 } be a (minimum) 2-cut of G and let H 1 , . . . , H t be the components of G \ C. Order components so that there is an integer 0 ≤ s ≤ t such that every vertex of H i is adjacent to both c 1 and c 2 for i > s. Note that if s = 0, we have all edges from C to G \ C so G contains a spanning 2-connected bipartite graph and by Theorem 7.1.7, pc(G) = 2. For each component H i with i ≤ s, define subsets H i,1 = N (c 1 ) ∩ H i and H i,2 = N (c 2 ) ∩ H i . Since each component is connected and C is a minimum cut, there must be an edge from H i,1 to H i,2 . Let e i = v i,1 v i,2 be one such edge in each component H i . Define the graph G 0 = G[C ∪ ( s i=1 {v i,1 , v i,2 })]. This graph is 2-connected and bipartite so pc(G 0 ) = 2 and notice that |G 0 | = 2 + 2s. Let G 1 be a subgraph of G obtained by adding a vertex to G 0 which has at least 2 edges into G 0 . Furthermore, let G i be a subgraph of G obtained by adding a vertex to G i-1 which has at least 2 edges into G i-1 . By Proposition 7.1.6, pc(G i ) = 2 for all i. We claim that there exists such a sequence of subgraphs of G such that G n-(2+2s) is a spanning subgraph of G. In order to prove this, suppose that G i is the largest such subgraph of G and suppose there exists a vertex v ∈ G \ G i . Certainly every vertex which is adjacent to both c 1 and c 2 is in G i . This means v ∈ H j for some 1 ≤ j ≤ s. Since H j is connected, there exists a path from v i,1 to v within H j . Let w be the first vertex on this path which is not in G i . Since diam(G) = 2, we know that w must be adjacent to at least one vertex of C. This means that d G i (w) ≥ 2 so we may set G i+1 = G i ∪ w for a contradiction. This completes the proof.

Finally we prove an upper bound for pc(G) for general graphs which is best possible as we saw before. -B i is bipartite or 3-connected : Then by Theorem 7.1.7 and Corollary 7.1.8, B i can be colored with 2 colors having the strong property. We color B i in such a way. -κ(B i ) = 2 : Then by Theorem 7.2.1, B i can be colored with 3 colors having the strong property. We color B i in such a way. It is easy to see that G is proper connected if there are no more uncolored edges in G since each B i has the strong property. Thus, suppose that there remain uncolored edges in G. It is clear that these edges induce a forest F in G. We color them as follows. Take one of the blocks, say B 1 , which contains a vertex v ∈ B 1 which is incident with some uncolored edges. Clearly, v is an endpoint of a bridge in G. We color these uncolored edges incident to v with different colors starting with color rd B 1 (v) + 1. Then, we have that rd G (v) ≤ ∆(G). We do the same for the rest of the vertices incident to bridges in B 1 . Then, we extend our coloring for each tree going out from B 1 in a Breadth First Search (BFS) way, coloring its edges with different colors (observe from Corollary 7.1.5 that rd G (w) ≤ ∆(G) ≤ ∆ for each vertex w in the interior of a tree) until we reach the rest of the blocks. And finally, for each of these blocks (in this order), we repeat the previous step. Before proving that this coloring makes G proper connected, it is important to mention that, if we reach a block B i with some color c ≥ rd B i (w) + 1, and the corresponding vertex, say w, of B i has more than crd B i (w) uncolored incident edges, then, when we color these edges, we do not repeat color c. Also, it is important to remark that, by coloring F in this way, we have that in any path that traverses some block from one tree in F to another, at least one of the colors before or after traversing the block is not used in the block.

We now prove that G is proper connected. Let v, w be vertices of G. It is clear that if both belong to the same block B i , then there exists a proper path between them and the same happens if they belong to the same tree outside the blocks. If v ∈ B i , w ∈ B j and B i ∩ B j = {u}, then there exist two paths P 1 , P 2 between v and u in B i , and two paths P 3 , P 4 between u and w in B j with the strong property. Suppose without losing generality that end(P 1 ) = start(P 3 ) and end(P 2 ) = start(P 4 ), then we obtain the paths P 1 P 3 and P 2 P 4 between v and w. It is clear that start(P 1 P 3 ) = start(P 2 P 4 ) and end(P 1 P 3 ) = end(P 2 P 4 ) since start(P 1 P 3 ) = start(P 1 ) = start(P 2 ) = start(P 2 P 4 ) and end(P 1 P 3 ) = end(P 3 ) = end(P 4 ) = end(P 2 P 4 ). Therefore, these paths are proper. Now, if B i ∩ B j = ∅ and there is a tree T in F such that B i ∩ T = {u 1 } and B j ∩ T = {u 2 }, we form a proper path between v and w as follows. Let P 1 be the unique (proper) path in the tree T between u 1 and u 2 . Let P 2 be the proper path in B i between v and u 1 such that end(P 2 ) = start(P 1 ). This path exists since we have the strong property in each block. Analogously, let P 3 be the proper path in B j between u 2 and w such that end(P 1 ) = start(P 3 ). Finally the path P = P 2 P 1 P 3 is proper between v and w. The same idea applies if v is in a block B i and w is in a tree T in F such that B i ∩ T = {u}. The idea also applies in the case that v is in a tree T i in F , w is in a tree T j in F and there is a block B such that T i ∩ B = {u 1 } and T j ∩ B = {u 2 }. Finally, the result holds by induction on the number of trees and blocks between vertices v and w using the remark stated before to guarantee the paths always traverse the blocks. Therefore, pc(G) ≤ ∆(G) if ∆(G) ≥ 3 and pc(G) ≤ 3 otherwise.

To end the section, based on the Theorem 7.2.1 and the previous section, we extend the Conjecture 7.1.1 to general graphs. -Case n = 2p for p ≥ 2 : Take a hamiltonian cycle C = v 1 , v 2 , . . . , v 2p of G and alternate colors on the edges using colors 1 and 2 starting with color 1. Color the rest of the edges using color 1. It is clear that there are p ≥ k edges with color 2. We will prove that this coloring gives us k proper paths between each pair of vertices of G. Take two vertices v, w such that c(v, w) = 2. This edge colored with color 2 is one proper path between v and w. Now, since there are at least other p -1 ≥ k -1 edges colored with color 2 and the rest of the edges are colored with color 1, we have at least k -1 proper paths between v and w using these edges. That is, for each vertices v ′ , w ′ such that c(v ′ , w ′ ) = 2 we form the proper path v, v ′ , w ′ , w. The case where c(v, w) = 1 is similar. Finally, color the rest of the edges with color 1. We now show that this coloring gives k proper paths between each pair of vertices v and w of G. First, take v = v 1 and w = v 2 (or similarly taking w = v 2p-1 ). We have the edge v 1 v 2 and the path v 1 , v 2p-1 , v 2 . Now since n = 2p -1 ≥ 2k we have at least (p -1) -2 ≥ k -2 edges in the cycle C with color 2 different from v 2 , v 3 and v 2p-2 , v 2p-1 and therefore we form the following k -2 proper paths between v 1 and v 2 of the form v 1 , v ′ , v 2 where v ′ is an endpoint of each of these edges such that c(v 1 , v ′ ) = 2. Now take v = v 1 and w = v 3 (analog taking w = v 2p-2 ). This case is similar to the previous except changing the second formed path to v 1 , v 2 , v 3 . Suppose now that v = v 1 and w = w ′ with w ′ / ∈ {v 2 , v 3 , v 2p-2 , v 2p-1 }. We take the edge v 1 w ′ and now, since there are at least i(p -1) -1 ≥ k -1 edges in the cycle C with color 2 with endpoints different from v ′ , we form the following k -1 proper paths between v 1 and w ′ of the form v 1 , v ′ , w ′ where v ′ is an endpoint of each of these edges such that c(v 1 , v ′ ) = 2. The rest of the cases are similar to those described before in the case n = 2p forming most of the proper paths with length 3. PSTP remains NP-Complete. Next, we focused on the PSTP and WSTP in the acyclic edge-colored graphs. We gave three different ways to solve the problem if any, PST or WST, exists, and a polynomial time algorithm that finds a Maximum Proper Tree (MPT) and Maximum Weak Proper Tree (MWT) in a given acyclic edge-colored graph. The first approach consists in finding the maximum matchings for each monochromatic subgraph, and then in taking the union of edges in all the matchings and as a result we find a PST, if any. It is obvious that a PST is also a WST. However, if no PST exists in a given edge-colored graph, we can not determine whether a WST exists. As we have shown, such an easy and intuitive approach doesn't apply in the case of WST. The second approach uses an algorithm based on the Breadth-First-Search. This algorithm finds a PST in a complete acyclic edge-colored graph. In the case of general acyclic edge-colored graphs, we presented similar algorithms based on the maximum matching algorithm for edgecolored graphs. These algorithms return the MPT and MWT in case where neither PST nor WST exist in the graph. The third approach is an algorithm based again on the maximum matching algorithm, though it returns the maximum proper forest (maximum in terms of edges). This can be interesting for designing an approximation algorithm for MSTP.

