Ayman El

Oriented Hamiltonian Cycles in Tournaments: a Proof of Rosenfeld's Conjecture

Keywords: , 2, 3, 4, 5}] Exc 5: [5B, (2, 1, 1), {1, 2, 3}, 45213, 51423]

Rosenfeld in 1974 conjectured that there is an integer N > 8 such that every tournament of order n ≥ N contains every non-directed cycle of order n. We prove that, with exactly 35 exceptions, every tournament of order n ≥ 3 contains each non-directed cycle of order m, 3 ≤ m ≤ n.

Preliminary study

In order to prove Rosenfeld's paths conjecture, Havet and Thomassé [7] introduced the concept of exceptions as follows: An exception is a pair (T ; P ), where T is a tournament illustrated in figures 1, 2, 3 and 4 and P is an outpath with |T | = |P | such that there exists two vertices x and y of T neither of which is an origin of P but s + (x, y) ≥ b 1 (P ) + 1. The dual of an exception is also an exception.

They established two categories of exceptions: the finite exceptions and the infinite families of exceptions. The notation of an exception is the following: [T ; P ; S; P 1 , ..., P k ] where T is a tournament illustrated in figures and , P is an outpath, S is the set of vertices of T which are not origin of P and the paths P 1 , ..., P k are the paths of T whose origins are precisely the vertices of V (T ) \ S.

Introduction

It was early proved by Camion [3] that a tournament has a Hamiltonian directed cycle if and only if it is strong. Concerning non-directed cycles, Thomassen [START_REF] Thomassen | Antidirected Hamiltonian cycles and paths in tournaments[END_REF], in 1973, proved that for any n ≥ 50, any tournament of order n contains a Hamiltonian antidirected cycle. Rosenfeld [START_REF] Rosenfeld | Antidirected Hamiltonian cycles in tournaments[END_REF], in 1974, after improving Thomassen's result to n ≥ 28, conjectured that there is an integer N ≥ 9 such that any tournament of order n ≥ N contains any nondirected Hamiltonian cycle. This conjecture was studied in different approach and many interesting result were successively established. At the beginning, Grünbaum [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF] proved the existence of cycles with block of length n -1, then, in 1983, the existence of cycles of two blocks was proved by Benhocine and Wojda [1]. In the same year, Petrović [START_REF] Petrović | Antidirected Hamiltonian circuits in tournaments, Graph Theory[END_REF] improved Thomassen's and Rosenfeld's bound by proving the same result for n ≥ 16. The first general proof of Rosenfeld's conjecture was established by Thomason [START_REF] Thomason | Paths and cycles in tournaments[END_REF] for tournaments of order n ≥ 2 128 . Thomason was contented only to indicate that it should be true for tournaments of order at least 9. Then, Havet [START_REF] Havet | Oriented Hamiltonian cycles in tournaments[END_REF] improved this bound by proving the conjecture for tournaments of order n ≥ 68 and for reducible tournaments of any order n ≥ 9. In this paper, we completely settle Rosenfeld's cycles conjecture: any tournament contains each Hamiltonian non-directed cycle with 30 exceptions, all of order less than 9 , that will be described explicitly together with the non-allowed cycles. Note that, Rosenfeld proposed the same conjecture for paths, which was proved by Havet and Thomassé [7]. The existence of Hamiltonian paths provides simply the existence of any non-Hamiltonian 1 path, but this is not applied for cycles. Our argument can be used to extend Rosenfeld's conjecture by showing that any tournament of order n contains any non-directed cycle of order m < n with exactly 5 exceptions, where m = n -1. A tournament is an orientation of a complete graph. If X is a subset of vertices of T , the subtournament induced by X is denoted by T (X). For abbreviation, we write x → y whenever (x, y) ∈ E(T ). Similarly, if X and Y are two subtournaments of T , we write X → Y if x → y for every pair (x, y) ∈ X × Y . A vertex x is said to be minimal in T if d -(x) = 0. Let P = x 1 ...x n be a path, x 1 is called the origin of P while x n is called its end. * P (resp. P * ) denotes the path x 2 ...x n (resp. x 1 ...x n-1 ), the path x n ...x 1 is denoted by P -1 . P is said to be an outpath(resp. inpath) if x 1 → x 2 (resp. x 1 ← x 2 ). P is said to be a directed outpath (resp. directed inpath) if x i → x i+1 (resp. x i ← x i+1 ) for all i, 1 ≤ i ≤ n -1. A block of P is a maximal directed subpath of P . In general, an oriented path is formed by successive blocks. The i th block of P is denoted by B i (P ) while its length is denoted by b i (P ). The type of an outpath P is the sequence +(b 1 (P ), b 2 (P ), ..., b s (P )) while that of an inpath Q is -(b 1 (Q), b 2 (Q), ..., b t (Q)). Occasionally, we write P = ±(b 1 (P ), ..., b s (P )) instead of P , if P is of type ±(b 1 (P ), ..., b s (P )). When two paths P and Q are of same type, we write P ≡ Q. Given a path P and a tournament T , we define the set Or(P, T ) = {v ∈ T ; v is an origin of a path Q in T such that Q ≡ P }. Let C = x 0 ...x n-1 x 0 be a cycle. In a similar way, we define the blocks B i (C) of C together with their lengths b i (C). The type of C is the sequence (b 1 (C), ..., b s (C)), we write C = (b 1 (C), ..., b s (C)). When two cycles C and C are of same type, we write C ≡ C . A rotation of a cycle C = a 0 a 1 ...a n-1 a 0 is an enumeration a 0 a 1 ...a n-1 a 0 of the vertices of C such that a 0 = a n-1 and a 1 = a 0 , after a rotation we re-write C = a 0 a 1 ...a n-1 a 0 , where a i = a i for all i ∈ {0, ..., n -1}. The dual of a digraph D is the digraph D such that (y, x) ∈ E(D) whenever (x, y) ∈ E(D). A tournament T is said to be strong if T contains an xy-directed outpath for every pair (x, y) ∈ T . Otherwise, T is said to be reducible. Let X be a set of vertices of T . The outsection generated by X, denoted by S + (X), is the set of vertices y such that T contains an xy-directed outpath for some x ∈ X. s + (X) denotes the cardinal of S + (X). Likewise, the insection generated by X, denoted by S -(X), is the set of vertices y such that T contains an xy-directed inpath for some x ∈ X. s -(X) denotes the cardinal of S -(X). If x 1 , ...x s are s vertices in T , we write S ± (x 1 , ..., x s ) instead of S ± ({x 1 , ..., x s }) and s ± (x 1 , ..., x s ) instead of s ± ({x 1 , ..., x s }). A vertex x of T is said to be an outgenerator (resp. ingenerator ) of T if S + (x) = V (T ) (resp. S -(x) = V (T )). In this paper, we say that a path P (resp. a cycle C) is contained in a tournament T if there exists a path Q (a cycle C ) lying in T such that Q ≡ P (resp. C ≡ C). If X is set of vertices of T , O X (resp. I X ) designate an arbitrary Hamiltonian directed outpath (resp. inpath)of T (X). Let x be a vertex in T and Y a subtournament of T , the set N + Y (x) (resp. N - Y (x)) denotes the set of outneighbors (resp. inneighbors) of x in Y . The cardinal of N + Y (x) (resp.

N - Y (x)) is denoted by d + Y (x) (resp. d - Y (x)).
The infinite families of exceptions are denoted by E i (n) = (F i (n); P ), where F i (n) is the tournament on n vertices illustrated in figures 5 and 6. For each E i (n), we define the set S of vertices of F i (n) which are not origin of P together with the conditions on the tournament. And finally, we give the paths P with origin x / ∈ S. Below the list of infinite families of exceptions:

Exception E 1 (n) = (F 1 (n), (1, n -2)); S = {1, 2, 3}. Conditions: |X| ≥ 1. Paths: for any u ∈ X, P = u132I X-u . Exception E 2 (n) = (F 2 (n), (2, n -3)); S = {3, 4}. Conditions: |X| ≥ 1. Paths: P = 1234I X , P = 2314I X and for any u ∈ X, P = u132I X-u . Exception E 3 (n) = (F 3 (n), (1, n-2)); S = {1, 3}. Conditions: N + (3) = {2} and 3 is an ingenerator of T (X) (in particular there exists a Hamiltonian directed inpath u3R 1 of T (X) and another 3vR 2 of T (X)). Paths: P = 21u3R 1 and for any y ∈ X \ {3}, P = y132I X-y . Exception E 4 (n) = (F 4 (n), (2, n -3)); S = {1, 4}. Conditions: N + (3) = {2} and 3 is an ingenerator of T (X) (in particular there exists a Hamiltonian directed inpath u3R 1 of T (X) and another 3vR 2 of T (X)). Paths: P = 214u3R 1 , P = 3241vR 2 and for any y ∈ X \ {3}, P = u132I X-u . Exception E 5 (n) = (F 5 (n), (1, n -2)); S = {1, 2}. Conditions: n ≥ 5, |Y | ≥ 2 and 2 is an ingenerator of T (X). Paths: for any x ∈ X \ {2}, P = x1I Y I X-x , and for any y ∈ Y , P = y1I Y -y I X . Exception E 6 (n) = (F 6 (n), (2, n -3)); S = {1, 3}. Conditions: |Y | ≥ 2 and 2 is an ingenerator of T (X). Paths: for any x ∈ X, P = xz13I Y -z I X-x (for a given z ∈ Y ) and for any y ∈ Y , P = y31I Y -y I X . Exception E 7 (n) = (F 7 (n), (1, 1, n -3)); S = {2, 3}. Conditions: T (Y ) is not 3-cycle and |Y | ≥ 3. Paths: since T (Y ) is not a 3-cycle, there is a path Q = - (1, n -5) in T (Y ) (this is clear if T (Y ) is not reducible; if T (Y ) is strong, there exists a vertex w ∈ Y such that T (Y ) -w is also strong, such a vertex is certainly an origin of Q since it has an inneighbor that is an origin of a Hamiltonian directed path of T (Y ) -w), we then have P = 1Q23 and for any y ∈ Y , P = y231O Y -y . Exception E 8 (n) = (F 8 (n), (n-4, 1, 1, 1)); S = X. Conditions: 3A is the 3-cycle, its set of vertices is {1, 2, 3}, we furthermore need |X| ≥ 2. Paths: P = 1O X-u 2u3, P = 2O X-u 1u3 and P = 3O X-u 2u1 for a given u ∈ X. Exception E 8 (n) = (F 8 (n), (n -4, 2, 1)); S = X. Conditions: 3A is the 3-cycle, its set of vertices is {1, 2, 3}, we furthermore need |X| ≥ 2. Paths: P = 1O X-u 32u, P = 2O X-u 13u and P = 3O X-u 21u for a given u ∈ X. Exception E 9 (n) = (F 9 (n), (n -6, 1, 1, 1, 1, 1)); S = X. Conditions: 5A is the 2-regular tournament, its set of vertices is {1, 2, 3, 4, 5}, we furthermore need |X| ≥ 2. Paths: P = 1O X-u 2u453, P = 2O X-u 3u514, P = 3O X-u 4u125, P = 4O X-u 5u231 and P = 5O X-u 1u342 for a given u ∈ X. Exception E 9 (n) = (F 9 (n), (n -6, 2, 1, 1, 1)); S = X. Conditions: 5A is the 2-regular tournament, its set of vertices is {1, 2, 3, 4, 5}, we furthermore need |X| ≥ 2. Paths: P = 1O X-u 32u45, P = 2O X-u 43u51, P = 3O X-u 54u12, P = 4O X-u 15u23 and P = 5O X-u 21u34 for a given u ∈ X. Exception E 10 (n) = (F 10 (n), (n -8, 1, 1, 1, 1, 1, 1, 1)); S = X. Conditions: 7A is the Paley tournament, its set of vertices is {1, 2, 3, 4, 5, 6, 7}, we furthermore need |X| ≥ 2. Paths: P = 1O X-u 2u45376, P = 2O X-u 3u56417, P = 3O X-u 4u67521, P = 4O X-u 5u71632, P = 5O X-u 6u12743, P = 6O X-u 7u23154 and P = 7O X-u 1u34265 for a given u ∈ X. Exception E 10 (n) = (F 10 (n), (n -8, 2, 1, 1, 1, 1, 1)); S = X. Conditions: 7A is the Paley tournament, its set of vertices is {1, 2, 3, 4, 5, 6, 7}, we furthermore need |X| ≥ 2. Paths: P = 1O X-u 32u4657, P = 2O X-u 43u5761, P = 3O X-u 54u6172, P = 4O X-u 65u7213, P = 5O X-u 76u1324, P = 6O X-u 17u2435 and P = 7O X-u 21u3546 for a given u ∈ X. Exception E 11 (n) = (F 11 (n), (1, 1, n -3)); S = {1, 2}. Conditions: |X| ≥ 2. Paths: P = 31O X 2 and for any u ∈ X, P = u1O X-u 23. Exception E 12 (n) = (F 1 2(n), (2, 1, n -4)); S = {1, 4}. Conditions: |X| ≥ 2. Paths: P = 231O X 4 and for any u ∈ X, P = u41O X-u 23. Exception E 13 (n) = (F 13 (n), (1, 1, n -3)); S = {1, 2}. Conditions: |X| ≥ 2. Paths: P = 3425O X 1, P = 4523O X 1, P = 5324O X 1 and for all u, v ∈ X, P = u1v2345O X\{u,v} .

1 2 3 X F 1 (n) 1 2 3 4 X F 2 (n) X 1 2 3 F 3 (n) X 1 2 4 F 4 (n) X 1 Y 2 F 5 (n) X 3 2 1 Y F 6 (n) Y 3 1 2 F 7 (n) 3A X F 8 (n) 5A X F 9 (n) 7A X F 10 (n) 1 X F 11 (n)
Exception E 14 (n) = (F 14 (n), (2, 1, n -4)); S = {1, 6}. Conditions: |X| ≥ 2. Paths: for every vertex u ∈ X, P = 24u53O X-u 61, P = 3u16245O X-u , P = 4u16253O + X -u, P = 5u16234O X-u and P = u61345O X-u 2.

They conclude the following: Theorem 2.1 Let T be a tournament of order n, P an outpath of order n and x, y two distinct vertices of T such that s + (x, y) ≥ b 1 (P ) + 1. Then at least one of the following holds:

(i) x or y is an origin of P in T .

(ii) (T ; P ) is an exception.

The following remark is due to Bou Hanna [START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF]: Remark 2.2 If T is a tournament of order 4, 6 or 8, there is at most 2 vertices x such that T -x may be 3A, 5A or 7A.

For the sake of our proof, we need to establish the following lemmas: Lemma 2.3 Let P be a non-directed outpath of order n and T a tournament of order n. If P is not directed and (T ; P ) is not a Grünbaum's exception then T contains P of origin of indgree at least 1 unless if (T ; P ) = (4B; (1,[START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF]).

Proof. This is clear if δ -(T ) ≥ 1. So we can assume that there exists x ∈ T of indegree zero. We can divide T -x into two sets of vertices V 1 and V 2 such that T (V 1 ) and T (V 2 ) is not 3A, 5A nor 7A and (1,1)) then T = 4B and P = (1,[START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF]. Lemma 2.4 Let P be a non-directed outpath of order n ≥ 3 and T a tournament of order n. If (T ; P ) is not the Grünbaum's exceptions then T contains P of origin of outdegree at least 2 unless if (T ; P ) = (4B; (1,1,1)) or (T ; P ) = (4B; [START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF]1)). In these cases, 1, 2 and 3 are origins of P . Moreover, if P is directed and T is distinct from 3A, there is an origin of P of outdegree at least 2. Else, every vertex of 3A is an origin of P .

Proof. Set X = {v ∈ T ; d + (v) ≥ 2}, we have |X| ≥ n -3 and s + (X) = n, since otherwise, there is a vertex x ∈ T -X such that x → X, a contradiction. If (T ; P ) is an exception, one can check the result. So we may suppose that (T ; P ) is not an exception, if |X| ≥ 2, the result follows by theorem 2.1. Otherwise, |X| = 1, if n = 3 then T is transitive while if n = 4 then T = 4B, the result follows.

Lemma 2.5 Let P be an outpath of order n having at least 3 blocks and T a tournament of order n. If (T ; P ) is not a Grünbaum's exception, then T contains a 1 ...a n ≡ P such that d + (a 2 ) ≥ 1 unless (T ; P ) = (4B; (1,1,1)) Definition 2.9 Let T be a tournament of order n, P a non-directed outpath of order n and x be a vertex of T such that d + (x) ≥ 2. The couple (T -x; * P ) is said to be a biexception with respect to x if x is not an origin of P in T , (T -x; * P ) is an exception and there exists a vertex y ∈ T such that x is the end of P whenever y is its origin in T . The notation of a biexception is Exc (k, i), where (T -x; * P ) is the biexception k, i = 0 when * P is an outpath and i = 1 when * P is an inpath. The set of vertices of T -x which are not origins of P in T when x is not its end is denoted by S. The biexceptions are the following: Exc (0, 0)(1): P = (2, 1), N + (x) = {1, 2, 3}, S = {1, 2, 3}. Exc (0, 0)( 2 Proof. We enumerate the finite exceptions and extend them in all possible ways by a vertex x of outdegree at least 2. To shorten the proof we use the following notation: Exc k: P [N + (x); S; P 1 ; ...; P t ], where P i are the paths of origin in T \ ({x} ∪ S) and of end distinct from x. Exc 0: [START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF]1) [{1, 2, 3}; {1, 2, 3}] [{1, 2}; {1, 2}; 3x21] The other possible outneighbors of x are equivalent Exc 1: (2,1,1) [{1,[START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF]∅;13x42;23x41;3x142;4x213] [{1,[START_REF] Camion | Chemins et circuits hamiltoniens des graphes complets[END_REF]{1,[START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF][START_REF] Camion | Chemins et circuits hamiltoniens des graphes complets[END_REF]4x312] 634x5271; 7x435261] Exc 49: - (1, 1, 1, 1, 1, 1, 1, 1) [{1, 2}; ∅; 1853426x7; 2853416x7; 3x4125768; 4x3125768; 5x3127846; 6x3127845; 7x3125648; 8x4125763 2,1,1,1,1,1) [{1 , 2}; ∅; 182367x54; 281367x54; 3x7216458; 4x8213675; 5x8214376; 6x7214358; 7x8216453; 2,1,2,1,1) [{2 , 8}; ∅; 1x4835267; 2738451x6; 3x4856172; 4x3852167; 5x3842167; 6x3842157; 7x3842156; 8732451x6] Lemma 2.12 Let T be a tournament on n vertices, P an outpath of order n and x a vertex of T of outdegree at least 2. If (T -x; * P ) belongs to one of the infinite families of exceptions, * P is an outpath and x is not an origin of P in T then any vertex of T distinct from x is an origin of P in T with end distinct from x unless if (T ; P ) is a biexception with respect to x.

Proof. We use the same notations of the previous lemmas and we denote by a and b a random vertices of X and Y respectively: Exc E 1 (n -1): (2, n -3) [{1, 2, 3}; ∅; 123xI X ; 231xI X ; l312xI X ; ax321I X-a ] [{1, 2}; {1}; 231xI X ; 312xI X ; ax132I X-a ] The other possible outneighbors of x are equivalent. Exc E 2 (n -1): (3, n -4) [{3, 4}; ∅; 12x34I X ; 2x314I X ; 3124xI X ; 4123xI X ; ax4321I X-a ] Exc E 3 (n -1): (2, n -3) [{1, 3}; {1}; 2x31vR 2 ; 321xvR 2 ; a21xI X-a ] Exc E 4 (n -1): (3, n -4) [{1, 4}; ∅; 13x42vR 2 ; 2x14u3R 1 ; 3x142vR 2 ; 43x12vR 2 ; ax142I X-a ] Exc E 5 (n -1): (2, n -3) [{1, 2}; {1}; ab1xI Y -b I X-a ; bx1I Y -b I X ] Exc E 6 (n -1): (3, n -4) [{1, 3}; ∅; 12x3I Y I X-2 ; 32x1I Y I X-2 ; ax3I Y I X-a ; bx31I Y -b I X ] Exc E 7 (n -1): (2, 1, n -4) [{2, 3}; ∅; 1x2O Y 3; 21xO Y 3; 31xO Y 2; bx2O Y -b 31] Exc E 8 (n -1): (n -4, 1, 1, 1) [N + (x) ⊆ X; X; 1O X 2x3; 2O X 1x3; 3O X 1x2] Exc E 8 (n -1): (n -4, 2, 1) [N + (x) ⊆ X; X; 1O X 32x; 2O X 13x; 3O X 21x] Exc E 9 (n -1): (n -6, 1, 1, 1, 1, 1) [N + (x) ⊆ X; X; 1O X 2x453; 2O X 3x514; 3O X 4x125; 4O X 5x231; 5O X 1x342] Exc E 9 (n -1): (n -6, 2, 1, 1, 1) [N + (x) ⊆ X; X; 1O X 32x45; 2O X 43x51; 3O X 54x12; 4O X 15x23; 5O X 21x34] Exc E 10 (n -1): (n -8, 1, 1, 1, 1, 1, 1, 1) [N + (x) ⊆ X; X; 1O X 2x45376; 2O X 3x56417; 3O X 4x67521; 4O X 5x71632; 5O X 6x12743; 6O X 7x23154; 7O X 1x34265] Exc E 10 (n -1): (n -8, 2, 1, 1, 1, 1, 1) [N + (x) ⊆ X; X; 1O X 32x4657; 2O X 43x5761; 3O X 54x6172; 4O X 65x7213; 5O X 76x1342; 6O X 17x2435; 7O

X 21x3546] Exc E 11 (n -1): (2, 1, n -4) [{1, 2}; {1}; 23xO X 1; 3x1O X 2; ax1O X-a 23] Exc E 12 (n-1): (3, 1, n-5) [{1, 4}; ∅; 12x43O X ; 2x413O X ; 3x12O X 4; 42x13O X ; a2x13O X-a 4] Exc E 13 (n -1): (2, 1, n -4) [{1, 2}; {1}; 2x1345O X ; 3x145O X 2; 4x153O X 2; 5x134O X 2; a2x345O X-a 1] Exc E 14 (n -1): (3, 1, n -5) [{1, 6}; ∅; 12x6345O X ; 2x61345O X ; 3x6145O X 2; 4x6153O X 2; 5x6134O X 2; a12x345O X-a 6]
Lemma 2.13 Let T be a tournament on n vertices, P an inpath of order n and x a vertex of T of indegree at least 2. If (T -x; * P ) belongs to one of the infinite families of exceptions, * P is an outpath and x is not an origin of P in T then any vertex of T distinct from x is an origin of P in T with end distinct from x unless if (T ; P ) is a biexception with respect to x.

Proof. We use the same notations of the previous lemma: [START_REF] Camion | Chemins et circuits hamiltoniens des graphes complets[END_REF]. Three cases may arise:

Exc E 1 (n -1): -(1, 1, n -3) [{1, 2, 3}; X; 13x2I X ; 21x3I X ; 32x1I X ] [{1, 2}; ∅; 1a3x2I X-a ; 2a3x1I X-a ; 32x1I X ; ax321I X-a ] Exc E 2 (n -1): -(1, 2, n -4) [{3, 4}; ∅; 1x234I X ; 2a31x4I X-a ; 34x21I X ; 4a12x3I X-a ; axa 3421I X\{a,a } ] If |X| ≥ 2 and if |X| = 1, a is not an origin of P . Exc E 3 (n -1): -(1, 1, n -3) [{1, 3}; ∅; 1a2x3R 1 ; 2r1u3(R 1 -r); 3v2x1R 2 ; ax231I X\{a,3} ] Where r is the end of R 1 . Exc E 4 (n -1): -(1, 2, n -4) [{1, 4}; ∅; 1v32x4R 2 ; 2r14u3(R 1 -r); 3x241I X-3 ; 4v32x1R 2 ; ax241I X-a ] Where r is the end of R 1 . Exc E 5 (n -1): -(1, 1, n -
If X = {2} and T (Y ) is not 3A, let Q = (1|Y |-2) in Y , [{1, 2}; {2}; 1Qx2; bxI Y -b 21]. If X = {2} and T (Y ) = 3A, [{1, 2}; {1, 2}; bxI Y -b 21].
If |X| ≥ 2, since 2 is an ingenerator of X, let 2Q be a Hamiltonian directed inpath of X with end v and let u be an inneighbor of 2 in X.

[{1, 2}; ∅; 1vI Y x2Q * ; 2uI Y x1I X\{2,u} ; axI Y 21I X\{a,2} ; bxI Y -b 21I X-2 ].
Exc E 6 (n -1): - (1, 2, n -4). Three cases may arise: {2} and Y {u → v}, [{1, 3}; {3, u}; 13xvu2; 2xu13v; vu12x3].

If X = {2} and |Y | ≥ 3, let u → v → w be three vertices of Y , [{1, 3}; ∅; 1uvw2x3I Y \{u,v,w} ; 2xb13I Y -b ; 3uvw2x1I Y \{u,v,w} ; bxb 13I Y \{b,b } 2]. If X =
If |X| ≥ 2, let u be an inneighbor of 2 in X, [{1, 3}; ∅; 1u2I Y x3I X\{2,u} ; 3u2I Y x1I X\{2,u} ; axb13I Y -b I X-a ; bxu213I Y -b I X\{2,u} ]. Exc E 7 (n -1): -(1, 1, 1, n -4). Let u → v be an arc of T (Y ), two cases may arise: If T (Y ) is not 3A, let Q = -(1, |Y | -2) be a Hamiltonian path of Y , [{2, 3}; ∅; 1xQ32; 2uvxO Y \{u,v} 31; 3uvxO X\{u,v} 21; axa 1I Y \{a,a } 32]. If T (Y ) is not 3A, S = {1}. Exc E 8 (n -1): -(1, n -5, 1, 1, 1). We have N -(x) ⊆ X. Let u → v be an arc of T (X).
Two cases may arise:

If |X| ≥ 4, let w ∈ N -(x).
[N -(x); ∅; 1x2O X\{u,v} 3vu; 2x3O X\{u,v} 1vu; 3x1O X\{u,v} 2vu; a1wx3O X\{a,u,v,w} 2vu] (The path O X\{a,u,v,w} may be empty).

If |X| = 3, say X = {u, v, w}. 1, 2 and 3 are origins as above. If u is an outneighbor of x, we have ux1v2w3 ≡ P . If u has an inneighbor v that is an inneighbor of x, we have uvx13w2. So the sole vertex that is not an origin of P is the inneighbor of x that is minimal in X.

If |X| = 2. [{u, v}; {u, v}; 1x32vu; 2x13vu; 3x21vu] Exc E 8 (n -1): - (1, n -5, 2, 1). Two cases may arise:

If |X| ≥ 3, let Q = (n -7, 1) be a path in T (X) and u = X -V (Q). [N -(x); ∅; 1x2Q3u; 2x3Q1u; 3x1Q2u; a1O X-a 2x3] If |X| = 2, say X = {u, v} [{u, v}; {1, 2, 3}; u1v2x3; v1u2x3] Exc E 9 (n -1): -(1, n -7, 1, 1, 1, 1, 1). Let u, v ∈ X. [N -(x) ⊆ X; ∅; 1x3O X\{u,v} 2u4v5; a1O X-a 23x54]. The other vertices of 5A are equivalent. Exc E 9 (n -1): -(1, n -7, 2, 1, 1, 1). Let u, v ∈ X. [N -(x) ⊆ X; ∅; 1x4O X\{u,v} 32u5v; a1O X-a 2x435]. The other vertices of 5A are equivalent. Exc E 10 (n -1): -(1, n -9, 1, 1, 1, 1, 1, 1, 1). Let u, v ∈ X. [N -(x) ⊆ X; ∅; 1x3O X\{u,v} 2u4v675; a1O X-a 23x7564]. The other vertices of 7A are equiv- alent. Exc E 10 (n -1): -(1, n -9, 2, 1, 1, 1, 1, 1). Let u, v ∈ X. [N -(x) ⊆ X; ∅; 1x4O X\{u,v} 32u5v67; a1O X-a 2x73546]. The other vertices of 7A are equiv- alent. Exc E 11 (n -1): -(1, 1, 1, n -4). Let u → v be an arc of T (X), [{1, 2}; ∅; 1xvu23O X\{u,v} ; 2uvx3O X\{u,v} 1; 3x1O X 2; a31xO X-a 2] Exc E 12 (n -1): -(1, 2, 1, n -5) [{1, 4}; ∅; 1a23xO X-a 4; 2x34O X 1; 3xa2O X-a 41; 4a23xO X-a 1; axa 1423O X\{a,a } ] Exc E 13 (n -1): -(1, 1, 1, n -4). Let u → v be an arc of T (X), [{1, 2}; ∅; 1342x5O X ; 2uvx345O X\{u,v} 1; 3x425O X 1; 4x523O X 1; 5x324O X 1; a34x5O X-a 12] Exc E 14 (n-1): -(1, 2, 1, n-5) [{1, 6}; ∅; 13455xO X 62; 2134x5O X 6; 3x254O X 61; 4x235O X 61; 5x243O X 61; 6345x) X 12; ax3425O X-2 61]
From now on, we call the lemmas 2.10, 2.11, 2.12 and 2.13 the building lemmas.

Main result

Below are all the exceptions (T ; C) where T does not contains C: The exceptions A i for 1 ≤ i ≤ 12 are found by Havet [START_REF] Havet | Oriented Hamiltonian cycles in tournaments[END_REF]. We show our main result in the other cases, for the Hamiltonian cycles, using the following theorem due to El-Sahili and Ghazo-Hanna [START_REF] Sahili | About the Number of Oriented Hammiltonian Paths and Cycles in Tournaments[END_REF], by discussing the minimal indegree of the tournament. The following theorem concerns the tournaments with minimal indegree at most 1 and the non-antidirected cycles. 4, 5, 6, 7, 8, 9, 10, 11, 12}. Proof. By theorem 3.1, we may suppose that δ -(T ) ≤ δ + (T ). Set δ = δ -(T ). Let v ∈ T be a vertex such that d -(v) = δ, u be its inneighbor if it exists and T 2 = T (N + (v)). Let B be a block of C of maximal length and set C = a 0 a 1 ...a n-1 a 0 such that B is forward and it ends at a δ+2 . If δ = 0, let s = min{i; a i → {a i-1 , a i+1 }}. Clearly, the path P = a i+1 a i+2 ...a n-1 a 0 ...a i-1 is contained in T 2 if and only if (T 2 ; P ) = (3A; (1,1)) and in this case (T ; C) = A 8 , we have vP v ≡ C. So we may suppose that δ = 1, we distinguish 3 cases: Case 1: If |B| ≥ 4, rotate C until a n-1 → a n-2 for the first time, then b 1 (a 0 a n-1 a n-2 ...a 3 ) = 1.

A 1 = (3A; (2, 1)), A 2 = (5A; (2, 1, 1, 1)), A 3 = (7A; (2, 1, 1, 1, 1, 1)), A 4 = (4C; (1, 1, 1, 1)), A 5 = (6M ; (1, 1, 1, 1, 1, 1)), A 6 = (8C; (1, 1, 1, 1, 1, 1, 1, 1)), A 7 = (6N ; (4, 2)), A 8 = (4B; (2, 2)), A 9 = (5C; (4, 1)), A 10 = (8D; (1, 1, 1, 1, 1, 1, 1, 1)), A 11 = (6O; (1, 1, 1, 1, 1, 1)), A 12 = (6P ; (1, 1, 1, 1, 1, 1)), A 13 = (6F ; (2, 1, 2, 1)), A 14 = (6D; (2, 2, 1, 1)), A 15 = (5A; (1, 1, 1, 1)), A 16 = (5F ; (1, 1, 1, 1)), A 17 = (7K; (1, 1, 1, 1, 1, 1)), 4C 5A 6M 7A 8C 6N 4B 5F 8D 6O 6P
Theorem 3.2 Let T be a tournament of order n such that δ -(T ) ≤ 1 and C a non-directed cycle of order n. If C is not antidirected then T contains C if and only if (T ; C) is not one of the exceptions A i , i ∈ {1,
(a) If d - T 2 (u) ≥ 2 then, by theorem 2.1, by taking two inneighbors of u in T 2 , u has an inneighbor w origin of a path P ≡ a 0 a n-1 ...a 3 in T 2 , and thus wuvP -1 ≡ C, or (T 2 ; a 0 a n-1 ...a 3 ) is an exception, the possible exceptions are: Dual Exc 0, 1,[START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF][START_REF] Sahili | About the Number of Oriented Hammiltonian Paths and Cycles in Tournaments[END_REF][START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF][START_REF] Havet | Oriented Hamiltonian cycles in tournaments[END_REF][START_REF] Petrović | Antidirected Hamiltonian circuits in tournaments, Graph Theory[END_REF][START_REF] Rosenfeld | Antidirected Hamiltonian paths in tournaments[END_REF][START_REF] Thomason | Paths and cycles in tournaments[END_REF]15,16,18,19,20,22,23,24,25,26,27,29,30,31,32,33,34,37,38,40,41,43,44,47,48,49. By checking each one, for Dual Exc 1,[START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF][START_REF] Havet | Oriented Hamiltonian cycles in tournaments[END_REF][START_REF] Petrović | Antidirected Hamiltonian circuits in tournaments, Graph Theory[END_REF][START_REF] Rosenfeld | Antidirected Hamiltonian paths in tournaments[END_REF][START_REF] Thomason | Paths and cycles in tournaments[END_REF]15,16,18,19,20,22,23,24,25,26,27,29,30,31,32,34,37,38,40,41,43,44,47,48,49

(b) If d - T 2 (u) = 1, set N -(u) = {w}. If T 2 -w contains a path P ≡ a 2 a 3 ...a n-2 , then wuP vw ≡ C. Otherwise, (T 2 -w; a 2 a 3 ...a n-2
) is a Grünbaum's exception, so by theorem 2.1 u has an outneighbor x origin of a path Q ≡ a 0 a 1 ...a n-3 in T 2 and so Qvux ≡ C, unless if T 2 = w → 3A and a 0 a 1 ...a n-3 is directed, which is the exception A 7 .

Case 2: If |B| = 3, denote by B and B" the blocks of C containing the arc a 0 → a n-1 and a 4 → a 3 respectively, without loss of generality, we can suppose that |B | ≥ |B"|, since otherwise, we can take C = a 0 a 1 ...a n-1 a 0 = a 3 a 2 ...a 4 a 3 , we have two cases:

(a) If d - T 2 (u) ≥ 2, suppose that s + T 2 (N - T 2 (u)) ≥ |B | + 1
, then u has an inneighbor w origin of a path P ≡ a 0 a n-1 ...a 3 and so wuvP -1 ≡ C or (T 2 ; a 0 a n-1 ...a 3 ) is an exception. By checking all the possible exceptions, u has an inneighbor w origin of a path Q ≡ a 3 a 4 ...a n-1 a 0 and thus vuQv ≡ a 1 a 2 ...a 1 = C unless if (T 2 ; a 0 a n-1 ...a 3 ) is one of the exceptions: 1, 20, 22, 23, 24, 28, 29, 30, 34, 38, 43, 46, 47, 48, 49. If it is 28, then u614523vu ≡ C. Otherwise, there exist w, t ∈ N - T 2 (u) and y ∈ N + T 2 (u) such that y is an origin of a path R ≡ a 2 a 3 ...a n-3 in T 2 \ {w, t} and w → t, thus wuRvtw ≡ C. So we may assume that s

+ T 2 (N - T 2 (u)) ≤ |B |, u has an outneighbor origin of a path R ≡ a n-1 a n-2 ...a 2 and thus uvR -1 u ≡ C, unless if (T 2 ; a n-1 ...a 2 ) = (4B; (2, 1)), in this case v41u32v ≡ C. (b) If d - T 2 (u) = 1, set N -(u) = {w}. Since d + T 2 (u) = |T 2 | -1, u has an outneighbor origin of a path R ≡ a n-1 a n-2 ...a 2 in T 2 and thus uvR -1 u ≡ C, unless if (T 2 ; a n-1 ...a 2 )
is one of the exceptions 0, 1, 7 and in these cases, by taking u in the place of v, we obtain an inneighbor of u having an indegree greater than 1 and the result follows.

- T 2 (N + T 2 (u)) = 2, let y ∈ N - T 2 (u) be an origin of a path R ≡ a 1 a 0 a n-1 ...a 6 (R may be an arc) in T (N -(u)) and {w, t} = N + T 2 (u), we have yuwvtR -1 ≡ a 1 a 2 ...a 1 = C. (b) If a 0 → a n-1
, then (T 2 ; a 0 a n-1 ...a 3 ) is one of the exceptions 6, 12, 18, 31, let

w, t ∈ N + T 2 (u) and y ∈ N - T 2 (u) such that y is an origin of a path R ≡ a 1 a 0 a n-1 ...a 6 in T 2 \ {w, t}, we have yuwtvR -1 ≡ a 1 a 2 ...a 2 = C. (c) If a n-1 → {a 0 , a n-2 },
without loss of generality we can suppose that a 4 → {a 3 , a 5 }, so (T 2 ; a 0 a n-1 ...a 3 ) is one of the exceptions 0, 4, 8, 9, 33, 37. There exist w ∈ N - T 2 (u), t ∈ N + T 2 (u) and y ∈ T 2 \{w, t} such that w → y and t is an origin of a path R ≡ a 3 a 4 ...a n-2 in T 2 \ {w, y}; we have ywuRvy ≡ C.

So we may assume that

d + T 2 (u) ≤ 1. If |T | ≥ 6, there exists w ∈ N - T 2 (u) such that T 2 -w has at least two origins of a path R ≡ a 4 a 5 ...a 0 and thus vwuRv ≡ a 1 a 2 ...a 1 = C, unless if T 2 = 4B and R is directed, in this case (T ; C) = A 13 . And if |T | ≤ 5, easily one can find C unless if (T ; C) = A 1 .
To complete the proof of the non-antidirected cycles, we consider the other cases in the following theorem: Theorem 3.3 Let T be a tournament of order n with δ -(T ) ≥ 2 and C a non-directed cycle of order n. If C is not antidirected then T contains C if and only if (T ; C) is not one of the exceptions A i , i ∈ {2, 3, 13}.