We also studied sufficient conditions for edge-colored graphs to have PST. We gave some sufficient conditions related to the number of vertices, number of colors and the connectivity. We also gave a similar relation between the minimum rainbow degree of a graph, the number of colors and the order of the graph. Next, we studied the same problem in the case of the edge-colored multigraphs. Since for 2edge-colored multigraphs PST are equivalent to Hamiltonian proper paths, we give sufficient conditions based on the number of edges and of vertices for a 2-edge-colored multigraph to have a Hamiltonian path. We have done the same work for c-edge-colored multigraphs. Finally, other parallel results were presented on the coloring of simple graphs in order to guarantee some proper connectivity.

Perspectives for future work

In order to complete the study of pst and wst problems, we would like to pursue our work in the following directions. First, we would like to develop algorithms for solving the pst and the wst problems in other special classes of graphs, such as planar graphs. Second, concerning the existence of pst and wst, we would like to propose sufficient conditions in the case of edge-colored multigraphs, and to give more sufficient conditions for the wst problem in edge-colored graphs. Also, we would like to find such conditions for Proper Hamiltonian cycles in edge-colored multigraphs, as well as for cycles partition and 1-tree-cycles partition in edgecolored graphs and multigraphs. Finally, the following are conjectures we formulated and we believe we can prove, which would allow to provide necessary conditions for the wst problem. ⌉. Then G c has a weak spanning tree.

Conjecture 8.2.1, if true, is the best possible. Indeed, consider a rainbow complete graph on nkj with suppose k = 1 and j = 2, and c ≥ (n-k-j)(n-k-j-1) 2 + 1. Since k = 1 the three vertices say v, w and z such that v are connected to the complete graph on nkj vertices and to w and z only to w by edges in one color. Then the resulting graph has no WST since if you choose as root a vertex in the complete graph you can not connect w and z, if you will choose as root v you can not connect z and if you will choose w as root you can not connect the complete graph. Conjecture 8.2.2, if true, is the best possible. Indeed, consider a rainbow complete graphs on n -3k -1 ≥ k + 1 vertices, one on k + 1 and 2 on k. Choose a vertex x and y in the graph on n -3k -1 and k + 1 vertices. Connect now the x with all vertices of one graph in k vertices, y with vertices of the second one and x and y between them by edges in new color. The resulting graph has c ≥ (n-3k-1)(n-3k-2) 2 + k(k -1) + k(k+1) 2 + 1, and the rainbow degree equal k. This graph has no WST since no matter in what of the 4 graphs we will choose the root, one of the 4 graphs can not be added to the WT.

In addition, the following Conjectures 8.2.3, 8.2.4, 8.2.5 have already been mentioned previously, concern the coloring of the graph in order to guarantee the proper connectivity. In conclusion, our future work and long term objective is to provide a complete study of the pst and wst problems in different kinds of graphs and multigraphs, including sufficient and necessary conditions for pst and wst to exist, as well as approximation algorithms with performance guarantee of at least logarithm of the optimal solution.

Figure 1 . 2 -

 12 Figure 1.2 -Proper Tree and Weak-proper Tree

Theorem 3 . 1 . 13 .

 3113 [START_REF] Feng | Characterization of edge-colored complete graphs with properly colored hamilton paths[END_REF] A K c n (c ≥ 2) has a properly edge-colored Hamiltonian path if and only if K c n contains a properly edge-colored spanning 1-path-cycle subgraph.

Theorem 3 . 1 . 16 .

 3116 [START_REF] Manoussakis | Minimal colorings for properly colored subgraphs[END_REF] If c ≥ 1 2 (n -1)(n -2) + 2, then every K c n has a properly edge-colored Hamilton cycle.

Corollary 3 . 1 . 21 .

 3121 [START_REF] Abouelaoualim | Cycles and paths in edge-colored graphs with given degrees[END_REF] Let G c be an c-edge-colored graph graph, c ≥ 3. If, for every vertex x, d i (x) ≥ d ≥ 1, i ∈ {1, 2, ..., c}, then G c has a properly edge-colored path of length 2⌊ c 2 ⌋d

Theorem 3 . 1 . 23 .

 3123 [START_REF] Abouelaoualim | Cycles and paths in edge-colored graphs with given degrees[END_REF] Let G c be a c-edge-colored multigraph of order n with minimum colored degree greater than or equal to ⌈ n+1 2 ⌉. I)If c = 2, then G c has a properly edge-colored Hamiltonian cycle when n is even, and a properly edge-colored cycle of length n -1, when n is odd. II) If c ≥ 3, then G c has a properly edge-colored Hamiltonian cycle.

Theorem 3 . 1 . 27 .

 3127 [START_REF] Bang-Jensen | Alternating cycles and paths in edge-coloured multigraphs : a survey[END_REF] One can construct a maximum cycle subgraph and a maximum 1-pathcycle subgraph, respectively in a c-edge-colored multigraph G on n vertices in time O(n 3 ).

Theorem 3 . 2 . 2 .

 322 [START_REF] Bang-Jensen | Alternating cycles and trails in 2-edge-coloured complete multigraphs[END_REF] A 2-edge-colored complete bipartite multigraph is Hamiltonian iff it is color-connected and has a spanning cycle subgraph. There is an algorithm for constructing a Hamiltonian alternating cycle in a color-connected 2-edge-colored complete bipartite multigraph on n vertices in time O(n 2,5 ) (if one exists).Another condition for a 2-edge-colored complete multigraph to contain a Hamiltonian alternating cycle was obtained by Chetwynd and Hilton[START_REF] Chetwynd | Alternating Hamiltonian cycles in two-colored complete bipartite graphs[END_REF] : Theorem 3.2.3. [27] A 2-edge-colored complete bipartite graph B with partite sets U and W (|U | = |W | = n) is Hamiltonian iff B has a spanning cycle subgraph and, for every k = 2, ..., n-1 and every pair of k-sets X and Y such that

Conjecture 3 . 3 . 7 .

 337 [START_REF] Brualdi | Multicolored trees in complete graphs[END_REF] If the complete graph K 2n has a coloring of factorization, then the edges of K 2n can be partitioned into n heterochromatic spanning trees.And they proved the following. Theorem 3.3.8.[START_REF] Brualdi | Multicolored trees in complete graphs[END_REF] If the complete graph K 2n (n ≥ 3) has a coloring of factorization, then there exist two edge-disjoint heterochromatic spanning trees.

Theorem 3 . 3 . 20 .

 3320 [START_REF] Albert | Multicoloured Hamilton cycles[END_REF] Let c < 1 64 . If n is sufficiently large and k ≤ ⌈cn⌉, then every k-bounded edge-colored complete digraph -→ K n contains a heterochromatic Hamiltonian directed cycle.

  is a subgraph composed by m proper edge colored trees and a collection of proper edge colored cycles, all vertex disjoint.

Contents

4. 1

 1 NP-completeness and non-approximability results . . . . . . . . . . . 36 4.1.1 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 4.2 Maximum colored trees in acyclic edge-colored graphs . . . . . . . . 43 4.2.1 pst . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4.2.2 wst . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 4.3 Proper spanning forests in acyclic graphs . . . . . . . . . . . . . . . . 49 4.4 Random graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 4.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Figure 4 . 1 -

 41 Figure 4.1 -Graph with Ω(n 2 ) colors

  Figure 4.1.