Proof. By theorem 3.1, we may suppose that δ -(T ) ≤ δ + (T ). Set δ = δ -(T ). Let v ∈ T be a vertex such that d -(v) = δ and denote by T 1 = T (N -(v)) and T 2 = T (N + (v)). Let B be a block of C of maximal length and set C = a 0 a 1 ...a n-1 a 0 such that B is forward and it ends at a δ+2 . We distinguish 4 cases:

Case 1: If a n-1 → a 0 → a 1 . Rotate C until a n-1 → a n-2 for the first time. (a) Suppose first that (T 1 ; a 1 ...a δ ) is a Grünbaum's exception, let a ∈ T 2 such that d - T 2 (a) = ∆ -(T 2 ). Set T 1 = T 1 ∪ {a} and T 2 = T 2 -a. We have two cases: (i) If a 1 → a 2 . Suppose that a n-3 → a n-2 . If T 1 = 4B then T 1 contains a path x 0 x 1 ...x δ ≡ a 0 a 1 ...a δ such that x 0 , x δ = a, if x 0 has an inneighbor in T 2 origin of a path P ≡ a n-1 a n-2 ...a δ+2 then x 0 ...x δ vP -1 x 0 ≡ C. Otherwise, since T 1 is regular, d - T 1 (x 0 ) = |T 1 | 2 ,
then, without loss of generality we may suppose that d - T 2 (x 0 ) ≥ 2; in fact, if δ = 5 or 7, it is clear, and if δ = 3, one can check that. By theorem 2.1, (T 2 ; a n-1 a n-2 ...a δ+2 ) is an exception. By the building lemmas, any vertex w of T 2 is origin of a path Q ≡ a 0 a n-1 ...a δ+2 in T 2 ∪ {x 0 } with end distinct from x 0 and thus wx 1 ...x δ vP -1 ≡ C, unless if (T 2 ; a n-1 a n-2 ...a δ+2 ) is one of the following biexceptions with respect to x 0 : (From now on, the possible biexceptions are determined by the number of blocks of the path, the lengths of some blocks and the outdegree and indegree of x 1 in T 2 ) Exc (0, 1):

Since d + T 2 (x 2 ) = 2 then x 1 = a. We have x 0 3x 2 x 3 v1a2x 0 ≡ C. Exc (4, 1): d - T 2 (x 1 ) = 1 so x 1 = a and d + T 2 (x δ ) ≤ 3 then x δ is an origin of a path P ≡ a 1 a 0 a n-1 ...a δ+2 in (T 2 ) ∪ {x 0 , x δ } with end distinct from x 0 so x δ x δ-1 ...x 1 vP -1 ≡ a 1 a 2 ...a 0 a 1 = C. Exc (8, 1): Clearly x 1 = a so if 3 → x δ it is solved as above and if x δ → 3 we have x 0 x 1 ...x δ 341v25x 0 ≡ C. Exc (9, 1)(1): If d + T 2 (x δ ) ≥ 1 then
it is clear that there is an outneighbor of x δ origin of a path P ≡ a δ+1 ...a n-1 in (T 2 ) ∪ {v} with end and inneighbor of x 0 , so x 0 ...x δ P x 0 ≡ C. Else, δ = 3, if x 1 = a, since 5 is an origin of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 ∪{x 0 } with end distinct from x 0 then 5x 3 x 2 x 1 vP -1 ≡ C, so x 1 = a and there is an inneighbor u of x δ origin of a path P ≡ a 0 a n-1 ...a δ+2 in (T 2 ) ∪ {x 2 } with end distinct from x 2 , so ux δ ax 0 vP -1 u ≡ C. Exc (9, 1)(2): x 1 = a, so, as above, we may suppose that V (T 2 ) -S ⊆ N + (x δ and then x 0 ...x δ 413v25x 0 ≡ C. Exc (9, 1)(3) and ( 4): x 1 = a, so, as above, we may suppose that V (T 2 ) -S ⊆ N + (x δ and then x 0 ...x δ 513v24x 0 ≡ C. Exc (15,1): x 1 = a, so, as above, we may suppose that V (T 2 ) -S ⊆ N + (x δ and then x 0 ...x δ 4532v61x 0 ≡ C. Exc (16,1)(1), ( 2) and ( 3): x 1 = a, so, as above, we may suppose that V (T 2 ) -S ⊆ N + (x δ and then x 0 ...x δ 123v654x 0 ≡ C. Exc (16, 1)(4): x 1 = a, so, as above, we may suppose that V (T 2 ) -S ⊆ N + (x δ and then x 0 ...x δ 123v456x 0 ≡ C. Exc (32,1): x 1 = a, so, as above, we may suppose that V (T 2 ) -S ⊆ N + (x δ and then x 0 ...x δ 136v425x 0 ≡ C. Exc (33,1): x 1 = a, so, as above, we may suppose that V (T 2 ) -S ⊆ N + (x δ and then x 0 ...x δ 463v7521x 0 ≡ C. Exc (E 8 (5), 1): x 1 = a, so, as above, we may suppose that V (T 2 ) -S ⊆ N + (x δ and then x 0 ...x δ 123vwux 0 ≡ C.

And if

T 1 = 4B (a 0 ...a 7 = (2, 1, 3, 1)), if there exists x ∈ T 1 such that d - T 2 (x) ≥ 2, let y 1 ...y 4 ≡ a 1 .
..a 4 be a path in T 1 of origin x, if y 1 has an inneighbor u origin of a path P ≡ a 0 a n-1 ...a 6 then uy 1 ...y 4 vP -1 ≡ C, otherwise, (T 2 ; a 0 a n-1 ...a 6) is an exception. By the building lemmas, we may assume that (T 2 ; a 0 a n-1 ...a 6 ) is one of the following biexceptions with respect to y 1 : (We can suppose that N - [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF], 0), (E 9 [START_REF] Havet | Oriented Hamiltonian paths in tournaments: a proof of Rosenfeld's conjecture[END_REF], 0), (E 10 (9), 0): Let P be an antidirected outpath in T 2 -u, we have uy 1 ...

y 4 P vu ≡ C. So d - T 2 (x) = 1 for all x ∈ T 1 , let u 1 u 2 u 3 be a directed outpath in T 1 , let w 1 ∈ N + T 2 (u 3 ), we have d + T 2 -w 1 (u 1 ) ≥ |T 2 -w 1 | -2 so u 1 has an outneighbor origin of a path P ≡ a n-2 a n-3 ...a 4 in T 2 -w 1 and thus u 2 u 3 w 1 vP -1 u 1 u 2 ≡ C unless if T 2 -w 1 is a transitive triangle, set V (T 2 ) = {a, w 1 , w 2 , w 3 } where w 2 → w 3 . If a → w 3 then u 2 u 3 w 1 vw 2 w 3 u 1 au 2 ≡ C, so w 3 → a and then we may suppose that w 1 → a. If w 3 → u 1 then u 2 u 3 w 1 vw 2 w 3 u 1 au 2 ≡ C, so u 1 → w 3 . Suppose that w 2 → w 1 then u 2 → w 1 since otherwise u 3 u 1 w 3 vw 2 w 1 u 2 au 3 ≡ C, and thus u 3 → w 3 since otherwise u 1 u 2 w 1 vw 2 w 3 u 3 au 1 ≡ C, but now d -(w) = 4 then w 1 → u 1 and so u 2 u 3 w 3 vw 2 w 1 u 1 au 2 ≡ C.
So we may assume that w 1 → w 2 and, without loss of generality, T (w 1 , w 2 , w 3 ) = 3A. We remark that

u i → w j implies that u i+1 → w j-1 , so if u 2 → w 3 then u 3 vaw 1 w 2 u 1 w 3 u 2 u 3 ≡ C and if u 2 → w 1 then u 3 vaw 3 w 1 u 1 w 2 u 2 u 3 ≡ C. So we can assume that a n-2 → a n-3 . Let x n-1 x 0 x 1 ...x δ-2 ≡ a n-1 a 0 a 1 ...a δ-2 in T 1 such that d + T 2 (x n-1 ) = max{d + T 2 (x); x ∈ T 1 }. If x n-1 has an outneighbor origin of a path P ≡ a δ-1 ...a n-3 in T 2 then x 0 x 1 ...x δ-2 P vx n-1 x 0 ≡ C. Otherwise, either d + T 2 (x δ-2
) ≥ 2 and (T 2 ; a δ-1 ...a n-3 ) is one of the following biexceptions with respect to x δ-2 : Dual Exc (E 1 (5), 1): We have v1w23ux δ-2 ...

x 0 x n-1 v ≡ C. Dual Exc (E 5 (5), 1): We have 13x 1 42x n-1 x 0 5v1 ≡ C. Or d + T 2 (x) = 1 for all x ∈ T 1 . Let a ∈ T 2 and x 0 x 1 ...x δ ≡ a 0 a 1 ...a δ in T 1 ∪ {a}, we have d - T 2 (x 0 ) ≥ |T 2 -a| -1 so
there is an inneighbor of x 0 origin of a path P ≡ a n-1 a n-2 ...a δ+2 in T 2 -a and thus x 0 ...x δ vP -1 x 0 ≡ C.

(ii) If a 2 → a 1 . Suppose that a n-3 → a n-2 , T 1 contains a path x 0 x 1 ...x δ ≡ a 0 a 1 ...a δ such that x 0 , x δ = a, if x 0 has an inneighbor in T 2 origin of a path P ≡ a n-1 a n-2 ...a δ+2 then x 0 ...x δ vP -1 x 0 ≡ C. Otherwise, either d - T 2 (x 0 ) ≥ 2 and (T 2 ; a n-1 a n-2 ...a δ+2 ) is one of the following biexceptions with respect to x 0 : Exc (1, 1): There is a vertex b ∈ T 2 such that d - T 2 (b) ≥ 2 and T 2 -b = 4A, so C can be found as above unless if T 1 ∪ {b} = 4B, in this case there is an origin of P = -(1, 1, 1) in T 1 ∪ {b} outneighbor of 3 and so v1423P v ≡ C. Exc (6, 1): We have x δ → 1 so x 0 ...x δ 142v35x 0 ≡ C. Exc (10, 1): Since T 2 cannot be 6F nor 6H, then if T 1 = 5A, let y 1 ...y 5 ≡ a n-1 a n-2 ...a δ+2 in T 1 , d - T 2 (y 1 ) ≥ 3 implies that there is an inneighbor u of y 1 origin of P ≡ a 0 a 1 ...a δ in T 2 and thus P vy 5 ...y 1 u ≡ C. So T contains C. Otherwise, T 1 = 3A, if x δ → 1 then x 0 ...x δ 12v453x 0 ≡ C, so we can assume that x 1 = a since otherwise, taking x δ ...x 1 ≡ a 1 ...a δ , (T 2 , a n-1 ...a δ+2 ) is not a biexception with respect to x 0 . T 2 \ {2} = 5E, so let y 0 ...y δ ≡ a 0 ...a δ in T 1 ∪ {2} such that 3, 4 or 5 is an inneighbor of y 1 , which is an origin of P = (1, 2, 1) in T 2 \ {2}, we have y 0 ...y δ vP -1 y 0 ≡ C. Exc [START_REF] Thomason | Paths and cycles in tournaments[END_REF]1): We have x δ → 3 so x 0 ...x δ 354v21x 0 ≡ C. Exc (22,1): There is y 0 ...y δ ≡ a n-δ-1 a n-δ ...a n-1 in T 1 having an inneighbor origin of P ≡ a n-δ-2 ...a 1 in T 2 , so vP -1 y 0 ...y δ v ≡ C. Exc (24, 1): We have x δ → 1 so x 0 ...x δ 1523v64x 0 ≡ C. Exc (26,1): Let y 1 ...y δ ≡ a 3 ...a δ+2 in T 1 , (T 2 ; a 1 a 0 a n-1 ...a δ+4 ) is not an exception so there is an inneighbor of y 1 origin of a path P ≡ a 1 a 0 a n-1 ...a δ+4 in T 2 , we have P y 1 ...

y δ+2 v ≡ C. Thus T contains C. Or T 1 = 4B and d - T 2 (x) = 1 for all x ∈ T 1 . If there exists v ∈ T such that T (N -(v )) is a transitive triangle, set T " 1 = T (N -(v )) = {u 1 , u 2 , u 3 } and T " 2 = T (N + (v )) such that u 1 → u 2 → u 3 , if u 1 has an inneighbor ori- gin of a path P ≡ a n-1 a n-2 ...a 4 in T " 2 then u 1 u 3 u 2 vP -1 u 1 ≡ C, otherwise, (T 2 ; a n-1 a n-2 ...a 4
) is one of the biexceptions Exc (4, 1), [START_REF] Petrović | Antidirected Hamiltonian circuits in tournaments, Graph Theory[END_REF]1), (15, 1), (16, 1) and (33, 1) with respect to u 1 , but in these cases (T 2 ; a n-4 a n-5 ...a 1 ) is not an exception so there exists an origin x of a path P ≡ a n-4 a n-5 ...a 1 in T 2 such that xu 1 ≡ a n-4 a n-3 , thus vP -1 u 1 u 3 u 2 v ≡ C. So we may assume that for any vertex x ∈ T of minimal indegree, T (N -(x)) = 3A, thus the inneighbors w 1 , w 2 and w 3 of x 0 , x 1 and x 3 are pairwasly distinct, so (T 2 ; a n-1 a n-2 ...a 5 ) is one of the exceptions 1, 18, 22, 24, 26, 34, 38 and 47, if

it is Exc 18 then | x∈T 1 N - T 2 (x)| ≥ 4
thus, without loss of generality, we may suppose that there exists an inneighbor of x 0 origin of a path P ≡ a 4 a 3 ...a n-2 in T 2 and therefore x 0 x 1 x 3 vP -1 x 0 ≡ a 5 a 6 ...a 5 = C. Otherwise, there exists x ∈ T 2 origin of a path P ≡ a n-5 a n-6 ...a 1 such that xx 3 ≡ a n-5 a n-4 and thus vP -1 x 3 x 2 x 1 x 0 v ≡ C. So we may assume that a n-2 → a n-3 , set T 1 = {u 1 , u 2 , u 3 }, if u 3 has an inneighbor origin of a path P ≡ a 2 a 3 ...a n-3 in T 2 then u 2 u 3 P vu 1 u 2 ≡ C, otherwise, without loss of generality, since there is no biexceptions such that

P = -(1, 3, Q), we may suppose that s + T 2 ( 3 i=1 N - T 2 (u i )) ≤ 3; if |N + T 2 (T 1 )| ≥ 2, let u, w ∈ N + T 2 (T 1
) such that w is an origin of a path P ≡ a 3 a 4 ...a n-3 in T 2 -u, we have

u 2 uu 3 P vu 1 u 2 ≡ C, otherwise, set {u} = N + T 2 (T 1 ), we have u 1 uu 2 u 3 vI T 2 -u u 1 ≡ C. (b) So we can assume that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception, let x 1 ...x δ ≡ a 1 ...a δ in T 1 such that d - T 2 (x 1 ) = max{d - T 2 (x); x ∈ Or(a 1 ...a δ ; T 1 )}.
If there is an inneighbor u of x 1 origin of P ≡ a 0 a n-1 ...a δ+2 in T 2 then ux 1 ...x δ vP 1 ≡ C. Otherwise, we have two cases:

(i) If a 1 → a 2 , either d - T 2 (x 1 ) ≥ 3 and (T 2 , a 0 a n-1 ...a δ+2
) is an exception. By the building lemmas we may suppose that (T 2 ; a 0 a n-1 ...a δ+2 ) is one of the following biexceptions with respect to x 1 : Dual Exc (0, 0)(1): We have

x δ → 3 so 12x 1 ...x δ 3v1 ≡ C. Dual Exc (1, 0)(3): 4 → x 2 , a contradiction. Dual Exc (4, 0)(2): If d + T 1 (x 1 ) = 2 then d + (x 1 ) = 5 and T (N + (x 1 )) = 5A since {x 2 , v} → {4, 5} that is solved unless if T (N + (x 1 )) = 5E
If d + (x 2 ) = 6 then N + (x 2 ) = 6H and if d + (x 2 ) = 5 then T (N + (6) = 6H; thus T contains C. Or d - T 2 (x 1 ) = 2, by Lemma 2.4, (T 1 ; a 1 ...a δ ) = (4B; (1, 1, 1)), (4B; (2, 1)), (3A; (2)) or |T 1 | = 2. If |T 1 | ≥ 3,
by the building lemmas we may assume that (T 2 ; a 0 a n-1 ... a δ+2 ) is one of the following biexceptions with respect to x 1 : Dual Exc (0, 0)(1):

d -(3) ≥ 4 is a contradiction. Dual Exc (1, 0): d -(4) ≥ 5 is a contradiction.
In the other possible biexceptions, N -(x 1 ) \ S = ∅ and so we can take x δ as origin of y 1 ...y δ ≡ a 1 ...a δ , where y 2 = x 1 , the result follows. And if

|T 1 | = 2, d -(x 1 ) = 2 and T (N + (x 1 )) = {v} → (N + (x 1 ) \ {v, x 2
}) which implies that (T 2 ; a 0 a n-1 ...a δ+2 ) can be only one of the following biexceptions with respect to x 1 : Dual Exc (31, 0), (48, 0): We have d -(x 2 ) = 2 and T (N + (x 2 )) is not one of the tournaments of these biexceptions. Dual Exc (E 8,9,10 (|T 2 |), 0) and (E 8,9,10 (|T 2 |), 0): There is an inneighbor of x 1

origin of P ≡ a n-1 a n-2 ...a + δ + 1 in T 2 so x 1 x 2 vP -1 x 1 ≡ C. (ii) If a 2 → a 1 , if d - T 2 (x 1 ) ≥ 2 and d - T 2 (x 2 ) = |T 2 | then (T 2 ; a 0 a n-1 ...a δ+2
) is an exception. If (T 2 ; a 0 a n-1 ...a δ+2 ) is not one of the dual of the exceptions 0, 1,[START_REF] Sahili | About the Number of Oriented Hammiltonian Paths and Cycles in Tournaments[END_REF][START_REF] Havet | Oriented Hamiltonian paths in tournaments: a proof of Rosenfeld's conjecture[END_REF]18,19,22,33 , E 1 (|T 2 |) andE 1 ( [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF] with X = 3A, then by Lemma 2.8, N + T 2 (x 1 ) contains an origin of w of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 and so wx 1 vQx 2 P -1 ≡ C in T , where Q ≡ a 3 ...a δ in T 1 \ {x 1 , x 2 }. So we may assume that (T 2 ; a 0 a n-1 ...a δ+2 ) is one of the following exceptions: Exc 4,19 or 33: Any vertex of T 2 is an origin of a path P ≡ a 1 a 0 ...a δ+2 in

Dual Exc 0: Set N - T 2 (x 1 ) = {1, 2}, we have x 1 3v1x 2 2x 1 ≡ C. Dual
T 2 ∪ {x 2 }, so if T 1 -x 2 contains a path y 2 ...y δ ≡ a 2 ...a δ , we have y 2 ...y δ vP -1 y 2 ≡ a 2 a 3 ...a 2 = C. Otherwise, T 1 -x 2 = 3A, let y 1 y 2 y 3 be a directed outpath in T 1 -x 2 and u ∈ N + T 2 (y 1 ), any vertex of (T 2 -u) ∪ {x 2 } is an origin of a path P ≡ x 6 x 7 ...x 1 in (T 2 -u) ∪ {x 2 }, thus vuy 1 y 2 y 3 P v ≡ a 2 a 3 ...a 2 = C. Dual Exc E 1 (4): If δ = 2 then x 1 4321vx 2 x 1 ≡ C. If δ = 3 then x 2 x 1 O T 2 vx 3 x 2 ≡ C. If δ = 4, let P ≡ a n-5 a n-6 ...a 1 be a path in T 2 and Q = (1, 2) be a path in T 1 of origin x 2 , we have QvP -1 x 2 ≡ a n-4 a n-5 ...a n-4 = C. Dual Exc E 1 (6): δ ≥ 4, if δ = 4, let P = -(1, 4) be a path in T 2 , we have x 1 P vx 4 x 3 x 2 x 1 ≡ C. If δ = 5, let Q = -(2, 2) be a path in T 1 , we have vQO T 2 v ≡ a n-4 a n-3 ...a n-4 = C. If δ = 6, let Q = -(1, 2, 1, 1
) be a path in T 1 of origin x and P = (2, 3) be a path in T 2 of origin an outneighbor of x, we have

QvP -1 x ≡ a n-2 a n-1 ...a n-2 = C. Otherwise, either d - T 2 (x 1 ) ≥ 2, d - T 2 (x 2 ) < |T 2 | and (T 2 , a 0 a n-1 ...a δ+2
) is an exception. By the building lemmas we may assume that it is one of the following biexceptions with respect to

= -(1, 1) in T 2 \ {i -1, i}, we have x 1 ...x δ i(i -1)vP x 1 ≡ a 1 ...a n-1 a 0 a 0 = C. So we can suppose that 1, 4, 5 ∈ N -(x δ ). If a δ-1 → a δ then 1x 1 ...x δ-1 v2x δ 3541 ≡ C. Otherwise, if x δ-1 ← 1 or 2, say 1, we have 2x 1 ...x δ-1 13x δ 45v2 ≡ C. Else, if 3 → x δ-1 we have x 1 vx 2 ...x δ-1 3514x δ 2x 1 ≡ C. Assume now that x δ-1 → 3. Suppose first that 4 → x δ-1 , if T (N + (4)) = 5A then T contains C; else, we have {5, x 2 } → x δ so T (N + (5)) = 5A. So we can assume that x δ-1 → 4, we have N + (x δ-1 ) = {1, 2, 3, 4, v} where T (N + (x δ-1 )) = 5A; so T contains C. Dual Exc (9, 0)(1): If x δ → 1 then 2x 1 ...x δ 1453v2 ≡ C. Otherwise, if d + T 2 (x δ ) ≥ 1, let u ∈ N + T 2 (x δ ), there exists x ∈ N - T 2 (x 1 ) origin of a path P = -(1, 1) in T 2 \ {u, w} where w ∈ N + T 2 (u); we have xx 1 ...x δ uwvP -1 ≡ C. So we can sup- pose that T 2 → x δ . We have N -(x 2 ) = 5E, so T contains C and thus T contains C. Dual Exc (9, 0)(2): If N + (x δ ) ∩ {2, 3, 4, 5} = ∅ then there exist u ∈ N + T 2 (x δ )
and x ∈ N - T 2 (x 1 ) such that x is an origin of a path P = -(1, 1) in T 2 \ {u, w} where w ∈ N + T 2 (u); we have xx 1 ...x δ uwvP -1 ≡ C. So we can assume that {2, 3, 4, 5} ⊆ N -(x δ ). If a δ-1 → a δ then 4x 1 ...x δ-1 v312x δ 54 ≡ C. Otherwise, if 1 or 2 → x δ-1 , say 1, then 4x 1 ...x δ-1 15x δ 32v1 ≡ C; and if 4 or 5 → x δ-1 , say 4, then 5x 1 ...x δ-1 43x δ 21v5 ≡ C. So we may suppose that N + (x δ-1 ) = {1, 2, 4, 5, v} so

T (N + (x δ-1 )) = ← - 5E thus T contains C. Dual Exc (15, 0): If x δ → 6 then 1x 1 ...x δ 63v5421 ≡ C and if N + (x δ )∩{1, 2, 3} = ∅, say 1, then 6x 1 ...x δ 13v2456 ≡ C. So we may suppose that {1, 2, 3, 6} ⊆ N -(x δ ), then T (N + (6)) = 6C since otherwise x δ ≡ 4 or 5 but {1, 2, 3} N + (x 1 ) nor N -(x 1 ). Therefore T contains C. Dual Exc (16, 0): If x δ → 6 then 4x 1 ...x δ 62v5314 ≡ C. If {1, 2, 3} ∩ N + (x δ ) = ∅, say 1 ∈ N + (x δ ), then 4x 1 ...x δ 13v6524 ≡ C. If {4, 5} ∩ N + (x δ ) = ∅, say 4 ∈ N + (x δ ), then 5x 1 ...x δ 46v3125 ≡ C. So we may assume that N -(x δ ) = T 2 , then a δ → a δ-1 .
One can find P ≡ a δ ...a n-1 a 0 in T 2 ∪ {x δ , v} with origin an inneighbor of x δ-1 and end an inneighbor of x 1 ; we have x 1 ...x δ-1 P x 1 ≡ C. Dual Exc (33,0)

: If {2, 3, ..., 7} ∩ N + (x δ ) = ∅ then there exist u ∈ N + T 2 (x δ ) and x ∈ N - T 2 (x 1 ) such that x is an origin of a path P ≡ a 0 a n-1 ...a δ+4 in T 2 \ {u, w} where w ∈ N + T 2 (u), then xx 1 ...x δ uwvP -1 ≡ C. So assume that {2, ..., 7} ⊆ N -(x δ ), if a δ-1 → a δ then there exist x ∈ N - T 2 (x 1 ) origin of P ≡ a 0 a n-1 ...a δ+1 in T 2 ∪ {x δ }; we have xx 1 ...x δ-1 vP -1 ≡ C.
Otherwise, one can find P ≡ a δ ...a n-1 a 0 in T 2 ∪ {x δ , v} with origin an inneighbor of x δ-1 and end an inneighbor of x 1 ; we have x 1 ...

x δ-1 P x 1 ≡ C. Dual Exc (E 1 (|T 2 |), 0): We have T (N + (2)) = F 1 nor F 5 , so T contains C. Dual Exc (E 3 (|T 2 |), 0): If x δ → X then if X = 3A, let Q = (1, |X| -2) be a
Hamiltonian path of X, we have 1x 1 ...x δ Q2v1 ≡ C and if X = 3A = {3, 4, 5} then 3x 1 ...x δ 5421v3 ≡ C. So we may suppose that X

N + (x)δ). If x δ → 2 then 1x 1 ...x δ 2I X v1 ≡ C. Assume now that 2 → x δ , we have T (N + (2)) = F 1 , F 3 nor F 5 , so T contains C. Dual Exc (E 5 (|T 2 |), 0): If x δ → 1 then 2x 1 ...x δ 1I T 2 \{1,2} v2 ≡ C. So assume that 1 → x δ . If N + (1) = F 1 nor F 5 then T contains C. Otherwise, x δ → x 1 , so 21x 2 ...x δ x 1 I T 2 \{1,2} v2 ≡ C. Dual Exc (E 8 (|T 2 |), 0), (E 9 (|T 2 |), 0) and (E 10 (|T 2 |), 0): Set X = {u → w}.
Without loss of generality, any vertex x of 3A, 5A and 7A respectively is an origin of a path P ≡ a δ+1 ...a n-1 a 0 in T 2 ∪ {v}; so if there is a such x ∈ N + (x δ ) then x 1 ...x δ P x 1 ≡ C. Now we may assume that 3A, 5A or 7A → x δ respectively. If a δ-1 → a δ then any vertex of 3A, 5A or 7A is an origin of a path P ≡ a δ+1 ...a n-1 a 0 in T 2 ∪ {x δ } with end w so wx 1 ...x δ-1 vP ≡ C. Otherwise, x δ-1 has an inneighbor x in T 2 -u. x is an origin of a path P ≡ a δ ...a n-1 a 0 in T 2 ∪ {x δ , v} with end u, so ux 1 ...x δ-1 P ≡ C. Or d - T 2 (x 1 ) = 1. By Lemma 2.3, (T 1 ; a 1 ...a δ ) = (4B; -(1, 2)) or T 1 is transitive and a 1 ...a δ is directed (it is clear that δ = 2 or 3). Suppose first that (T 1 ; a 1 ...a δ ) = (4B; - (1,[START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF]). Set C = u 0 u 1 ...u n-1 u 0 such that u 0 ≡ a δ+1 and u 1 ≡ a δ . We have u 1 ...u 4 = (2, 1), so let y 1 ...y 4 = (2, 1) in T 1 . We have d - T 2 (y 1 ) ≥ 3, so y 1 has an inneighbor u ∈ T 2 origin of a path P ≡ u 0 u n-1 ...a δ+2 in T 2 and so uy 1 ...y 4 vP -1 ≡ C, or (T 2 ; u 0 u n-1 ...u δ+2 ) is an exception. The building lemmas give the result unless if (T 2 ; u 0 u n-1 ...u δ+2 ) is a biexception with respect to y 1 . Since any vertex of T 1 distinct from x 1 can be an origin of a path [START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF]1), we may assume that |N -(y 1 )| ≥ 3 and |S| ≥ 3, and since the path u 1 u 0 ...u δ+2 has an even number of blocks with first block of length 2, there is no biexception verifying these conditions. Now if T 1 is a transitive triangle, if C has a block of length 1, let s = min {i; a i is contained in a block of length one}. Without loss of generality we can suppose that s = δ + 2, indeed, by considering C and C, we may assume that a s+1 → a s , do a i = a i-1 until a δ+2 = a s . Thus a δ+3 → a δ+4 , let w ∈ N + T 2 (x 2 ), there is an outneighbor of x 1 origin of a path P ≡ a 1 a 0 a n-1 ...a δ+4 in T 2 -w, and thus x 1 x 3 x 2 wvP -1 x 1 ≡ a 2 a 3 ...a 1 a 2 = C unless if |T 2 | = 3, we may suppose that T = 7K and so 13654721 = (2, 2, 2, 1) ≡ C. Assume now that C = (2, 2, ..., 2). If there exists w ∈ N + T 2 (x 2 ) such that d - T 2 (w) ≥ 1, let t ∈ T 2 be an inneighbor of w, and P ≡ a 1 a 0 a n-1 ...a δ+5 be a path in T 2 \ {w, t} of origin an outneighbor of x 1 (this is possible since

d + T 2 \{w,t} (x 1 ) ≥ |T 2 \ {w, t}| -1), we have x 1 x 3 x 2 wtvP -1 x 1 ≡ a 2 a 3 ...a 2 = C. Else, N + T 2 (x 2 ) = {w} and d + T 2 (w) = 0, we have d - T 2 (
x 3 ) ≥ 3 so there is an inneighbor t ∈ T 2 of x 3 origin of a path Q = (2, 1) containing w. Let P ≡ a δ+4 ...a n-2 be a path of origin an inneighbor of x 2 in T 2 -V (Q), we have vQ -1 x 3 x 1 x 2 P v ≡ a n-1 a 0 a 1 ...a n-1 = C. The last case is when |T 1 | = 2, let u be the inneighbor of x 1 in T 2 , if d - T 2 (u) ≥ 1, take x 1 in the place of v and set T 1 = T (N -(x 1 )) and T 2 = T (N + (x 1 )). Let y 1 y 2 ≡ a 1 a 2 in T 1 , and let w ∈ N - T 2 (y 1 ). Since T 2 = v → (T 2 -v), w is an origin of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 , and thus wy 1 y 2 x 1 P -1 ≡ C, unless if (T 2 \ {w, v}; a n-2 a n-3 ...a δ+2 ) is a Grünbaum's exception. If d - T 2 -v (w) ≥ 1, any inneighbor t of w in T 2 -v is an origin of a path P ≡ a n-1 a n-2 ...a δ+2 in T 2 -w, so wy 1 y 2 x 1 P -1 w ≡ C, else, any outneighbor t of w in T 2 -v is an origin of a path P ≡ a n-4 a n-5 ...a 1 in T 2 -w, so x 1 P -1 wy 1 y 2 x 1 ≡ C. So we may suppose that d -

T 2 (u) = 0. We have d - T 2 -u (x 2 ) ≥ 2 and so s + T 2 -u (N - T 2 (x 2 )) ≥ 3, since otherwise there is an inneighbor of x 2 of outdegree 1. If s = min {4 ≤ i ≤ n -2; d + C (a i ) = 2} exists, let P ≡ a n-1 a n-2 .
..a s+1 be a path in T 2 -u of origin an inneighbor of x 2 and Q ≡ a 3 a 4 ...4 s-1 be a Hamiltonian path of T 2 \(V (P )∪{u}), we have

x 2 x 1 uQvP -1 x 2 ≡ C. Else, C = (2, 2, 2, 1), let P ≡ a 1 a 2 .
..a n-3 be a Hamiltonian path of (T 2 -u) ∪ {v} with origin and end distinct from v, we have

x 1 P ux 2 x 1 ≡ C. Case 2: If a 1 → a 0 → a n-1 . Do a i = a i-1 until a n-2 → a n-1 for the first time.
(a) Suppose first that (T 1 ; a 1 ...a δ ) is a Grünbaum's exception, we have two cases:

(i) If a 2 → a 1 . For x i ∈ T 1 let P i ≡ a 0 a 1 ...a δ-1 in T 1 . If there is an outneighbor of an x i origin of a path P ≡ a n-1 a n-2 ...a δ+1 in T 2 then P i vP -1 x i ≡ C. Otherwise, either there exists x i ∈ T 1 such that d + T 2 (x i ) ≥ 2 or T 1 = 3A and d + T 2 (x i ) = 1 for all x i ∈ T 1 .
In the first case, denote by B the block of C containing the arc a n-2 → a n-1 and let Q ≡ a 2 a 3 ...a δ+1 be a path in T 1 of origin x i , say x 1 .

If |B | = 1,, (T 2 ; a n-1 a n-2 ...a δ+1 ) is an exception, so by the building lemmas it is one of the following biexceptions with respect to x 1 : Dual Exc (1, 1)(1), (2) and(4): x 2 has an outneighbor origin of a directed outpath P of order 3 in T 2 with end distinct from 1, so x 1 x 3 x 2 P v1x 1 ≡ C. Dual Exc (1,1)(3)

If |B | = 2, suppose that s - T 2 (N + T 2 (x 1 )) = 2 and set N + T 2 (x 1 ) = {u → w}. If N + (x 2 ) ∩ (T 2 \ {u, w}) = ∅, let t ∈ N + (x 2 ) ∩ (T 2 \ {u, w}), we have |T 2 | ≥ 4 so there exists P ≡ a n-2 a n-3 ...a δ+1 in T 2 -t of origin distinct from u and w, thus t * P 1 vP -1 x 1 t ≡ C. Otherwise, by induction on i, we may suppose that T 1 ← T 2 \ {u, w}. If d + T 1 (u) ≥ 1 or d + T 1 (w) ≥ 1, assume without loss of gen- erality that it is u, then there exists a path R ≡ a n-1 a 0 ...a δ-1 in T 1 ∪ {u} of origin x and end distinct from u, let S ≡ a n-2 a n-3 ...a δ+1 be a path in T 2 -u of origin distinct from w, we have RvSx ≡ a n-1 a 0 ...a n-1 = C. Otherwise, if T 2 \ {u, w} contains a path R ≡ a n-3 a n-4 ...a δ+1 then u * P 1 vR -1 x 1 wu ≡ C, else, let t ∈ T 2 \ {u, w}, R ≡ a n-3 a n-4 ...a δ+2 in T 2 \
{u, w, t} and S ≡ a 1 ...a δ in (T 1 -x 1 ) ∪ {t} with origin and end distinct from t, we have uSvR -1 x 1 wu ≡ C. So assume that s - T 2 (N + T 2 (x 2 )) ≥ 3 and (T 2 ; a n-1 a n-2 ...a δ+1 ) is an exception. By the building lemmas, it is one of the following biexceptions with respect to x 1 : Dual Exc [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF]1) and(7, 1

): We have 2vP -1 1 14352 ≡ C. If |B | = 3, if s - T 2 (N + T 2 (x 1
)) ≥ 4 then, since there is no a such biexception, T contains C. So we may suppose that s -

T 2 (N + T 2 (x 1 )) ≤ 3, if there exists u ∈ N + T 2 (x 1 ) ∩ N + T 2 (x 2 ), let w ∈ N + T 2 (x 1 ) -u, t ∈ N - T 2 (x 1 ) such that (T 2 \ {u, w, t}, a n-4 a n-5 ...a δ+1 ) is not a Grünbaum's exception and R ≡ a n-4 ...a δ+1 in T 2 \ {u, w, t}, we have u * P 1 vR -1 x 1 wtu ≡ C. Otherwise, let t ∈ N + T 2 (x 2 ), if N + T 2 (x 1 ) = {u, w}, then there exists R ≡ a n-4 a n-5 ...a δ+1 in T 2 \ {u, t} of origin distinct from w thus t * P 1 vR -1 x 1 ut ≡ C, so we may assume that s - T 2 (N + T 2 (x 1 )) = 3, then either t is an origin of a path R ≡ a n-1 a n-2 ...a δ+1 and thusP 2 vR -1 x 2 ≡ C or t is an origin of R -1 and thus vP -1 2 R -1 v ≡ a 1 a 2 ...a 1 = C. Assume now that T 1 = 3A and d + (x i ) = 1 for all x i ∈ T 1 . If |T 2 | = 3 then if T 2 is transitive, x 3 has an outneighbor origin of an antidirected outpath S in T 2 and thus vx 1 x 2 x 3 Sv ≡ a 1 a 2 ...a 1 = C, and if T 2 is a circuit then x 3 x 2 x 1 vO T 2 x 3 ≡ C. So |T 2 | ≥ 4. If the path R = a 2 a 3 ...a n-2 has at least three blocks, one can find R in T 2 ∪ {v} with origin x ∈ N -(x 1 ) and end y ∈ N -(x 3 ), thus x 2 x 1 Rx 3 x 2 ≡ C.
Otherwise, by taking R = a 2 a 1 ...a 6 , we can suppose without loss of generality that C = [START_REF] Camion | Chemins et circuits hamiltoniens des graphes complets[END_REF][START_REF] Camion | Chemins et circuits hamiltoniens des graphes complets[END_REF]1,1). and4 → T 1 . x 1 has an outneighbor in T 2 -4 origin of an antidirected inpath P , thus x 1 x 3 4x 2 vP -1 x 1 ≡ C. (ii) If a 1 → a 2 , for x i ∈ T 1 set P i ≡ a 3 a 4 ...a δ+2 be a path in T 1 of origin x i and end y i . There exists a P i such that for a

If N + T 2 (x i ) = {u} for all i, then T 2 = 4B, let V (T 2 ) = {u 1 , u 2 , u 3 }, we have x 1 u 1 vuu 2 x 3 u 3 x 2 x 1 ≡ C. Otherwise, T 2 = 4A
u i ∈ N + T 2 (x i ) we have N - T 2 -u i (y i ) ≥ 2. If there exists Q ≡ a δ+3 a δ+4 ...a n-1 a 0 or R ≡ a n-2 a n-3 ...a δ+1 in T 2 -u i of origin and inneighbor of y i , then vuP i Qv ≡ a 1 a 2 ...a 1 = C or vR -1 P -1 i v ≡ a δ a δ+1 ...a δ = C. Otherwise (T 2 -u i ; Q) is
an exception, so by the building lemmas, it is one of the following biexceptions with respect to y i : (in fact, let z i be the end of

P * i , if d + T 2 -u i (z i ) = 0
by taking the path P j of origin y i , we have y j = z i and since

d - T 2 -u j (z i ) ≥ |T 2 | -2, T contains C) Dual Exc (1, 0): (T 2 , a 2 a 1 a 0 a n-1 a n-2
) is not a biexception with respect to y i so z i is an origin of a path Exc (4,1), [START_REF] Havet | Oriented Hamiltonian cycles in tournaments[END_REF]1), [START_REF] Petrović | Antidirected Hamiltonian circuits in tournaments, Graph Theory[END_REF]1), (33, 1): (T 2 ; a 2 a 1 a 0 ...a δ+5 ) is not a biexception with respect to y i so z i is an origin of a path [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF], 1) with respect to z i so T contains C. Dual Exc (E 8 [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF], 0), (E 9 [START_REF] Havet | Oriented Hamiltonian paths in tournaments: a proof of Rosenfeld's conjecture[END_REF], 0) and (E 10 (9), 0): (T 2 -1; Q) is not one of the biexceptions Dual Exc (E 8 (5), 0)), (E 9 [START_REF] Havet | Oriented Hamiltonian paths in tournaments: a proof of Rosenfeld's conjecture[END_REF], 0) and (E 10 (9), 0) with respect to z i so T contains C. 1,[START_REF] Sahili | About the Number of Oriented Hammiltonian Paths and Cycles in Tournaments[END_REF]18,19,22,33

P ≡ a 3 a 2 a 1 a 0 a n-1 a n-2 in T 2 ∪ {y i } and thus z i x i vP -1 z i ≡ a 4 a 5 ...a 4 = C. Therefor T contains C. Exc (3, 1): (T 2 ; R) is the biexception Dual Exc (1, 0) with respect to y i so T contains C.
P ≡ a 4 a 3 a 2 ...a δ+4 in T 2 ∪ {y i }, thus * (P -1 i )u i vP -1 z i ≡ a 4 a 5 ...a 4 = C, therefore T contains C. Exc (E 8 (5), 1): (T 2 -1; Q) is not the biexception Exc (E 8
(b) Assume now that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception, let x 1 ...x δ ≡ a 1 ...a δ in T 1 such that d + T 2 (x 1 ) = max {d + T 2 (x); x ∈ Or(a 1 ...a δ ; T 1 )}. If there is an outneighbor u of x 1 origin of a path P ≡ a 0 a n-1 ...a δ+2 or an outneighbor w of x δ origin of a path Q ≡ a δ+1 a δ+2 ...a n-1 in T 2 , then ux 1 ...x δ vP -1 ≡ C or vx 1 ...x δ Qv ≡ C. Otherwise, two cases may arise: (i) If a 2 → a 1 , if d + T 2 (x 1 ) ≥ 2 and d + T 2 (x 2 ) = 0, then, by lemma 2.8 N - T 2 (x 1 ) contains an origin w of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 and so wx 1 vQx 2 P -1 ≡ C where Q ≡ a 3 ...a δ in T 1 \ {x 1 , x 2 } unless if (T 2 ; a 0 a n-1 ...a δ+2 ) is one of the exceptions Exc 0,
, E 1 (4), E 1 (6), so if T 1 -x 2 contains a path y 2 ...y δ ≡ a 2 ...a δ then every vertex of T 2 is an origin of a path R ≡ a δ+1 a δ+2 ...a 0 in T 2 ∪{x 2 } with end in T 2 , thus vy 2 ...y δ Rv ≡ a 1 a 2 ...a 1 = C. Otherwise, (T 1 -x 2 ; a 2 ...a δ )
is a Günbaum's exception, more precisely T 1 = 4B, then a 6 → a 7 and so the path a 0 a n-1 ...a δ+2 has an odd number of blocks, so (T 2 ; a 0 a n-1 ...a δ+2 ) is either one of the exceptions 4, 33, E 1 (4), E 1 [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF], let y 2 ...y 5 ≡ a 2 ...a 5 be a path in T 1 , y 2 has an outneighbor origin of a path R ≡ a 1 a 0 ...a 7 in T 2 and thus y 2 ...y 5 vR -1 y 2 ≡ a 2 a 3 ...a 2 = C; or the exception Exc 19, let y 3 ...y 6 ≡ a 3 ...a 6 be a path in T 1 , y 3 has an inneighbor origin of a path R ≡ a 2 a 1 ...a 8 in T 2 , thus y

3 ...y 6 vR -1 y 3 ≡ a 3 a 4 ...a 3 = C. Otherwise, if d + T 2 (x 1 ) ≥ 2 then (T 2 ; a 0 a n-1 ...a δ+2
) is an exception, so by the building lemmas we may assume that it is one of the following biexceptions with respect to x 1 : Exc (0,0)(2): T Exc (9,0), (15, 0), (16, 0), (26, 0)(27, 0), (31, 0), (33, 0) and (48, 0):

(N - T 2 (x 1 )) = 3A so T contains C and so T contains C. Exc (1, 0): We have δ ≥ 3 so x δ → 4 which is an origin of a path Q. A contradiction. Exc (2, 0): x δ → 1, a contradiction. Exc (4, 0): δ = 2 so 21x 1 v453x 2 2 ≡ C. Exc (6, 0): We have x δ-1 → x δ so d + T 2 (x δ ) ≥ 1 and 4 → (T 1 -x 1 ) so N + T 2 (x δ ) ∩ {1, 2, 3, 5} = ∅, a contradiction.
If d + T 2 (x δ ) ≥ 1, set s = min {i ≥ δ + 2; d + C (a s ) = 0} and let Q 1 ≡ a δ+1 a δ+2 ...a s-1 of origin x and Q 2 ≡ a 0 a n-1 ...a s+1 of origin a be two paths such that T 2 = V (Q 1 ) ∪ V (Q 2 ), x ∈ N + (x δ ) and a ∈ N + (x 1 ), we have ax 1 ...x δ Q 1 vQ -1 2 ≡ C. Otherwise, δ = 2, there exists R ≡ a 1 a 0 a n-1 ...a δ+2 in T 2 ∪ {x 2 }
of origin an outneighbor of x 1 and with end distinct from x 2 , we have

x 1 vP -1 x 1 ≡ a 2 a 3 ...a 2 = C. Exc (E 5 (|T 2 |), 0): We have T 2 → x δ , so d + (x 2 ) < δ if and only if δ = 2, then δ = 2
∈ N - T 2 (y 1 ) such that (T 2 -u; a δ+1 a δ+2 ...a n-2 ) = (5C; -(2, 1, 1)). Since d - T 2 -u (x δ ) ≥ |T 2 -u| -1, then there is an inneighbor w of x δ origin of a path P ≡ a δ+1 a δ+2 ...a n-2 in T 2 -u; we have ux 1 ...x δ P vu ≡ C. Thus T contains C. (ii) a 1 → a 2 , either d + T 2 (x 1 ) ≥ 2 and (T 2 ; a 0 a n-1 ...a δ+2
) is an exception. So by the building lemmas, it is one of the following biexceptions with respect to So w → 2 andthus x 1 wx 2

x 1 : Exc (1, 0): d -(4) < δ if and only if δ = 2. So δ = 2 and T (N + (4)) = 4A, then T contains C and thus T contains C. Exc (6, 0), (27, 0), (33, 0): (T (N + (x 2 )); a 0 a n-1 ...a δ+2 ) is not a biexception so T contains C and thus T contains C. Exc (26, 0): There is a block of length 3 so a δ-1 → a δ . If x 1 → 4 then T (N -(4)) = 6H so T contains C. Then we may assume that N + T 2 (x 1 ) = {5, 6}. Also if x 2 → 5 then T (N -(5)) = 6H and so T contains C. We can suppose that 5 → x 2 . We have {1, 2, 3, 5} → x δ since otherwise x δ has an outneigh- bor that is an origin of a path Q ≡ a δ+1 a δ+2 ...a n-1 which is a contradiction. Now if x δ → 6 we have 45x 2 ...x δ-1 v126x δ 3x 1 4 ≡ C and if 6 → x δ we have 45x 2 ...x δ-1 v6x δ 321x 1 4 ≡ C. Exc (E 13 (|T 2 |), 0): We have T (N -(1)) = F 13 (|T 2 |) so T contains C. Or d + T 2 (x 1 ) ≤ 1, set {u} = N + T 2 (x 1 ) if it exists, if there exist w, t, y ∈ T 2 such that w → {x 2 , t}, {t, y} → x 1 and y is an origin of a path P ≡ a n-3 a n-4 ...a δ+2 then twx 2 ...x δ vP -1 x 1 t ≡ C. Otherwise, suppose first that N + T 2 (x 1 ) = {u}, if there exists w ∈ N - T 2 (x 2 ) such that d + T 2 -u (w) ≥ 1, let w ∈ N + T 2 -u (w), then (T 2 \ {w, w }; a n-3 a n-4 ...a δ+2 ) is one of the following exceptions: Dual Exc 0: Set u ≡ 1, if 3 → w then x 1 wx 2 ...x δ vw 321x 1 ≡ C, so w → 3. If 2 → w then u is an origin of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 , a contradiction.
...x δ 3w 21x 1 ≡ C. Exc 1: u ≡ 4, if w → u, let {y 2 → y 1 } = T 2 \{u, w, 1, 2} then uwx 2 ...x δ vy 1 y 2 x 1 12u ≡ C. Otherwise, d + T 2 (u) ≥ 3 then T 2 -u = 5E, we have twx 2 ...x δ vu3x 1 12t ≡ C. Dual Exc 4: Set u ≡ 1. If w → u, let {y 2 → y 1 } = T 2 \ {w, u, 2, 4, 5}, we have uwx 2 ...x δ vy 1 y 2 x 1 452u ≡ C. Otherwise, since T 2 -u = 6D then u is an origin of P ≡ a 0 a n-1 ...a δ+2 in T 2 , a contradiction.
Exc 7: u ≡ 5, so there exists an outneighbor of w satisfying the conditions implying the result. Dual Exc 33: We have

d + T 2 (u) ≥ 3, a contradiction. Or {w} ⊆ N - T 2 (x 2 ) = {u, w} and d + T 2 -u (y) = 0 for all y ∈ N - T 2 (x 2 ). If u ∈ N - T 2 (x 2 ) then d + (u) < δ + (T ) a contradiction, so N - T 2 (x 2 ) = {w}. Take x 2 in the place of v and set T 2 = T (N -(x 2 )) and T 2 = T (N + (x 2 )
). Without loss of generality we may assume that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception, so let

y 1 ...y δ ≡ a 1 ...a δ be a path in T 1 . Obviously, y 1 or y δ is an inneighbor of v, so if (T 2 -v; a n-1 a n-2 ...a δ+2 ) is not a GRünbaum's exception, let Q ≡ a n-1 ...a δ+2 , if y 1 ∈ N -(v) then vy 1 ...y δ x 2 Q -1 v ≡ C and if y δ ∈ N -(v) then x 2 y 1 ...y δ vQ -1 x 2 ≡
C. So we may assume that (T 2 -v; a n-1 a n-2 ...a δ+2 ) is a Grünbaum's exception, which implies u is an origin of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 , a contradiction. Assume now that d + T 2 (x 1 ) = 0, then by lemma 2.3 (T 1 ; a 1 ...a δ ) = (4B; (1, 2)) or a 1 ...a δ is directed. Suppose first that there exists w ∈ N -

T 2 (x 2 ) such that d + T 2 (w) ≥ 1 and let w ∈ N + T 2 (w), then (T 2 \ {w, w }; a n-2 a n-3 ...a δ+2 ) is a Grünbaum's exception. If (T 1 ; a 1 ...a δ ) = (4B; (1, 2)), take C = u 0 u 1 ...u n-1 u 0 = a δ+1 a δ ...a δ+2 a δ+1 , x δ ...x 1 ≡ u 1 ...u δ with d + T 2 (x δ ) ≥ 1, the result follows. Else, a 1 ...a δ is directed, let P ≡ a δ+4 a δ+5 ...a 0 a 1 be a path in T 2 -w, we have vx δ ...x 2 wx 1 P v ≡ a 2 a 3 ...a 2 = C. So we may assume that N - T 2 (x 2 ) = {w} with d + T 2 (w) = 0, clearly d + T 2 (x δ ) ≥ 2, then by theorem 2.1 x δ has an outneighbor origin of a path P ≡ a δ+1 a δ+2 ...a n-1 in T 2 and so vx 1 ...x δ P v ≡ C. Case 3: If a 0 → {a 1 , a n-1 }. Denote by B the block of C containing the arc a 0 → a n-1 .
We distinguish seven cases:

(a) If |B | = 1 and a 1 → a 2 . Suppose first that (T 1 ; a 1 ...a δ ) is a Grünbaum's exception, if a δ+4 → a δ+3 , T contains C = u 0 u 1 ...u n-1 u 0 = a 2 a 3 ...a 2 by Case 3(d), so a δ+3 → a δ+4 . Let a ∈ T 2 such that d + T 1 (a) < |T 1 | and d - T 2 (a) = max {d - T 2 (x); x ∈ T 2 , d + T 1 (x) < |T 1 |}, set T 1 = T 1 ∪ {a} and T 2 = T 2 -a. Since T 1 = 4B, there exists x 0 ...x δ ≡ a 0 ...a δ such that a = x δ , if x 0 has an outneighbor origin of a path P ≡ a n-1 a n-2 ...a δ+2 in T 2 then x 0 ...x δ vP -1 x 0 ≡ C. Otherwise, if d + T 2 (x 0 ) ≥ 2 then (T 2 ; a n-1 a n-2 ...a δ+2 ) is
an exception, by the building lemmas we may assume that it is one of the following biexceptions with respect to x 0 : Dual Exc (0, 1): Since d - T 2 (x 0 ) ≥ 2 then a = x 2 and thus [START_REF] Petrović | Antidirected Hamiltonian circuits in tournaments, Graph Theory[END_REF]1): If x δ → 3 then x 0 has an outneighbor origin of an antidirected inpath P in T [START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF][START_REF] Sahili | About the Number of Oriented Hammiltonian Paths and Cycles in Tournaments[END_REF][START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF] and thus x 0 ...

x 3 → 3, therefore x 0 ...x 3 32v1x 0 ≡ C. Dual Exc (3, 1): If x δ → 2 then x 0 ...x δ 24v31x 0 ≡ C so 2 → x δ and then if x 1 = a x δ ...x 1 ≡ x 1 ...a δ , so T contains C, thus x 1 = a. One can check that x 0 → x δ and so x 0 x δ x...x 1 ≡ a 0 a 1 ...a δ , but x 1 → 2, so T contains C. Dual Exc (4, 1): If x 1 = a then N + T 2 (x 1 ) ⊆ N + T 2 (x δ ), thus x 0 ...x δ 54v132x 0 ≡ C. Oth- erwise, x 0 x δ ...x 1 54v132x 0 ≡ C. Dual Exc (8, 1): If x 1 = a then N + T 2 (x 1 ) ⊆ N + T 2 (x δ ), thus x 0 ...x δ 21v345x 0 ≡ C. Oth- erwise, x 0 x δ ...x 1 21v345x 0 ≡ C. Dual Exc
x δ 31vP -1 x 0 ≡ C, so 3 → x δ and x 1 = a, therefore x 0 x δ ...x 1 31vP -1 x 0 ≡ C. Dual Exc (15, 1): If x 1 = a then N + T 2 (x 1 ) ⊆ N + T 2 (x δ ), thus x 0 ...x δ 54v3261x 0 ≡ C. Otherwise, x 0 x δ ...x 1 54v3261x 0 ≡ C. Dual Exc (16, 1): If x 1 = a then N + T 2 (x 1 ) ⊆ N + T 2 (x δ ), thus if x 0 → 5, x 0 ...x δ 21v6435x 0 ≡ C and if x 0 → 4 x 0 ...x δ 21v6534x 0 ≡ C. Otherwise, if x 0 → 5, x 0 x δ ...x 1 21v6435x 0 ≡ C and if x 0 → 4 x 0 x δ ...x 1 21v6534x 0 ≡ C Dual Exc (32, 1): If x 1 = a then N + T 2 (x 1 ) ⊆ N + T 2 (x δ ), thus x 0 ...x δ 21v6435x 0 ≡ C. Otherwise, x 0 x δ ...x 1 21v6435x 0 ≡ C. Dual Exc (33, 1): If x 1 = a then N + T 2 (x 1 ) ⊆ N + T 2 (x δ ), thus x 0 ...x δ 43v67521x 0 ≡ C. Otherwise, x 0 x δ ...x 1 43v67521x 0 ≡ C. Dual Exc (E 8 (5), 1): If x 1 = a then N + T 2 (x 1 ) ⊆ N + T 2 (x δ ), thus x 0 ...x δ 21v3wux 0 ≡ C. Otherwise, x 0 x δ ...x 1 21v3wux 0 ≡ C. Dual Exc (E 8 (6), 1): If x δ → 1 then x 0 ...x δ 13v2O X x 0 ≡ C so 1 → x δ and then if x 1 = a x δ ...x 1 ≡ x 1 ...a δ , so T contains C, thus x 1 = a. One can check that x 0 → x δ and so x 0 x δ ...x 1 ≡ a 0 a 1 ...a δ , but x 1 → 1, so T contains C. Otherwise, d + T 2 (x 0 ) ≤ 1, if there exists w ∈ N - T 2 (x 1 ) ∩ N - T 2 (x 0 ) then if x 0 has an inneighbor origin of a path P ≡ a n-2 a n-3 ...a 5 in T 2 -w then wx 1 x 2 x 3 vP -1 x 0 w ≡ C and else (T -w; a n-3 a n-4 ...a 5 ) is one of the following exceptions: Exc 0: {1, 2} → x 0 , if w → 3 then wx 1 x 2 x 3 v21x 0 3w ≡ C, so we can suppose that 3 → w. If x 3 → 2 then wx 1 x 2 x 3 23v1x 0 w ≡ C and if x 3 → 3 then wx 1 x 2 x 3 31v2x 0 w ≡ C, so {2, 3} → x 3 and thus x 1 = a. If 1 → x 2 then vwax 0 x 3 x 2 123v ≡ C, so x 2 → {1, 2, 3} and thus v12x 2 x 0 x 3 3wav ≡ C. Exc 1: N + T 2 -w (x 0 ) = {4}, if w → 2 then wx 1 ...x δ v43x 0 12w ≡ C and if w → 4 then wx 1 ...x δ v21x 0 34w ≡ C, so we can suppose that {2, 4} → w since otherwise one can find C. If d + T 2 -w (x delta ) ≥ 1, set V (T 2 -w) = {x, y, x , y } such that x → {x 0 , y}, x → y and x δ → x , then wx 1 ...x δ x y vyxx 0 w ≡ C. Otherwise, if 3 → x 1 then 3x 1 ...x δ v14w2x 0 3 ≡ C so x 1 → 3 and x 1 = a and thus 1x δ ...x 1 34vw2x 0 1 ≡ C. Exc 7: N + T 2 -w (x 0 ) = {5}
, in a similar way we can suppose that {1, 2, 3} → w. If

d + T 2 -w (x δ ) ≥ 1, set V (T 2 -w) = {x, y, z,
x , y } such that xyz and x y are two directed outpaths, x → x 0 and x δ → x , then wx 1 ...x δ x y vzyxx 0 w ≡ C. Otherwise, if there exists w ∈ N -(x 1 ) ∩ {1, 2, 3}, say w = 3, then w x 1 ...x δ vw1254x 0 w ≡ C, so x 1 → {1, 2, 3} and x 1 = a and thus x 0 x δ ...x 1 1w243v5x 0 ≡ C. Exc 4 and 33: In a similar way we can suppose that (T 2 -w) → w.

If N + (x δ ) ∩ (T 2 - w) = ∅ then there exist x ∈ N + T 2 -w (x δ ), y ∈ N + T 2 -w (x
) and P ≡ a n-3 a n-4 ...a δ+4 in T 2 \ {w, x, y} of origin an inneighbor of x 0 , thus wx 1 ...x δ xyvP -1 x 0 w ≡ C. So we may assume that (T 2 -w) → x δ and so

x 1 = a. Let x ∈ N + T 2 (x 1 ), w ∈ N - T 2 (x 0 ) ∩ N - T 2 (x δ ) and P ≡ a n-3 a n-4 ...a δ+4 be a path in T 2 \ {w, w , x } of origin an inneighbor of x 0 , we have w x δ ...x 1 x wvP -1 x 0 w ≡ C. Otherwise, d - T 2 (x 1 ) ≤ 1, then x 1 = a, in fact, suppose that x 1 = a, if |T 1 | = 3 then x 1 ∈ T 1 with d - T 2 (x 1 ) ≤ 1, a contradiction, and if |T 1 | = 5 then x 1 ∈ T 1 with d - T 2 (x 1 ) ≤ 2, a contradiction. Suppose first that T 1 = 3A, if x 3 → x 1 then y 0 ...y 4 = x 2 x 0 x 1 x 3 ≡ a 0 ...a 4 with d - T 2 (y 1 ) ≥ 2 then T contains C, so x 1 → x 3 and thus d - T 2 (x 1 ) = 1, set N - T 2 (x 1 ) = {u}. We also have N + T 2 (x 0 ) = {u}. If a n-2 a n-3 a n-4 is directed then by T contains C = u 0 u 1 ...u 0 = a n-2 a n-1 ...a n-2 by Case 1 or 2, so a n-2 a n-3 a n-4 is not directed. If a n-2 a n-3 a n-4 = (1, 1)
, since the number of blocks of a 4 a 5 ...a n-3 is odd then T 2 -u contains a path P ≡ a 4 a 5 ...a n-3 and so

x 0 x 3 x 2 x 1 P vux 0 ≡ C, and if a n-2 a n-3 a n-4 = -(1, 1), if C = (2, 1, 2, 1, 1, 1, 2, 1), by taking C = u 0 u 1 ...u 0 ≡ a 5 a 6 ...a 5 it is solved in Case 4(c)(ii), otherwise, let u ∈ N - T 2 (x 3
) and P ≡ a 3 a 4 ...a n-4 be a path in T 2 -u of origin in N -(x 0 ), we have x 1 x 2 x 0 P vu x 3 x 1 ≡ C. Assume now that T 1 = 5A, if a n-3 → a n-2 then T contains C = u 0 u 1 ...u 0 ≡ a 4 a 5 ...a 4 by Case 4(c)(i), so we may suppose that a n-2 → a n-3 . Let y 0 ...y 4 ≡ a 0 ...a 4 be a path in T 1 with end x 0 and let w ∈ N + T 2 (y 0 ), since d - T 2 (x 0 ) ≥ |T 2 -w|-2 then by theorem 2.1 x 0 has an inneighbor origin of a path P ≡ a 5 a 6 ...a n-3 in T 2 -w and so y 0 ...y 4 P vwy 0 ≡ C, unless if (T 2 -w; a 5 ...a n-3 ) = (5C; [START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF]1,1)), this case is solved in Case 4(c)(i) by taking C = u 0 u 1 ...u n-1 u 0 ≡ a 8 a 9 ...a 8 .

Assume now that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception and let x 1 ...x δ ≡ a 1 ...a δ be a path in

T 1 such that d - T 2 (x 1 ) = max {d - T 2 (x); x ∈ Or(a 1 ...a δ , T 1 )}, if x 1 has an inneighbor u origin of a path P ≡ a 0 a n-1 ...a δ+2 then ux 1 ...x δ vP -1 ≡ C. Otherwise, (T 2 ; a 0 a n-1 ...a δ+2 ) is one of the following biexceptions with respect to x 1 : Exc (1, 1): If d - T 2 (x 1 ) = 2 then, by Lemma 2.4, (T 1 ; a 1 ...a δ ) = (4B; (2, 1)), (4B; (1, 1, 1)), (3A; (2)) or |T 1 | = 2. If it is (4B; (2, 1)) then the block a 0 ...a 3 is greater than B, a contradiction. If it is (4B; (1, 1, 1)) or (3A; (2)) then any vertex of {x 1 , x 2 , x δ } may play the role of x 1 , so if N - T 2 (x 1 ) = {1, 3} then d -(x 2 ) < δ, a contradiction and if N - T 2 (x 1 ) = {1, 2} then N - T 2 (x 2 ) = {1, 3}, a contradiction and if N - T 2 (x 1 ) = N - T 2 (x 2 ) = {2, 3} then d -(x 1 ) = δ with T (N -(x 1 )) = 3A nor 4B, the result follows. So we may assume that δ = 2. If x 2 → 1 then x 1 x 2 123v4x 1 ≡ C, else, x 1 v4231x 2 x 1 ≡ C. So we may assume that N + T 2 (x 1 ) = {1, 2, 3} and {2, 4} ⊆ N + (x 2 ). Since d -(2) = 4 then 2 → x δ . By supposing that d + T 2 (x δ ) ≥ 1 one can find C so we may assume that T 2 → x δ and so a δ → a δ-1 and d + T 1 (x δ = δ -1. x δ is an origin of a path Q ≡ a 1 ...a δ so 4Qv3124 ≡ C. Exc (6, 1): Since d - T 2 (x 1 ) = 2 then δ = 2 and thus 5x 1 x 2 142v35 ≡ C. Exc (10, 1): d -(2) ≥ 6 a contradiction. Exc (12, 1): Since d - T 2 (x 1 ) = 2 then δ = 2 and thus 1x 1 x 2 354v21 ≡ C. Exc (22, 1): If x δ → 4 then 1x 1 ...x δ 46v3521 ≡ C, so we may assume that {4, 5, 6} → x δ . If a δ-1 → a δ then 2x 1 ...x δ-1 v5x δ 41632 ≡ C, so a δ → a δ-1 . If 1 → x δ-1 then 2x 1 ...x δ-1 15x δ 46v32 ≡ C, so x δ-1 → {1, 2, 3}. Then {4, 5, 6}∩N -(x δ-1 ) = ∅, without loss of generality we may suppose that 4 → x δ-1 then 3x 1 ...x δ-1 46x δ 52v13 ≡ C. Exc (24, 1): By supposing that d + T 2 (x δ ) ≥ 1 one can find C, otherwise, x δ is an origin of a path Q ≡ a 1 ...a δ in T 1 and so 6Qv524316 ≡ C. Exc (26, 1): T (N -(4)) = 6H so the result follows. (b) If |B | = 1 and a 2 → a 1 . Suppose first that (T 1 ; a 1 ...a δ ) is a Grünbaum's exception, if a δ+4 → a δ+3 (in this case |T 2 | ≥ 5), for x i ∈ T 1 let P i = y i 3 y i 4 ...y i δ+2 ≡ a 3 a 4 .
..a δ+2 be a path of origin x i in T 1 . If there exists x i ∈ T 1 having an outneighbor origin of a path P ≡ a 2 a 1 ...a δ+4 in T 2 then P i vP -1 x i ≡ a 3 a 4 ...a 4 = C. Otherwise, if δ = 5 or 7, then d - T 2 (x i ) ≥ 2 and b i (P i ) = 1 and so (T 2 ; a 2 a 1 ...a δ+4 ) is an exception. By the building lemmas we may suppose that it is a biexception with respect to x i ; the sole biexception verifying that a 3 a 2 ...

a δ+4 = +(1, 1, 1, 1, Q) and T 2 -S ⊆ N + (x i ) is Dual Exc (22, 1): d -(1) = 8,
a contradiction. So we may suppose that δ = 3, if there exists

x i ∈ T 1 such that d + T 2 (x i ) ≥ 2 then (T 2 ; a 2 a 1 ...a δ+4
) is an exception, by the building lemmas we may suppose that it is one of the following biexceptions with respect to an x i : Dual Exc (4, 1), [START_REF] Havet | Oriented Hamiltonian cycles in tournaments[END_REF]1), [START_REF] Petrović | Antidirected Hamiltonian circuits in tournaments, Graph Theory[END_REF]1), (24, 1), (33, 1): (T 2 ; a 6 a 7 ...a 3 ) is not a biexception with respect to y i 4 , thus y i 5 is an origin of a path Q ≡ a 5 a 6 ...a 3 in T 2 ∪ {y i 4 , y i 5 } with end in T 2 , thus vy i 3 Qv ≡ a 4 a 5 ...a 4 = C. Dual Exc (22,1), (E 8 (5), 1): (T 2 ; a 8 a 9 ...a 3 ) is not a biexception with respect to x i , thus

y i 4 is an origin of a path Q ≡ a 6 a 7 ...a 3 in T 2 ∪ {y i 4 , x i } with end in T 2 , thus vy i 4 Qv ≡ a 4 a 5 ...a 4 = C. Otherwise, d + T 2 (x i ) = 1 for all x i ∈ T 1 , if a 7 → a 8 , there exists u ∈ N - T 2 (y i 5 ) such that T 2 -u = 4A with y i 3 → 4, so (T 2 -u; a 2 a 1 ...a 8 ) = (4A; (1, 1, 1
)) with y i 3 → 4 and thus y i 3 has an inneighbor origin of a path P ≡ a 2 a 1 ...a 8 in T 2 , we have

y i 3 y i 4 y i 5 uvP -1 y i 3 ≡ a 3 a 4 ...a 3 = C. Else, a 8 → a 7 , let u ∈ N - T 2 (y i 3 ) and w ∈ N + T 2 (u), T 2 \ {u
, w} has at least 2 origins of a path a 6 a 7 ...a n-1 so y i 5 has an inneighbor origin of a path P ≡ a 6 a 7 ...a n-1 in T 2 \ {u, w}, we have vwuy i 3 y i 4 y i 5 P v ≡ C. So we may assume that a δ+3 → a δ+4 , if there exists x ∈ T 1 such that d + T 2 (x) = 1 then T 1 = 3A, set N + T 2 (x) = {u} and let y 3 y 4 y 5 ≡ a 3 a 4 a 5 be a path in T 1 of end x and w, t ∈ N -(y 3 ) such that w → t. T 2 \ {w, t} contains a path Q ≡ a 6 a 7 ...a n-1 of origin an inneighbor of x and then vtwy 3 y 4 y 5 Qv ≡ C unless three cases:

(i) If |T 2 | = 4, set V (T 2 ) = {u, w, t
Now d + T 2 (x) ≥ 2 for all x ∈ T 1 , let a ∈ T 2 such that d + T 2 (a) = ∆ + (T 2 ) and set T 1 = T 1 ∪ {a} and T 2 = T 2 -a. Let x n-1 x 0 ...x δ-1 ≡ a n-1 a 0 ...a δ-1 be a path in T 1 such that d - T 2 (x n-1 ) = max {d - T 2 (x); x ∈ Or(a n-1 a 0 ...a δ-1 )}. If x n-1 has an in- neighbor origin of a path P ≡ a n-2 a n-3 ...a δ+1 in T 2 then x 0 ...x δ-1 vP -1 x n-1 x 0 ≡ C. Otherwise, if d - T 2 (x n-1 ) ≥ 2 then (T 2 ; a n-2 a n-3 ...a δ+1
) is an exception. By the building lemmas we may assume that it is one of the following biexceptions with respect to Dual Exc (E 9 [START_REF] Havet | Oriented Hamiltonian paths in tournaments: a proof of Rosenfeld's conjecture[END_REF], 0), (E 10 (9), 0): Let Q ≡ a n-4 a n-3 ...a δ-5 be a path of origin x n-1 in T 1 , since d - T 2 (x n-1 ) ≥ 5 and (T 2 ; a δ-5 a δ-4 ...a n-6 ) is not an exception then x n-1 has an inneighbor origin of a path R ≡ a δ-5 a δ-4 ...a n-6 in T 2 and thus vQRv ≡ a n-6 a n-5 ...a n-6 = C. Dual Exc (E 9 (8), 0), (E 10 (10), 0): Let Q ≡ a n-5 a n-4 ...a δ-6 be a path of origin x n-1 in T 1 , since d + T 2 (x n-1 ) ≥ 3 and (T 2 ; a δ-6 a δ-5 ...a n-7 ) is not an exception then x n-1 has an outneighbor origin of a path R ≡ a δ-6 a δ-5 ...a n-7 in T 2 and thus vQRv ≡ a n-7 a n-6 ...a n-7 = C. Exc (0, 1): We have Exc (16,1): If δ = 3 then it is solved in Case 1 by taking C = a 0 a 1 ...a 0 = a 7 a 6 ...a 7 . If δ = 5 or 7, there exists x ∈ T 1 such that 6 → x (x can be x n-1 or x 0 ), if T 2 = 7C then, by Theorem 2.1, x has an inneighbor origin of an antidirected Hamiltonian inpath P in T 2 , also if T 2 = 7C, 6 is an origin of P , let Q ≡ a 4 a 5 ...a δ+3 be a path in T 1 of origin x, we have QvP -1 x ≡ a 4 a 5 ...a 4 = C. Exc (32, 1): Similar to Exc (16,1). Exc (33,1):

x n-1 : Dual Exc (1, 0): We have x δ-2 x δ-3 ...x n-1 3421vx δ-1 x δ-2 ≡ C. Dual Exc (7, 0): If δ = 3 it
d + T 2 (x n-1 ) = 2 then x n-1 → a and thus T 1 → a → T 2 , x 0 → 1 or 2, say 1, we have a3x n-1 x 2 x 0 1v2a ≡ C. Exc (4, 1): d - T 2 (x 0 ) = 4 then x 0 = a and so d - T 2 (x δ-1 ) = 4, by Theorem 2.1 x 0 or x δ-1 is an origin of a path y n-1 y 0 ...a δ-1 ≡ a n-1 a 0 ...a δ-1 in T 1 with d - T 2 (y n-1 ) = 4 then T contains C. Exc (8, 1): d - T 2 (x 0 ) ≥ 3 then x 0 = a and thus {2, 3, 4} → x δ-1 , if x δ-1 → 1 then x 0 ...x δ-1 134v25x n-1 x 0 ≡ C and if x δ-1 → 5 then x 0 ...x δ-1 541v32x n-1 x 0 ≡ C. Exc (9, 1): For all u ∈ N + T 2 (x δ-1 ), there exist u 1 , u 2 , w 1 , w 2 ∈ T 2 such that uu 1 u 2 and w 1 w 2 are two directed outpaths, w 1 ∈ N - T 2 (x n-1 ) and V (T 2 ) = {u, u 1 , u 2 , w 1 , w 2 }, thus x 0 ...x δ-1 uu 1 u 2 vw 2 w 1 x n-1 x 0 ≡ C.
d - T 2 (x 0 ) = 6 then x 0 = a and so d - T 2 (x δ-1 ) = 6, by Theorem 2.1 x 0 or x δ-1 is an origin of a path y n-1 y 0 ...a δ-1 ≡ a n-1 a 0 ...a δ-1 in T 1 with d - T 2 (y n-1 ) = 6 then T contains C. Exc (E 8 (5), 1), (E 8 (6), 1): d - T 2 (x 0 ) ≥ 3 then x 0 = a and thus {1, 2, 3} → x δ-1 . So d + X (x δ-1 ) ≥ 1 and then x δ-1 has an outneighbor origin of a path P ≡ a n-1 a n-2 ...a δ+2 in T 2 , therefore x δ-1 x δ-2 ...x n-1 vP -1 x δ-1 ≡ C. So we may assume now that d - T 2 (x n-1 ) = 1, and thus T 1 = 3A, if x 7 → x 8 it is solved
in Case 4(c) by taking C = a 0 a 1 ...a 0 = a 7 a 6 ...a 7 , so a 8 → a 7 . Let y 1 y 2 y 3 be a directed outpath in T 1 such that y 1 = x n-1 , and let u ∈ N + T 2 (y 3 ). Set P = a 2 a 1 a 0 ...a 7 , we have P = +(1, 1, 1, Q, 1) with an odd number of blocks (P can be +(1, 1, 1)) and

d + T 2 -u (y 1 ) ≥ |T 2 -u| -2
, so by Theorem 2.1 y 1 has an outneighbor origin of a path * P in T 2 -u and thus y 1 y 2 y 3 wv( * P ) -1 y 1 ≡ a 2 a 3 ...a 2 = C unless if (T 2 ; * P ) is an exception. By the building lemmas we may assume that (T 2 -u; * P ) is a biexception with respect to y 1 , and the sole possible biexceptions are: Dual Exc (4, 1)( 2 Assume now that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception and let x 1 ...x δ ≡ a 1 ...a δ be a path in

T 1 such that d - T 2 (x 1 ) = max {d - T 2 (x); x ∈ Or(a 1 ...a δ , T 1 )}, if x 1 has an inneighbor u origin of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 then ux 1 ...x δ vP -1 ≡ C. Oth- erwise, if d - T 2 (x 1
) ≥ 2 and d + T 2 (x 2 ) = 0 then, by lemma 2.8 (T 2 ; a 0 a n-1 ...a δ+2 ) is one of the exceptions 1, 18 or 22. If T 1 -x 2 contains a path y 2 ...y δ ≡ a 2 ...a δ then every vertex of T 2 is an origin of a path P ≡ a 1 a 0 ...a δ+2 in T 2 ∪ {x 2 } with end distinct from x 2 so y 2 ...y δ vP -1 y 2 ≡ a 2 a 3 ...a 2 = C; and if (T 1 -x 2 ; a 2 a 3 a 4 ) = (3A; (1, 1)), then without loss of generality we may suppose that Or(a 0 a n-1 ...a δ+2 ; T 2 ) ⊆ N + (x) for all x ∈ T 1 -x 2 , so if (T 2 ; a 0 a n-1 ...a δ+2 ) is the exception Exc 18 then 14365x 2 x 3 x 4 x 1 v21 ≡ C and if it is the exception Exc 22 then one can check that there exist w