Figure 4 . 2 -

 42 Figure 4.2 -Redution 3-SAT formula B = (x 1 ∨ x 2 ∨ x 3 ) ∧ ( x2 ∨ x 3 ∨ x 4 ) to wst (fig.(a)) and pst (fig.(b)).

Theorem 4 . 1 . 3 .

 413 The maximum proper tree problem on G c is NP-hard even for c = Ω(n 2 ).

Lemma 4 . 1 . 5 .

 415 Opt(G) = Opt(B) + 2s + 1.

Lemma 4 . 1 . 7 .

 417 Opt(G) = Opt(B) + 2s

Figure 4 .

 4 Figure 4.3 -A Complete graph on 3 colors

Theorem 4 . 1 . 9 .

 419 The pst is NP-complete for complete graphs K c n , colored with c ≥ 3 colors.Démonstration. Let G c = (V, E) be an instance of the pst problem with n nodes and c ≥ 2 colors. Construct a new colored complete graph H c+1 on 2n vertices and (c+1)-colors, as follows. Add all edges in the complement graph G c using a new color, and retain the edges of G c with their original color. Also, use the extra color to form a complete graph on a new set of n vertices as well as a complete bipartite graph between the old vertices and new vertices.

Theorem 4 . 2 . 1 .

 421 An acyclic graph G c has a pst iff the union of maximum matchings of each of the colors in χ c contains exactly (n -1) edges.

Lemma 4 . 2 . 3 .

 423 If an acyclic edge-colored graph contains a non-rainbow-yeo vertex, then it has no pst.

Figure 4 . 4 -

 44 Figure 4.4 -Acyclic Complete graph decomposition

  end if 10: end for 11: return the pst yeo-vertex and the corresponding component. The root of the subtrees are likewise computed recursively. Thus we get an auxiliary rooted edge-colored tree T .

Figure 4 . 5 -

 45 Figure 4.5 -Construction of a Maximum Proper Spanning Forest

Theorem 4 . 3 . 1 .

 431 Let G c be an acyclic c-edge-colored graph. Then, the maximum proper spanning forest problem can be solved in polynomial time in the order of G c .

Theorem 4 . 3 . 2 .

 432 Let G c be a c-edge-colored acyclic graph and d : V (G c ) → {0, 1, . . . , n -1}, an integer function. In addition, consider 0

Theorem 4 . 4 . 1 .

 441 If p = λ log n

Contents 5 . 1

 51 Some general results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 5.2 Minimum number of colors in terms of the connectivity . . . . . . . 58 5.3 Minimum number of colors in terms of the rainbow degree . . . . . 61 5.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

Figure 5 . 1 -

 51 Figure 5.1 -Rainbow partition

Figure 5 . 2 -

 52 Figure 5.2 -K-connected case

Theorem 5 . 2 . 3 .

 523 Let G c be a c -edge-colored simple k-connected graph such that |V (G c )| = n. If n ≥ k 2 + 3k + 4 2 and c ≥ (nk -1) (nk -2) 2 + k + 1, then G c hasa properly edge-colored spanning tree. Démonstration. By Lemma 5.1.2 the number of bridges on E (H 0 ) is less or equal to kp + 1. The result then holds directly from Proposition 5.2.1 and Proposition 5.2.2.

Figure 5 . 3 -

 53 Figure 5.3 -Rainbow degree case

Proposition 5 . 3 . 2 .

 532 Let G c be a c-edge-colored connected graph with rainbow degree rd (G c ) = k such that |V (G c )| = n with n ≥ k 2 + 3k + 4 2 and c ≥

Contents 6 . 1

 61 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 6.2 Preliminary results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 6.3 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 6.4 Summary and remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

Lemma 6 . 2 . 2 .

 622 Suppose that for every vertex x in G c , c = 2, rd(x) = 2 and n ≥ 14. If m ≥ (n -3)(n -4) + 3n -2, then G c has two matchings M r and M b in colors, say red and blue, such that |M r | = ⌊ n 2 ⌋ and |M b | ≥ ⌈ n-2 2 ⌉. Démonstration. Let us denote by E r (G c ) and E b (G c ) the set of edges colored in red and blue, of sizes |E r (G c )| = m r and |E b (G c )| = m b , respectively. Observe that, since for every vertex x in G c , rd(x) = 2, we have d i (x) ≥ 1 for i ∈ {r, b}.

2 -

 2 1) n is even, |M | = n 2 and 2p < n -(1a) If P is not a proper cycle, then there are at least (n -2 + pn -2p 2 )(c -1) missing edges in G c different from red and the minimum value of this function is (2n -4)(c -1) for p = n-2 2 . -(1b) If P is a proper cycle, then there are at least (2pn -4p 2 )(c -1) missing edges in G c different from red and the minimum value of this function is(2n -4)(c -1) for p = n-2 2 . -(2) n is odd, |M | = n-12 and 2p < n -1 -(2a) If P is not a proper cycle, then there are at least (n-3-p+pn-2p 2 )(c-1) missing edges in G c different from red and the minimum value of this function is (2n -6)(c -1) for p = n-3 2 . -(2b) If P is a proper cycle, then there are at least (2pn -2p -4p 2 )(c -1) missing edges in G c different from red and the minimum value of this function is(2n -6)(c -1) for p = n-3 2 . -(3) n is even, |M | = n-22 and 2p < n -(3a) If P is not a proper cycle, then there are at least (n-4-2p+pn-2p 2 )(c-1) missing edges in G c different from red and the minimum value of this function is (2n -8)(c -1) for p = n-42 . -(3b) If P is a proper cycle, then there are at least (2pn -4p -4p 2 )(c -1) missing edges in G c different from red and the minimum value of this function is (2n -8)(c -1) for p = n-4 2 .

Lemma 6 . 2 . 5 . 2 + 3 ,

 62523 Let G be a connected non-colored simple graph on n vertices, n ≥ 9. If m ≥ (n-2)(n-3) then G has a matching M of the size |M | = ⌊ n 2 ⌋. Démonstration. By a theorem of [23], a 2-connected graph on n ≥ 10 vertices and m ≥ (n-2)(n-3) 2

Lemma 6 . 3 . 2 .

 632 Let G c be a 2-edge-colored multigraph. Assume that G c , contains a proper cycle C, of length at most n -2 and that there exists a red edge xy in G c -C. If d b C (x) + d b C (y) > |C|, then G c has a proper cycle of length |C| + 2 containing xy.

2 and d b (x) ≥ n+1 2 , 2 - 1 .

 2221 then G c has a proper hamiltonian cycle for n even and a proper cycle of length n -1 for n odd. Let us suppose therefore that for some vertex, say x, and for some color, say red, d r (x) ≤ n+1 Notice now that d r (x) > 0 and d b (x) > 0, otherwise, for example if d r

2 -

 2 again a contradiction. Thus we may conclude that there are two distinct neighbors, say y and z, of x such that c(xy) = r and c(xz) = b in G c . Replace now the vertices x, y, z by a new vertex s such that N b (s) = N b G c -{x,y,z} (y) and N r (s) = N r G c -{x,y,z} (z). The obtained graph, say G ′ , has n -2 vertices and at least(n-1)(n-2)+n-[(n-1)+ n+1 1+2(n-2)] = n 2 -11n 2 +8 > f 0 (n-2) = n 2 -6n+8 edges. By the induction hypothesis, G ′ has a proper hamiltonian cycle for n-2 even and a proper cycle of length n -3 otherwise. If G ′ has a proper hamiltonian cycle then coming back to G c we may easily find a proper hamiltonian cycle in G c . Assume now that n -2 (and thus n) is odd. Let C be a proper cycle of length n -3 in G ′ . If s belongs to C, then as previously we may easily find a proper cycle of length n -1 in G c . Assume therefore that s does not belong to C. By Lemma 6.3.2, if d b C (x) + d b C (y) > |C| or d r C (x) + d r C (z) > |C|) then we may integrate the edge xy (respectively xz) in C in order to obtain a cycle of length n -1. Assume therefore that d b C b(x) + d b C (y) ≤ |C| and d r C (x) + d r C (z) ≤ |C|. But then the number of edges of G c is at most n(n -1) -(n -3) -(n -3) = n 2 -3n + 6 < f 0 (n) = n 2 -2n + 2, again a contradiction. This completes the argument and the proof.

Theorem 6 . 3 . 4 .

 634 Let G c be a 2-edge-colored multigraph. For n ≥ 14, if m ≥ f 1 (n) = n 2 -3n+4, then G c has a proper hamiltonian path. Démonstration. By Theorem 6.3.1, as n 2 -3n + 4 ≥ (n -3)(n -4) + 3n -2, it suffices to prove that for every vertex x of G c , rd(x) = 2, i.e., d r (x) = 0 and d b (x) = 0. Assume therefore that either d r (x) = 0 or d b (x) = 0 but not both, as d r (x) = 0 and d b (x) = 0 can be easily checked as not possible. In that case G cx has at least n2 -3n + 4 -(n -1) = n 2 -4n + 5 = [(n -1) -1][(n -1) -2] + (n -1)edges. By Lemma 6.3.3, G cx has either a proper hamiltonian cycle or a cycle of length n -2. If G cx has a proper hamiltonian cycle then in a trivial maner we join x and the cycle in order to obtain a proper hamiltonian path. Assume therefore that n -1 is odd and that G cx has a proper cycle, say C, of length n -2. Let y be the vertex outside C in G cx. By using arguments similar to those of Lemma 6.3.2 we may show that d r (x) + d r (y) ≤ |C| and d b (x) + d b (y) ≤ |C|. It follows that the number of edges of G c is at most n

Theorem 6 . 3 . 6 . 2 + 1 ,

 63621 Let G c be a c-edge-colored multigraph on n vertices, n ≥ 3 and c ≥ 3. If m ≥ c(n-1)(n-2) then G c has a proper hamiltonian path.

Theorem 6 . 3 . 7 .

 637 Let G c be a connected 3-edge-colored multigraph on 9 vertices. If m ≥ 72, then G c has a proper hamiltonian path.

Theorem 6 . 3 . 8 .

 638 Let G c be a connected 3-edge-colored multigraph on 10 vertices. If m ≥ 94, then G c has a proper hamiltonian path. Proof. Before beginning the proof we state the following. CLAIM : For any two colors, say r and b, |M r,b | ≥ 4 Let us assume |M r,b | < 4. We consider that M r,b is maximum. Let q be the number of edges of M r,b , then M r,b = {e 1 , • • • , e q } with 1 ≤ q ≤ 3. So there are 10 -2q vertices without edge of color r and b between them. Let I be the set of those vertices. Now we are going to count the number of missing edges of G c . Since any arbitrary couple of vertices in I can not have more than two incident edges with color r and b with e i , then the number of missing edges of color r or b between M r,b and I is at least 4q 10-2q 2 10 -2q -1 = 2q(10 -2q). This number is justified by the number of combinations of 2 among |I| = 10 -2q and each vertex of I in this number is counted 10 -2q -1 times. Inside of I, the number of missing edges of color r and b is (10 -2q)(10 -2q -1)

  2.5, a perfect matching. In this matching, all edges are in color r except one. We cannot have at the same time |M b | ≤ 3 and |M g | ≤ 3, otherwise we have a contradiction about the number of missing edges. To see this, we have to analyze and calculate like the first step. Let us assume that |M r,b | = n 2 . If |M r,g | = n 2 , then the missing edges in color g are 2n -4 obtained from the lemma 6.2.4, and one edge more not parallel to the edges of M r,b since M g is not perfect. By |M r,g | = n 2 , the missing edges in color b are 2n -3 and the missing edges in color r are n -1 since d r (z) = 0. Adding all these numbers lead to 5n -7 missing edges. Let us now suppose that |M r,g | < n

2 .

 2 Then in the contraction process we delete n -2 edges in color b at x and n -2 in color b at z ′ and n+1 2 at z in color r. In total we have (5n-7) 2 deleted edges that is less than 3n -8 for n > 9. -d b (x) ≤ n -3 Let us suppose that d b (x) = n -3. Otherwise we can use any vertex z ∈ N (x) for the contraction process, since we delete at most n -4 edges in color b at x and 2n -4 in color b and say r at z, that is enough to respect the induction hypothesis.

  is part of the path with c(us) = r and c(sv) = b. To obtain the proper hamiltonian path for the initial graph we replace usv in the path with uyxzv such that c(uy) = r, c(yx) = b, c(xz) = r and c(zv) = b. Now, if we have usv with c(us) = g and c(sv) = r (the case that c(sv) = b is similar) then we replace it with uzxyv such that c(uz) = g, c(zx) = r, c(xy) = b and c(yv) = r. The case that s is either at the beggining or at the end of the path is easy to observe.

  (c) |P | = n -2 Let us suppose that M b = n 2 . By the same observations as above we can use one vertex in M b \ (P ∩ M b ) as x for the contraction process. Let us suppose that M b = (n-2)

  (b) Let us suppose that M b = (n-2) 2

  is part of the path with c(us) = r and c(sv) = b. To obtain the proper hamiltonian path for the initial graph we replace usv in the path with uyxzv such that c(uy) = r, c(yx) = b, c(xz) = r and c(zv) = b. Now, if we have usv with c(us) = g and c(sv) = r (the case that c(sv) = b is similar) then we replace it with uzxyv such that c(uz) = g, c(zx) = r, c(xy) = b and c(yv) = r. The case that s is either at the beggining or at the end of the path is easy to observe. Suppose without losing generality that |E b | ≥ |E r | ≥ |E g |, then |E b | ≥ 3(n-2)(n-3) 2 + 3, otherwise we have contradiction with the number of edges in G c . Since the subgraph in color b is connected because of the rainbow degree of the vertices, we have by Lemma 6.2.5 that there is a matching |M b | = n 2 for n even and |M b | = (n-1) 2

Contents 7 . 1

 71 Bipartite graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 7.2 General graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 7.3 Minimum degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 7.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

Theorem 7 . 1 . 4 .

 714 Let G = K n,m , m ≥ n ≥ 2k for k ≥ 1. Then pc k (G) = 2Démonstration. Take the bipartition of G = A ∪ B. Then split each set A and B into the setsA 1 , A 2 , B 1 , B 2 such that |A i |, |B i | ≥ k for i = 1, 2. This is clearly possible since |A|, |B| ≥ 2k. Now color the graph in the following way. Put c(v, w) = 1 for all v ∈ A 1 and w ∈ B 1 , and for all v ∈ A 2 and w ∈ B 2 . Finally put color 2 to the rest of the edges, that is, c(v, w) = 2 for all v ∈ A 1 and w ∈ B 2 , and for all v ∈ A 2 and w ∈ B 1 (see Figure7.1). Now we prove that this coloring produces k proper paths between each pair of vertices of G. First, consider two vertices v, w ∈ A 1 (an identical argument holds for pairs in other sets). Since the cardinality of each set is at least k, we formk proper paths v, b 1 , a 2 , b 2 , w choosing b 1 ∈ B 1 , a 2 ∈ A 2 and b 2 ∈ B 2 . If v ∈ A 1and w ∈ A 2 (similarly for v ∈ B 1 and w ∈ B 2 ) we have at least 2k proper paths formed as v, b, w for each choice of b ∈ B. The final case is when v ∈ A 1 and w ∈ B 1 (that is, v and w are adjacent). Here we have at least k + 1 proper paths, as follows. One path is simply the edge vw while the k that remain are of the form v, b 2 , a 2 , w for each choice of b 2 ∈ B 2 and a 2 ∈ A 2 . This completes the proof.

Figure 7 . 1 -

 71 Figure 7.1 -Coloring of K 4,5 . Normal edges represent color 1 and bold edges color 2.

Corollary 7 . 1 . 5 .

 715 If G is a tree then pc(G) = ∆.We present now the following proposition. Proposition 7.1.6. If pc(G) = 2 then pc(G ∪ v) = 2 as long as d(v) ≥ 2.