∈ N + T 2 (x 4 ) and w ∈ N + T 2 (w) such that T 2 \ {w, w } contains a path P ≡ a 0 a n-1 ...a δ+4 of origin an inneighbor x of x 1 then xx 1 ...x 4 ww vP -1 ≡ C. If d - T 2 (x 1 ) ≥ 1 and d + T 2 (x 2 ) ≥ 1 then (T 2 ; a 0 a n-1 ...a δ+2
) is an exception. By the building lemmas we may assume that it is one of the following biexceptions with respect to x 1 :

Exc (1, 1)(1): If x δ → 3 then 1x 1 ...x δ 34v21 ≡ C and if x δ → 4 then 2x 1 ...x δ 41v32 ≡ C, so {3, 4} → x δ . If a δ-1 → a δ then 1x 1 ...x δ-1 v3x δ 421 ≡ C so a δ → a δ-1 . If δ = 2 then v43x 2 x 1 12v ≡ C. If δ = 3,
we can assume that T 1 is transitive since otherwise x δ is an origin of a directed inpath which has 4 as inneighbor, 2 ← x 2 or x 3 since otherwise d + (2) = 5, without loss of generality suppose that x 2 → 2, we have x 1 x 3 v3241x 2 x 1 ≡ C. If δ = 4 then a 1 ...a 4 is antidirected and T 1 = 4A and thus

x 2 x 4 x 3 x 1 = -(1, 2). We have v14x 2 x 4 x 3 x 1 23v ≡ C. Exc (1, 1)(2): By supposing that N + (x δ ) ∩ {1, 3, 4} = ∅ one can find C, and since d -(2) = 4 then T 2 → x δ . If δ = 2 then v43x 2 x 1 12v ≡ C, else, N -(x 1 ) = 4A, so T contains C. Exc (1, 1)(3): By supposing that N + (x δ )∩{1, 4} = ∅ one can find C so {1, 4} → x δ . If a δ-1 → a δ then 2x 1 ...x δ-1 v4x δ 132 ≡ C so a δ → a δ-1 . If δ = 2 then v14x 2 x 1 23v ≡ C. If δ = 3 then T 1 is transitive, if 2 → x 2 then x 2 x 3 v143x 1 2x 2 ≡ C and if x 2 → 2 then x 1 x 3 v3142x 2 x 2 ≡ C. If δ = 4 then a 1 ...a 4 is transitive and T 1 = 4A. We have 4 → x 2 or x 3 , say x 2 , then v14x 2 x 4 x 3 x 1 23v ≡ C. Exc (1, 1)(4): By supposing that N + (x δ ) ∩ {1, 3, 4} = ∅ one can find C so {1, 3, 4} → x δ . Similarly we can assume that a δ → a δ-1 . If δ = 2 or 3 it is solved as Exc (1, 1)(1). If δ = 4, since d - T 1 (x 1 ) = 1 then T 1 = 4A
and so x 2 or x 4 is an origin of an antidirected inpath in T 1 which has 4 as inneighbor and the result follows. Exc [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF]1):

If x δ → 1 then 5x 1 ...x δ 142v35 ≡ C, so {1, 2, 3} → x δ . If a δ-1 → a δ , if x δ → 4 then 5x 1 ...x δ-1 v14x δ 235 ≡ C and if 4 → x δ then 5x 1 ...x δ-1 v1x δ 4235 ≡ C, so a delta → a δ-1 . If δ = 2 then 241x 2 x 1 5v32 ≡ C. If δ = 3 then 34vx 1 x 2 x 3 2513 ≡ C. If δ = 4 then a 1 ...a 4 is antidirected and T 1 = 4A, we have N + (x 4 ) ∩ {4, 5} = ∅ so let x ∈ N + T 2 (x 4 ), x 1 has an outneighbor in T 2 origin of a path Q = -(2, 1) in T 2 -x then vxx 4 ...x 1 Qv ≡ C. If δ = 5, if a 1 ...
+ (x δ ) ∩ {1, 2, 3} = ∅ one can find C so {1, 2, 3} → x δ . If a δ-1 → a δ then 5x 1 ...x δ-1 v1x δ 3425 ≡ C, so a δ → a δ-1 . If δ = 2 then 4213x 1 x 2 v54 ≡ C. If δ = 3 then 5243x 3 x 1 x 2 v15 ≡ C. If δ = 4 then x 2 x 1 35v421x 4 x 3 x 2 ≡ C. If δ = 5, if a 1 ...a 5 = -(2,
+ (x δ ) ∩ {3, 4, 5} = ∅ one can find C so {3, 4, 5} → x δ . If a δ-1 → a δ then 1x 1 ...x δ-1 v3x δ 5421 ≡ C, so a δ → a δ-1 . If δ = 2 then v54x 2 x 1 213v ≡ C. If δ = 3 then x 3 x 1 1v4532x 2 x 3 ≡ C. If δ = 4 then T 1 = 4A, we have 453x 2 x 4 x 3 x 1 1v24 ≡ C. If δ = 5, if a 1 ...a 5 = -(2,
x δ → 4 then 1x 1 ...x δ 46v3521 ≡ C, so {4, 5, 6} → x δ . If a δ-1 → a δ then 3x 1 ...x δ-1 v6x δ 52413 ≡ C, so a δ → a δ-1 . If 1 → x δ-1 then 2x 1 ...x δ-1 15x δ 46v32 ≡ C and if 4 → x δ-1 then 1x 1 ...x δ-1 46x δ 53v21 ≡ C. Exc (24, 1): By supposing that d + T 2 (x δ ) ≥ 2 one can find C so T 2 → x δ and thus a δ → a δ-1 . If 1 → x δ-1 then 3x 1 ...x δ-1 15x δ 62v43 ≡ C. If 4 → x δ-1 then 1x 1 ...x δ-1 46x δ 35v21 ≡ C. If 6 → x δ-1 then 2x 1 ...x δ-1 61x δ 54v32 ≡ C.
Exc (26, 1)( 1), ( 2): By supposing that N + (x δ ) ∩ {1, 2, 3, 6} = ∅ one can find C, so {1, 2, 3, 6} → x δ and then 5x 1 ...x δ-1 v6x δ 32145 ≡ C. Exc (26,1)(3): By supposing that N + (x δ ) ∩ {1, 2, 3, 5} = ∅ one can find C, so {1, 2, 3, 5} → x δ . If x δ → 4 then T (N -(4)) = 6H and so T contains C, thus 4 → x δ and 6x 1 ...x δ-1 v1x δ 45236 ≡ C. So we may suppose that d - T 2 (x 1 ) = 1, set {u} = N - T 2 (x 1 ),so by Lemma 2.3 three cases may arise:

(i) If T 1 = 4B and a 1 ...a 4 = -(1, 2). We can suppose that u is minimal in T 2
and T 2 -u = 3A, 5A or 7A, since otherwise by taking x 1 in the place of v, v is minimal in T (N + (x 1 )). If T 2 = 4B then x 2 has an inneighbor origin of a path P ≡ a n-2 a n-3 ...a δ in T 2 and thus

x 4 x 1 x 3 vP -1 x 2 x 4 ≡ C. Otherwise, set V (T 2 ) = {u, u 1 , u 2 , u 3 }, we have x 2 x 1 uu 1 x 4 vu 3 u 2 x 3 x 2 ≡ C.
(ii) If T 1 is a transitive triangle and a 1 a 2 a 3 is directed, if a δ+4 → a δ+3 it is solved in case 1 by taking C = a 0 a 1 ...a 0 = a δ+3 a δ+2 ...a δ+3 , so we may assume that a δ+3 → a δ+4 . Similarly we may assume that a n-2 → a n-3 since otherwise T contains C = a n-1 a 0 ...a n-1 . Suppose that there exists w

∈ N + T 2 -u (x 2 ), if x 1 has an outneighbor origin of a path P ≡ a n-1 a n-2 ...a δ+2 in T 2 -w then x 1 wx 2 x 3 vP -1 x 1 ≡ C, otherwise (T 2 -w; a n-1 ...a δ+2
) is a Grünbaum's exception and then w is an origin of a path P ≡ a n-2 a n-3 ...a δ in T 2 and thus x 1 x 3 vP -1 x 2 x 1 ≡ C. So we may assume that N + T 2 (x 2 ) = {u}, and thus x 3 → u (to obtain T (N -(x 1 )) a transitive triangle). We have |T 2 | ≥ 5 since otherwise T (N -(x 3 )) = 3A, then a n-1 a n-2 ...a 5 has at least four blocks, let s = max {i;

d + C (a i ) = 0}, let w ∈ N + T 2 (u), Q ≡ α n-1 a n-2 .
..a s+1 and Q ≡ a s-1 a s-2 ...a 5 be two paths such that

V (Q) ∪ V (Q ) = V (T 2 ) \ {u, w}, we have x 1 wux 3 vQ -1 x 2 Q -1 x 1 ≡ C. (iii) If δ = 2, if |T 2 | = 2 then (T ; C) = A 2
which is an exception, so we may assume that |T 2 | ≥ 4. If there exists w ∈ N + T 2 -u (x 2 ), since (T 2 -w; a 4 a 5 ...a n-1 ) = (4B; -(1, 2)) (in fact, otherwise, it is solved in Case 1 by taking C = a 0 a 1 ...a 0 = a 3 a 4 ...a 3 ), x 1 has an outneighbor origin of a path P ≡ a 4 a 5 ...a n-1 in T 2 -w and thus vwx 2 x 1 P v ≡ C. So we can suppose that (T 2 -u) → x 2 , there exist w 1 , w 2 ∈ T 2 -u and P ≡ a 5 a 6 ...a n-1 a path in T 2 \ {w 1 , w 2 } of origin an inneighbor of x 2 and thus vw

1 x 1 w 2 x 2 P v ≡ C unless if T 2 = 4B, in this case set V (T 2 ) = {u, u 1 , u 2 , u 3 }, we have vu 2 u 1 x 2 x 1 uu 3 v ≡ C. (c) If |B | = 2 and a 1 → a 2 . Without loss of generality, we can suppose that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception. In fact, if it is, set C = a 0 a 1 ...a n-1 a 0 = a δ a δ-1 ...a δ+1 a δ . So let x 1 ...x δ ≡ a 1 ...a δ in T 1 such that d - T 2 (x 1 ) = max {d - T 2 (x); x ∈ Or(a 1 ...a δ , T 1 )}. We can suppose that if s + T 2 (N - T 2 (x 1 )) = 2 then T (N + T 2 (x 1 ) = 3A, 5A, 7A since oth- erwise, set N - T 2 (x 1 )
= {u → w}, in T , taking w in the place of v, T (N + (w)) has v as a maximal vertex and u as a minimal vertex. So let Q ≡ a n-2 a n-3 ...a δ+2 be a path in T (N + T 2 (x 1 )), we have uwx 2 ...x δ Q -1 x 1 u ≡ C. Otherwise, if x 1 has an inneighbor u origin of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 then ux 1 ...x δ vP -1 ≡ C. Else, (T 2 ; a 0 a n-1 ...a δ+2 ) is an exception, so by the building lemmas we may assume that it is one of the following biexceptions with respect to x 1 : Exc (5, 1): T (N + (x 2 )) = 5B, 5C. Exc [START_REF] Havet | Oriented Hamiltonian paths in tournaments: a proof of Rosenfeld's conjecture[END_REF]1)

: We have {1, 2} ⊆ N -(x 1 ), if d + T 2 (x δ ) ≥ 1 one can find Q ≡ a δ+1 a δ+2 .
..a 0 in T 2 ∪ {v} of origin an outneighbor of x δ and end an inneighbor u of x 1 , thus ux 1 ...x δ Q ≡ C. So we may assume that T 2 → x δ and a δ → a δ-1 . If a 1 ...a δ is antidirected then 5x δ ...x 1 v21435 ≡ C. So δ = 4 and a 1 ...a 4 = (1, 2), we have

x 4 v32145x 1 x 2 x 3 x 4 ≡ C. (d) If |B | = 2 and a 2 → a 1 . Suppose first that (T 1 ; a 1 ...a δ ) is a Grünbaum's exception. If d + T 2 (x) ≥ 2 for all x ∈ T 1 , let a ∈ T 2 such that d + T 2 (a) = ∆ + (T 2 ) . Set T 1 = T 1 ∪ {a} and T 2 = T 2 -a and let x 0 ...x δ ≡ a 0 ...a δ be a path in T 1 such that d + T 2 (x δ ) = max {d + T 2 (x);
x ∈ Or(a δ ...a 0 , T 1 )}, if x δ has an outneighbor in T 2 origin of a path P ≡ a δ+1 a δ+2 ...a n-2 then x 0 ...x δ P vx 0 ≡ C. Otherwise, by Lemma 2.4, d + T 2 (x δ ) ≥ 2 then (T 2 ; a δ+1 ...a n-2 ) is an exception. By the building lemmas we may suppose that it is one of the following biexceptions with respect to x δ : Exc (1, 0): If δ = 5, there exists y 0 y n-1 ...y δ+2 ≡ a 0 a n-1 ...a δ+2 such that y 0 has an inneighbor origin of a path Q ≡ a 1 a 2 ...a δ in T 2 , we have y 0 Qvy δ+2 ... Exc (27,0): Taking the block a δ+3 a δ+4 a δ+5 a δ+6 of maximal length, T contains C. Exc (48,0), (E 8 , 0), (E 9 , 0) and (E 10 , 0): There is an outneighbor of x δ of outdegree greater than ∆ + (T ) which is a contradiction. Suppose now that there exists x ∈ T 1 such that d + T 2 (x) = 1, let u 1 a such vertex and u 1 u 2 u 3 be a directed outpath of T 1 . There exist u ∈ N + T 2 (u 1 ) and an inneighbor of u 3 origin of a path Q ≡ a 6 a 7 ...a n-1 in T 2 -u and thus vuu 1 u 2 u 3 Qv ≡ C, unless if T is the Paley tournament which contains C or (T 2 -u; a 6 a 7 ...a n-1 ) = (4B; (1,[START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF]), set T 2 -u = y 1 → {y 2 , y 3 , y 4 }, by supposing that u i → y j j = 2, 3 or 4 one can find C so we can assume that {y 2 , y 3 , y 4 } → T 1 and u → {y 2 , y 3 , y 4 }; we have vy 3 y 2 u 1 u 2 u 3 uy 4 y 1 v ≡ C. Assume now that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception and let x 1 ...x δ ≡ a 1 ...a δ be a path in

y 0 ≡ C, thus T contains C. Assume now that δ = 3, let u 1 u 2 u 3 be a directed outpath of T 1 such that u 1 → a. If u 3 ← 1 or 2, say 1, then u 1 av4231u 3 u 2 u 1 ≡ C. So assume that {1, 2} → u 3 , if d - T 2 (a) ≥ 1, say a → 1, then 1au 1 u 2 u 3 2v431 ≡ C. So we may suppose that d + T 2 (a) = 4, if u 1 has an outneighbor in T 2 -a, say 1, we have v1u 1 u 2 u 3 2a43v ≡ C. So T 2 -a → u 1 , we have v21u 1 u 2 u 3 34av ≡ C.
T 1 such that d - T 2 (x 1 ) = max {d - T 2 (x); x ∈ Or(a 1 ...a δ , T 1 )}. If x 1 has an inneighbor u origin of a path P ≡ a 0 a n-1 ...a δ in T 2 then ux 1 ...x δ vP -1 ≡ C. Otherwise, suppose that d - T 2 (x 1 ) ≥ 2, if s + T 2 (N - T 2 (x 1 )) = 2 set N - T 2 (x 1 )
= {u → w}, taking w in T in the place of v, then T (N + (w)) has u as a minimal vertex and v as a maximal vertex. So we may suppose that T 2 has a minimal vertex x. If

x 2 → x, if |T 2 | ≥ 4 let Q ≡ a n-1 a n-2 ...a δ+2 in T 2 -
x of origin distinct from u and w, we have x 1 xx 2 ...x δ+2 vQ -1 x 1 ≡ C and if |T 2 | = 3, we may assume that |T | = 6, since, otherwise, an another origin of a path y 1 y 2 y 3 ≡ a 1 a 2 a 3 in T 1 is an outneighbor of x which is an origin of a path R ≡ a 0 a 6 a 5 in T 2 , and so xy 1 y 2 y 3 vR -1 ≡ C; thus (T ; C) = A 14 . So x → x 2 , taking x in the place of w, the maximal component has at least 3 vertices, u, w and x 2 . So we can assume that s + T 2 (N - T 2 (x 1 )) ≥ 3 and so if d + T 2 (x 2 ) = 0 then, by lemma 2.8 (T 2 ; a 0 a n-1 ...a δ+2 ) is the exception 7, if T 1 -x 2 contains a path y 2 ...y δ ≡ a 2 ...a δ then every vertex of T 2 is an origin of a path P ≡ a 1 a 0 ...a δ+2 in T 2 ∪ {x 2 } and so y 2 ...y δ vP -1 y 2 ≡ a 2 a 3 ...a 2 = C, otherwise, without loss of generality we may assume that x 4 → 5 and 1 → x 1 , thus 31x 1 ...x δ 52v23 ≡ C. And if d + T 2 (x 2 ) ≥ 1 then (T 2 ; a 0 a n-1 ...a δ+2 ) is an exception. By the building lemmas we may suppose that it is one of the following biexceptions with respect to x 1 : Exc (5, 0), [START_REF] Havet | Oriented Hamiltonian paths in tournaments: a proof of Rosenfeld's conjecture[END_REF]0)

: If d + T 2 (x δ ) ≥ 1 one can find Q ≡ a δ+1 a δ+2 .
..a 0 in T 2 ∪ {v} of origin an outneighbor of x δ and end an inneighbor u of x 1 , thus ux 1 ...

x δ Q ≡ C. So we may assume that T 2 → x δ and |T 1 | = 2 since otherwise T (N + (1)) = T ({x 1 , x 2 , x δ , 2, 4}) = 5B. Thus v25x 2 x 1 143v ≡ C. So we may assume that d - T 2 (x 1 ) = 1, set N - T 2 (x 1 )
= {u}, and, by Lemma 2.3, either (T 1 ; a 1 ...a δ ) = (4B; -(1, 2)), in this case we can suppose that (T 1 -x 1 ) → u since otherwise T (N -(x 1 )) = 4B, then d + T 2 (u) ≥ 3 and so (T 2 ; a n-1 a n-2 ...a δ+2 ) is one of the exceptions 18,24,26,34,38,47, and in these ones, u has an outneighbor origin of a path Q ≡ a n-2 a n-3 ...a δ+1 , thus

x 2 x 1 x 3 x 4 vQ -1 x 2 ≡ C; or |T 1 | = 2, if x 2 → u then, taking x 1 in the place of v we can suppose that d - T 2 (u) = 0 which is a contradiction, so assume that u → x 2 . Also taking x 1 in the place of v we can suppose that d - T 2 (u) = 1, thus either (T 2 -u; a n-1 a n-2 ...a δ+2 ) = (4A; (1, 1, 1)
+ T 2 (N - T 2 (T 1 )) ≤ 3 so s + T 2 (N + T 2 (T 1 )) = |T 2 |
, therefore, for y 2 y 3 ...y δ+1 ≡ a 2 a 3 ...a δ+1 , there is an outneighbor u ∈ T 2 of y 2 origin of a path P ≡ a 1 a 0 ...a δ+3 and thus y 2 ...y δ+1 vP -1 y 2 ≡ a 2 a 3 ...a 2 = C. So we can suppose that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception. Let x 1 ...

x δ ≡ a 1 ...a δ in T 1 such that d + T 2 (x 1 ) = max {d + T 2 (x); x ∈ Or(a 1 ...a δ ; T 1 )}. If x 1 has an outneighbor u ∈ T 2 origin of a path P ≡ a 0 a n-1 ...a δ+2 then ux 1 ...x δ vP -1 ≡ C. Otherwise, if d + T 2 (x 1 ) ≥ 2 and d + T 2 (x 2 ) = 0 then, since a 0 a n-1 .
..a δ+2 has an even number of blocks with first block of length 1 and last block of length 2, by lemma 2.8, (T 2 ; a 0 a n-1 ...a δ+2 ) is one of the exceptions 19 or

E 1 (4). If a δ-1 → a δ , let y 0 ...y δ-1 ≡ a 0 ...a δ-1 be a path in T 1 such that d + T 2 (y 0 ) ≥ 2, T 2 has at least |T 2 | -1 origin of a path P ≡ a n-1 a n-2 .
..a δ+1 , so there is an outneighbor of y 0 origin of a path P in T 2 and thus y 0 ...y δ-1 vP -1 y 0 ≡ C. Otherwise, a δ → a δ-1 , then T 1 has at least 3 origins of a 1 ...a δ , and so Or(a 1 ...a δ ; T 1 ) ∪ {x 2 } ⊆ N + (x) for all x ∈ Or(a 0 a n-1 ...a δ+2 ; T 2 ), then d + (x) > ∆ + (T ) for all x ∈ Or(a 0 a n-1 ...a δ+2 ; T 2 ), a contradiction. Then d + T 2 (x 2 ) ≥ 1 and (T 2 ; a 0 a n-1 ...a δ+2 ) is an exception, by the building lemmas we may assume that it is one of the following biexceptions with respect to x 1 : Exc (2, 0): If x δ has an outneighbor in T 2 then it is an origin of a directed outpath, we have vx 1 ...

x δ O T 2 v ≡ C. Else, δ = 2, we have 1vx 2 x 1 3421 ≡ C. Exc (E 1 (|T 2 |), 0): T (N -(2)) = F 1 , F 3 , F 5 , so T contains C. Exc (E 3 (|T 2 |), 0): T (N -(1)) = F 1 , F 3 , F 5 , so T contains C. Exc (E 5 (|T 2 |), 0): T (N -(1)) = F 1 , F 3 , F 5 , so T contains C. Exc (E 11 (|T 2 |), 0): T (N + (3)) = F 11 , F 13 , so T contains C. Exc (E 13 (|T 2 |), 0): T (N -(1)) = F 13 , so T contains C. Assume now that d + T 2 (x 1 ) = 1, so by Lemma 2.3, (T 1 ; a 1 ...a δ ) = (4B; -(1, 1, 1)), (4B; -(2, 1)) or (3A; -(2)) or |T 1 | = 2. First, if T 1 = 4B, we can assume that N + T 2 (x 1 , x 2 , x 4 ) = {u} since otherwise there is a vertex x of minimal indegree with T (N -(x)) = 4B. We have d + T 2 (u) ≥ 3, so if a 1 ...a δ = -(1, 1, 1) then (T 2 -u, a n-1 a n-2 ... a δ+2 ) = (6D; -(1, 2, 2)), we have 62ux 1 x 4 x 2 x 3 v43516 ≡ C. And if a 1 ...a δ = -(2, 1) then (T 2 -u; a n-1 a n-2 ...a δ+ ) = (6H; -(1, 1, 3), (7B; -(2, 1, 3)) or (7C; -(2, 1, 3)). In this case d + T 2 -u (x 4 ) ≥ 4 so it has an outneighbor w ∈ T 2 -u origin of a path P ≡ a n-1 a n-2 ...a δ+1 in T 2 . We have x 4 x 1 x 2 x 3 vP -1 x 4 ≡ C. Now if T 1 = 3A, we can suppose that N + T 2 (x i ) are pairwaisly different, which is a contradiction. So as- sume that |T 1 | = 2, let u ∈ N + T 2 (x 2
) and P ≡ a n-1 a n-2 ...a 3 in T 2 -u of origin and inneighbor of x 1 , we have

x 2 uvP -1 x 1 x 2 ≡ C. (f) If |B| = |B | = 3. Do a i = a i-1 until a δ → a δ-1 . Suppose first that (T 1 ; a 1 ...a δ ) is a Grünbaum's exception. If d + T 2 (x) ≥ 2 for all x ∈ T 1 , let a ∈ T 2 such that d + T 2 (a) = ∆ + (T 2 ) and set T 1 = T 1 ∪ {a} and T 2 = T 2 -a. Let x 0 x 1 ...x δ ≡ a 0 a 1 ...a δ in T 1 such that d + T 2 (x 0 ) ≥ 2,
this is possible by Lemma 2.4. If T 2 contains a path P ≡ a n-1 a n-2 ...a δ+2 of origin an outneighbor of x 0 then x 0 ...x δ vP -1 x 0 ≡ C. Otherwise (T 2 ; a n-1 ...a δ+2 ) is an exception, so by the building lemmas we may assume that it is one of the following biexceptions with respect to x 0 : Exc (0, 0), (4, 0) and (33, 0): x δ has an outneighbor origin of a path Q ≡ a δ+1 a δ+2 ...a n-2 in T 2 so x 0 ...x δ Qvx 0 ≡ C. Exc (9, 0): a = x 1 so x δ ← {1, 3, 4}. In fact, if not, the path x 1 ...x δ ≡ x δ ...x 1 . If x δ → 2 then x 0 ...x δ 23541vx 0 ≡ C. So we may assume that N - T 2 = {1, 2, 3, 4}, thus x 0 ...x δ 513v24x 0 ≡ C. Exc (15, 0): a = x 1 so {2, 4, 5, 6} ⊆ N -(x δ ). Then x δ has an outneighbor in {1, 3} which is an origin of a path Q ≡ a δ+1 ...a n-2 , we have x 0 ...x δ Qvx 0 ≡ C. Exc (16, 0): x δ → 6 so x 0 ...x δ 651423vx 0 ≡ C. Suppose now that there exists x ∈ T 1 such that d + T 2 (x) = 1 so T 1 = 3A. Let u 3 be a such vertex and u 1 u 2 u 3 be a directed outpath of T 1 . There exist w, t ∈ T 2 such that w → {u 1 , t} and u 3 has an inneighbor origin of a path Q ≡ a 7 a 8 ...a 0 in T 2 \ {w, t} and thus vtwu 1 u 2 u 3 Qv ≡ a 1 a 2 ...a 2 = C unless two case:

(i) (T 2 \ {w, t}; a 7 a 8 ...a 0 ) = (4B; (1,[START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF]), where u 3 → {u} → {y 1 , y 2 , y 3 }, in this case we can suppose that {u, y 1 , y 2 , y 3 } → {w, t} since otherwise one can find C, so there exist w , t such that w → {t , u 3 } and Q ≡ a 7 a 8 ...a 0 in T 2 \ {w , t } with origin an inneighbor of u 2 , thus vt w u 3 u 1 u 2 Qv ≡ C.

(ii) T 2 \ {w, t} = {y 1 → y 2 } where u 3 → y 1 , if w or t → y 1 one can find C so we can suppose that d + T 2 (y 1 ) = 3. Then if there exists u i ← y 1 so T contains C.

Assume now that T 1 → y 1 , if T 2 -y 1 = 3A then there is an outneighbor of a u i origin of a directed outpath in

T 2 -y 1 thus u i+2 y 1 u i+1 u i O T 2 -y 1 vu i+2 ≡ C, else,
there is an outneighbor of a u i origin of an antidirected outpath R in T 2 -y 1 thus u i y 1 u i+2 u i+1 vR -1 u i ≡ C.

So we may suppose that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception. Let

x 1 ...x δ ≡ a 1 ...a δ in T 1 such that d + T 2 (x 1 ) = max {d + T 2 (x); x ∈ Or(a 1 ...a δ ; T 1 )}. If d + T 2 (x 1 ) ≥ 2 then s + T 2 (N + T 2 (x 1
)) ≥ 3 and so (T 2 ; a 0 a n-1 ...a δ+2 ) is an exception, by the building lemmas we may suppose that it is one of the following biexceptions with respect to

x 1 : Exc (7, 0), (E 8 (|T 2 |), 0), (E 9 (|T 2 |), 0) or (E 10 (|T 2 |), 0). In these exceptions, we have respectively T (N + (x 2 )) = 5C, N -(u) = F 8 , F 9 or F 10 , where u ∈ X; thus T contains C. Without loss of generality, we can suppose that d + T 2 (x 1 ) ≤ 1 and d + T 2 (x δ ) ≤ 1, so assume that N + T 2 (x 1 ) = {u}, if d + T 2 (u) ≥ 2 then (T 2 -u; a n-1 a n-2 ...a δ+2
) is one of the exceptions: 0, [START_REF] Sahili | About the Number of Oriented Hammiltonian Paths and Cycles in Tournaments[END_REF][START_REF] Havet | Oriented Hamiltonian cycles in tournaments[END_REF][START_REF] Petrović | Antidirected Hamiltonian circuits in tournaments, Graph Theory[END_REF]15,16,32,33,37, E 8 , E 9 or E 10 . It is easy to check that there is an inneighbor w of x 2 origin of a path P ≡ a 1 a 0 ...a δ+2 in T 2 ∪ {x 1 } and thus x 2 ...x δ vP -1 x 2 ≡ a 2 a 3 ...a 2 = C. Otherwise, d + T 2 (u) = 1, take u in the place of v in T and set T 1 = T (N -(u)) and T 2 = T (N + (u)), x 1 and v are origins of two paths P ≡ a 0 a n-1 ...a δ+2 and P ≡ a δ+1 a δ+2 ...a n-1 in T 2 . Let y 1 ...y δ ≡ a 1 ...a δ in T 1 , y 1 or y δ is an inneighbor of x or v, thus y 1 ...y δ uP y 1 ≡ a 1 a 2 ...a 1 = C or uy 1 ...y δ P u ≡ C. Therefore T contains C and so T contains C.

(g) If |B| = |B | ≥ 4. Do a i = a i-1 until a δ → a δ-1 . (T 1 ; a 1 ...a δ ) is not a Grünbaum's ex-
ception since its first block has the same direction of its last one. Let x 1 ...x δ ≡ a 1 ...a δ in T 1 such that d + T 2 (x 1 ) = max {d + T 2 (x); x ∈ Or(a 1 ...a δ ; T 1 )}. If there is an outneighbor u of x 1 origin of P ≡ a 0 a n-1 ...a δ+2 in T 2 , then ux 1 ...x δ vP -1 ≡ C. Otherwise, if d + T 2 (x 1 ) ≥ 2 then (T 2 ; a 0 a n-1 ...a δ+2 ) is an exception, and since a 1 a 0 ...a δ+2 = (3, ..., 2) then it is not a biexception with respect to x 1 , so by the building lemmas T contains C. Else, d + T 2 (x 1 ) = 1, then by Lemma 2.4, (T 1 ; a 1 ...a δ ) = (4B; -(1, 1, 1)) and N + T 2 (x) = {u} for all x ∈ Or(a 1 ...a δ ; T 1 ) and so d + T 2 (u) ≥ 3, since for x ∈ Or(a 1 ...a δ ; T 1 ) and d + T 2 (x) ≥ 2 we have s + T 2 (N + T 2 (x)) ≥ 3 and (T 2 ; a 0 a n-1 ...a δ+2 ) is not an exception. We have a 0 a n-1 ...a δ+3 = +(2, ..., 1) with an even number of blocks; so there is an outneighbor w of u in T 2 origin of a path P ≡ a 0 a n-1 ...a δ+3 and thus wux 1 x 4 x 2 x 3 vP -1 ≡ C. ..a δ+2 be a path in T 1 , since d - T 2 (x 3 ) ≥ 2 then x 3 has an inneighbor origin of a path P ≡ a 2 a 1 ...a δ+4 and thus x 3 ...x δ+2 vP -1 x 3 ≡ a 3 a 4 ...a 3 = C unless if (T 2 ; a 2 a 1 ...a δ+4 ) is an exception and by the building lemmas we may assume that it is a biexception with respect to x 3 , but since any vertex of T 1 is an origin of a path Q ≡ a 3 ...a δ+2 then if N - T 2 (x 3 ) ∪ S = V (T 2 ), one can find C, thus we are in one of the following biexceptions with respect to x 3 :

Exc (22, 1): We have |T 1 | = 3, set V (T 1 ) = {u 1 , u 2 , u 3 } we have T 1 → {4, 5, 6} and {1, 2, 3} → T 1 , thus u 1 1u 3 u 2 v6452u 1 ≡ C. Exc (24, 1): d -(4) > δ a contradiction. So we may assume that a δ+3 → a δ+4 . If there exists x ∈ T 1 such that d + T 2 (x) = 1 then T 1 = 3A = {x 1 , x 2 , x 3 }, without loss of generality, suppose that d + T 2 (x 1 ) = 1. Let u ∈ N + T 2 (x 2 ), since d - T 2 (x 1 ) ≥ |T 2 -u| -1 then x 1
has an inneighbor origin of a path P ≡ a δ+3 a δ+4 ...a 0 in T 2 -u and thus vux 2 x 3 x 1 P v ≡ a 1 a 2 ...a 2 = C unless four cases:

(i) If T 2 -u = {u 1 → u 2 } with x 1 → u 1 . If x 3 → u then u 2 vux 3 x 1 x 2 u 1 u 2 ≡ C, so u → x 3 . If u → u 2 then u 2 vu 1 x 1 x 2 x 3 uu 2 ≡ C, so u 2 → u and without loss of generality x 3 → u 2 . If u 1 → u then uvu 2 x 3 x 1 x 2 u 1 u ≡ C, so u → u 1 and thus T = 7A which contains no (2, 1, 1, 1, 1, 1), (T ; C) = A 3 . (ii) If T 2 -u = {u 1 , u 2 , u 3 } is transitive such that x 1 → u 1 → {u 2 , u 3 }.
We can suppose that u 1 → x 2 and thus u → x 3 , we have

x 1 ux 3 x 2 vu 3 u 1 u 2 x 1 ≡ C. (iii) If (T 2 -u; a 6 ...a 0 ) = (4A; (1, 1, 1)) and N + T 2 (x 1 ) = {4}. If 1 → u then 21ux 2 x 3 x 1 34 v2 ≡ C, if 3 → u then 43ux 2 x 3 x 1 12v4 ≡ C and if 4 → u then 14ux 2 x 3 x 1 23v1 ≡ C; so u → {1, 3, 4}. If 2 → x 3 then 1v4x 1 x 2 x 3 23u1 ≡ C and if x 3 → 2 then x 1 2x 3 x 2 v1u43x 1 ≡ C.
(iv) If (T 2 -u; a 6 ...a 0 ) = (5C; (2, 1, 1)) and N + T 2 (x 1 ) = {5}. We can suppose that 5 → x 2 since otherwise x 1 has an inneighbor origin of a path P = (2, 1, 1) in T 2 -5 and thus v5x 2 x 3 x 1 P v ≡ a 1 a 2 ...a 2 = C. If N -(u) ∩ {1, 2, 3} = ∅, say 1 → u, then 21ux 2 x 3 x 1 345v2 ≡ C, so u → {1, 2, 3}. If 4 → u then 54ux 2 x 3 x 1 123v5 ≡ C, so u → 4. If 5 → u then 15ux 2 x 3 x 1 234v1 ≡ C, so u → 5, and thus x 3 → u. Therefore 3vux 3 x 1 x 2 51423 ≡ C.

Now we can suppose that

d + T 2 (x) ≥ 2 for all x ∈ T 1 . Let a ∈ T 2 such that d + T 2 (a) = ∆ + (T 2
) and set T 1 = T 1 ∪ {a} and T 2 = T 2 -a. Let x 0 ...x δ ≡ a 0 ...a δ be a path in T 1 , if x 0 has an inneighbor origin of a path P ≡ a n-1 a n-2 ...a δ+2 in T 2 then x 0 ...x δ vP -1 x 0 ≡ C. Otherwise, if d - T 2 (x 0 ) ≥ 2 then (T 2 ; a n-1 a n-2 ...a δ+2 ) is an exception. By the building lemmas we may assume that it is one of the following biexceptions with respect to x 0 : Exc (1, 1)( 1), ( 2), ( 4): By definition of a, a cannot be x 1 , so {2, 4} → x δ , since otherwise by taking x δ ...x 1 ≡ a 1 ...a δ the result follows. If

x δ → 3 then x 0 ...x δ 34v21x 0 ≡ C, so 3 → x δ . Since d + T 2 (x δ ) ≥ 2 thenx δ → 1, if 3 → x 1 then x 0 ...x δ 12v43x 0 ≡ C, so x 1 → 3. If δ = 5
, T 1 has at least 4 origins of an antidirected Hamiltonian inpath and since so d + (a) ≥ 6, a contradiction, in fact, d - T 2 (x) ≥ 3 for all x ∈ T 1 so we can suppose that any origin of an antidirected Hamiltonian inpath in T 1 is an outneighbor of a. Thus δ = 3 and so 3vax 1 x 0 x 3 4213 ≡ C. Exc (1, 1)(3): As above wen can assume that δ = 3. If x δ → 4 then x 0 ...x δ 41v32x 0 ≡ C and if 4 → x δ then 3vax 1 x 0 x 3 4213 ≡ C. Exc [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF]1): By definition of a, a cannot be x 1 , so x δ → 5 and thus x 0 ...x δ 51v432x 0 ≡ C. Exc (7, 1): Similarly we can suppose that {1, 2, 3} → x δ . If x δ → 4 then x 0 ...x δ 45v321 x 0 ≡ C and if x δ → 5 then x 0 ...x δ 51v432x 0 ≡ C. Exc [START_REF] Rosenfeld | Antidirected Hamiltonian paths in tournaments[END_REF]1): If δ = 5, since T 1 contains at least 4 origin of an antidirected Hamiltonian inpath and one of them is an outneighbor of 2, so the result follows and thus δ = 3. Let y 1 y 2 y 3 be a directed outpath of T 1 of origin x 0 , one can find an outneighbor u ∈ T 2 of x 0 such that y 3 has an inneighbor in T 2 -u origin of a path Q = (2, 1, 1) and thus vuy 1 y 2 y 3 Qv ≡ a 1 a 2 ...a 1 = C. Exc (22, 1): If T 2 = 7A then there exists a ∈ T 2 such that there is a path x 0 ...x δ ≡ a 0 ...a δ in T 1 ∪ {a } with x 0 has an inneighbor origin of a path P ≡ a n-1 a n-2 ...a δ+2 in T 2 -a and the result follows, so d + T 2 (a) ≥ 4. There exists a ∈ T 2 such that T 2 -a = 6F, 6H and the result follows; in fact, the condition

d + T 2 (a ) ≥ 1 is sufficient instead of d + T 2 (a ) = ∆ + (T 2 ). Exc (24, 1): By definition of a, a cannot be x 1 so {1, 2, 3, 6} → x δ . So N + T 2 (x δ ) ⊆ {4, 5}, say x δ → 4, we have x 0 ...x δ 46v1253x 0 ≡ C. Else, set N - T 2 (x 0 ) = {u}, since d - T 2 (x) ≥ 2 for all x ∈ T 1 then a ≡ x 1 . Since u is not an origin of a path P ≡ a n-1 a n-2 ...a δ+2 then N + T 2 (a) ∩ N + T 2 (x 0 ) = ∅, let w ∈ N + T 2 (a) ∩ N + T 2 (x 0 ). If T 2 -w contains a path P ≡ a n-2 a n-3 .
..a δ+2 of origin distinct from u then wx 1 ...x δ vP -1 x 0 w ≡ C. Otherwise, two cases may arise:

(i) If (T 2 -w; a n-2 ...a δ+2 ) = (4A; (1, 1, 1)), there exists w ∈ N + T 2
(a) such that T 2 -w contains a path P and the result follows. In fact, otherwise, u is an origin of a path Q = (2, 1, 1) in T 2 and thus x 0 ...x δ vQ -1 x 0 ≡ C.