Figure 7 . 2 -

 72 Figure 7.2 -Smallest 2-connected graph with pc(G) = 3

Figure 7 . 3 -

 73 Figure 7.3 -Placement of vertices.

Figure 7 . 4 -

 74 Figure 7.4 -Placement of vertices.

Figure 7 . 5 -

 75 Figure 7.5 -Placement of vertices.

Theorem 7 . 2 . 3 .

 723 If diam(G) = 2 and G is 2-connected, then pc(G) = 2.

Theorem 7 . 2 . 4 .

 724 Let G be a connected graph. Consider ∆(G) as the maximum degree of a vertex which is an endpoint of a bridge in G.Then pc(G) ≤ ∆(G) if ∆(G) ≥ 3 and pc(G) ≤ 3 otherwise.Démonstration. Let B 1 , B 2 , . . . B s be the blocks of G with at least 3 vertices. For each block of B i we have the following cases.

Conjecture 7 . 2 . 5 .

 725 If G is a 2k-connected graph with k ≥ 1, then pc k (G) ≤ 3.This conjecture is proved for k = 1 in Theorem 7.2.1. Now we prove a stronger result for complete graphs.

Theorem 7 . 2 . 6 .

 726 Let G = K n , n ≥ 4, and k > 1. If n ≥ 2k then pc k (G) = 2Démonstration.

Figure 7 . 6 -

 76 Figure 7.6 -Coloring of K 13 . Normal edges represent color 1 and bold edges color 2.

Conjecture 8 . 2 . 1 . 2 + 2 ,

 82122 Let G c be be a k-connected edge-colored graph such that n ≥ (k+j) 2 +3(k+j)-2 2 and c ≥ (n-k-j)(n-k-j-1) with j = ⌈ 1+ √ 1+4k 2

Conjecture 8 . 2 . 2 . 2 + 2 + 2 ,

 822222 Let G c be a c-edge-colored connected graph with rainbow degree rd(G c ) = k and n ≥ 4k + 2. If c ≥ (n-3k-1)(n-3k-2) k(k -1) + k(k+1) then G c has a weakly proper edge-colored spanning tree.

Conjecture 8 . 2 . 3 .

 823 If κ(G) = 2 and δ(G) ≥ 3, then pc(G) = 2. Conjecture 8.2.4. If G is a 2k-connected bipartite graph with k ≥ 1, then pc k (G) = 2. Conjecture 8.2.5. If G is a 2k-connected graph with k ≥ 1, then pc k (G) ≤ 3.

Table des matières

 des Heterochromatic subgraphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Vertex Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Edge Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.3 Heterochromatic Hamiltonian Cycle . . . . . . . . . . . . . . . . . . . . . 3.3.4 Heterochromatic Spanning Tree . . . . . . . . . . . . . . . . . . . . . . . . 3.3.5 Proper trees and weak proper trees in edge colored graphs . . . . . . . . . 3.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 Introduction 2 Notation and Terminology 2.1 Notation and terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Notions and definition of Complexity . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 The notion of Complexity . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Class P . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.3 Class NP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.4 NP-complete problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.5 Class coNP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.6 Best, worst and average case complexity . . . . . . . . . . . . . . . . . . . 2.2.7 Approximability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.8 Non-approximability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.9 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Related Work 3.1 Proper edge-colored Hamiltonian cycles and paths . . . . . . . . . . . . . . . . . 3.2 Bipartite multigraphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 4 Algorithmics aspects of PST and WST problems 4.1 NP-completeness and non-approximability results . . . . . . . . . . . . . . . . . . 4.1.1 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Maximum colored trees in acyclic edge-colored graphs . . . . . . . . . . . . . . . 4.2.1 pst . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v Chapitre 1

  Notation and terminology . . . . . . . . . . . . . . . . . . . . . . . . .

	Chapitre 2
	Notation and Terminology
	Contents
	2.1

6 2.2 Notions and definition of Complexity . . . . . . . . . . . . . . . . . . 14 2.2.1 The notion of Complexity . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.2.2 Class P . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.2.3 Class NP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.2.4 NP-complete problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.2.5 Class coNP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.2.6 Best, worst and average case complexity . . . . . . . . . . . . . . . . . . 18 2.2.7 Approximability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.2.8 Non-approximability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.2.9 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Table 2 .

 2

	Size of		Complexity function
	problem	n	n 2	2 n
	10	0.01µs 0.1µs	1.024µs
	20	0.02µs 0.4µs	1.049 ms
	30	0.03µs 0.9µs	1.074 s
	40	0.04µs 1.6µs 18.3 minutes
	50	0.05µs 2.5µs	13.0 days
	60	0.06µs 3.6µs	36.6 years
	70	0.07µs 4.9µs 374 centuries

1 -Comparison of complexity functions

  Proper edge-colored Hamiltonian cycles and paths . . . . . . . . . . 22 3.2 Bipartite multigraphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 3.3 Heterochromatic subgraphs . . . . . . . . . . . . . . . . . . . . . . . . 29 3.3.1 Vertex Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3.3.2 Edge Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 3.3.3 Heterochromatic Hamiltonian Cycle . . . . . . . . . . . . . . . . . . . . 32 3.3.4 Heterochromatic Spanning Tree . . . . . . . . . . . . . . . . . . . . . . . 33 3.3.5 Proper trees and weak proper trees in edge colored graphs . . . . . . . . 34 3.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

	Chapitre 3
	Related Work
	Contents
	3.1

This is the joint work done with Wenceslas Fernandez de La Vega, Yannis Manoussakis, Carlos Martinhon, Rahul Muthu, and Rachid Saad.

This is a joint work with Raquel Agueda, Yannis Manoussakis, Gervais Mendy and Rahul Muthu.

This is a joint work with Raquel Agueda Mate, Yannis Manoussakis, Gervais Mendy and Leandro Montero.

+ 4 + 4 = 12 + 8 = 20 missing edges.-No vertex has more than one neighbor in the clique.Then, for any vertex of I, we have 1 more missing edge and that give us

more edges.

This is a joint work with Shinya Fujita, Aydin Gerek, Colton Magnant, Yannis Manoussakis, Leandro Montero, Zsolt Tuza.

Acknowledgments

In the first place I would like to thank the committee members, as well as the committee chair. I gratefully thank Prof. Li Hao and Prof. Dominique Barth, for the interest expressed in my the-

Let us now suppose that |M r | = n 2 and |M b,g | = n-2 2 .

(1) So, in the case that P r and P b,g are not cycles, we use again lemma 6.2.4 to get missing edges

The minimum value of this function is 6n -16 and is greater than 5n -9 for all n ≥ 8. Since G c has no hamiltonian path, then p ′ = n-2 2 . Since P r is a not a cycle, there is at least 2(n -2 + pn -2p 2 ) missing edges in G c in color b and g. By lemma 6.2.3 there are also at least n -4 missing edges of color r. By considering the matching M r , we know there are at least 2 missing edges of color b and g, parallel to the edges of the matching . Finally by considering the matching M b,g , we add 2 missing edges of color r at the extremities of P b,g , otherwise it will be a proper hamiltonian path. By addition, we obtain a function with one variable

2 , the latter function is greater than 5n -9 because its minimum value is 5n -8.

(2) P r is a cycle and P b,g is a not a cycle. By lemma 6.2.4, we have g 1 (p, p ′ ) = 2(2pn -4p 2 )+(n-4+np ′ -2p ′ -2p ′2 ), for 1 ≤ p ≤ n-2 2 , 1 ≤ p ′ ≤ n-4 2 and g 1 (p, p ′ )-f (p, p ′ ) = 2np -2n -4p 2 + 4 ≥ 0. For p ′ = n-2 2 , we add n -4 (n -4 from the lemma 6.2.3) to 2(2pn-4p 2 ) There are also 2 missing parallel edges of color r and b for the extremities. In total we n -2 for the color r and b for the extremities to 2(2pn -4p 2 ) and the minimum value is 5n -8.

(3) P r is a not a cycle and P b,g is a cycle. Then g 2 (p, p ′ ) = 2(n -2 + pn -2p 2 ) + (2p ′ n -4p ′ -4p ′2 ), for 1 ≤ p ≤ n-2 2 , 1 ≤ p ′ ≤ n-4 2 and g 1 (p, p ′ )f (p, p ′ ) ≥ 0. For p ′ = n-2 2 , when we consider the cycle P b,g , for each edge of color r of P b,g , there are 2 missing edges of color r between this edge and the 2 vertices out of P b,g . In total there are n -2 missing edges in color r in which we add 2 missing parallel edges of color b and g from M r . The resultant function is 2(n -2 + pn -2p 2 ) + n and its minimum value is 5n -8.

(4) P r is a cycle and P b,g is a cycle. g 3 (p, p ′ ) = 2(2pn -4p 2 ) + (2p ′ n -4p ′ -4p ′2 ) for 1 ≤ p ≤ n-2 2 , 1 ≤ p ′ ≤ n-4 2 and we have g 3 (p, p ′ )f (p, p ′ ) = (g 1 (p, p ′ )f (p, p ′ )) + (g 2 (p, p ′ )f (p, p ′ )) ≥ 0. For p ′ = n-2 2 ,as previous, we add n to 2(2pn -4p 2 ) and the total has 5n -8 as minimum value. The contradiction about the number of edges means that G c has a proper hamiltonian path in this case.

color. The monochromatic subgraph in color b has at least m b ≥ (n-2)(n-3) 2 + ⌈ n 3 ⌉. We distinguish two possibilities. If this subgraph is connected, then by Lemma 6.2.5 there is a perfect matching in color b of the size n 2 for n even and n-1 2 if n is odd. Otherwise, if the subgraph is disconnected, then there is a matching of size n-2 2 for n even and n-1 2 if n is odd.

Let M b denote the maximum matching in E b . We deal with the longest proper compatible path P with M b .

-n is even, and M b ≥ n-2 2 . Let P = x 1 y 1 x 2 y 2 . . . x p y p be the longest proper compatible path with M b . It is easy to check that |P | ≥ 4, otherwise there is no enough edges in C c . Suppose now that P is also a proper cycle. Then we distinguish 2 cases :

Observe that between the vertices in the cycle and the vertices in G c \ P there are no edges in color r and g, otherwise a longest proper path can be found. Take the triplet x, y and z from the vertices of the cycle (consecutive). We count the edges incident to x. There is at most n -1 edges in color b. Observe that the maximum number of edges in color r and color g, is when the vertices of the cycle form a complete multigraph in color r and g. In the worst case we delete at x at most 2(n -9) edges in color r and g and n -1 in color b. We can see that there is at most 2n -18 + n -1 = 3n -19 edges incident at x in this case. Using this triplet we can do the contraction. Let us suppose that M b = (n-2)

2 . In this case as in the previous one, between the vertices in the cycle and the vertices in M b \ (P ∩ M b ), there is no edges in color r and g. As before, we take x, y and z from the vertices of the cycle. Observe that the maximum number of edges in color r and color g is when the vertices of the cycle form a complete multigraph in color r and g and then there are edges in both colors to the two unmatched vertices in G c \ M b . Then we delete at x as previous at most n -1 edges in color b plus 2(n -9) + 4 = 2n -14 edges in color r and g. Using this triplet we can do the contraction.

As previously, there is no edges in color r and g between the vertices in cycle and the vertices in G c \ P . Let us take the triplet x, y and z from the vertices of M b \ (P ∩ M b ). We count the edges incident to x. There is at most n -1 edges in color b. Observe that the maximum number of edges in color r and color g, is when the vertices of M b \ (P ∩ M b ) form a complete multigraph in color r and g. Since |P | ≥ n -6, there are 6 vertices in G c \ P , therefore in the worst case we delete at x at most 2(5) edges in color r and g and n -1 in color b. We can see that there is at most n -9 edges incident to x in this case. Using this triplet we can do the contraction.

Let us see first the case when we contract 2 vertices, v and his neighbor, say y. In this case let us suppose that after the contraction of v and y a set of vertices S is disconnected. Taking in consideration the total number of edges that has the resulting multigraph |S| is equal to one. Denote the vertex in S by q. Then there can be at most two parallel edges incidents to q. Take then q and contract with y, there is another vertex disconnected q ′ . This implies that G c has at most 3 (n-2)(n-3) 2 + 4 edges. A contradiction.

Let us suppose that after the contraction of v, y and z, where v is connected to y and to z by edges in different colors, there is set of |S| vertices disconnected from the rest of the multigraph. Taking in consideration the total number of edges that has the resulting multigraph |S| is equal to one.

Denote by w the vertex in S. Clearly, there can be at most 3 edges between w and vertices y and z, there can be at most 3 edges between w and v. Since 1 < d r,g,b (w) ≤ 6 and 1 < N r,g,b (w) ≤ 3, we can take the triplet w, y ′ and z ′ (where y ′ and z ′ in {v, y, z}) and contract it. Suppose there is one disconnected vertex w ′ . Then if w ′ = z = y = v then there are at most 3 incident edges to w ′ . Otherwise there can be at most 6 incident edges to w ′ but there are 3 of them connected to w. In conclusion the multigraph G c can have at most 3 (n-2)(n-3) 2 + 6 + 3 edges. A contradiction. Now as the connectivity is proved the theorem holds. Theorem 6.3.10. Let G c be a c-edge-colored multigraph on n vertices, n ≥ 11 and c ≥ 3. Assume that for every vertex

Proof. By Lemma 6.3.5 it is enough to prove the theorem for c = 3. As m ≥ f (n) = 3(n-2)(n-3) 2 + 7 then E( Ḡc ) ≤ 6n -16. The proof will be done by construction of a proper hamiltonian path or if it is not possible, by reduction to Theorem 6.3.9, i.e., a connected c-edge-colored multigraph on n ′ ≥ 9 and m ′ ≥ g(n ′ ) = 3(n ′ -2)(n ′ -3)

2

+ n ′ has a proper hamiltonian path. We will do this reduction by contracting 2 or 3 vertices depending if there exists a vertex x in G c , such that |N r,g,b (x)| = 1 or not.

-There exist x ∈ G c such that N r,g,b (x) = 1. Let denote the neighbor of x by y. We will replace x and y by a new vertex s, such that N r (s) = N r (y) and N b (s) = N g (s) = ∅.

Clearly, we will delete 3 edges between x and y and at most 2(n -2) edges in color b and g incident to y. This gives us a total of 2n -1 deleted edges. For the reduction hypothesis we can delete at most f (n)g(n -1) = 3(n-2)(n-3) 2 + 7 -3(n-3)(n-4)

2

n + 1 = 2n -1 edges, where n ′ = n -1. Since we delete exactly 2n -1 edges the reduction hypothesis holds.

Suppose now that there is one edge in color say r (resp. g) between x p+1 y p+1 , then using edges x p+1 x i , y p+1 y i or y p+1 x i , x p+1 y i in color b, we can insert the edge x p+1 y p+1 in color r into the cycle. Since this will contradict our hypothesis we can conclude that there are 2 missing edges in color b from x p+1 y p+1 to each edge x i y i in P . In the worst case there will be one edge missing in color b for x p+1 and one for y p+1 , for each edge x i y i . Since there are (2p) 2 edges x i y i we conclude that vertex

2 is less than or equal to 2n -6. Suppose that there are no edges in color r and g between x p+1 y p+1 . Clearly p ≤ (n-4) 2 , otherwise for p = (n-2) 2 by rainbow degree condition we have both edges and we are in the previous case. Then in the worst case

we can choose a different matching replacing x p+1 y p+1 and x p+2 y p+2 by x p+1 x p+2 and y p+1 y p+2 . Since there are no edges in color r and g between x p+1 x p+2 (otherwise previous case applies) we conclude that the degree in all the colors of d r,g,b (x p+1 ) ≤ 2(n -2p -3) + n -1 and this is less than or equal to 2n -6 for n ≥ 11.