(ii) If (T 2 -w; a n-2 ...a δ+2 ) is a Grünbaum's exception, if there exists w ∈ T 2 -w such that T 2 -w = 3A, 5A, 7A, the result follows, so we can assume that

N + T 2 -u (a) ⊆ {x ∈ T 2 -u/T 2 -x = 3A, 5A, 7A}. By completing T 2 under the conditions |N + T 2 -u (a)| ≤ 2 and d + T 2 (a) = ∆ + (T 2 )
, u is an origin of a path P ≡ a n-1 a n-2 ...a δ+2 in T 2 which is a contradiction. Assume now that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception and let x 1 ...x δ be a path in

T 1 such that d + T 2 (x 1 ) = max {d + T 2 (x); x ∈ Or(a 1 ...a δ , T 1 )}. If x 1 has an outneighbor u origin of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 then ux 1 ...x δ vP -1 ≡ C. Otherwise, if d + T 2 (x 1 ) ≥ 2 then (T 2 ; a 0 a n-1 ...a δ+2
) is an exception. By the building lemmas we may assume that it is one of the following biexceptions with respect to x 1 : Dual Exc (0, 1):

T (N -(x 2 )) = T (x 1 , 1, 2) = 3A then T contains C. Dual Exc (2, 1): T (N -(1)) = 4A, F 1 (4) so T contains C.
Dual Exc (4, 1), (9, 1)( 2), ( 3), ( 4), (15, 1), (16, 1)( 2), (33, 1) and (E 5 , 1)(2): Since T (N + (x 2 )) has v as minimal vertex then (T (N + (x 2 )); a 0 a n-1 ...a δ+2 ) is not an exception, unless if it is E 8 [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF], this case will treated below. Dual Exc [START_REF] Havet | Oriented Hamiltonian cycles in tournaments[END_REF]1): If d + T 2 (x) = 4 for an x ∈ T 1 then T (N + (x)) = 5A, 5D, 5E, so we may assume that d - T 2 (x) ≥ 2 for all x ∈ T 1 . If x δ → 2 then 5x 1 ...x δ v21v345 ≡ C, if x δ → 3 then 5x 1 ...x δ 3124v5 ≡ C and if x δ → 4 then 5x 1 ...x δ 43v125 ≡ C, so {2, 3, 4} → x δ and thus x 2 = x δ i.e. δ ≥ 2. If δ = 3 then 35x 2 x 1 v12x 3 43 ≡ C. If δ = 4 then (T 2 ; a 0 a 1 ...a 4 ) is not an exception, so let y 1 ...y 4 ≡ a 6 ...a 9 , since d - T 2 (y 4 ) ≥ 2 then y 4 has an inneighbor u origin of a path P ≡ a 0 ...a 4 in T 2 and thus P vy 1 ...y 4 u ≡ C. So δ = 5 and x δ ...x 1 ≡ x 1 ...x δ since otherwise there is a block of length 3. Since d - T 2 (x 4 ) ≥ 2, one can find an inneighbor of x 4 origin of a path P ≡ a 1 a 0 ...a 7 in T 2 ∪ {x 5 } of end distinct from x 5 and thus x 4 x 3 ... [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF], 1): If N + (x δ )∩{1, 2, 3} = ∅, say 1 ∈ N + (x δ ), we have ux 1 ...x δ 12v3w u ≡ C, so {1, 2, 3} → x δ and thus x 2 = x δ i.e. δ ≥ 3. If T 1 = 3A, 5A or 7A, x 2 is an origin of a path y 1 ...y δ ≡ a 1 ...a δ and so 1y 1 ...y δ v3u2w1 ≡ C, so T 1 = 3A, 5A, 7A. If a δ-1 → a δ , let u 0 ...u δ-1 ≡ a 0 ...a δ-1 be a path in T 1 , any vertex of T 2 is an origin of a path P ≡ a n-1 ...a δ+1 so u 0 ...u δ-1 vP -1 u 0 ≡ C, so a δ → a δ-1 . If N -(x δ-1 ) ∩ {u, w} = ∅, say u ∈ N -(x δ-1 ), we have wx 1 ...x δ-1 u1x δ 23vw ≡ C so x δ-1 → {u, w} and thus x δ-1 = x 2 i.e. δ ≥ 4. If a 1 ...a δ is symmetric then (T 2 ; a 0 a n-1 ...a δ+2 ) is not a biexception with respect to x δ , so by taking x δ ...x 1 ≡ a 1 ...a δ one can find C. So we may assume that δ = 4, let u 1 ...u 4 ≡ a 6 ...a 9 , every vertex of T 2 is an origin of a path P ≡ a 0 a 1 ...a 4 , thus

C = a 0 a 1 ...a 0 = a s+δ+2 a s+δ+1 ... a s+δ+2 . Dual Exc (E 5 , 1)(1): If x δ → 1 then wx 1 ...x δ 12v3uw ≡ C, so we may assume that {1, 2, 3} → x δ . If a δ-1 → a δ then wx 1 ...x δ-1 v1x δ 23uw ≡ C, so a δ → a δ-1 . Then δ = 2 and thus uw3vx 2 x 1 12u ≡ C. Dual Exc (E 8
u 4 P vu 1 ...u 4 ≡ a 9 a 0 ...a 9 = C. Else, d + T 2 (x 1 ) ≤ 1, if there exist w ∈ N - T 2 (x 1 ) ∩ N - T 2 (x 2 ) and P ≡ a n-1 a n-2 ...a δ+2 in T 2 -w of origin an inneighbor of x 1 then x 1 wx 2 ...a δ vP -1 x 1 ≡ C. Otherwise, since (T 2 -w; a n-1 a n-2 ...a δ+2 ) is not a Grünbaum's exception, d + T 2 (x 1 ) = 1, set N + T 2 (x 1 ) = {u}.
Two cases may arise:

(i) If d - T 2 (x 2 ) ≥ 2 then either (T 2 -w; a n-1 a n-2 ...a δ+2 ) = (4A; (1, 1, 1)) or (5C; (2, 1, 1 )) or u is minimal in T 2 and a n-1 a n-2 ...a δ+2 is directed. If (T 2 -w; a n-1 ...a δ+2 ) = (4A; (1, 1, 1)), u ≡ 4, if 4 → x 2
then, without loss of generality suppose that 3 → w, 24x 2 ...x δ vw3x 1 12 ≡ C; so x 2 → 4. Now we suppose that for any w ∈ N - T 2 (x 2 ) T 2 -w = 4A with u ≡ 4. So if 3 → x 2 then u is an origin of a path P ≡ a 0 a n-1 ...a δ+2 , a contradiction. So 4 → w and w → 3, if there is a path P = (1, 1) in T (1, 2, w) of origin an inneighbor of (2,1,1)

x 2 then x 2 ...x δ v43x 1 P -1 x 2 ≡ a 2 a 3 ...a 2 = C, so T (1, 2, w) = 3A. If δ = 2 then x 2 x 1 v4213wx 2 ≡ C, if δ = 3 and T 1 = 3A then 3x 2 x 3 x 1 v241w3 ≡ C, if δ = 3 and T 1 is transitive, if a 1 a 2 a 3 is antidirected then 2v4x 1 x 3 x 2 w312 ≡ C and if a 1 a 2 a 3 is directed then x 2 x 1 x 3 v4213wx 2 ≡ C, finally if δ = 4,
), if N + (w) ∩ {1, 2, 3} = ∅, say w → 1, then 1wx 2 ...x δ v43x 1 251 ≡ C so {1, 2, 3} → w. If 5 → x 2 then 15x 2 ...a δ vw243x 1 1 ≡ C so x 2 → 5. If N -(x 2 ) ∩ {1, 2, 3} = ∅, say 1 → x 2 , then T 2 -1 = 5C
and the result follows, so x 2 → {1, 2, 3} and thus 4 → x 2 and w → 5.

If N + (x δ ) ∩ {1, 2, 3} = ∅, say x δ → 1, then 5x 1 ...x δ 142w3v5 ≡ C, so {1, 2, 3} → x δ and thus x 2 = x δ i.e. δ ≥ 3. If a δ-1 → a δ then x 1 wx 2 ...x δ-1 v1x δ 2354x 1 ≡ C, so a δ → a δ-1 . If x δ → 5 then x 1 wx 2 ...a δ 51v432x 1 ≡ C, so 5 → x δ . If (T 1 ; a 1 ...a δ ) = (4B; (1, 2)) then x 1 x 3 x 4 x 2 ≡ a 1 ...
2 -u = {u 1 → u 2 }, if |T 1 | = 2 then u 2 → x 2 and so x 2 → u 1 thus x 1 x 2 u 1 u 2 vux 1 ≡ C, and if |T 1 | = 3 then T 1 is transitive and T 2 → x 3 , we have x 3 u 2 x 2 x 1 vu 1 ux 3 ≡ C. (b) If |B | = 1 and a 2 → a 1 . Suppose first that (T 1 ; a 1 ...a δ ) is a Grünbaum's exception
and for x i ∈ T 1 let y i 0 ...x i δ-1 ≡ a 0 ...a δ-1 be a path of origin x i in T 1 . If an x i has an inneighbor origin of a path P ≡ a n-1 a n-2 ...a δ+1 in T 2 then y i 0 ...y i δ-1 vP -1 y i 0 ≡ C. Otherwise, (T 2 ; a n-1 a n-2 ...a δ+1 ) is an exception, so by the building lemmas it is one of the biexceptions (0, 1), [START_REF] Camion | Chemins et circuits hamiltoniens des graphes complets[END_REF]1), [START_REF] Sahili | About the Number of Oriented Hammiltonian Paths and Cycles in Tournaments[END_REF]1), [START_REF] Havet | Oriented Hamiltonian cycles in tournaments[END_REF]1), [START_REF] Petrović | Antidirected Hamiltonian circuits in tournaments, Graph Theory[END_REF]1), (15, 1), (16, 1), (32, 1), (33, 1), (E 8 [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF], 1) and (E 8 (6), 1) with respect to x i . For Exc(0, 1), T = 7A and thus 13465721 ≡ C, for Exc(4, 1) and Exc(33, 1), (T 2 ; a n-1 a n-2 ...a δ+1 ) is not a biexception with respect to y i 1 . And for the remaining biexceptions, there is an outneighbor of x i origin of a path P ≡ a δ+2 a δ+3 ...a 0 and therefore vy i δ-1 y i δ-2 ...y i 0 P v ≡ a 1 a 2 ...a 1 = C. Assume now that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception and let x 1 ...x δ ≡ a 1 ...a δ be a path in T 1 such that d + T 2 (x 1 ) = max {d + T 2 (x); x ∈ Or(a 1 ...a δ ; T 1 )}. If x 1 has an outneighbor u origin of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 then ux 1 ...x δ+2 vP -1 ≡ C.

Otherwise, if d +

T 2 (x 1 ) ≥ 2 and d + T 2 (x 2 ) = 0 then, by lemma 2.8, (T 2 ; a 0 a n-1 ...a δ+2 ) is one of the following exceptions: Dual Exc 0, 4, 19, 33: If T 1 -x 2 contains a path y 2 ...y δ ≡ a 2 ...a δ then every vertex of T 2 is an origin of a path P ≡ a 1 a 0 ...a δ+2 in T 2 ∪ {x 2 } of end distinct from x 2 and thus y 2 ...y δ vP -1 y 2 ≡ a 2 a 3 ...a 2 = C; and if (T 1 ; a 2 ...a δ ) = (3A; (1,1)), let a ∈ T 2 and y 1 ...y 4 ≡ a 1 ...a 4 be a path in (T 1 -x 2 ) ∪ {a}, a = y 4 , every vertex of T 2 -a is an origin of a path P ≡ a 0 a n-1 ...a δ+2 in (T 2 -a) ∪ {x 2 }, thus y 1 ... T 2 (x 1 ) ≥ 2 and d + T 2 (x 2 ) ≥ 1 then (T 2 ; a 0 a n-1 ...a δ+2 ) is an exception, so by the building lemmas we may assume that it is one of the following biexceptions with respect to x 1 : (When |T 2 | = 5 and a δ → a δ-1 we can suppose that δ = 2. In fact, if δ = 4 or 5 take C = a 0 a 1 ...a 0 = a 0 a n-1 ...a 0 and it is solved in an another case, and if δ = 3 then a 0 a 1 ...a 3 is a block greater than B.) [START_REF] Sahili | About the Number of Oriented Hammiltonian Paths and Cycles in Tournaments[END_REF]1): By the building lemmas, if N + (x δ ) ∩ {1, 2, 5} = ∅ then x δ has an outneighbor origin of a path P ≡ a δ+1 a δ+2 ...a 0 in T 2 ∪ {x 1 } and thus vx 2 ...x δ P v ≡ a 1 a 2 ...a 1 = C, so we may assume that {1, 2, 5} → x δ . If a δ-1 → a δ then 1x 1 ...x δ-1 v435 x δ 21 ≡ C, so a δ → a δ-1 . Then δ = 2 and thus 421vx 2 x 1 534 ≡ C. Dual Exc [START_REF] Havet | Oriented Hamiltonian cycles in tournaments[END_REF]1): By the building lemmas, if N + (x δ ) ∩ {2, 3, 4} = ∅ then x δ has an outneighbor origin of a path P ≡ a δ+1 a δ+2 ...a 0 in T 2 ∪ {x 1 } and thus vx 2 ...x δ P v ≡ a 1 a 2 ...a 1 = C, so we may assume that {2, 3, 4} → x δ . If a δ-1 → a δ then 5x 1 ...x δ-1 v213 x δ 45 ≡ C, so a δ → a δ-1 . Then δ = 2 and thus 253vx 2 x 1 142 ≡ C. Dual Exc (9, 1)( 1), ( 3), ( 4): By the building lemmas, if N + (x δ ) ∩ {3, 5} then x δ has an outneighbor origin of a path P ≡ a δ+1 a δ+2 ...a 0 in T 2 ∪ {x 1 } and thus vx 2 ...x δ P v ≡ a 1 a 2 ...a 1 = C, so we may suppose that {3, 5} → x δ . If a δ-1 → a δ then 4x 1 ...x δ-1 v5x δ 31 24 ≡ C, so a δ → a δ-1 . Then δ = 2 and thus 253vx 2 x 1 142 ≡ C. Dual Exc [START_REF] Petrović | Antidirected Hamiltonian circuits in tournaments, Graph Theory[END_REF]1)(2): By the building lemmas, if N + (x δ ) ∩ {3, 4} then x δ has an outneighbor origin of a path P ≡ a δ+1 a δ+2 ...a 0 in T 2 ∪ {x 1 } and thus vx 2 ...x δ P v ≡ a 1 a 2 ...a 1 = C, so we may suppose that {3, 4} → x δ . If a δ-1 → a δ then 5x 1 ... ). For any u ∈ T 2 , one can find a path P ≡ a δ a δ+1 ...a 0 in T 2 ∪ {x δ , v} ending by an outneighbor of x 1 , so in both cases, a δ-1 → a δ and a δ → a δ-1 , x δ-1 is an origin of a path Q ≡ a δ-1 a δ ...a 0 ending by an outneighbor of x 1 and thus T 2 (u), t ∈ T 2 \ {u, w} and P ≡ a n-3 a n-4 ...a δ+2 be a path in T 2 \ {u, w, t}, we have

y 4 vP -1 y 1 ≡ a 1 a 2 ...a 1 = C. Dual Exc E 1 (4): If δ = 2 then x 1 4321vx 2 x 1 ≡ C. If δ = 3 then x 3 x 2 O T 2 vx 1 x 3 ≡ C. If δ = 4,
Dual Exc (0, 1): If δ = 2 then x 2 v123x 1 x 2 ≡ C. If δ = 3 then x 2 → 3 and thus x 3 v123x 2 x 1 x 3 ≡ C. Dual Exc (2, 1): If δ = 2 then 1vx 2 x 1 4231 ≡ C. Else, we have x δ → 2 so 4x 1 x 2 213v4 ≡ C. Dual Exc
x 1 x 2 ...x δ-2 P x 1 ≡ a 1 a 2 ...a 1 = C unless if T 2 → x δ-1 , in this case T (N -(x δ ) = 7A. Dual Exc (E 8 (5), 1): If x δ → 1 then wx 1 ...x δ 12v3uw ≡ C, so we may assume that {1, 2, 3} → x δ . If a δ-1 → a δ then wx 1 ...x δ-1 v1x δ 23uw ≡ C, so a δ → a δ-1 . Then δ = 2 and thus uw3vx 2 x 1 12u ≡ C. Else, by lemma 2.4, either T 1 = 4B or δ = 2. If T 1 = 4B, set T 1 = {u 1 , u 2 , u 3 } → u 4 , we have N + T 2 (u i ) = {u} for all i = 1, 2,
u 2 wuu 4 u 1 vP -1 u 3 tu 2 ≡ C. If δ = 2 and N + T 2 (x 1 ) = {u}, if |T 2 | > 3, let P ≡ a n-2 a n-3 ...a 2 be a path in T 2 of origin distinct from u, we have x 2 vP -1 x 1 x 2 ≡ C. Otherwise, |T 2 | = 3, set T 2 -u = {w → t}. If T 2 = 3A, i.e. t → u → w, then u → x 2 , if w → x 2 then x 2 ux 1 vtwx 2 ≡ C and if x 2 → w then x 1 wx 2 vutx 1 ≡ C.
So we may assume that T 2 is a transitive triangle, also, for every vertex x ∈ T , T (N + (x)) and

T ( N -(x)) are transitive, since, other- wise, T or T contains C. Thus {w, t} → x 2 , if u → x 2 then x 2 ux 1 vtwx 2 ≡ C, and if x 2 → u then (T ; C) = A 14 . (c) If |B | = 2.
We distinguish two cases:

(i) If a 2 → a 1 , suppose first that (T 1 ; a 1 ...a δ ) is a Grünbaum's exception, every vertex x i ∈ T 1 is an origin of a path P i ≡ a 0 a 1 ...a δ-1 and d - T 2 (x i ) ≥ 2, denote by y i the origin of * P i . If x i has an inneighbor origin of a path P ≡ a n-1 a n-2 ...a δ+1 in T 2 then P i vP -1 x i ≡ C. Otherwise, (T 2 ; a n-1 a n-2 ...a δ+1 ) is an exception, so by the building lemmas it is an biexception with respect to x i (Considering that a n-1 a n-2 ...a δ+1 = -(1, Q, 1) with odd number of blocks and that d - T 2 (y i ) ≥ 2), in any possible biexception except Exc(0, 1), there is an origin w of a path P ≡ a n-1 a n-2 ...a δ+1 in T 2 -S which is an outneighbor of y i , let x j = y i , we have P j vP -1 y i ≡ C. And if it is the biexception Exc(0, 1) with respect to x i , in this case, by induction on i, we may assume that 1 → T 1 , so d + (1) = 4, a contradiction. So we may assume that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception, let x 1 ...

x δ ≡ a 1 ...a δ in T 1 such that d + T 2 (x 1 ) = max {d + T 2 (x); x ∈ Or(a 1 ...a δ , T 1 )}. If x 1 has an outneighbor u origin of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 then ux 1 ...x δ vP -1 ≡ C. Otherwise, if d + T 2 (x 1 ) ≥ 2 and s - T 2 (N + T 2 (x 1 )) ≥ 3 then (T 2 ; a 0 a n-1 ...a δ+2 ) is an exception. If d +
T 2 (x 2 ) = 0, then, by lemma 2.8, T contains C. So we may assume that d + T 2 (x 2 ) ≥ 1 and thus, by the building lemmas, we can suppose that (T 2 ; a 0 a n-1 ...a δ+2 ) is one of the following biexceptions with respect to x 1 : Dual Exc (3, 1): T (N -(1)) = 4A, then T contains C or it will be solved. Dual Exc (11,1): If N + (x δ ) ∩ {1, 2, 4, 5}, one can find C, so we may assume that {1, 2, 4, 5} → x δ and thus if δ > 2 then T (N + (2)) = 5E, so we may assume that δ = 2, therefore 13v2x 5), then T contains C or it will be solved. Dual Exc (E 8 [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF], 1): T (N -(1)) = F 8 (6), then T contains C or it will be solved.

2 45x 1 ≡ C. Dual Exc (E 2 (5), 1): T (N -(x 2 )) = F 2 (
If d + T 2 (x 1 ) ≥ 2 and S - T 2 (N + T 2 (x 1 )) = N + T 2 (x 1 ) = {u → w}, if there exists t ∈ N + T 2 \{u,w} (x 2 ), then |T 2 | ≥ 4, so there exists Q ≡ a n-1 a n-2 .
..a δ+2 in T 2 -t of origin distinct from u, w, we have x 1 tx 2 ...x δ vQ -1 x 1 ≡ C and if x 2 → u, since (T 2 ; a n-2 a n-3 ...a δ+2 ) is not a Grünbaum's exception, there exists Q ≡ a n-3 ...a δ+2 in T 2 \ {u, w}, we have wux 2 ...x δ vQ -1 x 1 w ≡ C. So we may assume that N + T 2 (x 2 ) ⊆ {w}, if δ ≥ 3 then x δ → u, let P ≡ a δ+2 a δ+3 ...a n-2 be a path in T 2 \ {u, w}, we have wvx 2 ...x δ uP x 1 w ≡ C, and if δ = 2, set s = min {i ≥ 3; d + C (a i ) = 2} and let P ≡ a 2 a 3 ...a s-1 be a path in T 2 such that u, w ∈ V (P ) and T 2 -V (P ) contains a path Q ≡ a s+1 a s+2 ...a n-1 , we have

x 1 x 2 P vQx 1 ≡ C. Assume now that d + T 2 (x 1 ) = 1, set N + T 2 (x 1 ) = {u}, by lemma 2.4, either T 1 = 4B or δ = 2. In the first case, set T 1 = {u 1 , u 2 , u 3 } → u 4 , we can suppose that {u 1 , u 2 , u 3 } → {u} and d - T 2 (u) = 0 since otherwise, by taking u 1 in the place of v the result follows. If a 1 ...a 4 = -(2, 1), there exist w, t ∈ T 2 -u such that t is an origin of a path Q ≡ a n-2 a n-3 ...a δ+2 in T 2 \ {u, w}, we have u 4 uu 2 wu 1 vQ -1 u 3 u 4 ≡ C, else, a 1 ...a 4 = -(1, 1, 1), if there exists w ∈ N - T 2 (u 4 ), then there exist a path Q ≡ a n-1 a n-2 ...a δ+2 in T 2 -w of origin distinct from u, thus u 3 u 2 wu 4 u 1 vQ -1 u 3 ≡ C, otherwise, let w, t ∈ T 2 -u such that t is an origin of a path Q ≡ a δ+2 a δ+3 ...a n-2 in T 2 \ {u, w}, we have u 4 wu 2 u 1 uu 3 Qvu 4 ≡ C, then T contains C. In the second case, suppose that there exists w ∈ N + T 2 (x 2 ) ∩ N - T 2 (x 1 ), if T 2 -w contains a path P ≡ a n-1 a n-2 .
..a δ+2 of origin an inneighbor of x 1 then x 1 wx 2 vP -1 x 1 ≡ C, else, without loss of generality, we can suppose that (T 2 -w; a n-1 ...a δ+2 ) is a Grünaum's exception, in this case, one can find a path P ≡ a δ+1 a δ+2 ...a 0 a 1 in T 2 ∪ {v} with origin w and end u,

so x 1 x 2 P x 2 ≡ a 1 a 2 ...a 1 = C. So we can assume that (T 2 -u) → x 2 , if u → x 2
then T 2 -u contains a path P ≡ a 5 a 6 ...a 0 we have vx

1 ux 2 P v ≡ a 1 a 2 ...a 1 = C. Now if x 2 → u, if there exists w ∈ N - T 2 (x 1 ) such that T 2 -w contains a path P ≡ a 2 a 3 ...a n-3 of origin distinct from u so x 1 x 2 P vwx 1 ≡ C, otherwise, |T 2 | = 4 and d - T 2 (u) = 0, set T 2 = {u, w 1 , w 2 , w 3 } we have w 1 w 2 vw 3 x 1 x 2 uw 1 ≡ C. (ii) If a 1 → a 2 , suppose first that (T 1 ; a 1 ...a δ ) is a Grünbaum's exception, let a ∈ T 2 such that d + T 1 (a) = min {d + T 1 (x); x ∈ T 2 } and set T 1 = T 1 ∪ {a} and T 2 = T 2 -a. Let x 0 ...x δ ≡ a 0 ...a δ be a path in T 1 such that d - T 2 (x 0 ) = max {d - T 2 (x); x ∈ Or(a 0 ...a δ ; T 1 )}, if x 0 has an inneighbor origin of a path P ≡ a n-1 a n-2 ...a δ+2 in T 2 then x 0 ...x δ vP -1 x 0 ≡ C or an inneighbor origin of a path Q ≡ a δ+1 a δ+2 ...a n-2 then x δ ...x 0 Qvx δ ≡ C. Otherwise, if T 2 = 3A, there exists x ∈ T 1 such that x → a, thus vaO T 1 O T 2 v ≡ a 1 a 2 ...a 1 = C. Assume now that T 2 = 3A, if d - T 2 (x 0 ) ≥ 2 then (T 2 ; a n-1 a n-2 ...a δ+2 )
- T 2 (x) ≥ 3 for all x ∈ T 1 . So let y 3 ...y δ+2 ≡ a 3 ...a δ+2 be a path in T 1 such that d - T 2 (y δ+2 ) ≥ 3 and let u ∈ N + T 2 (y 3 ), d - T 2 -u (y δ+2 ) ≥ 2 so y δ+2 has an inneighbor origin of a path P = (1, 1, 2) in T 2 , thus vuy 3 ...y δ+2 P v ≡ a 1 a 2 ...a 1 = C. Dual Exc 8: Since d - T 2 (x 0 ) = 2 then T 1 = 3A, if x 3 → 2 then x 0 ...x 3 21v345x 0 ≡ C, so 2 → x 3 . If x 3 → 3 then x 0 ...x 3 3124v5x 0 ≡ C, so 3 → x 3 . If x 3 → 4 then x 0 ...x 3 4351v2x 0 ≡ C, so 4 → x 3 . Thus T 1 = 4A, if x 2 = a then x 0 x 1 x 2 v4x 3 3125x 0 ≡ C, so x 1 = a and thus d + T 1 (x)
≥ 2 for all x ∈ T 2 . So 4 → x 2 and thus x 0 x 1 x 2 43x 3 21v5x 0 ≡ C. Dual Exc 9: If N + (x 0 ) ∩ {4, 5} = ∅ then by the building lemmas we may assume that T 2 → x 1 and thus x 1 = a. x 1 has an inneighbor origin of a path P ≡ a 6 a 7 ...a 0 in T 2 -3, thus v3x 0 (x 0 ) = 2 so δ = 3. By the building lemmas we may assume that (T 2 -X) → x 1 so T 2 = 4A. d - T 1 (w) = 2 so x 1 = a. Let x → y be an arc in T 2 -X and P ≡ a 6 a 7 ...a n-3 be a path in T 2 -(X ∪ {x, y}), we have

x 3 x 1 P v ≡ a 1 a 2 ...a 1 = C. So we may suppose that {4, 5} ⊆ N -(x 0 ), if x δ → 2 then x 0 ...x δ 21v354x 0 ≡ C, so 2 → x δ . If x δ → 3 then x 0 ...x δ 32v514x 0 ≡ C, so 3 → x δ . If x δ → 4 then x 0 ...x δ 43v125x 0 ≡ C, so {4, 5} → x δ . If δ = 5 then 2 → x 0 and x δ → 1, thus x 0 ...x δ 1453v2x 0 ≡ C, so δ = 3 and T 1 = 4A. If x 2 = a then x 0 x 1 x 2 v3x 3 2415x 0 ≡ C, so x 1 = a. Thus 1 → {x 2 , x 3 }, we have x 0 x 1 x 2 14x 3 32v5x 0 ≡ C.
yxx 1 x 3 x 2 x 0 uP vwy ≡ C. Suppose now that d - T 2 (x 0 ) = 1, if T 2 is
a transitive triangle, we may assume that T 1 = 4A, and thus Assume now that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception, let x 1 ...x δ ≡ a 1 ...a δ be a path in T 1 , if x 1 has an outneighbor w origin of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 then wx 1 ...x δ vP -1 ≡ C and if there exists w ∈ N -

d + T 1 (a) ≥ 1; set V (T 2 ) = {u 1 , u 2 , u 3 } such that u 1 u 2 u 3 is a directed outpath and V (T 1 ) = {y 1 , y 2 , y 3 }. Without loss of generality sup- pose that y 1 → a, if a → u 3 , T 1 ∪ {u 1 } contains a path z 0 ...z 3 ≡ a 0 ...a 3 and thus z 0 ...z 3 vau 2 u 3 z 0 ≡ C or z 0 ...z 3 vu 2 au 3 z 0 ≡ C, so u 3 → a. If u 1 → y 3 then u 3 vay 1 y 2 y 3 u 1 u 2 u 3 ≡ C, so y 3 → y 3 . There exist i ∈ {1, 2, 3} and {j, k} = {1, 2} such that y i → u j and u k → y i-1 then avu j y i y i+1 y i-1 u k u 3 a ≡ C. If |T 2 | = 4 then by taking C = a 4 a 5 ...a 4 it is solved in Case 1, so we may assume that |T 2 | ≥ 5. Set {u} = N - T 2 (x 0 ), if a = x 1 then (T 2 -u) → {x 2 , x 3 }, let P ≡ a 6 a 7 ...a n-3 be a path in T 2 -u and {x, y} = V (T 2 -u) -V (P ), we have x 0 x 1 x 2 xyx 3 P vux 0 ≡ C, so x 2 = a. If a 7 → a 6 ,
T 2 (x 1 ) ∩ N - T 2 (x 2 ) such that T 2 -w contains a path Q ≡ a n-1 a n-2 .
..a δ+2 with origin an inneighbor of x 1 then x 1 wx 2 ...x δ vP -1 x 1 ≡ C. Otherwise, we distinguish 4 cases:

(i) If d + T 2 (x 1 ) ≥ 2 and s - T 2 (N + T 2 (x 1
)) ≥ 3 and so (T 2 ; a 0 a n-1 ...a δ+2 ) is an exception. By the building lemmas, we may assume that it is one of the following biexceptions with respect to x 1 : Dual Exc (3, 1): d -(1) ≥ 5, a contradiction. Dual Exc [START_REF] Rosenfeld | Antidirected Hamiltonian cycles in tournaments[END_REF]1): T (N -(4)) = 5E nor F 2 , so T contains C or it will be solved. Dual Exc (E 2 (5), 1):

T (N + (x 2 )) = F 2 nor 5E, so T contains C or it will be solved. Dual Exc (E 8 (6), 1): d -(1) ≥ 7, a contradiction. (ii) If d + T 2 (x 1 ) = s - T 2 (N + T 2 (x 1 )) = 2, set N + T 2 (x 1 ) = {u → w}, clearly N - T 2 (x 2 ) = {w}. If |T 2 | ≥ 5, since otherwise there is a vertex in T 2 of indegree greater than 4. If |T 2 | = 5 then T 2 \ {u, w} = 3A, let t 1 t 2 t 3 = O T 2 \{u,w} , we have t 3 wx 2 ...x δ vux 1 t 1 t 2 t 3 ≡ C. So δ ≥ 6, let a ∈ T 2 \ {u, w} such that T 2 \ {u, w, a} = 3A, 5A, 7A, and let y 1 ...y δ ≡ a 1 ...a δ be a path in (T 1 -x 2 ) ∪ {a}, y 1 has an outneighbor t origin of a path P ≡ a 0 a n-1 ...a δ+2 in (T 2 -a) ∪ {x 2 } of end distinct from x 2 , thus ty 1 ...y δ vP -1 ≡ C, unless if |T | = 9 and a 4 ...a 7 is directed, in this case w is an origin of a path P ≡ a 8 a 7 ...a 3 = -(1, 3, 1) in T 2 and thus x 2 x 1 vP -1 x 2 ≡ C. (iii) If N + T 2 (x 1 ) = {u}, suppose that there exists w ∈ N - T 2 (x 1 ) ∩ N - T 2 (x 2
) so (T 2 -w; a n-1 a n-2 ...a δ+2 ) is a Grünbaum's exception, suppose that u ≡ 1, if T 2 -w = 5A or 7A then N + (w) ∩ {2, 3} = ∅ and so w is an origin of a path P ≡ a 1 a 2 ...a δ+2 in T 2 ∪ {x 1 } with end distinct from x 1 and thus

x 2 ...x δ vP -1 x 2 ≡ a 2 a 3 ...a 2 = C. If T 2 -w = 3A, then w → 2 and so 3 → w. If w → u then uwx 2 ...x δ v32x 1 u ≡ C, so u → w. If δ = 2, by taking C = u 0 u 1 ...u 0 ≡ a 3 a 4 ...a 3 it is solved in Case 1.
If T 1 is a transitive triangle then x 1 is an origin of an antidirected outpath Q in T 1 and x 1 has an inneighbor origin of a path P ≡ a 3 a 2 a 1 a 0 , thus QvP -1 x 1 ≡ a 4 a 5 ...a 4 = C. If 3, which is an origin of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 , has an inneighbor in T 1 that is an origin of a path a 1 ...a δ in T 1 then T contains C. So we may assume that |Or(a 1 ...a δ ; T 1 )|leq2 and 3 → Or(a 1 ...a δ ; T 1 ). If a δ → w then 1x 1 ...x δ w2v31 ≡, so we may assume that

N + T 1 (w) = {x 1 , x 2 , x 4 }. We have d + T 1 (x 1 ) = 2 so T 1 = 4A. If a 1 ...a 4 is antidirected then d + T 2 (x 4 ) = 2,
{2, 3} → {x 1 , x 3 } = Or((2, 1); T 1 ), thus x 3 3x 1 x 4 x 2 vw13x 3 ≡ C. Assume now that N - T 2 (x 2 )
= {u}, we may suppose that d - T 2 (u) = 0 since otherwise, by taking x 2 in the place of v one can find C. If a δ+2 a δ+3 ...a n-1 has at least 4 blocks, then (T 2 -u) ∪ {x 1 } contains a path P ≡ a 0 a n-1 ...a δ+2 with origin w and end in T 2 , thus wux 2 ...x δ vP -1 ≡ C, so a δ+2 ...a n-1 has only two blocks. If δ = 2, let x → y be an arc in T 2 -u, we have yxx 1 vI T 2 \{u,x,y} ux 2 y ≡ C. If δ = 3, let P ≡ a 3 a 4 ...a n-3 be a path in T 2 -u and suppose that x 1 x 3 x 2 = -(1, 1), we have x 2 x 3 x 1 P vux 2 ≡ C. So δ = 4, if a 1 ...a 4 is antidirected, let P ≡ a 5 a 6 ...a n-1 be a path in T 2 , thus x 4 ... (2, 1, 1) in T 2 andthus vx 4

...x 1 P v ≡ a 3 a 4 ...a 3 = C. (iv) If d + T 2 (x 1 ) = 0, so x 1 is the unique origin of the path a 1 ...a δ in T 1 . Let w ∈ N - T 2 (x 2 ), we have (T 2 -w; a n-1 a n-2 ...a δ+2
) is a Grünbaum's exception. x δ has an outneighbor in T 2 -w, but any vertex in T 2 -w is an origin of a path P ≡ a δ+1 a δ+2 ...a n-3 in T 2 -w, so let P be a path of origin an outneighbor of x δ and set {a} = V (T 2 -w) -V (P ), we have x 1 wx 2 ...

x δ P vax 1 ≡ C. (d) If |B| > |B | ≥ 3. Do a i = a i-1 until a n-1 → a n-2
for the first time. Suppose first that (T 1 ; a 1 ...a δ ) is a Grünbaum's exception and for x i ∈ T 1 , let P i ≡ a 3 a 4 ...a δ+2 be a Hamiltonian path of T 1 of origin x i . If an x i has an outneighbor origin of a path

P ≡ a 2 a 1 ...a δ+4 in T 2 then P i vP -1 x i ≡ a 3 a 4 ...a 4 = C. Otherwise, if (T 2 ; a 2 a 1 ...a δ+4 )
is an exception so it is one of the duals of the exceptions (E 8 ; (∆ -4,

, then |N - T 2 (T 1 )| ≥ 3; else, (T 2 ; a 2 a 1 ...a δ+4 ) is not an exception, so s - T 2 ( x∈T 1 N + T 2 (x)) ≤ 3, then s - T 2 (N - T 2 (T 1 )) = |T 2 | and |N - T 2 (T 1 )| ≥ 3 unless if |T 2 | = 5 and T 1 = 3A. In the first case, |N - T 2 (T 1 )| ≥ 3, let u, w, t ∈ N - T 2 (T 1 ) 1, 1, 1)), (E 8 ; (∆, -6, 1, 1, 1, 1, 1)) or (E 10 ; (∆ -8, 1, 1, 1, 1, 1, 1, 1))
, since (T 2 -u; a 1 a 0 a n-1 ...a δ+4 ) and (T 2 -u; a 0 a n-1 ...a δ+3 ) cannot be two exceptions in the same time, then w or t is an origin of a path Q ≡ a 1 a 0 ...a δ+4 or R ≡ a 0 a n-1 ...a δ+3 in T 2 -u and thus

x i u * P i vQ -1 x i ≡ a 2 a 3 ...a 2 = C or P i vR -1 x i ≡ a 1 a 2 ...a 1 = C, where P i ≡ a 1 a 2 ...a δ+1 is a path in T 1 ∪ {u} with origin x i . Now if |T 2 | = 5 and T 1 = 3A, we have N + T 2 (T 1 ) → N - T 2 (T 1 ) where N + T 2 (T 1 ) = {u 1 , u 2 , u 3 } = 3A and N - T 2 (T 1 ) = {b 1 , b 2 }, thus u 1 u 2 b 2 b 1 P i u 3 va 1 ≡ C. Assume now that (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception and let x 1 ...x δ ≡ a 1 ...a δ be a path in T 1 such that d - T 2 (x 1 ) = max {d - T 2 (x); x ∈ Or(a 1 ...a δ , T 1 )}. If x 1 has an inneighbor u origin of a path P ≡ a 0 a n-1 ...a δ in T 2 then ux 1 ...x δ vP -1 ≡ C. Oth- erwise, since d - T 2 (x 1 ) ≥ 2, (T 2 ; a 0 a n-1 ...a δ
) is one of the following biexceptions with respect to x 1 : Dual Exc (6, 0), (31, 0), (E 11 , 0) and (E 13 , 0): In this case, d - T 2 (x 1 ) = 2, so by Lemma 2.4, since δ ≥ 3 and a 1 ...a δ is not directed, T 1 = 4B and a 1 ...a δ has 3 origins, which implies that there exists a common inneighbor z of the origins of a ) is not a Grünbaum's exception, so let x 0 ...x δ-1 ≡ a 0 ...a δ-1 be a path in T 1 such that d - T 2 (x 0 ) = max {d - T 2 (x); x ∈ Or(a 0 ...a δ-1 ; T 1 )}, if x 0 has an inneighbor origin of a path P ≡ a n-1 a n-2 ...a δ+1 in T 2 then x 0 ...x δ-1 vP -1 x 0 ≡ C, otherwise, we distinguish three cases:

1 ...a δ in T 2 d + (z) > ∆ + (T ), a contradiction. Dual Exc (26, 0): If x δ → 1 then 4x 1 ...x δ 16523v4 ≡ C.
(i) If d - T 2 (x 0 ) = 2 and s - T 2 (N - T 2 (x 0 )) = 2, set N - T 2 (x 0 ) = {u → w}.
Suppose that x 1 has an outneighbor t ∈ T 2 \ {u, w}, if T 2 \ {u, w, t} contains a path P ≡ a n-4 a n-5 ...a δ+1 then tx 1 ...x δ-1 vP -1 uwx 0 t ≡ C, else, if t has an inneighbor t ∈ T 2 \ {u, w} then T 2 \ {u, w, t, t } contains a path P ≡ a n-5 a n-6 ...a δ+1 and thus tx

1 ...x δ-1 vP -1 wux 0 t t ≡ C. Otherwise, t → (T 2 \ {u, w, t}, if x δ-1 has an outneighbor x in T 2 \ {u, w, t} then |T 2 | ≥ 8 since otherwise d -(x) > ∆ + (T ),
let xyz be a direccted outpath in T 2 and P ≡ a n-4 a n-5 ...a δ+4 be a path in T 2 \ {u, w, t, x, y, z}, we have tx

1 ...x δ-1 xyzvP -1 uwx 0 t ≡ C, else, (T 2 \ {u, w, t}) → x δ-1 , if t → x δ-2
, let x → y be an arc in T 2 \{u, w, t} and P ≡ a n-4 a n-5 ...a δ+3 be a path in T 2 \{u, w, t, x, y}, we have x 0 ...x δ-2 txyx δ-1 P -1 vuwx 0 ≡ C and if there exists x ∈ T 2 \ {u, w, t} an inneighbor of x δ-2 , let P ≡ a n-4 a n-5 ...a δ+2 be a path in T 2 \ {u, w, t, x}, we have tx 1 ...x δ-2 xx δ-1 vP -1 uwx 0 t ≡ C. So we may assume that N - T 2 (x δ-2 ) = {w}, let xy be an arc in T 2 \ {u, w, t} and P ≡ a δ+3 a δ+4 ...a n-4 in T 2 \ {u, w, x, y}, we have tx 1 ...x δ-2 wxyx δ-1 P vux 0 t ≡ C. So we can suppose now that (T 2 \ {u, w}) → x 1 , we have to mention that

T 1 ∪ {u} contains a path Q ≡ a n-1 a 0 ...a δ-1 of origin x 1 and end distinct from u, in fact, if a 2 → a 1 then x 1 x 0 ux 2 ...x δ-1 ≡ Q and if a 1 → a 2 , T 1 -x 1 contains a path Q ≡ a 2 .
..a δ-1 of origin distinct from x 0 and thus x 1 uQ ≡ Q, let P ≡ a δ+1 a δ+2 ...a n-4 be a path in T 2 -w, and set {t} = V (T 2 ) -V (P ), we have QvP wtx 1 ≡ a n-1 a 0 ...a n-1 = C.