-P is not a proper cycle By Lemma 6.2.4 we can check that |P | ≥ n -6. Clearly, there can not be edges in color r and g between M b \ (M b ∩ P ) and the extremities of P . Also, there can be at most 4 edges in color r and g from each edge in M b \ (M b ∩ P ) and edges y i-1 x i in P . Then we can conclude that in the worst case there are at most 2 (2p-2) 2 edges in color r and g, between x p+1 and vertices in P . Suppose now that there is one edge in color say r (resp. g) between x p+1 and y p+1 , then using edges x p+1 x i , y p+1 y i or y p+1 x i , x p+1 y i in color b, we can insert the edge x p+1 y p+1 in color r into P . Since this will contradict our hypothesis we can conclude that there are 2 missing edges in color b from x p+1 y p+1 to each edge x i y i in P . There will be one edge missing in color b between each vertex x p+1 , y p+1 and each edge x i y i . For |P | = n -6 and |P | = n -4 we already have a contradiction with the number of edges. Now, since there are (2p) 2 edges x i y i we conclude that the vertex

this is less than or equal to 2n -6. Suppose there are no edges in color r and g between x p+1 and y p+1 . Then the degree in color r and g at

we have already contradiction as before.

If |P | = n -4 there are 6n -20 missing edges in color r and g. Since there are no edges in color r and g between x p+1 and y p+1 , x p+2 and y p+2 we conclude 4 missing edges more in color r and g. Observe that if we can replace the edges x p+1 y p+1 and x p+2 y p+2 in M b by x p+1 x p+2 and y p+1 y p+2 , then there are missing the 4 parallel edges in color r and g. A contradiction. Otherwise if not, there are 2 more missing edges and again a contradiction.

If |P | = n-2, since there are no parallel edges x p+1 y p+1 we have that d r,g,b (x p+1 ) ≤ 2n -5 and d r,g,b (y p+1 ) ≤ 2n -5. We can use one of these vertices unless both inequalities become equalities. In this case we will try to replace the edges x i y i in P by x i x p+1 y i and x j y p+1 y j , i = j. Suppose that we have the edges x p+1 x i in color r and g, x p+1 y i in color b, y p+1 x j in color r and g, and y p+1 y j in color b. Then we have a proper hamiltonian path P ′ = x 1 y 1 . . . x i x p+1 y i x i+1 . . . x j y p+1 y j . . . x p y p .

Otherwise at least one of those edges is missing and therefore either d r,g,b (x p+1 ) ≤ 2n -6 or d r,g,b (y p+1 ) ≤ 2n -6.

(2) n is odd.

-P is a proper cycle. By Lemma 6.2.4 we can check that

Clearly, there are no edges in colors r and g between M b \ (M b ∩ P ) and P . As before we can see that d r,g (x p+1 ) ≤ 2(n -2p -1).

If there is an edge x p+1 y p+1 in color r or g, then we can use the edges in color b between x p+1 y p+1 and x i y i to extend P . Since it is not possible there are at most Suppose |P | < n -1. There can not be edges in colors r and g to the extremities of P and there can be only 4 edges in colors r and g between M b \ (M b ∩ P ) and y i-1 x i in P . Suppose now that there exists an edge x p+1 y p+1 in color r or g, then we can use edges in color b between x p+1 y p+1 and x i y i to extend P . Since it is not possible we conclude that there are only

Denote by w the vertex in X. Clearly, there can be at most 3 edges between w and vertices y and z, there can be at most 3 edges between w and v. We will distinguish 3 cases :

(1) Let us suppose that there exist 2 new triplets v, y ′ , z ′ and v, y ′′ , z ′′ , where y = y ′ = y ′′ and z = z ′ = z ′′ . Then suppose that after each contraction there is a vertex w ′ and w ′′ disconnected respectively. Clearly w = w ′ = w ′′ , otherwise we contradict the fact that the resulting multigraph after the contraction was disconnected. Since there can be at most 6 edges incident at each vertex w, w ′ and w ′′ , we have a contradiction with the number of edges in G c .

(2) Let us suppose that there exists only 1 new triplet v, y ′ , z ′ . This case can appear when say N r (v) ≤ n -1 and N b,g (v) ≤ 2 or when N r,g,b (v) ≤ 5 that implies that there are at most n -1 + 4 = n + 3 or 15 incident edges to v. For each triplet there exist w and w ′ and there can be at most 6 edges between w, w ′ and y, z and y ′ , z ′ . Take now the triplet v, y and z ′′ , such that

Then there exist a third vertex w ′′ with at most 6 incident edges to it. We conclude that G c has at most 3 (n-3)(n-4) 2 + 6 + 4 + 6 or 3 (n-4)(n-5)

2

+ n + 14 edges. A contradiction.

(3) Suppose there are no new triplets. Then N r (v) ≤ n -1 and N g,b = 1 or N r,g,b (v) ≤ 3, that implies that there are at most n-1+2 = n+1 or 9 edges incident to v. Suppose N r (v) ≥ 4.

Then after the contraction of v, y and z there is vertex w that is disconnected and there can be at most 4 edges between w and the vertices of the triplet. Take then the triplet v, y and z ′ , where z = z ′ . Suppose there is another vertex w ′ = w what is disconnected with at most 4 incident edges at it. This implies that G c has 3 (n-3)(n-4)

2

+ n + 9 edges. A contradiction. Suppose that N r,g,b (v) = 3. Then take the triplet v, y and z ′ = z and make the contraction. There is one more vertex w ′ disconnected. If w, w ′ ∈ N r,g,b (v) then there are at most 6 edges incident to each vertex, otherwise there are at most 3 edges incident to each of them if w, w ′ / ∈ N r,g,b (v). This implies that G c can have at most 3 (n-3)(n-4) 2

There is one more case to study when N r,g,b (v) ≤ 2. Then w / ∈ N r,g,b (v) and since d r,g,b (w) = 3 we can use the triplet w, z and y for the contraction. If there is one vertex more disconnected, then there can be 2 possibilities first w ′ = v and second w ′ = v. If w ′ = v then there are at most 3 edges incident to v and this implies that G c can have at most 3 (n-2)(n-3) 2 + 6 edges. A contradiction. If w ′ = v then G c can have at most 3 (n-3)(n-4) 2 + 6 + 3 + 3 edges. A contradiction. Now as the connectivity is proved the theorem holds.

Bipartite graphs

In this section, we study proper connection numbers in bipartite graphs. We state a general conjecture for pc k (G) where G is a bipartite graph with some specific connectivity that depends on k. Following that, we show that this conjecture is best possible in the sense of connectivity. Later, we prove some results for specific classes of graphs such as complete bipartite graphs with lower connectivity assumptions than that which is required for the conjecture. Then, we prove that the conjecture is true for complete bipartite graphs. Finally, we study the case k = 1 and obtain results for trees and other graphs depending on their connectivity. We end the section by obtaining, as main result, the proof of the conjecture for the special case k = 1 and some corollaries stemming from it.

If true, Conjecture 7.1.1 is the best possible in the sense of connectivity. In the following we present a family of bipartite graphs which are (2k -1)-connected with the property that pc k (G) > 2. It is also clear that we cannot exchange the vertex connectivity for edge connectivity since it is easy to find graphs with connectivity 1 which have edge connectivity 2k.

Consider the complete bipartite graph G = K p,q with p = 2k -1 (k ≥ 1) and q > 2 p where G = V ∪ W , V = {v 1 , v 2 , . . . , v p } and W = {w 1 , w 2 , . . . , w q }. Clearly, G is (2k -1)-connected. We will show that pc k (G) > 2.

Proposition 7.1.2. Let p = 2k -1 (k ≥ 1) and q > 2 p . Then pc k (K p,q ) > 2.

Démonstration. Suppose that pc k (G) = 2 and consider a k-proper connected coloring of G with 2 colors. For each vertex w i ∈ W , there exists a p-tuple C i = (c 1 , c 2 , . . . , c p ) so that c(v j , w i ) = c j for 1 ≤ j ≤ p. Therefore, each vertex w i ∈ W has 2 p different ways of coloring its incident edges using 2 colors. Since q > 2 p , there exist at least two vertices w i , w j ∈ W such that C i = C j . As pc k (G) = 2, there exist k internally disjoint proper paths in G between w i , w j . Using this, we will arrive to a contradiction. First, observe that one of these paths between w i , w j (say P ) must have only one intermediate vertex v l ∈ V since otherwise, if all the paths have at least two intermediate vertices in V , we would have |V | ≥ 2k, which is a contradiction. Hence, as C i = C j we have c(v l , w i ) = c(v l , w j ) and therefore the path P is not properly colored, leading to a contradiction.