(ii) If d - T 2 (x 0 ) ≥ 2 and s - T 2 (N - T 2 (x ) ) ≥ 3 then (T 2 ; a n-1 a n-2 ...a δ+1
) is an exception, so, by the building lemmas, we may assume that it is one of the following biexceptions with respect to x 0 : Dual Exc [START_REF] Havet | Oriented Hamiltonian paths in tournaments: a proof of Rosenfeld's conjecture[END_REF]0)

: If d + T 2 (x δ-1 ) ≥ 1 then there exist xyz and ab two directed outpaths such that V (T 2 ) = {a, b, x, y, z}, x ∈ N + (x δ-1 ) and b ∈ N -(x 0 ), thus x 0 ...x δ-1 xyzvabx 0 ≡ C, so T 2 → x δ-1 . Let u ∈ N - T 2 (x δ-2 ), there ex- ists P ≡ a n-1 a n-2 ...a δ+2 in T 2 -u of origin an inneighbor of x 0 , we have x 0 ...x δ-2 ux δ-1 vP -1 x 0 ≡ C. Dual Exc (E 8 (6), 1), (E 9 (8), 1), (E 10 (10), 1): If there exists x ∈ N + T 2 -X (x δ-1 ) then |T 2 | ≥ 8, let xyz be a directed outpath in T 2 -X and P ≡ a n-1 a n-2 ...a δ+4 be a path in T 2 \{x, y, z} of origin an inneighbor of x 0 , we have x 0 ...x δ-1 xyzvP -1 x 0 ≡ C, so (T 2 -X) → x δ-1 . If there exists x ∈ N - T 2 -X (x δ-2 ), let P ≡ a n-1 a n-2 .
..a δ+2 be a path in T 2 -x of origin an inneighbor of x 0 , we have x 0 ...x δ-2 xx δ-1 vP -1 x 0 ≡ C, so x δ-2 → (T 2 -X) and thus |T 2 | ≥ 8. If x δ-2 has an inneighbor x ∈ X such that there is an inneighbor y ∈ X -x of x 0 of positive indegree in T 2 -x, set {z → y} = X -x and let a, b ∈ T 2 -X and P ≡ a δ+3 a δ+4 ...a n-4 be a path in T 2 \ {x, y, z, a, b}, we have x 0 ...x δ-2 xabx δ-1 P vzyx 0 ≡ C, so we may suppose that T 2 = {u, w, t} is transitive such that u → w → t, N - T 2 (x 0 ) = {u, t} and N - T 2 (x δ-2 ) = {t}. Let ab be an arc in T 2 -X and P ≡ a n-1 a n-2 ...a δ+4 be a path in T 2 \{a, b} with origin an inneighbor of x 0 , we have x 0 ...

x δ-3 vx δ-2 abx δ-1 P -1 x 0 ≡ C. (iii) If d - T 2 (x 0 ) = 1, set N - T 2 (x 0 ) = {u}, we have x 0 is maximal in T 1
. By lemma 2.3, a 0 ...a δ-1 is directed. If T 1 is transitive with δ = 3 or 4, then x 0 is an origin of a path Q ≡ a 1 ...a δ in T 1 and thus x 0 has an outneighbor w origin of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 , we have wQvP -1 ≡ C. If T 1 = 4B, we can suppose that, without loss of generality, {x 1 , x 2 , x 3 } → u, so d - T 2 -u (x 1 ) ≥ 3 and thus x 1 has either an inneighbor origin of a path P ≡ a n-2 a n-3 ...a δ+1 in T 2 -u and thus x 0 ux 2 x 3 vP -1 x 1 x 0 ≡ C, or (T 2 -u; a n-2 a n-3 ...a δ+2 ) is one of the dual of the exceptions 22, 24, 38 or 47, in this case x 1 has an outneighbor origin of a path Q ≡ a δ+4 a δ+5 ...a 1 in T 2 -u and thus vx 2 x 3 ux 0

x 1 x 0 P v ≡ a 2 a 3 ...a 2 = C. Now if |T 1 | = 2,
we can suppose that a 5 → a 6 since otherwise by Case 2 one can find C = a 0 a 1 ...a 0 = a)a 5 a 4 ...a 5 , also we can suppose that a n-5 → a n-4 since otherwise one can find C = a n-4 a n-3 ...a n-4 . The blocks containing the arcs a n-5 → a n-4 and a 5 → a 6 cannot be of length 2 since otherwise it is solved in Case 4(c). Obviously u is not maximal in T 2 so let w ∈ N + T 2 (u), if d - T 2 \{u,w} (x 1 ) ≥ 2 then x 1 has an inneighbor origin of a path P ≡ a n-3 a n-4 ...a 3 in T 2 \ {u, w} and thus wux 0 P -1 x 1 vw ≡ C unless if (T 2 \ {u, w}, a n-3 a n-4 ...a 3 ) is one the exceptions Exc 0, 4, 8, 9, 33, 37, E 8 (6), E 9 [START_REF] Havet | Oriented Hamiltonian paths in tournaments: a proof of Rosenfeld's conjecture[END_REF] or E 10 (9), if it is the exception 0 then if d + T 2 (x 1 ) ≥ 2 then T 2 contains a path P ≡ a 0 a n-1 ...a 3 of origin an outneighbor x of x 1 and thus xx

1 x 0 vP -1 ≡ C unless if s - T 2 (N + T 2 (x 1 )) ≤ 3, but since T 2 contains a circuit triangle with at least two inneighbors of x 1 then T 2 = {u, w} → {1, 2, 3} and N + T 2 (x 1 ) = {u, w}, thus vx 0 w12u3x 1 v ≡ C, so d - T 2 (x 1 ) ≥ 4, let t ∈ N +
T 2 (x 0 ), x 1 has an inneighbor origin of a path P = (1, 2) in T 2 -t, we have x 1 x 0 tvP -1 x 1 ≡ a 2 a 3 ...a 2 = C. Otherwise, one can check that (T 2 ∪ {x 1 }) \ {u, w} contains a path P ≡ a n-1 a 0 ...a n-6 of origin and end in T 2 , thus vwux 0 P v ≡ a n-5 a n-4 ...a n-5 = C. So we may assume that d - T 2 \{u,w} (x

1 ) ≤ 1, in other terms d + T 2 (x 1 ) ≥ |T 2 | -3, if s - T 2 (N + T 2 (x 1
)) ≥ 4 then x 1 has an ouitneighbor x origin of a path P ≡ a 0 a n-1 ...a 3 in T 2 and thus xx 1 x 0 vP -1 ≡ C unless if (T 2 ; a 0 a n-1 ...a 3 ) = E 8 [START_REF] Havet | Oriented Hamiltonian paths in tournaments: a proof of Rosenfeld's conjecture[END_REF], in this case x 1 has an outneighbor origin of a path

P ≡ a n-4 a n-3 ...a 0 in T 2 , then vP -1 x 1 x 0 v ≡ a n-1 a 0 ...a n-1 = C, otherwise, s - T 2 (N + T 2 (x 1 ) ≤ 3, then |T 2 | = 5 and N + T 2 (x 1 ) → N - T 2 (x 1
), in this case u and x 1 have a common outneighbor w then x 1 has an inneighbor x origin of a path

P = (1, 1) in T 2 \ {u, w }, we have w ux 0 P -1 x 1 vw ≡ C. (f) If |B| = |B | ≥ 4. Do a i = a i-1 until a δ → a δ-1 . If (T 1 ; a 1 ...a δ ) is not a Grünbaum's exception, there exists x 1 ...x δ ≡ a 1 ...a δ such that d - T 2 (x 1 ) ≥ 2 and s - T 2 (N - T 2 (x 1
)) ≥ 3, then, since (T 2 ; a 0 a n-1 ...a δ+2 ) is not an exception, by theorem 2.1 there is an inneighbor u of x 1 origin of a path P ≡ a 0 a n-1 ...a δ+2 in T 2 and thus ux 1 ...

x δ vP -1 ≡ C. Assume now that (T 1 ; a 1 ...a δ ) is a Grünbaum's exception, let a ∈ T 2 such that d - T 2 (a) = ∆ -(T 2 ) and set T 1 = T 1 ∪{a} and T 2 = T 2 -a. If there exists x 0 ...x δ ≡ a 0 ...a δ in T 1 such that x δ = a then d - T 2 (x 0 ) ≥ 2,
so there is an inneighbor of x 0 origin of a path P ≡ a n-1 a n-2 ...a δ+2 in T 2 unless if (T 2 ; a n-1 ...a δ+2 ) is one of the following biexceptions with respect to x 0 : Dual Exc (6, 0): x 1 = a so {x 1 , x δ } → 5, then by taking 5 in the place of v the result follows. Dual Exc (31,1): In a similar way take 6 in the place of v. Dual Exc (E 11 [START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF], 0): In a similar way take 3 in the place of v. Otherwise, T 1 = 4B, if there is an inneighbor u of a vertex y i ∈ T 1 origin of a path P ≡ a 0 a n-1 ...a δ+3 in T 2 then uy i ay i+1 y i+2 vP -1 ≡ C. Else, if there is a vertex

y i ∈ T 1 such that d - T 2 (y i ) = 1, we have s + T 2 (N + T 2 (y i )) ≥ |T 2 |-1 so there exists u ∈ N + T 2 (y i ) origin of a path P ≡ a 2 a 1 ...a δ+4 in T 2 and thus y i y i+1 y i+2 vP -1 y i ≡ a 3 a 4 ...a 3 = C. So we can assume that d - T 2 (y i ) ≥ 2 for all y i ∈ T 1 and s - T 2 (N - T 2
(T 1 )) ≥ 3 and (T 2 ; a 0 a n-1 ...a δ+3 ) is one of the following exceptions: Dual Exc 5,[START_REF] Havet | Oriented Hamiltonian paths in tournaments: a proof of Rosenfeld's conjecture[END_REF]51 w). If u has an inneighbor origin of an antidirected outpath P in T 2 \ {w, t} then uP vtwu ≡ C. Otherwise, (T 2 \ {w, t}; (1, 1, ..., 1)) is one of the following biexceptions with respect to u:

, E 8 (|T 2 |), E 9 (|T 2 |) and E 10 (|T 2 |): There exists x ∈ N + T 2 (y i ) origin of a path P ≡ a 2 a 1 ...a δ+4 in T 2 so y i y i+1 y i+2 vP -1 y i ≡ C.
-(T ) ≤ δ + (T ). Set δ = δ -(T ). Let v ∈ T be a vertex such that d -(v) = δ and denote by T 1 = T (N -(v)) and T 2 = T (N + (v)).
; C) = A 10 . (b) If d - T 2 (u) ≥ 4, let w ∈ N - T 2 (u) such that d + T 2 (w) ≥ 2 and t ∈ N + T 2 (
Exc (1, 1)(4): We can suppose that T (w, t, 1, 2) = 4A since otherwise it contains P = (1, 1, 1) with origin an inneighbor of u, so u → t and t → {1, 2}. If t → 4 then u34t1vw2u ≡ C, else, u124tv3wu ≡ C if w → 3 and u124tvw3u ≡ C if 3 → w. Exc (1, 1)( 1), ( 2), ( 3): There exist w , t ∈ T 2 such that w → {t , u} and d - T 2 \{w ,t } (u) ≥ 3, so T contains C. Exc (24,1): By taking T 2 \ {4, 6}, we can suppose that T (w, t, 1, 2, 3, 5) = 6H and u → t, so {1, 2, 3} → w and thus u6541w23vtu ≡ C. Exc (22,1): Taking T 2 \ {1, 5}, we obtain d - T 2 \{1,5} (u) ≥ 4, and the result follows.

(c) If d + T 2 (u) ≤ 2 and d - T 2 (u) ≤ 3, then |T 2 | ≤ 4. If |T 2 | = 2 then T = 3A ∪ {x} which is the exception A 4 . So we may assume that |T 2 | = 4, set V (T 2 ) = {w 1 , w 2 , w 3 , w 4 }, if N + T 2 (u) = {w 1 } then T contains C if
such that d - T 2 (x 1 ) = max {d - T 2 (x); x ∈ Or(-(1, 1, ..., 1); T 1 )} and d + T 2 (x δ ) ≥ 1 (This is possible unless if δ = 2). Let u ∈ N + T 2 (x δ ), if x 1 has an inneighbor origin of an antidirected outpath P in T 2 -u then x 1 ...x δ uvP -1 x 1 ≡ C. Otherwise, suppose first that d - T 2 -u (x 1 ) ≥ 2 then (T 2 -u; (1, 1, ..., 1)
) is one of the following exceptions: Exc 1: We distinguish four cases:

(i) If N - T 2 (x 1 ) = {1, 2, 3} then by the building lemmas we may suppose that {2, 4} ⊆ N -(x 2 ). By supposing that there exists i ∈ {1, 3, 4} ∩ N -(u) one can find C, so we may assume that u → {1, 3, 4}. If T 1 = 4A then x 1 x 3 vx 4 x 2 41u32x 1 ≡ C, so T 1 = 4A. If there exists u ∈ N + T 2 -u (x δ ) the result follows so

N + T 2 (x δ ) = {u}. If δ = 4 then x δ is an origin of an antidirected inpath in T 1 with d - T 2 (x δ ) > d - T 2 (x 1 ), a contradiction. If δ = 2, if x 1 → u then ux 1 43x 2 12vu ≡ C, and if u → x 1 then 4ux 1 1x 2 23v4 ≡ C. (ii) If N - T 2 (x 1 ) = {1, 2}
then by the building lemmas we may suppose that {2, 4} ⊆ N -(x 2 ). By supposing that there exists i ∈ {3, 4} ∩ N -(u) one can find C, so we may assume that u → {3, 4} and thus 3 → {x 2 , ..., x δ }. If there exists u ∈ N + T 2 -u (x δ ) the result follows so

N + T 2 (x δ ) = {u}. If δ = 4, if u → 2 then x 1 vx 4 x 3 x 2 34u21x 1 ≡ C and if 2 → u then 43x 3 x 2 x 1 1x 4 2uv4 ≡ C. If δ = 2, if x 1 → u then ux 1 43x 2 12vu ≡ C and if u → x 1 then 4ux 1 1x 2 23v4 ≡ C. (iii) If N - T 2 (x 1 ) = {1, 3}
then by the building lemmas we may suppose that {2, 3, 4} ⊆ N -(x 2 ). By supposing that there exists i ∈ {1, 3, 4} ∩ N -(u) one can find C, so we may assume that u → {1, 3, 4}. If x δ → 2 then x 1 ...x δ 2v4u31x 1 ≡ C, if x δ → 3 then x 1 vx 2 ...x δ 3u142x 1 ≡ C and if x δ → 4 then x 1 vx 2 ...x δ 4u312x 1 ≡ C, so {2, 3, 4} → x δ . If δ = 4, suppose that x 1 → x 4 , since 2 and 4 are origins of an antidirected outpath P with end in {2, 4, u} then if x 3 has an inneighbor in {2, 4, u} we obtain x 4 x 1 vx 2 x 3 P -1 x 4 ≡ C, so x 3 → {2, 4, u} and thus 2 → u. If 1 → x 2 then 4x 1 x 4 3x 2 12x 3 uv4 ≡ C and if x 2 → 1 then 1x 2 x 1 3x 4 2ux 3 4v1 ≡ C. So x 4 → x 1 , and we can suppose that x 3 → 4. x 2 or x 4 is an origin of an antidirected inpath y 1 ...y 4 in T 1 such that y 4 ∈ {x 1 , x 3 } and thus y 1 ...y 4 4v1u32y

1 ≡ C. If δ = 2, if x 2 → 1, if u → 2 then x 1 x 2 1v2u43x 1 ≡ C, so 2 → u; if u → x 1 then 1ux 1 3x 2 42v1 ≡ C and if x 1 → u then T -v = 7A and (T ; C) is the exception A 6 . So we may assume that 1 → x 2 , if x 1 → u then ux 1 21x 2 34vu ≡ C and if u → x 1 then 24x 2 1x 1 u3v2 ≡ C. (iv) If N - T 2 (x 1 ) = {2, 3}
. By supposing that there exists i ∈ {1, 4} ∩ N -(u) one can find C, so we may assume that u → {1, 4}. If

x δ → 2 then x 1 ...x δ 2v1u43x 1 ≡ C, so 2 → x δ .
Suppose first that u → 3, by supposing that there exists i ∈ {1, 4} ∩ N + (x δ ) one can find C, so we may assume that x δ → {1, 4}. Suppose that

x δ → 3 then 3 → {x 2 , x 3 } if it exist. If u → x 1 then x 1 ...x δ 3v241ux 1 ≡ C, so x 1 → u. If u → 2 then x 1 vx 2 ...x δ 312u4x 1 ≡ C, so 2 → u. If δ = 4 then if x 2 → 2 and so x 2 x 3 x 4 vx 1 4u312x 2 ≡ C and if 2 → x 2 , without loss of generality suppose that x 4 → x 2 , so 1x 1 43x 3 x 4 x 2 2uv1 ≡ C. If δ = 2 then 31x 2 2ux 1 4v3 ≡ C. Other- wise, 3 → x δ , suppose that δ = 4, if u → x 3 then 3ux 3 x 2 x 1 2x 4 41v3 ≡ C, so x 3 → u. If u → x 1 , let a ∈ N - T 2 (x 3 ), b ∈ N + T 2 -a (u) and {c → d} = T 2 \ {a, b, u}, we have bux 1 x 2 x 3 ax 4 cdvb ≡ C, so x 1 → u. Suppose that x 1 → x 3 , if x 2 → x 4 then ux 1 x 3 x 2 x 4 4213vu ≡ C, so x 4 → x 2
and thus, similarly x 2 → 4. So 4 → x 3 and since d + (x 1 ) = 5 then x 4 → x 1 and so ux 4 x 1 x 2 x 3 4213vu ≡ C. So we may assume that x 3 → x 1 , if x 1 → x 4 then if x 3 has an inneighbor origin of an antidirected outpath P in T 2 of end distinct from u then x 4 x 1 vx 2 x 3 P x 4 ≡ C, otherwise, 2 → u, N - T 2 (x 3 ) = {1, 3} and u → x 2 . If x 2 → 3 then x 2 vx 3 ux 1 x 4 4213x 2 ≡ C, otherwise, x 2 3x 4 x 1 vx 3 241ux 2 ≡ C. So x 4 → x 1 , if 4 → x 2 then x 3 x 1 x 4 x 2 4213vux 3 ≡ C and if x 2 → 4 then 4 → x 3 and thus x 4 x 1 x 2 x 3 4213vux 4 ≡ C. Now if δ = 2 then ux 1 43x 2 12vu ≡ C. So we may assume that 3 → u, if δ = 2 then if 4 → x 2 then u3x 1 2x 2 41vu ≡ C, so x 2 → 4 and thus u → 2 and x 1 → u since otherwise we are in the case (i) or (ii). If x 2 → 3 then x 1 → u and thus 3x 2 ux 1 142v3 ≡ C and of 3 → x 2 then ux 1 43x 2 12vu ≡ C. So δ = 4, suppose that x 4 → x 1 , similarly we may assume that {1, 2} → x 2 , also x 3 → 4 and thus 4 → {x 2 , x 4 }. If

x 3 → x 1 then x 1 → u and x 2 x 4 x 1 x 3 or x 4 x 2 x 1 x 3 is an antidirected inpath with d - T 2 (x 2 ) ≥ d - T 2 (x 1 ) and d - T 2 (x 4 ) ≥ d - T 2 (x 1
), a contradiction, so x 1 → x 3 . Similarly we may suppose that x 4 → x 2 . We have

x 1 x 3 x 4 x 2 12vu34x 1 ≡ C, so x 1 → x 4 . If u → x 3 then 31x 4 2x 1 x 2 x 3 uv3 ≡ C, so x 3 → u. Suppose that u → x 1 . If there exists u ∈ N + T 2 (x 4
) such that x 1 has an inneighbor in T 2 -u of outdegree at least 2 in T 2 -u then T contains C, so we may assume that {1, 2, 3, 4} → x 4 . Similarly, by taking x 4 in the place of v, we may assume that x 2 → 4, so 4 → x 3 and thus 1ux 1 x 2 x 3 4x 4 23v1 ≡ C. So x 1 → u and then u → {2, x 2 } and x 3 → x 1 . And thus T 1 cannot contains a vertex of outdegree 0 in T 1 , so without loss of generality, if there exists x ∈ T such that d -(x) = 4 and T (N -(x)) contains a vertex of outdegree 0 then T contains C, by duality, if d + (x) = 4 and T (N + (x)) contains a vertex of indegree 0 then T contains C, so we may assume that x 2 → 4 and thus 4 → {x 3 , x 4 }. We have u3x 1 x 2 x 3 4x 4 12vu ≡ C.

Exc 20: One can check that for any i ∈ {3, 4, 5, 6}, 1 or 2 is an origin of an antidirected outpath P in T 2 \ {i, i + 1, u}, so if there exists i ∈ N -(u) ∩ {3, 4, 5, 6} then x 1 ...x δ ui(i + 1)vP -1 x 1 ≡ C, thus u → {3, 4, 5, 6}. If there exists u ∈ N + T 2 -u (x δ ), 1 or 2 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x δ u vP -1 x 1 ≡ C, so N + T 2 (x δ ) = {u}. There exist w ∈ N - T 2 (x δ-1 ), w ∈ N + T 2 (w) and a ∈ {1, 2} pairwaisly distinct such that x δ has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x δ ax 1 ...x δ-1 ww vP -1 x δ ≡ C. Exc 22: Without loss of generality we can suppose that {1, 2} ⊆ N -(x 1 ). One can check that for any i ∈ {4, 5, 6}, there exists an outneighbor j of i such that 1 or 2 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u) ∩ {1, 2, 3, 4, 5} then x 1 ...x δ uijvP -1 x 1 ≡ C, thus u → {4, 5, 6}. If there exists u ∈ N + T 2 -u (x δ ) distinct from 4, 1 or 2 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x δ u vP -1 x 1 ≡ C, and if 4 ∈ N + (x δ ) then x 1 ...x δ 4536v21x 1 ≡ C, so N + T 2 (x δ ) = {u}. There exist w ∈ N - T 2 (x δ-1 ), w ∈ N + T 2 (w) and a ∈ {4, 6} pairwaisly distinct such that x δ has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x δ ax 1 ...x δ-1 ww vP -1 x δ ≡ C. Exc 23: One can check that for any i ∈ {1, 2, 3, 4, 5}, there exists an outneighbor j of i such that 4 or 6 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u) ∩ {1, 2, 3, 4, 5} then x 1 ...x δ uijvP -1 x 1 ≡ C, thus u → {1, 2, 3, 4, 5}. If there exists u ∈ N + T 2 -u (x δ ), 4 or 6 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x δ u vP -1 x 1 ≡ C, so N + T 2 (x δ ) = {u}. There exist w ∈ N - T 2 (x δ-1 ), w ∈ N + T 2 (w) and a ∈ {4, 6} pairwaisly distinct such that x δ has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x δ ax 1 ...x δ-1 ww vP -1 x δ ≡ C. Exc 24: Without loss of generality we may suppose that 1 ∈ N -(x 1 ). One can check that for any i ∈ {2, 4, 5, 6}, there exists an outneighbor j of i such that 1 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u) ∩ {2, 4, 5, 6} then x 1 ...x δ uijvP -1 x 1 ≡ C, thus u → {2, 4, 5, 6}. If 3 → u then x 1 ...x δ u3456v21x 1 ≡ C, so u → 3. If there exists u ∈ N + T 2 -u (x δ ), 1 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x δ u vP -1 x 1 ≡ C, so N + T 2 (x δ ) = {u}. There exist w ∈ N - T 2 (x δ-1 ), w ∈ N + T 2 (w) and a ∈ N - T 2 (x 1 ) ∩ N - T 2 (x δ ) pairwaisly distinct such that x δ has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x δ ax 1 ...x δ-1 ww vP -1 x δ ≡ C. Exc 29 :One can check that for any i ∈ {2, 3, 5}, there exists an outneighbor j of i such that 4 or 6 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists

i ∈ N -(u)∩{2, 3, 5} then x 1 ...x δ uijvP -1 x 1 ≡ C, thus u → {2, 3, 5}. Similarly if 1 ∈ N -(u) then x 1 ...x δ u1453v26x 1 ≡ C, thus u → 1. If there exists u ∈ N +
T 2 -u (x δ ), 4 or 6 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x δ u vP -1 x 1 ≡ C, so N + T 2 (x δ ) = {u}. There exist w ∈ N - T 2 (x δ-1 ), w ∈ N + T 2 (w) and a ∈ {4, 6} pairwaisly distinct such that x δ has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x δ ax 1 ...x δ-1 ww vP -1 x δ ≡ C. Exc 30: One can check that for any i ∈ {3, 4, 5}, there exists an outneighbor j of i such that 4 or 6 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u) ∩ {3, 4, 5} then x 1 ...x δ uijvP -1 x 1 ≡ C, thus u → {3, 4, 5}. Similarly if there exists

i ∈ N -(u) ∩ {1, 2, 6}, set {u 1 → u 2 } = T (1, 2, 6) -i, we have x 1 ...x δ i5u 1 u 2 v43x 1 ≡ C, thus u → {1, 2, 6}. If there exists u ∈ N +
T 2 -u (x δ ), 4 or 6 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x δ u vP -1 x 1 ≡ C, so N + T 2 (x δ ) = {u}. There exist w ∈ N - T 2 (x δ-1 ), w ∈ N + T 2 (w) and a ∈ {4, 6} pairwaisly distinct such that x δ has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x δ ax 1 ...x δ-1 ww vP -1 x δ ≡ C. Otherwise, d - T 2 -u (x 1 ) ≤ 1. If δ ≥ 4 then, by lemma 2.6, T 1 ∈ {4A, 4B}. If T 1 = 4B, taking x 3 x 2 x 4 x 1 as antidirected inpath in T 1 with u ∈ N + T 2 (x 1 ) such that d + T 2 -u (x 3 ) ≥ 2 the result follows. So we may suppose that T 1 = 4A ≡ 4A, N - T 2 (x 1 ) = {u, w} and {u} ⊆ N + T 2 (x δ ) ⊆ {u, w}. If there exists x ∈ N + T 2 (x 1 ) ∩ N + T 2 (x 2 ) then x 1 has an outneighbor origin of an antidirected inpath P in T 2 \{u, x} and thus

x 1 xx 2 x 3 x 4 uvP -1 x 1 ≡ C unless if T 2 \{u, x} = 3A, set T 2 \ {u, x} = {w, w 1 , w 2 }, if w → x then xx 2 x 3 x 4 uvw 2 x 1 w 1 wx ≡ C so x → w. If x 2 → w 1 , without loss of generality suppose that w 2 → x, then w 1 x 2 x 3 x 4 uvww 2 xx 1 w 1 ≡ C, so w 1 → x 2 . If w → u then ux 4 x 3 x 2 xvw 2 x 1 w 1 wu ≡ C, so u → w. If w 2 → x then x 2 x 3 x 4 ux 1 w 1 vww 2 xx 2 ≡ C so x → w 2 and thus w 1 → x (Since otherwise w is an origin of an antidirected outpath in T 2 -u). If x 2 → w 2 then x 2 x 3 x 4 ux 1 w 1 vwxw 2 xx 2 ≡ C so w 2 → x 2 and if x 2 → w then x 2 x 3 x 4 ux 1 xw 1 w 2 vwx 2 ≡ C so w → x 2 . If x 4 → w then T (N -(x 4 )) = T (x 1 , w 1 , w 2 , x) = 4A so T contains C, thus w → x 4 . If x 2 → u then T (N + (x 4 )) = 4A so u → x 2 . If x 3 → x then T (N + (x 2 )) = 4A so x → x 3 . If u → w 2 then x 1 x 2 x 3 xx 4 ww 1 vw 2 ux 1 ≡ C and if w 2 → u then x 1 x 2 x 3 xx 4 w 2 uvw 1 wx 1 ≡ C. So we may assume that N + T 2 (x 1 ) → x 2 . If x 2 →
w, since w is not an origin of an antidirected Hamiltonian outpath in T 2 -u then if w is an origin of an antidirected inpath P of order 3 in T 2 -u, let Q be an antidirected Hamiltonian inpath of T 2 \ ({u} ∪ V (P )), we have x 2 x 3 x 4 uvQ -1 x 1 P -1 x 2 ≡ C, and if not then w → (T 2 \ {u, w}) and T 2 \ = 3A, 5A or 7A, since {x 3 , u} ⊆ N + (x 4 ) ∩ N -(x 1 ) then there exists u 1 u 2 u 3 u 4 ≡ x 1 x 2 x 3 x 4 in T (x 1 , x 3 , x 4 , u) such that u 1 = x 1 and u 4 = x 4 , x 2 has an inneighbor origin of an antidirected outpath P in T 2 \{u, w} of order |T 2 |-3, let {w } = V (T 2 )-(V (P )∪{u, w}), thus u 1 u 2 u 3 u 4 x 2 P vw wx 1 ≡ C, so w → x 2 and thus N + T 2 (x 2 ) = {u}. So without loss of generality we may assume that N + T 2 (x 4 ) = {u} since otherwise we can take x 4 x 3 x 2 ≡ x 2 x 3 x 4 and we find C. If d - T 2 (u) = 0 then N -(u) = {x 2 , x 3 , x 4 , v} with T (N -(u)) = 4A and thus T contains C so we may assume that d -

T 2 (u) ≥ 1. If d + T 2 -u (x 3 ) ≥ 2, let u ∈ N - T 2 (u), w ∈ N + T 2 -u (x 3
) and P be an antidirected outpath in T 2 \ {u , w }, we have ux

2 x 1 x 3 w vP -1 u u ≡ C. Otherwise, x 3 → u so d - T 2 -u (x 3 ) ≥ 2, let u ∈ N + T 2 (u), w ∈ N - T 2 -u (x 3
) and P be an antidirected outpath in

T 2 \ {u , w }, we have x 3 x 2 x 1 uu vP -1 w x 3 ≡ C. Now if δ = 2, if d - T 2 -u (x 1 ) = 1 set {w} = N - T 2 -u (x 1 ). Suppose that d + T 2 -w (x 2 ) ≥ 2, then |T 2 | ≥ 5 since otherwise d -(x 2 ) ≤ 1, let u ∈ N + T 2 -w (x 2
) distinct from u, we may assume that x 1 → u. If x 1 has an outneighbor origin of an antidirected inpath P in T 2 \ {u, u } then x 1 u x 2 uvP -1 x 1 ≡ C, otherwise, (T 2 \ {u, u }; - (1, 1, ..., 1)) is a Grünbaum's exception and d + T 2 (x 2 ) = 2. Since w is not an origin of an antidirected outpath in T 2 -u then there exists w ∈ N - T 2 -u (u ), let P be an antidirected outpath in T 2 \ {u, u , w }, we have x 2 w u x 1 uvP -1 x 2 ≡ C. So we may assume that N + T 2 -w (x 2 ) = {u}. If |T 2 | ≥ 7, one can find u 1 , u 2 , u 3 ∈ T 2 \ {u, w} such that T 2 \ {w, u, u 1 , u 2 , u 3 } is not 4A nor 6H, thus x 2 has an inneighbor origin of an antidirected outpath P in T 2 \ {w, u, u 1 , u 2 , u 3 }, suppose that u 3 → u 2 , therefore x 1 u 2 u 3 x 2 P vu 1 x 1 ≡ C. So we may suppose that |T 2 | ∈ {3, 5} and we have 3 cases:

(i) If T 2 \{u, w} is a transitive triangle, set T 2 = T 2 \{u, w} = {u 1 , u 2 , u 3 } such that u i → u i+1 for i ∈ {1, 2}. Obviously {u 2 , u 3 } → w. If x 2 → w then x 2 wu 2 u 3 x 1 u 1 vux 2 ≡ C, so w → x 2 . If w → u 1 , if x 1 → u then x 2 wu 1 vux 1 u 3 u 2 x 2 ≡ C so u → x 1 , if u has an outneighbor u i in T 2 then x 1 uu i vu i+2 u i+1 x 2 wx 1 ≡ C so T 2 → u and thus x 1 x 2 uu 2 u 3 vu 1 wx 1 ≡ C; so u 1 → w. If u has an inneighbor u i in T 2 then x 1 vx 2 uu i wu i+1 u i+2 x 1 ≡ C, so u → T 2 . If x 1 → u then ux 1 u 2 u 1 x 2 u 3 wvu ≡ C, so u → x 1 and thus u 1 ux 1 wx 2 u 2 u 3 vu 1 ≡ C. (ii) If T 2 \ {u, w} is a circuit triangle, set T 2 = T 2 \ {u, w} = {u 1 , u 2 , u 3 }. If x 2 → w, if w has an inneighbor u i in T 2 then x 2 wu i u i+1 x 1 u i+2 vux 2 ≡ C, and if w → T 2 , if d - T 2 (u) ≥ 1, say u 1 → u, then x 1 x 2 uu 1 u 2 vu 3 wx 1 ≡ C and if d - T 2
(u) = 0 then (T ; C) = A 10 , so we may assume that w → x 2 . If x 1 → u, if T 2 → w then ux 1 u 2 u 1 x 2 u 3 wvu ≡ C and otherwise, by taking x 1 in the place of v one can find C; so u → x 1 . If u has an outneighbor u i in T 2 then u i ux 1 wx 2 u i+1 u i+2 vu i ≡ C, so T 2 → u. If w has an outneighbor u i in T 2 then u i wx 1 x 2 uu i+1 u i+2 vu i ≡ C, so T 2 → w. We have

x 1 vx 2 uu 1 wu 2 u 3 x 1 ≡ C. (iii) If T 2 \ {u, w} = {z}. We have T -u = 5A then (T ; C) = A 5 . If d - T 2 -u (x 1 ) = 0 i.e. N - T 2 (x 1 ) = N + T 2 (x 2 ) = {u}, if there exists u 1 , u 2 , u 3 , u 4 ∈ T 2 -u such that u 4 is an origin of an antidirected outpath P in T 2 \ {u 1 , u 2 , u 3 }, suppose that u 3 → u 2 , then u 1 x 1 u 2 u 3 x 2 P vu 1 ≡ C. Otherwise, if |T 2 | = 5,
. If d + T 2 (x 1 ) ≥ 3, let {u 1 , u 2 , u 3 } ⊆ N + T 2 (x 1
) such that u 2 → u 3 and let P be an antidirected outpath in

T 2 \ {u 1 , u 2 , u 3 }, we have u 1 x 1 u 3 u 2 x 2 P vu 1 ≡ C, so d + T 2 (x 1 ) ≤ 2. We have d - T 2 (x 1 ) ≥ 3, let w 1 , w 2 , w 3 ∈ N - T 2 (x 1
) such thatw 2 → w 3 and let P be an antidirected outpath in T 2 \ {w 1 , w 2 , w 3 }, we have w 3 w 2 x 1 w 1 x 2 P vw 3 ≡ C.

Case 4: If δ ≥ 3, δ is odd and, without loss of generality, T 1 = 3A, 5A, 7A. Let x 1 ...x δ be a Hamiltonian antidirected outpath of T 1 such that x δ → x 1 and d - T 2 (x 2 ) = max {d - T 2 (y 2 ); y 1 ...y δ is an antidirected outpath in T 1 with y δ → y 1 }. We distinguish two cases:

(a) If d + T 2 (x 1 ) ≥ 3. Suppose first that d + T 2 (x δ ) ≥ 1 and let u ∈ N + T 2 (x δ ). Obviously d + T 2 -u (x 1 )
≥ 2 so there is an outneighbor of x 1 origin of an antidirected inpath P in T 2 -u and thus x

1 ...x δ uvP -1 x 1 ≡ C; unless if (T 2 -u; -(1, 1, ..., 1
)) is one of the following biexceptions with respect to x 1 : Dual Exc (4, 1): If there exists u ∈ N + T 2 -u (x 1 ) such that u is an origin of an antidirected inpath P in T 2 -u then u x 1 ...x δ P vu ≡ C, so we may assume that u → {1, 5}. If 3 → x 2 , 3 is an origin of an antidirected outpath P in T 2 \ {4, 5} then 45x 1 x δ ...x 2 P v4 ≡ C. Otherwise, 4 → x 2 and we may assume that u → 3, 3 is an origin of an antidirected outpath P in T 2 \ {2, 4} so 24x 2 ...x δ x 1 P v2 ≡ C. Dual Exc [START_REF] Havet | Oriented Hamiltonian cycles in tournaments[END_REF]1) 4): If 2 → u then 4x 1 ...x δ u21v354 ≡ C and if 5 → u then 4x 1 ...x δ u53v124 ≡ C, so u → {2, 5} and thus 34x 2 ...x δ x 1 15u2v3 ≡ C. Dual Exc (33, 1): In 7A, any vertex i is an origin of a path

: If 2 → u then 5x 1 ...x δ u21v345 ≡ C so u → 2. If 3 → u then 5x 1 ...x δ u3124v5 ≡ C and if 4 → u then 2x 1 ...x δ u4351v2 ≡ C, so u → {3, 4}. If 1 → x 2 , if u → 5 then 23x 1 x δ ...x 2 14u5v2 ≡ C and if 5 → u then 41x 2 ...x δ u5x 1 32v4 ≡ C. And if 5 → x 2 then 15x 2 ...x δ x 1 32u4v1 ≡ C. Dual Exc (9, 1)(1): If 1 → u then 2x 1 ...x δ u2453v2 ≡ C, if 2 → u then 4x 1 ...x δ u21v354 ≡ C, if 3 → u then 4x 1 ...x δ u31v524 ≡ C, if 4 → u then 5x 1 ...x δ u43v125 ≡ C and if 5 → u then 4x 1 ...x δ u53v124 ≡ C, so u → {1, 2, 3, 4, 5}. If 1 or 2 → x 2 , say 1, then 23x 1 x δ ...x 2 14u5v2 ≡ C and if 4 → x 2 then 51x 1 x δ ...x 2 43u2v5 ≡ C. Dual Exc (9, 1)(2): If 4 → u then 5x 1 ...x δ u43v125 ≡ C and if 5 → u then 2x 1 ...x δ u54v132 ≡ C, so u → {4, 5} and thus 23x 1 x δ ...x 2 14u5v2 ≡ C. Dual Exc (9, 1)(3), (
P i = -(1, 1, 1, 1, 2) of end i + 1, so if there exists i ∈ {1, 4, 6, 7} such that i → u then (i + 1)x 1 ...x δ uP * i v(i + 1) ≡ C, so u → {1, 4, 6, 7}. Thus 62x 2 ...x δ x 1 4537v1u6 ≡ C. Dual Exc (E 8 (5), 1): Set X = {u 1 , u 2 }. If 1 → u then u 1 x 1 ...x δ u13u 2 2vu 1 ≡ C so u → {1, 2, 3}. Without loss of generality suppose that u 1 → x 2 and u 2 → u, we have 31x 1 x δ ...x 2 u 1 2u 2 uv3 ≡ C. Or (T 2 -u; (1, 1, .., 1)) is an exception with N - T 2 (x 2 ) = N + T 2 (x δ ) = {u},
so T 1 is a transitive triangle (easy to prove) and N + T 2 (x δ ) = {u}. If u is an origin of an antidirected inpath P in T 2 -z for a z ∈ N + (x 1 ) then zx 1 x 2 x 3 P vz ≡ C, otherwise, let u 1 , u 2 ∈ N + T 2 -u (x 1 ) such that u 1 → u 2 , if T 2 \ {u 1 , u 2 } contains an antidirected outpath P of origin and end in T 2 -u then u 2 x 1 vx 2 P -1 x 3 u 1 u 2 ≡ C, otherwise, T 2 \ {u 1 , u 2 } = 4A and y 1 ...y 4 is the sole antidirected outpath in T 2 \ {u 1 , u 2 }. If u = y 4 , since u is not an origin of an antidirected inpath in T 2 -u 1 then u 1 → y 1 and thus u

2 x 1 x 2 x 3 y 4 y 3 y 2 vy 1 u 1 u 2 ≡ C. Else, u = y 1 , since u 2 u 1 y 4 y 3 y 2 is not antidi- rected y 4 → u 1 . If y 3 → u 2 then u 2 x 1 x 2 x 3 y 1 y 4 u 1 vy 2 y 3 u 2 ≡ C so u 2 → y 3 and thus u 1 x 1 vx 2 y 2 y 1 y 3 u 2 x 3 y 4 u 1 ≡ C. Assume now that N -(x δ ) = T 2 , let a ∈ N + T 2 (x 1 ), b ∈ N - T 2 (x δ-1
) and P be an antidirected oupath in T 2 \ {a, b}, we have ax

1 ...x δ-1 bx δ P va ≡ C. (b) If d + T 2 (x 1 ) ≤ 2 i.e. d - T 2 (x 1 ) ≥ |T 2 | -2. Let u ∈ N - T 2 (x 2 ), if d + T 2 (u) = 0 and N - T 2 (x 2 ) = {u}, let a ∈ N - T 2 -u (x 1
) and P be an antidirected inpath in T 2 \ {a, u}, we have x 1 x δ ...x 3 vx 2 P uax 1 ≡ C. So we may suppose that there exists w ∈ N + T 2 (u), if x 1 has an inneighbor y origin of an antidirected outpath P in T 2 \ {u, w} then yx 1 x δ ...x 2 uwvP -1 ≡ C, otherwise, either (T 2 \ {u, w}; (1, 1, ..., 1)) is one of the following exceptions: Exc

1: If V (T 2 \ {u, w}) = {v 1 , v 2 , v 3 , v 4 } such that v 2 ∈ N -(x 1 ) ∩ N -(v 1 ) and v 4 ∈ N -(w) ∩ N -(v 3 ) then x 1 x δ ...x 2 uwv 3 v 4 vv 1 v 2 x 1 ≡ C. So we may assume that w → 4. Suppose first that {1, 2} ⊆ N -(x 1 ), so w → 3. If u → 1 or 2, say 1, then x 1 x δ ...x 2 u1v4w32x 1 ≡ C, so {1, 2} → u. If x 2 → 4 then x 1 x δ ...x 3 vx 2 4w32u1x 1 ≡ C so 4 → x 2 .
Without loss of generality suppose that u → 4, since 1 is an origin of an antidirected outpath P in T 2 \ {u, 4}, then x 1 x δ ..x 2 u4vP -1 x 1 ≡ C. So we may assume that |N -(x 1 ) ∩ {1, 2}| = 1, and thus {u, w, 3} ⊆ N -(x 1 ). Suppose that 1 ∈ N

-(x 1 ), so w → {1, 3}, if u → 2 then x 1 x δ ...x 2 u2v4w31x 1 ≡ C, so 2 → u. If u → 3 then x 1 x δ ...x 2 u3v241wx 1 ≡ C, so 3 → u. If u → 4 then x 1 x δ ...x 2 u4v213wx 1 ≡ C, so 4 → u. If 1 → x 2 then x 1 x δ ...x 2 12vu34wx 1 ≡ C and if x 2 → 1 then x 1 x δ ...x 3 vx 2 14u23wx 1 ≡ C.
So we may assume that N + T 2 (x 1 ) = {1, 4} and thus 4 → {1, 4}. If u → 2, without loss of generality suppose that w → 3, then

x 1 x δ ...x 2 u241v3wx 1 ≡ C, so 2 → u. If u → 3 then x 1 x δ ...x 2 u3v241wx 1 ≡ C, so 3 → u. If 1 → x 2 , since d - T 2 \{1,2} (4) ≥ 2 then T 2 \ {1, 2} contains an antidirected outpath of origin distinct from 4, thus x 1 x δ ...x 2 12vP -1 x 1 ≡ C and if x 2 → 1 then x 1 x δ ...x 3 vx 2 1w43u2x 1 ≡ C. Exc 24: If d - T 2 (w) ≥ 2,
one can find and inneighbor u 1 of w in T 2 -u having an outneighbor u 2 in T 2 \{u, w} such that T 2 \{u, w, u 1 , u 2 } contains an antidirected outpath P of origin an inneighbor of x 1 and thus x 1 x δ ...x 2 uwu 1 u 2 vP -1 x 1 ≡ C, so we may assume that d

- T 2 (w) = 1. If d + T 2 (u) ≥ 2, let w ∈ N + T 2 -w (u), T 2
\{u, w } contains an antidirected outpath P of origin an inneighbor of x 1 and thus

x 1 x δ ...x 2 uw vP -1 x 1 ≡ C, so d + T 2 (u) = 1. We may suppose that x 2 → (T 2 -u), let a ∈ N - T 2 \{u,w} (x 1 ), b ∈ N +
T 2 \{u,w} (x 2 ) and P be an antidirected inpath in T 2 \ {u, w, a, b}, we have

x 1 x δ ...x 2 bwP uax 1 ≡ C. Or T 2 \ {u, w} = {u 1 → u 2 } such that u 1 ∈ N + (x 1
). Suppose first that u → u 1 , then w → u 2 and x 1 → w. And since d + (u) = 4 then x 3 → u and thus, without loss of generality suppose that w → u 1 , x 1 x 2 x 3 uvu 2 wu 1 x 1 ≡ C. So we may assume that u 1 → u. If u 1 → x 2 , if u → x 1 then x 1 x 3 x 2 u 1 u 2 vwux 1 ≡ C and if w → x 1 then, without loss of generality suppose that w → u 2 , we have x 1 x 3 x 2 u 1 uvu 2 wx 1 ≡ C, so we can assume that x 2 → u 1 . If T 2 → x 3 , without loss of generality we may assume that u 2 → w, thus x 1 x 2 ux 3 u 2 wvu 1 x 1 ≡ C. So we may assume that d

+ T 2 (x 3 ) = 1. Suppose that x 3 → w, if u → u 2 then x 1 x 2 x 3 wuu 2 vu 1 x 1 ≡ C, so u 2 → u. If x 1 → u then x 1 x 2 x 3 wvu 2 u 1 ux 1 ≡ C, so u → x 1 . If u 2 → w then x 1 x 2 x 3 wu 2 uvu 1 x 1 ≡ C, so w → u 2 . If x 1 → w then u 2 u 1 x 3 ux 2 x 1 wvu 2 ≡ C, so w → x 1 .
If u 1 → w then there exists an outneighbor of x 2 in {u 2 , w} origin of an antidirected inpath P of end in {u, w, u 2 } and thus x 1 x 3 vx 2 P x 1 ≡ C, so w → u 1 and thus x 2 → {u 2 , w}. Therefore uu 2 x 1 x 3 wx 2 u 1 vu ≡ C, so we can assume that w → x 3 . Suppose that

x 3 → u 1 , if u 2 → x 2 then wux 2 u 2 x 1 x 3 u 1 vw ≡ C, so x 2 → u 2 . If u → x 1 then x 2 u 2 u 1 wux 1 x 3 vx 2 ≡ C, so x 1 → u and thus w → x 1 . If x 2 → w then w → u 2 and u 2 → u, so x 1 x 3 vx 2 wu 1 uu 2 x 1 ≡ C, so w → x 2 . If w → u 2 then x 1 vx 3 x 2 wu 2 u 1 ux 1 ≡ C, so u 2 → w. If u 2 → u then x 1 x 2 x 3 u 1 vwu 2 ux 1 ≡ C, so u → u 2 . Therefore u 2 ux 2 wx 1 x 3 u 1 vu 2 ≡ C. Now suppose that x 3 → u 2 , if x 1 → w then x 1 x 2 x 3 u 2 u 2 uvwx 1 ≡ C, so w → x 1 . If u 1 → w then x 2 x 1 vx 3 u 2 u 1 wux 3 ≡ C, so w → u 1 then x 2 → {w, u 2 } and thus u 2 → w; if u → x 1 then wux 1 x 3 u 2 x 2 u 1 vw ≡ C, so x 1 → u and then T -v = 7A, therefore (T ; C) = A 6 . Suppose now that x 3 → u. We have 2 cases: (i) If u 1 → w. If x 1 → w then x 1 x 2 x 3 uvu 2 u 1 wx 1 ≡ C, so w → x 1 . Suppose that u 2 → u then u → x 1 ; if u 2 → w then x 1 x 2 x 3 uu 2 wvu 1 x 1 ≡ C, so w → u 2 . If x 2 → u 2 then x 1 x 3 vx 2 u 2 u 1 wux 1 ≡ C, so u 2 →
→ u 2 , if u → x 1 then x 2 vx 3 x 1 uwu 1 u 2 x 2 ≡ C and if x 1 → u then wx 3 u 1 u 2 vux 1 x 2 w ≡ C. (ii) If w → u 1 . If w → u 2 then x 1 x 2 x 3 uvu 2 wu 1 x 1 ≡ C, so u 2 → w. If u 2 → u then x 1 x 2 x 3 uu 2 wvu 1 x 1 ≡ C, so u → u 2 . And thus x 1 → w, therefore x 1 x 2 x 3 uu 1 u 2 vwx 1 ≡ C. Case 5: If δ = 3 and for all x ∈ T such that d -(x) = 3 we have T (N -(x)) = 3A. Set V (T 1 ) = {y 1 , y 2 , y 3 }. If d + T 1 (x) ≥ 2 for any x ∈ T 2 . If |T 2 | = 4 then d + T 2 (x) ≤ 2
for all x ∈ T 2 , and so T 2 = 4A. Also, d + T 1 (x) = 1 for all x ∈ T 2 . Without loss of generality we may suppose that y 1 → 3. If 4 → y 3 then y 1 vy 2 y 3 4213y 1 ≡ C and if

y 3 → 4 then 4 → {y 1 , y 2 }. Since T (N + (2)) = 3A then N + (2) = {3, y 1 , y 3 }. Thus T -1 = 7A and (T ; C) = A 6 . Now if |T 2 | ≥ 6, without loss of generality suppose that d + T 2 (y 1 ) = max{d + T 2 (y i ), i ∈ {1, 2, 3}}. Let a ∈ N - T 2 (y 2 ) ∩ N - T 2 (y 3 ) and set T 1 = T 1 ∪ {a} and T 2 = T 2 -a. We have x 1 x 2 x 3 x 4 = y 3 ay 2 y 1 = -(1, 1, 1). If d + T 2 (x 4 ) = 0 then there exists a ∈ T 2 such that d + T 1 (a ) = 3, by taking x 1 a x 3 x 4 ≡ x 1 ...x 4 we obtain d + T 2 -a (x 4 ) = 1. So we may suppose that there exists u ∈ N + T 2 (x 4 ) with d - T 2 -u (x 1 ) ≥ 2. In fact, otherwise, d + T 2 (x 1 ) ≥ |T 2 | -3 with N + T 2 (x 1 ) ⊆ N - T 2 (x 4 ), thus d - T 2 (x 4 ) ≥ |T 2 | -3 and so d + T 2 (x 4 ) ≤ |T 2 | -(|T 2 | -3), then |T 2 | -3 ≤ 3, but |T 2 | ≥ 6, so |T 2 | = 6; let a , u ∈ N + T 2 (x 1 ), we have x 3 a x 4 x 1 = -(1, 1, 1) with d - T 2 \{a ,u } (x 3 ) ≥ 2.
So if x 1 has an inneighbor origin of an antidirected outpath P in T 2 -u then x 1 ...x 4 uvP -1 x 1 ≡ C. Otherwise, (T 2 -u; (1, 1, ..., 1)) is an exception, and it will be treated below. Otherwise, there exist y ∈ T 2 such that d + T 1 (y) ≤ 1, suppose that, without loss of generality, {y 1 , y 3 } → y. Set T 1 = T 1 ∪ {y} and T 2 = T 2 -y. We have

x 1 ...x 4 = y 2 y 1 yy 3 = -(1, 1, 1). Since T (x 2 , x 3 , v) = 3A then d + T 2 (x 4 ) ≥ 1, so let u ∈ N + T 2 (x 4 ). If x 1 has an inneighbor origin of an antidirected outpath P in T 2 -u then x 1 ...x 4 uvP -1 x 1 ≡ C. Otherwise, if d - T 2 -u (x 1 ) ≤ 1, if |T 2 | = 4, set T 2 \ {y, u} = {w 1 → w 2 }. We have T (N -(y 3 )) = T (y 2 , w 1 , w 2 ) = 3A. If y 1 → w 1 then T (y 3 , w 2 , u) = 3A and so u → w 2 . Since T (w 1 , u, v) = 3A then y → w 2 .
Similarly, we find that T -v = 7A and (T ; C) = A 6 . So we may suppose that w 1 → y 1 . Similarly, we find that T -v = 7A and (T ; C) = A 6 . Now if |T 2 | ≥ 6, two cases may arise:

(i) If N - T 2 -u (x 1 ) = {w}. Suppose first that x 1 → u, then N -(x 1 ) = {w, x 2 , x 3 } and thus w ∈ N -(x 2 ). We have x 2 x 4 ux 1 = -(1, 1, 1), if d - T 2 -u (x 2 ) ≥ 2 then there exists u ∈ N + T 2 -u (x 1
) such that d - T 2 \{u,u } (x 2 ) ≥ 2 and the result follows, so N - T 2 (x 2 ) = {u, w} and thus w ∈ N -(x 4 ). We have T 2 \{u, w} ⊆ N + (x 1 )∩N + (x 2 ) with |T 2 \{u, w}| ≥ 3 and d - T 2 \{u,w} (x 4 ) ≥ 1 so we may find u" ∈ T 2 \ {u, w} such that d - T 2 \{u",x 3 } (x 4 ) ≥ 2 with x 4 x 1 u"x 2 = -(1, 1, 1) and the result follows. So we may assume that u → x 1 , and so N +

T 2 (x 1 ) ⊆ N -(x 4 ). Set T + = T (N + T 2 (x 1 )
). If x 2 has an outneighbor a ∈ T + then x 4 x 1 ax 2 = -(1, 1, 1) with x 2 → x 3 and d - T 2 \{a,x 3 } (x 4 ) ≥ 2, so the result follows. Thus T + ⊆ N -(x 2 ). If there exists an antidirected outpath P in T + such that x 1 has an outneighbor origin of an antidirecrted inpath P of order |T + | -1 in T 2 V (P ) then x 3 x 4 x 2 u 1 u 2 x 1 P vx 3 ≡ C. Otherwise, suppose that u has an outneighbor u 1 ∈ T + , then T (u, u 1 , w) = 3A and thus T + → w and w → u. If u 1 has an inneighbor u 2 ∈ T + , let P be an antidirected outpath in

T 2 \ {u, u 1 , u 2 }, then x 3 x 2 x 1 uu 1 u 2 x 4 P vx 3 ≡ C. Else, N + T + (u) = {u 1 }, if |T + | = 5 or 7, let P be an antidirected outpath in T + -u 1 of order |T + | -3, there exists and antidirecte outpath Q in T 2 -V (P ) of origin in T + , thus x 3 x 4 x 2 P x 1 Qvx 3 ≡ C. If |T + | = 3, set {u 2 → u 3 } = T + -u 1 , if u → x 2 then x 3 x 4 x 2 uu 1 x 1 u 3 u 2 wvx 3 ≡ C, and if x 2 → u, x 3 has an outneighbor u i distinct from u 1 , in fact, d + T 2 (x 3 ) ≥ 2, so if {u 1 , w} ⊆ N + (x 3 ) then ux 4 x 2 u 2 u 3 x 1 u 1 x 3 wvu ≡ C and if N + T 2 (x 3 ) = {u, w} or N + T 2 (x 3 ) = {u, u 1 } then T (N + (x 3 )) = 3A
, so without loss of generality we may suppose that x 3 → u 2 and thus ux 2 x 1 x 3 u 2 u 1 x 4 u 3 wvu ≡ C. So we can suppose that T + → u, if w has an outneighbor u 1 ∈ T + , let P be an antidirected outpath in T + -u 1 , we have x 1 ...x 4 uP vu 1 wx 1 ≡ C, so T + → w. Since T (x 1 , x 2 , x 3 ) = 3A then u → w. If |T + | = 5 or 7, let P be an antidirected outpath in

T + of order |T + | -3, there exists an antidirected outpath Q in T 2 -V (P ) of origin in T + , thus x 3 x 4 x 2 P x 1 Qvx 3 ≡ C. Now if |T + | = 3, set V (T + ) = {u 1 , u 2 , u 3 } such that u 1 → u 2 , since T (w, x 1 , x 3 ) = 3A then x 2 ∈ N + (u), thus x 3 x 4 x 2 uwu 1 u 2 x 1 u 3 vx 3 ≡ C.
(ii) If N -(x 1 ) = {x 2 , x 3 , u}, then x 3 → u and u → x 2 . We may assume that d +

i) If {1, 2} ⊆ N -(x 1 ) (
, by the building lemmas we can suppose that {2, 4} →

x 2 . If N -(u) ∩ {3, 4} = ∅ one can find C, so u → {3, 4}. So by supposing that N + (x 4 ) ∩ {1, 2, 3, 4} = ∅ we can find C, so {1, 2, 3, 4} → x 4 . Thus d - T 2 (x 3 ) ≥ 3. If x 4 → x 2 then x 1 ux 3 x 4 x 2 42v31x 1 ≡ C and if x 2 → x 4 , since T 2 -4 = 4A and d - T 2 -4 (x 3
) ≥ 2 then x 3 has an inneighbor origin of an antidirected outpath P in T 2 -4 and thus

x 3 x 2 x 4 x 1 4vP -1 x 3 ≡ C. (ii) If N - T 2 (x 1 ) = {1, 3}
, by the building lemmas we can suppose that {2, 3, 4} → x 2 . By supposing that there exists i ∈ {1, 3, 4} ∩ N -(u) one can find C, so we may suppose that u → {1, 3, 4}. If

x 4 → 2 then x 1 ...x 4 2v4u31x 1 ≡ C, if x 4 → 3 then x 1 vx 2 x 3 x 4 3u142x 1 ≡ C and if x 4 → 4 then x 1 vx 2 x 3 x 4 4u312x 1 ≡ C, so {2, 3, 4} → x 4 . If x 4 → x 2 then x 1 ux 3 x 4 x 2 42v31x 1 ≡ C and if x 2 → x 4 , since T 2 -4 = 4A and d - T 2 -4 (x 3
) ≥ 2 then x 3 has an inneighbor origin of an antidirected outpath P in T 2 -4 and thus x

3 x 2 x 4 x 1 4vP -1 x 3 ≡ C. (iii) If N - T 2 (x 1 ) = {2, 3}
. By supposing that there exists i ∈ {1, 4} ∩ N -(u) one can find C, so we may assume that u → {1, 4}. If there exists u ∈ {1, 4} ∩ N + (x 4 ) then d - T 2 -u (x 1 ) ≥ 3 and the result follows, and if there exists u ∈ {2, 3} ∩ N + (x 4 ) then d - T 2 -u (x 1 ) ≥ 2 with an inneighbor of outdegree at least 2 in T 2 -u , so the result follows. Thus {1, 2, 3, 4} → x 4 . We have y 1 ...y 4 = x 3 1x 4 x 1 = -(1, 1, 1) with d - T 2 \{1,4} (y 1 ) ≥ 2, so the result follows unless if T (x 2 , u, 2, 3) = 4A with N + T 2 \{1,4} (y 1 ) ∩ N -(4) = ∅, so the result follows.

Otherwise, we distinguish 3 cases: (ii) If N - T 2 -u (x 1 ) = {1, 3}, by the building lemmas we can suppose that {2, 3, 4} ⊆ N -(x 2 ). By supposing that there exists i ∈ {1, 3, 4} ∩ N -(u) one can find C, so we may suppose that u → {1, 3, 4}. If x 4 → 2 then x 1 ...x 4 2v4u31x 1 ≡ C, if x 4 → 3 then x 1 vx 2 x 3 x 4 3u142x 1 ≡ C and if x 4 → 4 then x 1 vx 2 x 3 x 4 4u312x 1 ≡ C, so {2, 3, 4} → x 4 . If x 1 → x 3 , since T (x 1 , x 3 , v) = 3A then x 2 has an outneighbor x ∈ {1, u}, but x 4 has an inneighbor origin of an antidirected outpath P in T 2 -x so x 4 x 1 x 3 x 2 xvP -1 x 4 ≡ C, so x 3 → x 1 . If 4 → x 3 , if 2 → u then 14x 3 x 2 x 1 3x 4 2uv1 ≡ C and if u → 2 then x 3 x 2 x 1 3x 4 41u2vx 3 ≡ C, thus x 3 → 4 and so ux 4 x 2 34x 3 x 1 13vu ≡ C.

(iii) If N - T 2 -u (x 1 ) = {2, 3}. By supposing that there exists i ∈ {1, 4} ∩ N -(u) one can find C, so we may assume that u → {1, 4}. If x 4 → 2 then x 1 ...x 4 2v1u43x 1 ≡ C, so 2 → x 4 . If 3 → x 4 then x 3 x 2 x 1 2x 4 34u1vx 3 ≡ C, so x 4 → 3. If u → x 1 then 241ux 1 ...x 4 3v2 ≡ C, so x 1 → u. If 2 → x 2 then x 3 x 4 x 2 2x 1 34u1vx 3 ≡ C and if x 2 → 2 then 312x 2 x 3 x 4 ux 1 4v3 ≡ C.

Exc 20: One can check that for any i ∈ {3, 4, 5, 6}, 1 or 2 is an origin of an antidirected outpath P in T 2 \ {i, i + 1, u}, so if there exists i ∈ N -(u) ∩ {3, 4, 5, 6} then x 1 ...x 4 ui(i + 1)vP -1 x 1 ≡ C, thus u → {3, 4, 5, 6}. If there exists u ∈ N + T 2 -u (x 4 ), 1 or 2 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x 4 u vP -1 x 1 ≡ C, so N + (w) and a ∈ {1, 2} pairwaisly distinct such that x 4 has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x 4 ax 1 ...x 3 ww vP -1 x 4 ≡ C. Otherwise, since T (x 1 , x 3 , v) = 3A then d + T 2 (x 2 ) ≥ 1, thus there exist x ∈ N + T 2 (x 2 ) and P an antidirected outpath P in T 2 -x of origin distinct from u, therefore x 4 x 1 x 3 x 2 xvP -1 x 4 ≡ C. Exc 22: Without loss of generality we can suppose that {1, 2} ⊆ N -(x 1 ). One can check that for any i ∈ {4, 5, 6}, there exists an outneighbor j of i such that 1 or 2 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u) ∩ {1, 2, 3, 4, 5} then x 1 ...x 4 uijvP -1 x 1 ≡ C, thus u → {4, 5, 6}. If there exists u ∈ N + T 2 -u (x 4 ) distinct from 4, 1 or 2 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x 4 u vP -1 x 1 ≡ C, and if 4 ∈ N + (x 4 ) then x 1 ... (w) and a ∈ {4, 6} pairwaisly distinct such that x 4 has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x 4 ax 1 ...x 3 ww vP -1 x 3 ≡ C. Otherwise, since T (x 1 , x 3 , v) = 3A then d + T 2 (x 2 ) ≥ 1, thus there exist x ∈ N + T 2 (x 2 ) and P an antidirected outpath P in T 2 -x of origin distinct from u, therefore x 4 x 1 x 3 x 2 xvP -1 x 4 ≡ C. Exc 23: One can check that for any i ∈ {1, 2, 3, 4, 5}, there exists an outneighbor j of i such that 4 or 6 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u) ∩ {1, 2, 3, 4, 5} then x 1 ...x 4 uijvP -1 x 1 ≡ C, thus u → {1, 2, 3, 4, 5}. If there exists u ∈ N + T 2 -u (x 4 ), 4 or 6 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x 4 u vP -1 x 1 ≡ C, so N + (w) and a ∈ {4, 6} pairwaisly distinct such that x 4 has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x 4 ax 1 ...x 3 ww vP -1 x 4 ≡ C. Otherwise, since T (x 1 , x 3 , v) = 3A then d + T 2 (x 2 ) ≥ 1, thus there exist x ∈ N + T 2 (x 2 ) and P an antidirected outpath P in T 2 -x of origin distinct from u, therefore x 4 x 1 x 3 x 2 xvP -1 x 4 ≡ C. Exc 24: Without loss of generality we may suppose that 1 ∈ N -(x 1 ). One can check that for any i ∈ {2, 4, 5, 6}, there exists an outneighbor j of i such that 1 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u) ∩ {2, 4, 5, 6} then x 1 ...x 4 uijvP -1 x 1 ≡ C, thus u → {2, 4, 5, 6}. If 3 → u then x 1 ...x 4 u3456v21x 1 ≡ C, so u → 3. If there exists u ∈ N + T 2 -u (x 4 ), 1 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x 4 u vP -1 x 1 ≡ C, so N + (x 4 ) pairwaisly distinct such that x 4 has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x 4 ax 1 ...x 3 ww vP -1 x 4 ≡ C. Exc 29 :One can check that for any i ∈ {2, 3, 5}, there exists an outneighbor j of i such that 4 or 6 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u) ∩ {2, 3, 5} then x 1 ...x 4 uijvP -1 x 1 ≡ C, thus u → {2, 3, 5}. Similarly if 1 ∈ N -(u) then x 1 ...x 4 u1453v26x 1 ≡ C, thus u → 1. If there exists u ∈ N + T 2 -u (x 4 ), 4 or 6 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ... (w) and a ∈ {4, 6} pairwaisly distinct such that x 4 has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x 4 ax 1 ...x 3 ww vP -1 x 4 ≡ C. Otherwise, since T (x 1 , x 3 , v) = 3A then d + T 2 (x 2 ) ≥ 1, thus there exist x ∈ N + T 2 (x 2 ) and P an antidirected outpath P in T 2 -x of origin distinct from u, therefore x 4 x 1 x 3 x 2 xvP -1 x 4 ≡ C. Exc 30: One can check that for any i ∈ {3, 4, 5, }, there exists an outneighbor j of i such that 4 or 6 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u) ∩ {3, 4, 5} then x 1 ... T 2 \{y,u } (y 1 ) = {1, 3}, by the building lemmas, we may assume that {2, 3, 4} → y, a contradiction. So we can suppose that N - T 2 \{y,u } (y 1 ) = {2, 3}. If 1 → u then P u 12v43y 1 ≡ C and if 4 → u then P u 41v32y 1 ≡ C, so u → {1, 4}. Suppose that y → y 4 then P = y 5 yy 4 y 2 y 3 y 1 = -(1, 1, ..., 1). We have {1, 4} → y 5 , if 2 → u then P 4vu 231y 5 ≡ C, so u → 2 and thus 3 → u (since ∆ ( T 2 ) = 3), we have P 43u v21y 5 ≡ C. So we may assume that y 4 → y and so y → y 3 , thus y 2 → y. So y 5 y 4 y 1 yy 3 y 2 = -(1, 1, ..., 1), but 4 → y 5 , so y 5 y 4 y 1 yy 3 y 2 u v3124y 5 ≡ C.

Exc 20: One can check that for any i ∈ {3, 4, 5, 6}, 1 or 2 is an origin of an antidirected outpath P in T 2 \ {i, i + 1, u}, so if there exists i ∈ N -(u) ∩ {3, 4, 5, 6} then x 1 ...x 6 ui(i + 1)vP -1 x 1 ≡ C, thus u → {3, 4, 5, 6}. If there exists u ∈ N + T 2 -u (x 6 ), 1 or 2 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x 6 u vP -1 x 1 ≡ C, so N + (w) and a ∈ {1, 2} pairwaisly distinct such that x 6 has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x 6 ax 1 ...x 5 ww vP -1 x 6 ≡ C. Exc 22: Without loss of generality we can suppose that {1, 2} ⊆ N -(x 1 ). One can check that for any i ∈ {4, 5, 6}, there exists an outneighbor j of i such that 1 or 2 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u) ∩ {1, 2, 3, 4, 5} then x 1 ...x 6 uijvP -1 x 1 ≡ C, thus u → {4, 5, 6}. If there exists u ∈ N + T 2 -u (x 6 ) distinct from 4, 1 or 2 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x 6 u vP -1 x 1 ≡ C, and if 4 ∈ N + (x 6 ) then x 1 ...x 6 4536v21x 1 ≡ C, so N + (w) and a ∈ {4, 6} pairwaisly distinct such that x 6 has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x 6 ax 1 ...x 5 ww vP -1 x 6 ≡ C. Exc 23: One can check that for any i ∈ {1, 2, 3, 4, 5}, there exists an outneighbor j of i such that 4 or 6 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u) ∩ {1, 2, 3, 4, 5} then x 1 ...x 6 uijvP -1 x 1 ≡ C, thus u → {1, 2, 3, 4, 5}. If there exists u ∈ N + T 2 -u (x 6 ), 4 or 6 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x 6 u vP -1 x 1 ≡ C, so N + (w) and a ∈ {4, 6} pairwaisly distinct such that x 6 has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x 6 ax 1 ...x 5 ww vP -1 x 6 ≡ C. Exc 24: Without loss of generality we may suppose that 1 ∈ N -(x 1 ). One can check that for any i ∈ {2, 4, 5, 6}, there exists an outneighbor j of i such that 1 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u) ∩ {2, 4, 5, 6} then x 1 ...x 6 uijvP -1 x 1 ≡ C, thus u → {2, 4, 5, 6}. If 3 → u then x 1 ...x 6 u3456v21x 1 ≡ C, so u → 3. If there exists u ∈ N + T 2 -u (x 6 ), 1 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x 6 u vP -1 x 1 ≡ C, so N + (x 6 ) pairwaisly distinct such that x 6 has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x 6 ax 1 ...x 5 ww vP -1 x 6 ≡ C. Exc 29 :One can check that for any i ∈ {2, 3, 5}, there exists an outneighbor j of i such that 4 or 6 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u)∩{2, 3, 5} then x 1 ...x 6 uijvP -1 x 1 ≡ C, thus u → {2, 3, 5}. Similarly if 1 ∈ N -(u) then x 1 ...x 6 u1453v26x 1 ≡ C, thus u → 1. If there exists u ∈ N + T 2 -u (x 6 ), 4 or 6 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x 6 u vP -1 x 1 ≡ C, so N + (w) and a ∈ {4, 6} pairwaisly distinct such that x 6 has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x 6 ax 1 ...x 5 ww vP -1 x 6 ≡ C.

Exc 30: One can check that for any i ∈ {3, 4, 5, }, there exists an outneighbor j of i such that 4 or 6 is an origin of an antidirected outpath P in T 2 \ {i, j, u}, so if there exists i ∈ N -(u) ∩ {3, 4, 5} then x 1 ...x 6 uijvP -1 x 1 ≡ C, thus u → {3, 4, 5}. Similarly if there exists i ∈ N -(u)∩{1, 2, 6}, set {u 1 → u 2 } = T (1, 2, 6)-i, we have x 1 ...x 6 i5u 1 u 2 v43x 1 ≡ C, thus u → {1, 2, 6}. If there exists u ∈ N + T 2 -u (x 6 ), 4 or 6 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x 6 u vP -1 x 1 ≡ C, so N + Finally, it is easy to check that there exists x ∈ T such that T -x / ∈ {5A, 5F, 7A, 7K}, so for all m ≤ n -2, T contains C.