Based on the previous result we prove the following.

Démonstration. It is easy to check that pc 2 (G) = 2 for 3 ≤ n ≤ 6 and pc 2 (G) = 3 for 7 ≤ n ≤ 8. Now let n ≥ 9. We will give a 2-proper coloring of G using c = ⌈ 3 √ n⌉ colors and we will also

General graphs

We begin this section by studying pc(G) for a general graph G. We show some easy results for specific classes such as complete graphs and cycles. Following this, we prove a result analogous to that obtained in the previous section for 2-connected graphs but using 3 colors instead of 2. We also show that this bound is sharp by presenting a 2-connected graph for which 2 colors are not enough to make it proper connected. As a main result of the section, we state an upper bound for pc(G) for general graphs that can be possibly reached as we saw in the previous section. Based on the results of 2-connected graphs we extend Conjecture 7.1.1 to general graphs and finally we prove this for complete graphs. By Vizing's Theorem [START_REF] Vizing | On an estimate of the chromatic class of a p-graph[END_REF], we have that the edge chromatic number of any graph is at most ∆ + 1 and therefore ∆ + 1 is a trivial upper bound for pc(G) for any graph G. First we present some easy results. By using alternating colors, it is easy to see that any path of length at least 2 and any cycle of length at least 4 has proper connection number 2. Also it is clear that the addition of an edge to G cannot increase pc k (G). Fact 7.2.0.2. For n ≥ 3, pc(P n ) = 2 and if n ≥ 4, pc(C n ) = 2. Furthermore, pc k is monotone decreasing with respect to edge addition.

The following theorem improves the Vizing's ∆ + 1 upper bound whenever the graph is 2-connected. This result is a natural extension of Theorem 7.1.7.

Theorem 7.2.1. Let G be a graph. If G is 2-connected, then pc(G) ≤ 3 and there exists a 3-edge-coloring of G that makes it proper connected with the following strong property. For any pair of vertices v, w there exist two paths P 1 , P 2 between them (not necessarily disjoint) such that start(P 1 ) = start(P 2 ) and end(P 1 ) = end(P 2 ).

As in Theorem 7.1.7, we note that an edge-connected version of this result is immediate from the proof.

Démonstration. Suppose G is a 2-connected graph and consider a spanning minimally 2-connected subgraph (meaning that the removal of any edge would leave G 1-connected). For the sake of simplicity, we call this subgraph G. This proof is by induction on the number of ears in an ear decomposition of G. The base case of this induction is when G is simply a cycle and we properly color the edges with at most 3 colors.

Let P be the last ear added in an ear decomposition of G and let G 1 be the graph after removal of the internal vertices of P . Since G is assumed to be minimally 2-connected, we know that P has at least one internal vertex. Let u and v be the vertices of P ∩G 1 so P = uu 1 u 2 . . . u p v. By induction, there is a 3-coloring of G 1 which is proper connected with the strong property. Color the edges of G 1 as such.

Minimum degree

In this section, we prove the following result concerning minimum degrees. The minimum degree condition is best possible. To see this, we construct the following graph. Let G i be a complete graph with n/4 vertices for i = 1, 2, 3, 4, and take a vertex v i ∈ G i for each 1 ≤ i ≤ 4. Let G be a graph obtained from G 1 ∪ G 2 ∪ G 3 ∪ G 4 by joining v 1 and v j with an edge for each 2 ≤ j ≤ 4. Then the resulting graph G is connected and it has δ(G) = n/4 -1 and pc(G) = 3. To prove Theorem 7.3.1, we will make use of the following theorems.

2 then G is panconnected meaning that, between any pair of vertices in G, there is a path of every length from 2 up to n -1. For this statement, we use the following notation. For a path

(1) The graph H 1 obtained from a path P with |P | ≥ 2 and m ≥ 0 isolated vertices v 1 , . . . , v m by joining each v i for (i ≤ m) within P with at least two edges.

(

The graph H 2 obtained from a path P with |P | ≥ 1 and even cycle C by identifying exactly one vertex (i.e., |P ∩ C| = 1).

(3) The graph H 3 obtained from H 2 and m ≥ 0 isolated vertices v 1 , . . . , v m by joining each v i for (i ≤ m) with at least two edges to either P -C or C -P in H 2 .

(4) The graph H 4 obtained from an even cycle C and two paths P 1 and P 2 by identifying an end of each path to a vertex of C. As in H 3 , we may also join vertices each with at least 2 edges to either a path P i or C.

(5) The graph H 5 obtained from the union of two disjoint cycles which are connected by two disjoint paths to form a 2-connected graph. Furthermore, we may also add vertices each with at least 2 edges to this structure. [START_REF] Albert | Multicoloured Hamilton cycles[END_REF] The graph H 6 obtained from H 5 by removing an edge from one of the cycles. Again we may add vertices each with at least 2 edges to this structure.

Démonstration. One can easily get a 2-coloring of H i which forces pc(H i ) = 2 for i = 1, 2, . . . , 6. For example, as for H 1 , by Fact 7.2.0.2 and Proposition 7.1.6, there is a 2-coloring of H 1 that is properly connected.

Proof of Theorem 7.3.1: If κ(G) ≥ 3, then by Corollary 7.1.8, we have pc(G) = 2. So we may assume that κ(G) = 1 or 2. We divide the proof into two cases according to the value of κ(G). By the choice of P 2 , note that P 2 ∩ S = ∅. Let P be a path joining P 1 and P 2 at the common vertex v. Then, utilizing P and the assumption δ(G) ≥ n/4, we will find a spanning subgraph which has a property of H 1 in Lemma 7.3.6. In order to show this, we need only show that each vertex in G \ P has at least 2 edges to P . As previously discussed, we know that all vertices in C 1 have at least 2 edges to P 1 so we need only check vertices x ∈ C 2 \ P If C 1 and C 2 are both 3-connected, then by Corollary 7.1.8, there is a 2-coloring of each with that strong property. Along with these colorings, we also color all edges between {u, v} and C i with color i. This coloring clearly shows that P C(G) = 2 so we may assume that at least one component C i has 1 ≤ κ(C i ) ≤ 2. Next we will suppose that 1 ≤ κ(C i ) ≤ 2 for both i = 1, 2. In this case, by the minimum degree condition and the fact that G is 2-connected, we may easily show that each component is hamiltonian connected (since n is large) so G is hamiltonian. This means pc(G) = 2. Finally, if we suppose C 1 is 3-connected while 1 ≤ κ(C 2 ) ≤ 2, each possible case contains a large (almost spanning) subgraph with the properties of H 4 from Lemma7.3.6, meaning that pc(G) = 2. This completes the proof of Theorem 7.3.1. In the present thesis, we studied proper spanning trees and weak-proper spanning trees in edge-colored graphs and multigraphs. Our main objective was to focus on particular families of edge-colored graphs, and to contribute in the following aspects :

Chapitre 8

Conclusion and Perspectives

(1) Demonstrate the NP-completeness and give some nonapproximability bounds for both PST and WST problems,

(2) Focus on the edge-colored complete graphs and show that the problem remains NPcomplete,

(3) Provide methods to solve the PST and WST problems, and devise algorithms for a more general problem of finding Maximum Proper Tree and Maximum Weak Proper tree in the case of edge-colored acyclic graphs.

(4) Give sufficient conditions for edge-colored graphs and multigraphs to have PST.

We believe that the results presented in the thesis contribute significantly to the understanding of the PST and WST problem in edge-colored graphs. In what follows, we recall briefly the results presented in this thesis and conclude with a few remarks on the future work and perspectives.

Contribution summary

As we have seen in Chapter 4, both problems PST and WST as well as their optimization versions are NP-Complete. We gave some nonapproximability bounds for both problems. For this reason we have studied these problems for some particular families of edge-colored graphs. We have seen that in the case of the edge-colored complete graphs the PST Problem (PSTP) remains NP-Complete. However, in the case of WST Problem (WSTP), it becomes trivial. We have also seen that in the case of the edge-colored graphs with many colors of order Ω(n 2 ), the