Figure 5 :Figure 6 :

 56 Figure 5: The infinite families of exceptions -1-

  |V 1 | = b 1 + b 2 -1 unless if (T -x; (b 1 , b 2 -1) = (3A; (1, 1)). Let a 1 ...a t = (b 1 , b 2 -1) in T (V 1 )and a t+2 ...a n = (b 3 -1, ..., b s ) in T (V 2 ). We have a 1 ...a t xa t+2 ...a n ≡ P and d + (a 1 ) ≥ 1. If (T -x; (b 1 , b 2 -1)) = (3A;

  ): P = (2, 1), N + (x) = {1, 2}, S = {1, 2}. Exc (1, 0)(1): P = (2, 1, 1), N + (x) = {1, 3}, S = {1, 2, 3}. Exc (1, 0)(2): P = (2, 1, 1), N + (x) = {2, 3}, S = {1, 2}. Exc (1, 0)(3): P = (2, 1, 1), N + (x) = {1, 2, 3}, S = {1, 2}. Exc (2, 0): P = (2, 2), N + (x) = {3, 4}, S = {3}. Exc (4, 0)(1): P = (2, 1, 1, 1), N + (x) = {1, 2}, S = {5}. Exc (4, 0)(2): P = (2, 1, 1, 1), N + (x) = {1,2, 3}, S = {1, 2}. Exc (6, 0): P = (2, 1, 2), N + (x) = {4, 5}, S = {4}. Exc (7, 0)(1): P = (3, 1, 1), N + (x) = {1, 2, 4}, S = {1}.

Figure 7 :

 7 Figure 7: Cycle's exceptions

Theorem 3 . 1

 31 Let T be a tournament and P a path such that |T | = |P |. The number of paths P in T is equal to that of P in T .

  one can find an inneighbor of u origin of a path Q ≡ a n-1 a n-2 ...a 2 in T 2 and thus uvQ -1 u ≡ C. While for Dual Exc 0, 4, 6, 33, if d + T 2 (u) ≥ 1, u has an outneighbor origin of a path R ≡ a n-2 a n-3 ...a 1 and so vR -1 uv. Otherwise, we have: Dual Exc 0: (T ; C) is the exception A 9 . Dual Exc 4: 543u21v5 ≡ C. Dual Exc 6: 15uv3241 ≡ C. Dual Exc 33: 765u4231v7 ≡ C.

Case 3 :

 3 If |B| = 2, if u has an outneighbor w origin of a path P ≡ a 0 a n-1 ...a 3 or Q ≡ a 4 a 5 ...a 1 then wuvP -1 ≡ C or uQvu ≡ C. Otherwise, suppose first that d + T 2 (u) ≥ 2, we have two cases: (a) If C = (2, 1, 2, 1, ..., 2, 1) then either (T 2 ; a 0 a n-1 ...a 3 ) is the dual of the exception Exc 3 or 45, in these cases there exist w, t ∈ N + T 2 (u) and y ∈ T 2 \ {w, t} such that w → y and t is an origin of a path R ≡ a 2 a 3 ...a n-3 in T 2 \ {w, y}, we have wuRvtw ≡ C, or s

T 2 (y 4 )

 24 ⊆ S, since otherwise, y 4 has an inneighbor origin of a path P ≡ a 1 a 0 ...a 6 in T 2 ∪ {y 1 } with end distinct from y 1 , so y 4 y 3 y 2 vP -1 y 4 ≡ a 2 a 3 ...a 2 = C) Dual Exc (0, 0): d -(3) = 5, a contradiction. Dual Exc (4, 0)(1): We have 15y 2 ay 1 v342y 4 1 ≡ C. Dual Exc (4, 0)(2): We have 43y 1 ay 4 v125y 2 4 ≡ C. Dual Exc (9, 0): We have 54y 1 ay 4 v3y 2 125 ≡ C. Dual Exc (15, 0): We have 21y 1 ay 4 v5346y 2 2 ≡ C. Dual Exc (33, 0): We have 21y 1 ay 4 v46573y 2 2 ≡ C. Dual Exc (E 1 (4), 0): We have 21y 1 ay 4 v43y 2 2 ≡ C. Dual Exc (E 8

  which is the next biexception; in any case T contains C. Otherwise T 1 = 3A and then 4 is an inneighbor of vertex in T 1 , so T contains C; or |T 1 | = 2 and then T (N -(4)) = 5A, so T contains C. Dual Exc (9, 0)(1), (15, 0)(1), (16, 0)(1) and (33, 0): We have d -(x 2 ) = δ and (T (N + (x 2 )), a 0 a n-1 ...a δ+2 ) is not an exception since T (N + (x 2 )) has v as a minimal vertex. Thus T contains C. Dual Exc (26, 0)(1):

x 1 :

 1 Dual Exc (0, 0)(2): If there is a vertex x ∈ T such that d -(x) = 3 and T (N -(x)) = 3A, or d + (x) = 3 and T (N + (x)) = 3A, then T contains C. Otherwise (T ; C) = (6Z; (2, 1, 2, 1)). Dual Exc (2, 0): We have N -(1) = 4A so T contains C and thus T contains C. Dual Exc (4, 0): If x δ → 1, 5 or 4 (say i) then 1 or 2 is an origin of a path P

  : It is not possible for all x i since otherwise d -(2) = 5 contradiction. Dual Exc (6, 1): , we have 14Qv2351 ≡ C. Dual Exc (10, 1): x 2 → 2 contradiction. Dual Exc (12, 1): 4 is an outneighbor of an x i contradiction. Dual Exc (22, 1): We have x i ← 4, 5, 6 for all i, so 315Q64v23 ≡ C. Dual Exc (24, 1): We have x i ← 5, 6 for all i, so 315Q64v23 ≡ C.

  and thus the first block of the path a δ+2 a δ+3 ...a 0 is at most one which is a contradiction. Otherwise, by lemma 2.4, (T 1 ; a 1 ...a δ ) = (4B; -(1, 1, 1)), (4B; -(2, 1)) or (3A; -(2)) or |T 1 | = 2: We have d + T 2 (x) = 1 for all x ∈ Or(a 1 ...a δ ; T 1 ). So there exist y 1 ...y δ ≡ a 1 ...a δ in T 1 and u

  is solved in Case 1 by taking C = a 0 a 1 ...a 0 = a 4 a 5 ...a 4 and if δ = 5, let Q ≡ a 5 a 6 ...a 9 be a path in T 1 of origin x 0 , since T 2 = 6H and d - T 2 (x 0 ) = 4 then x 0 has an inneighbor origin of a path R ≡ a 4 a 3 ...a n-1 in T 2 , thus QvR -1 x 0 ≡ a 5 a 6 ...a 5 = C. Dual Exc (48, 0): Since d + T 2 (x 0 ) = 2 then δ = 7. If δ = 3 then it is solved in Case 1 by taking C = a 0 a 1 ...a 0 = a 7 a 6 ...a 7 and if δ = 5 then it is solved in Case 4(c) by taking C = a 0 a 1 ...a 0 = a 9 a 8 ...a 9 .

  ): a ≡ 4 or 5 and a → u (since a has a maximal outdegree in T 2 ), if a ≡ 4 then 21y 1 y 2 y 3 u43v52 ≡ C and if a ≡ 5 then 32y 1 y 2 y 3 u54v13 ≡ C. Dual Exc (4, 1)(3): a ≡ 3 or 5, if a ≡ 3 then 14y 1 y 2 y 3 u32v51 ≡ C and if a ≡ 5 then 32y 1 y 2 y 3 u54v13 ≡ C. Dual Exc (9, 1)(1): a ≡ 1 or 3, if a ≡ 1 then v2y 1 y 2 y 3 u1453v ≡ C and if a ≡ 3 then v5y 1 y 2 y 3 u3124v ≡ C. Dual Exc (15, 1)(1): a ≡ 4 or 5, say a ≡ 4, then 21y 1 y 2 y 3 u46v352 ≡ C.

  a 5 = -(2, 1, 1) then 53vx 5 ...x 1 4125 ≡ C and if a 1 ...a 5 = -(1, 1, 2) then by Theorem 2.1 x 2 or x 5 is an origin of a path Q = -(1, 1, 2) and thus 1Qv43521 ≡ C. Exc (10, 1): By supposing that N

  1, 1) then 41vx 5 ...x 1 3524 ≡ C and if a 1 ...a 5 = -(1, 1, 2) then by Theorem 2.1 x 2 or x 5 is an origin of a path Q = -(1, 1, 2) and thus 1Qv43521 ≡ C. Exc (12, 1): By supposing that N

  1, 1) then by taking C = a 0 a 1 ...a n-1 a 0 = a 6 a 5 ...a 7 a 6 it is solved in case 1, and if a 1 ...a 5 = -(1, 1, 2) then by Theorem 2.1 x 2 or x 5 is an origin of a path Q = -(1, 1, 2) and thus 1Qv43521 ≡ C. Exc (22, 1): If

  ) and this is a contradiction since d + (3) = 1 or T 2 = {u, w, t} which is similar to the above case (d - T 2 (x 1 ) = 2). (e) If |B| > |B | ≥ 3. Do a i-1 = a i until a n-2 → a n-1 for the first time. Obviously we have a 2 → a 1 . Assume first that (T 1 ; a 1 ...a δ ) is a Grünbaum's exception. If there exists x 3 x 4 ...x δ+2 ≡ a 3 a 4 ...a δ+2 such that x 3 has an inneighbor u ∈ T 2 origin of a path P ≡ a 2 a 1 ...a δ+4 then x 3 x 4 ...x δ+2 vP -1 x 3 ≡ a 3 a 4 ...a 3 = C. Otherwise, since there is no a such exception, we have s

Case 4 :

 4 If {a 1 , a n-1 } → a 0 . Denote by B the block of C containing the arc a n-1 → a 0 . We distinguish six cases: (a) If |B | = 1 and a 1 → a 2 . Suppose first that (T 1 ; a 1 ...a δ ) is a Grünbaum's exception. If a δ+4 → a δ+3 , let x 3 ...a δ+2 ≡ a 3 .

  since d -(3) = 5 then 3 → x 4 , and if N + (x 4 ) ∩ {1, 2, w} = ∅, say x 4 → 1, then 4x 1 ...x 4 132wv4 ≡ C so {1, 2, w} → x 4 and thus T (N + (1)) = T (x 1 , x 2 , x 4 , w, 3) = T 2 and the result follows. Now if (T 2 -w; a n-1 a n-2 ...a δ+2 ) = (5C;

  a 4 with x 2 → 1 so T contains C. Otherwise, since (T 1 ; a 1 ...a δ ) = (4A;(1, 1, 1)) nor (5C;[START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF] 1, 1)) then there exists y 1 ...y δ ≡ a 1 ...a δ such that y 1 = x 1 , and if y 2 is distinct from x 1 and x 2 then d -(5) ≥ 7, a contradiction. If y 2 = x 1 then 1 → y 2 and so T contains C, soy 2 = x 2 . But T (N -(5)) = T 2 ,and the result follows. Assume now that u is minimal in T 2 -w and a n-1 a n-2 ...a δ+2 is directed. If |T 2 | ≥ 4, let a s be the origin of B, then T contains C = a 0 a 1 ...a 0 = a s+δ+2 a s+δ+1 ...a s+δ+2 . Otherwise, if δ = 2 and T 2 is transitive, let P = (1, 1) be a path in T 2 of origin an inneighbor of x 2 , we have x 2 x 1 vP -1 x 2 ≡ C, if δ = 2 and T 2 = 3A, then vx 1 O T 2 x 2 v ≡ C, and if δ = 3 then T 1 = 3A since otherwise d + (x 1 ) ≥ 4, and thus T = 7A where 12457631 ≡ C. (ii) If d - T 2 (x 2 ) = 1 then we may assume that u is minimal in T 2 and a n-1 a n-2 ...a δ+2 is directed, this is possible by taking x 2 in the place of v. If |T 2 | ≥ 4, let a s be the origin of B, then T contains C = a 0 a 1 ...a 0 = a s+δ+2 a s+δ+1 ...a s+δ+2 . Otherwise, set T

  let y 1 ...y 4 = (2, 1) be a path in T 1 such that y 1 = x 2 , y 1 has an outneighbor origin of a path P ≡ a 5 a 4 a 3 a 2 in T 2 and thus y 4 vP -1 y 1 ...y 4 ≡ C. Dual Exc E 1 (6): Every vertex of T 2 is an origin of a path P ≡ a 2 a 1 ...a n-3 = -(2, 1, 2), so let y 3 ...y δ+2 ≡ a 3 ...a δ+2 be a path in T 1 , y 3 = x 2 , thus y 3 ...y δ+2 vP -1 y 3 ≡ a 3 a 4 ...a 3 = C. And if d +

  x δ-1 v4x δ 31 25 ≡ C, so a δ → a δ-1 . Then δ = 2 and thus 253vx 2 x 1 142 ≡ C. Dual Exc (15, 1): If d + T 2 (x δ ) ≥ 1, one can find an outneighbor of x δ origin of a path P ≡ a δ+1 a δ+2 ...a 1 in T 2 ∪{v} of end and outneighbor u of x 1 and thus ux 1 ...x δ vP ≡ C. So we may suppose that d + T 2 (x δ ) = 0 and so a δ → a δ-1 . If δ = 5 or 6 take C = a 0 a 1 ...a 0 = a 0 a n-1 ...a 0 and it is solved in an another case; if δ = 4 then x 4 v435261x 1 x 2 x 3 x 4 ≡ C; if δ = 3 then a 0 a 1 ...a 3 is a block greater than B and if δ = 2 then 4vx 2 x 1 162534 ≡ C. Dual Exc (16, 1): If d + T 2 (x δ ) ≥ 1, one can check that x δ has an outneighbor origin of a path P ≡ a δ+1 a δ+2 ...a 0 in T 2 ∪ {v} of end 4 and so 4x 1 ...x δ P ≡ C, so a δ → a δ-1 . Let u ∈ N - T 2 (x δ-1 ), u is an origin of a path P = (2, 1) in T 2 \ {w, t} where w ∈ N + T 2 (x 1 ) and t ∈ T 2 , thus wx 1 ...x δ-1 P x δ tvw ≡ C. Dual Exc (32, 1): If d + T 2 (x δ ) ≥ 1, one can check that x δ has an outneighbor origin of a path P ≡ a δ+1 a δ+2 ...a 0 in T 2 ∪ {v} of end 5 or 6 and thus T contains C. Otherwise, let u ∈ N - T 2 (x δ-1 ), as above one can find C. Dual Exc (33, 1): By the building lemmas, N + T 2 (x 1 ) = {1, 2, 3, 6} and {1, 2, 4, 5, 6, 7} ⊆ N - T 2 (x δ ) (since a δ+1 a δ+1 ...a 0 ≡ a 0 a n-1 ...a δ+1

  3 and u → u 4 . Suppose that a 1 ...a 4 = -(2, 1), if a n-2 → a n-3 , let P ≡ a 3 a 4 ...a n-3 be a path in T 2 of origin distinct from u, we have u 4 u 2 u 1 P vu 3 u 4 ≡ C; else, let w ∈ N + T 2 (u), t ∈ T 2 \{u, w} and P ≡ a n-3 a n-4 ...a δ+4 be a path in T 2 \{u, w, t} (this is possible since d + T 2 (u) ≥ 3), we have u 2 u 1 wuu 4 vP -1 u 3 tu 2 ≡ C. Assume now that a 1 ...a 4 = -(1, 1, 1), if |T 2 | = 4, let T 2 -u = {y 1 , y 2 , y 3 } we have y 3 vu 1 u 4 uy 1 u 3 u 2 y 2 y 3 ≡ C; else, let w ∈ N +

  is one of the following exceptions: Dual Exc 2: If δ = 3 by taking C = a 0 a 1 ...a 0 = a 4 a 5 ...a 4 it is solved in Case 1. If δ = 5, since a 0 ...a δ has at least 3 origins in T 1 then d -(1) ≥ 6, a contradiction. Dual Exc 4: If δ = 3, there exists x ∈ T 1 such that d - T 2 (x) ≥ 3, since otherwise, by the building lemmas, we may assume that {1, 2, 4, 5} → x 1 , if x 1 = a, by definition of a we deduce that {3, 4, 5} → {x 2 , x 3 }. And if δ = 5, obviously d

  Dual Exc 15: If δ = 3 then by taking C = a 7 a 8 ...a 7 it is solved in Case 1, so δ ≥ 5. So T 1 contains at least 3 origins of an antidirected inpath, thus d -(4) ≥ 8, a contradiction. Dual Exc 33: If δ = 3, let P = -(1, 1, 1, 2, 1) be a path in T 2 of end an inneighbor of x 0 , we have x 0...x 3 vP x 0 ≡ a 7 a 8 ...a 7 = C, so δ ≥ 5. Let y δ-2 ...y 2δ-2 ≡ a δ-2 ...a 2δ-2 be a path in T 1 and u ∈ N + T 2 (y δ-2 ), d - T 2 -u (y 2δ-2 ) ≥ 2 so T 2 -u contains a path P ≡ a 2δ-1 a 2δ ...a δ-5 of origin an inneighbor of y 2δ-2 , thus y δ-2 ...y 2δ-2 P vuy δ-2 ≡ a δ-2 a δ-1 ...a δ-2 = C. Dual Exc 37: If δ = 3 then 5264731x 0 ...x 3 v5 ≡ C, so δ ≥ 5. Let y 3 ...y δ+2 ≡ a 3 ...a δ+2 be a path in T 1 and u ∈ N + T 2 (y 3 ), d - T 2 -u (y δ+2 ) ≥ 2 so y δ+2 has an outneighbor origin of a path P ≡ a δ+3 a δ+4 ...a 0 in T 2 -u, thus vuy 3 ...y δ+2 P v ≡ a 1 a 2 ...a 2 = C. Dual Exc E 1 (4): If δ = 3 then by taking C = a 4 a 5 ...a 5 it is solved in Case 1, so δ = 5. Let y 5 ...y 9 ≡ a 5 ...a 9 be a path in T 1 , by Theorem 2.1 y 5 has an outneighbor origin of a path P ≡ a 4 a 3 ...a 0 in T 2 , thus y 5 ...y 9 vP -1 y 5 ≡ a 5 a 6 ...a 5 = C. Dual Exc E 8 , E 9 , E 10 : d - T 2

2 ( 2 (x 3 ) 2 (

 2232 then every vertex x of T 1 has an inneighbor origin of a path P ≡ a 2 a 1 ...a 7 in T 2 and thus I X vP -1 x ≡ a 3 a 4 ...a 3 = C, so a 6 → a 7 . Similarly we may assume that a n-4 → a n-3 . Let x, y ∈ N - T 2 (x 1 ) such that x → y, if x 0 has an outneighbor origin of a path P ≡ a n-3 a n-4 ...a 4 in T 2 \ {x, y} then x 1 xyvP -1 x 0 x 3 x 1 ≡ C, otherwise, (T 2 \ {x, y}; a n-3 a n-4 ...a 4 ) is one of the following exceptions: Dual Exc 1: If d + T 2 -u (x 3 ) ≥ 2 then x 3 has an outneighbor origin of a path P ≡ a 4 ...a 7 in T 2 -u and thus x 0 ...x 3 P vux 0 ≡ C unless if (T 2 -u; a 4 ...a 7 ) = (4A;(1, 1, 1)), in this case there exists w ∈ N + T x 0 ) such that x 3 has an outneighbor origin of a path P ≡ a 0 a 1 0a 8 a 7 = -(2, 1) in T 2 -w and thus x 3 ...x 0 wvP -1 x 3 ≡ a 1 a 2 ...a 1 = C. So we may assume that d - T ≥ 3, so x 3 has an inneighbor origin of a path P ≡ a 6 a 7 ...a 0 = (1, 1, 2) in T 2 and thus vax1 x 0 x 3 P v ≡ a 1 a 2 ...a 1 = C. Dual Exc 5, 7: If d + T 2 -u (x 3 ) ≥ 3then x 3 has an outneighbor origin of a path P ≡ a 4 a 5 ...a n-3 = (1, 1, 2) in T 2 -u and thus x 0 ...x 3 P vux 0 ≡ C. Otherwise, let w ∈ N + T x 0 ) such that d - T 2 -w (x 3 ) ≥ 3, x 3 has an inneighbor origin of a path P ≡ a n-5 a n-6 ...a 2 = (1, 2, 1) in T 2 -w and thus wvP -1 x 3 ...x 0 w ≡ C. Dual Exc 24: We have 541x 0 x 3 x 1 xyv2365 ≡ C.

so x 4

 4 has an outneighbor x in T 2 such that T 2 -x contains a path P = -(1, 1) of origin an inneighbor of x 1 and thus xvP -1 x 1 ...x 4 x ≡ C. If a 1 ...a 4 = (1, 2) then Or((1, 2); T 1 ) = T 1 -Or((2, 1); T 1 ) and thus there exists y 4 ...y 7 ≡ a 4 ...a 7 = (2, 1) such that y 4 → 3 therefore 21w3y 4 ...y 7 v2 ≡ C. If a 1 ...a 4 = (2, 1) then

  x 1 vP x 4 ≡ C. So a 1 ...a 4 = (1, 2), if |T 2 | = 4 then u is an origin of a path P = (2, 1, 1) in T 2 ∪{x 1 } with end in T 2 , thus vx 4 x 3 x 2 P v ≡ a 3 a 4 ...a 3 = C; and if |T 2 | = 5 then x 1 has an inneighbor origin of a path P =

  In a similar way, we can assume that {1, 2, 3, 4, 5} → x δ , then T (N + (4) = 6H, thus T contains C. (e) If |B| = |B | = 3. Since the first and the last arcs of the path a 0 ...a δ-1 have the same direction then (T 1 ; a 0 ...a δ-1

Theorem 3 . 4

 34 Let T be a tournament of order n = 2k ≥ 4, T contains an antidirected Hamiltonian cycle C if and only if (T ; C) is not one of the exceptions.Proof. By theorem 3.1, we may suppose that δ

Case 1 :

 1 If δ = 0, T contains C if and only if T 2 contains an antidirected inpath P , in this case vP v ≡ C. So the sole exceptions are A 4 , A 5 and A 6 . Case 2: If δ = 1, set T 1 = {u}. 3 cases may arise: (a) If d + T 2 (u) ≥ 3, let w ∈ N + T 2 (u) such that d - T 2 (w) ≥ 1. If u has an outneighbor origin of an antidirected inpath P in T 2 -w then uP vwu ≡ C, otherwise, (T 2 -w; -(1, 1, ..., 1)) is one of the following biexceptions with respect to u: (We may suppose that w → (T 2 \ {S ∪ {w}}) since, otherwise, let R = (1, 1, ..., 1) be a path in (T 2 -w) ∪ {u} of origin in T 2 \ {S ∪ {w}} and end in T 2 , we have wRvw ≡ C) Dual Exc (0, 1): If 1 → w then uw13v2u ≡ C and if w → w then u1w3v2u ≡ C. Dual Exc (4, 1): We have u1w5v342u ≡ C. Dual Exc (8, 1): If w → 1 then u2w1v345u ≡ C and if w → 5 then u5w4v132u ≡ C. So we may assume that {1, 5} → w, in this case T -v = 7A and thus (T ; C) = A 6 . Dual Exc (9, 1)(1), (3): We have u415w3v2u ≡ C. Dual Exc (9, 1)(2): We have u514w3v2u ≡ C. Dual Exc (9, 1)(4): We have u5w231v4u ≡ C. Dual Exc (33, 1): We have u1w62435v7u ≡ C. Dual Exc (E 8 (5), 1): We have (T

  then d - T 2 (u) = 0 and thus (T ; C) = A 10 . If |T 2 | = 3 then T -v = 5A and thus (T ; C) = A 5 . So we may assume now that T 2 → x 2 . If |T | = 6 then there exists u ∈ T 2 such that T -u = 5A and thus (T ; C) = A 5 , so we may suppose that |T | ≥ 8

2 (x 2 ) 2 (x 1 )

 2221 = 0. There exist a, b ∈ T 2 -u such that x 2 has an inneighbor origin of an antidirected outpath P in T 2 \ {a, b}, thus ax 1 x 4 bx 2 P va ≡ C.So we may assume that d -T 2 -u (x 1 ) ≥ 2 and then (T 2 -u; (1, 1, ..., 1)) is an exception. For the both cases, if d - T ≥ 2 and (T 2 -u;(1, 1, ..., 1)) is an exception then it is one of the following ones: Exc 1: If d + T 1 (x) ≥ 2 for all x ∈ T 2 , 3 cases may arise:

  (i) If {1, 2} ⊆ N -(x 1 ). If 3 → u then x 1 ...x 4 u34v21x 1 ≡ C and if 4 → u then x 1 ...x 4 u41v32x 1 ≡ C, so u → {3, 4}. If x 4 → 1 or 2,say 1, then x 1 ...x 4 1v4u32x 1 ≡ C, so {1, 2} → x 4 and thus x 3 x 2 x 1 1x 4 23u4vx 3 ≡ C.

T 2 (x 4 ) 2 (x 3 ) 2 (x 3 )

 242323 = {u}. If d - T ≥ 1 then there exist w ∈ N - T , w ∈ N + T 2

2 (x 4 ) 2 (x 3 ) 2 (x 3 )

 242323 x 4 4536v21x 1 ≡ C, so N + T = {u}. If d - T ≥ 1, then there exist w ∈ N - T , w ∈ N + T 2

T 2 ( 2 (x 3 )

 223 x 4 ) = {u}. If d - T ≥ 1, then there exist w ∈ N - T 2 (x 3 ), w ∈ N + T 2

T 2 (x 4 ) 2 (x 3 ) 2 (x 3 ) 2 (x 1 )

 24232321 = {u}. If d - T ≥ 1, then there exist w ∈ N - T , w ∈ N + T 2 (w) and a ∈ N - T ∩ N - T 2

2 (x 4 ) 2 (x 3 ) 2 (x 3 )

 242323 x 4 u vP -1 x 1 ≡ C, so N + T = {u}. If d - T ≥ 1, the there exist w ∈ N - T , w ∈ N + T 2

2 (x 4 ) 2 9x 3 ) 2 (x 3 ) 2 ( 2 (x 2 ) 2 (x 2 2 (x 6 )

 2423232222226 x 4 uijvP -1 x 1 ≡ C, thus u → {3, 4, 5}. Similarly if there exists i ∈ N -(u)∩{1, 2, 6}, set {u 1 → u 2 } = T(1,[START_REF] Hanna | Paths in tournaments, a simple proof of Rosenfeld's Conjecture[END_REF][START_REF] Grünbaum | Antidirected Hamiltonian paths in tournaments[END_REF]-i, we have x 1 ...x 4 i5u 1 u 2 v43x 1 ≡ C, 74 thus u → {1, 2, 6}. If there exists u ∈ N + T 2 -u (x 4 ), 4 or 6 is an origin of an antidirected ouptath P in T 2 -u and thus x 1 ...x 4 u vP -1 x 1 ≡ C, so N + T = {u}. If d - T ≥ 1, then there exist w ∈ N - T , w ∈ N +T w) and a ∈ {4, 6} pairwaisly distinct such that x 4 has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x 4 ax 1 ...x 3 ww vP -1 x 4 ≡ C. Otherwise, since T (x 1 , x 3 , v) = 3A then d + T ≥ 1, thus there exist x ∈ N + T ) and P an antidirected outpath P in T 2 -x of origin distinct from u, thereforex 4 x 1 x 3 x 2 xvP -1 x 4 ≡ C. Case 6: If δ = 5 and for all x ∈ T such that d -(x) = 5 we have T (N -(x)) = 5A. Let a ∈ T 2 such that d - T 2 (a) ≥ 1 and set T 1 = T 1 ∪ {a} and T 2 = T 2 -a. Let x 1 ...x 6 = -(1, 1, ..., 1) be a path in T 1 and let u ∈ N + T . If x 1 has an inneighbor origin of an antidirected outpath P in T 2 -u then x 1 ...x 6 uvP -1 x 1 ≡ C. Otherwise, if d - T 2 -u (x 1 ) ≥ 2 then (T 2 -u; (1, 1, ..., 1)) is one of the following exceptions: Exc 1: Set V (T 1 ) = {y 1 , y 2 , ..., y 5 } such that y i ≡ i in 5A, we distinguish two cases:(i) If there exists y ∈ T 2 and i ∈ {1, 2, 3, 4, 5} such that {y i , y i+1 } → y, without loss of generality suppose that i = 1. Let u ∈ N + T 2 -y (y 1 ) and suppose that T 2 \ {u , y} = 4A. We haveP = y 5 y 3 y 4 y 2 yy 1 = -(1, 1, ..., 1). Since T (N + (x)) = 5A for all x ∈ T such that d + (x) = 5 then d + T 2 -y (y 1 ) ≥ 2. If {1, 2} ⊆ N -(y 5 ). If 3 → u then P u 34v21y 5 ≡ C and if 4 → u then P u 41v32y 5 ≡ C, so u → {3, 4}. If N + (y 1 ) ∩ {3, 4} = ∅ then d -(y 1 ) = 5 with T (N -(y 1 )) = 5A, a contradiction, so y 1 → 1 or 2, say 1, then P 1v4u 32y 5 ≡ C. If N - T 2 \{u ,y} (y 5 ) = {1, 3}, by the building lemmas we can suppose that {2, 3, 4} → y 3 . If 3 → u then P u 34v21y 5 ≡ C, if 4 → u then P u 42v31y 5 ≡ C and if 1 → u then P u 12v43y 5 ≡ C, so u → {1, 3, 4}. If y 1 → 2 then P 2v4u 31y 5 ≡ C, if y 1 → 3 or 4, say 3, then P 3u 4v21y 5 ≡ C, so N -(y 1 ) = {y 4 , y 5 , 2, 3, 4} and thus {2, 4} → y 4 and y 4 → 3. We have 21y 5 3y 3 4y 4 y 2 yy 5 u v2 ≡ C. Now if N - T 2 \{u ,y} (y 5 ) = {2, 3}, if 1 → u then P u 12v43y 5 ≡ C and if 4 → u then P u 41v32y 5 ≡ C, so u → {1, 4}. If y 1 → 2 then P 2v1u 43y 5 ≡ C, so 2 → y 1 . If 3 → y 1 then T (N -(y 1 )) = 5A so y 1 → 3 and thus {1, 4} → y 1 . But T (N -(y 1 )) = 5A so y 4 → 4 and {1, 2} → y 4 . In any case T (N -(y 2 )) = 5A and T (N + (y 2 )) = 5A, a contradiction. (ii) If for all x ∈ T 2 and for all i, N + (x) ∩ {y i y i+1 } = ∅. Let y ∈ T 2 such that d + T 2 (y) = ∆ + (T 2 ), without loss of generality suppose that y → {y 1 , y 5 }. We have P = y 1 yy 5 y 3 y 4 y 2 = -(1, 1, ..., 1). Let u ∈ N + T 2 -y (y 2 ), we may assume that T 2 \ {y, u } = 4A. If {1, 2} ⊆ N -(y 1 ), by the building lemmas we can suppose that {2, 4} → y, and since y 1 → 4 then 4 → y 5 and thus 31y 1 2y4y 5 y 3 y 4 y 2 u v3 ≡ C. If N -

T 2 ( 2 (x 5 )

 225 x 6 ) = {u}. There exist w ∈ N - T , w ∈ N + T 2

T 2 ( 2 (

 22 x 6 ) = {u}. There exist w ∈ N - T

T 2 (x 6 )

 26 = {u}. There exist w ∈ N - T 2 (x 5 ), w ∈ N + T 2

T 2 (x 6 )

 26 = {u}. There exist w ∈ N -

T 2 (

 2 x 6 ) = {u}. There exist w ∈ N - T 2 (x 5 ), w ∈ N + T 2

T 2 (x 6 ) 2 (x 5 ) 2 ( 2 (x 1 ) 2 (y 6 ) 2 (x 1 ) 2 (x 8 ) 2 (x 8 2 ( 2 (x 8 )

 262522126212828228 = {u}. There exist w ∈ N - T , w ∈ N + T w) and a ∈ {4, 6} pairwaisly distinct such that x 6 has an inneighbor origin of an antidirected outpath P in T 2 \ {a, w, w }, thus x 6 ax 1 ...x 5 ww vP -1 x 6 ≡ C. So we may assume that d + T ≥ 3 and by Theorem 2.1, x 1 is always an end of an antidirected inpath y 1 ...y 6 in T 1 and thus we may find u∈ N + T such that d - T 2 -u (y 1 ) ≥ 2, therefore T contains C.Case 7: If δ = 7 and for all x ∈ T such that d -(x) = 7 we haveT (N -(x)) = 7A, let a ∈ T 2 such that d + T 2 (a) = ∆ + (T 2 ) and set T 1 = T 1 ∪ {a} and T 2 = T 2 -a, there exists x 1 ...x 8 = -(1, 1, ..., 1) such that d - T ≥ 3 and d + T ≥ 2. Let u ∈ N + T 2 (x 8 ),T 2 -u has an antidirected Hamiltonian outpath P of origin an inneighbor of x 1 and thusx 1 ...x 8 uvP -1 x 1 ≡ C unless if (T 2 -u; (1, 1, ..., 1)) is one of the following biexceptions with respect to x 1 : Exc (22, 1): If N -(u) ∩ {4, 5, 6} = ∅, one can find C. Otherwise, let u ∈ N + T ), T 2 -uis not an exception and the result follows.Exc (24, 1): If d - T u) ≥ 1 one can find C, otherwise, let u ∈ N +T distinct from u, T 2 -u is not an exception, so the result follows. Corollary 3.5 Let T be a tournament on n ≥ 3 vertices and C be a non-directed cycle of order m, where 3 ≤ m ≤ n. T contains C if and only if (T ; C) is not one of the exceptions A i , 1 ≤ i ≤ 18.Proof. If m = n, theorems3.2, 3.3 and 3.4 give the result. If m = n -1, if there exists x ∈ T such that (T -x; C) is not an exception, the result follows. Otherwise, we discuss according to the values of n:(i) If n = 5, if T = 5A then (T ; C) = A 15 . Otherwise, let x ∈ T such that d + (x) = 3and let y be its inneighbor. it is easy to check that d -(y) ≥ 2 with T \ {x, y} = 3A and so (T ; C) = A 16 . (ii) If n = 7, if T -x = 6F for all x ∈ T then (T ; C) = A 18 . If T -x = 6M for all x ∈ T then (T ; C) = A 17 . It is easy to check that if there exists x ∈ T such that T -x = 6M then T -y / ∈ {6O, 6P } for all y ∈ T . Also, one can check that |{x ∈ T ; T -x = 6O}| ≤ 2 and |}x ∈ T ; T -x = 6P }| ≤ 3, so if T = 7K then there exists x ∈ T such that T -x / ∈ {6O, 6P } and the result follows.

  (ii) If (T 2 \ {w, t}; a 6 a 7 ...a n-1 ) = (4A;(1, 1, 1)), there exists w , t ∈ T 2 \ {w, t} such that T 2 \ {w , t } = 4A, thus y 3 has an inneighbor origin of a path P = (1, 1, 1) in T 2 \ {w , t }, therefore y 3 y 4 y 5 w t vP -1 y 3 ≡ a 3 a 4 ...a 3 = C.(iii) If (T 2 \ {w, t}; a 6 a 7 ...a n-1 ) = (5C; (2, 1, 1)) or T 2 \ {w, t} is a transitive triangle and a 6 a 7 a 8 is directed, both cases are solved in Case 1 by taking C = a 0 a 1 ...a 0 = a 7 a 6 ...a 7 .

, y}, if y ← w or t, say w, then ty 5 y 4 vuywy 3 t ≡ C, so y → {w, t}. Without loss of generality we may suppose that u → T 1 , so y 4 ← w or y, say w, we have y 3 uy 5 vytwy 4 y 3 ≡ C.

  x 1 vP -1 x 4 ≡ a 2 a 3 ...a 2 = C. Dual Exc[START_REF] Petrović | Antidirected Hamiltonian circuits in tournaments, Graph Theory[END_REF] 1)(1): If N + (x δ )∩{2, 3, 4, 5} = ∅, one can find w, t ∈ T 2 and P = -(1, 1) of origin an outneighbor u of x 1 in T 2 \ {w, t} such that x δ → w → t and thus ux 1 ...x δ wtvP -1 ≡ C, and ifx δ → 1 then 2x 1 ...x δ 1453v2 ≡ C. Dual Exc (16, 1): If d + T 2 (x δ ) ≥ 1, one can check that x δ has an outneighbor origin of a path P ≡ a δ+1 a δ+2 ...a 0 in T 2 ∪ {v} of end 4 and so 4x 1 ...x δ P ≡ C, so a δ → a δ-1 . Let u ∈ N - T 2 (x δ-1 ), uis an origin of a path P = (2, 1) in T 2 \ {w, t} where w ∈ N + T 2 (x 1 ) and t ∈ T 2 , thus wx 1 ...x δ-1 P x δ tvw ≡ C. Dual Exc (32, 1): If d + T 2 (x δ ) ≥ 1, one can check that x δ has an outneighbor origin of a path P ≡ a δ+1 a δ+2 ...a 0 in T 2 ∪ {v} of end 5 or 6 and thus T contains C. Otherwise, let u ∈ N - T 2 (x δ-1 ), as above one can find C. Dual Exc (E 1 , 1): Let a s be the origin of B, T contains

  and only if d - T 2 (w 1 ) ≥ 1, in fact, without loss of generality, suppose that w 2 ∈ N -(w 1 ) we have uw 2 w 1 vw 3 w 4 u ≡ C; so C is not contained in T if and only if (T ; C) is the exception A 11 . If N + T 2 (u) = {w 1 , w 2 }, say w 3 → w 4 , then T contains C if and only if w 1 or w 2 → w 3 , so C is not contained in T if and only if (T ; C) is the exception A 12 . Case 3: If δ is a positive even number. Let x 1 ...x δ be a Hamiltonian antidirected inpath of T 1

  x 2 and thus x 2 → w, therefore x 2 vx 3 x 1 u 2 uu 1 wx 2 ≡ C. So we may assume that u → u 2 . If w → u 2 , then u 2 → x 2 and thus x 2 → w. If u → x 1 then x 2 vx 3 x 1 uu 2 u 1 wx 3 ≡ C, and if x 1 → u then u 2 x 3 u 1 wvux 1 x 2 u 2 ≡ C. So we can suppose now that u 2 → w. Then w → x 2 and thus x 2

Lebanese University, KALMA Laboratory, Beirut, Lebanon

University Savoie Mont Blanc, LAMA Laboratory, Chambéry, France

Acknowledgments

77

I would like to express my gratitude to Professor Amine El Sahili and Doctor Maydoun Mortada for their insightful comments.