S D Adhikari

L Boza

Shalom Eliahou

M P Revuelta

M I Sanz

Equation-regular sets and the Fox-Kleitman conjecture

Keywords: 2010 Mathematics Subject Classification. 05D10, 11B75 Partition regularity, degree of regularity, monochromatic solution, discrete derivative

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Introduction

A Diophantine equation L is said to be n-regular, for some positive integer n, if for every n-coloring of N + = {1, 2, . . . }, there is a monochromatic solution to L. Further, L is said to be regular if it is n-regular for all n ≥ 1. Of course, (n + 1)regularity implies n-regularity. The degree of regularity of L, denoted dor(L), 1 is defined to be infinite if L is regular, or else, it is the largest n such that L is n-regular [START_REF] Rado | Studien zur Kombinatorik[END_REF]. Determining the degree of regularity of a given Diophantine equation is difficult in general, even if L is linear.

In this paper, we focus on a particular linear equation proposed by Fox and Kleitman in [START_REF] Fox | On Rado's Boundedness Conjecture[END_REF]. Given positive integers k, b, we shall denote by L k (b) the 2k-variable Diophantine equation

k ∑ i=1 (x i -y i ) = b.
Fox and Kleitman [START_REF] Fox | On Rado's Boundedness Conjecture[END_REF] showed that this equation is never 2k-regular, i.e., that

dor(L k (b)) ≤ 2k -1 (1)
for all b ∈ N + . Moreover, they conjectured that this upper bound is best possible.

Conjecture 1.1 ( [START_REF] Fox | On Rado's Boundedness Conjecture[END_REF]). Let k ≥ 1. Then there exists an integer b ≥ 1, depending on k, such that dor(L k (b)) = 2k -1.

The case k = 2 of the conjecture was recently settled in [START_REF] Adhikari | On a conjecture of Fox and Kleitman on the degree of regularity of a certain linear equation[END_REF], where it is shown that dor(L 2 (b)) = 3 for all b ≡ 0 mod 6. More generally, the authors determined dor(L 2 (b)) for all b ≥ 1, as follows. A reduced 3-variable version of the 4-variable equation L 2 (b) had already been studied in [START_REF] Bialostocki | On the degree of regularity of some equations[END_REF]. Indeed, they considered the equation x 3x 2 = x 2x 1 + b, which can be obtained from the equation (x 1y 1 ) + (x 2y 2 ) = b by setting y 1 = y 2 and renaming the resulting three variables x 1 , x 2 , y 2 as x 1 , x 3 , x 2 , respectively.

Finally, as pointed out by a referee, the conjecture has recently been fully settled, using sophisticated methods of additive combinatorics [ST].

The functions ν and f

We now introduce two functions ν, f : N + → N + which are somehow related to dor(L k (b)). In particular, they will provide nice reformulations of the above formulas for dor(L 2 (b)). For n ∈ N + , define ν(n) = the smallest non-divisor of n, i.e. the least integer m ≥ 2 such that m does not divide n. For instance, we have ν(n) = 2 if n is odd, and ν((p -1)!) = p if p is prime. Still for n ∈ N + , define f (n) = the largest r such that r! divides n.

Thus for instance, we have f (n) = 1 if n is odd, and f (n) = 2 if and only n is even and not divisible by 6, i.e. n ≡ 2, 4 mod 6.

As is easily seen, Theorem 1.2 is equivalent to the formulas

Contents

In Section 2, we provide basic lemmas regarding the behavior of dor(L k (b)). In Section 3 we introduce L-regular and L-singular sets with respect to a Diophantine equation L, and we provide tools to determine such sets for the equation L k (b). In Section 4 we give a new proof of the above formula for dor(L 2 (b)). In Section 5 we determine dor(L 3 (b)) for all b ≥ 1, with some computer assistance in the specific case b = 120, and we independently verify the Fox-Kleitman conjecture for k = 3. In Section 6 we briefly discuss the case k = 4. Finally, in Section 7 we determine dor(L k (b)) whenever k ≥ b.

Basic lemmas

We shall use the following notation from additive combinatorics. For subsets A, B of an abelian group G, we write

A + B = {a + b | a ∈ A, b ∈ B}, -A = {-a | a ∈ A} and kA = A + • • • + A, the k-fold sum of A with itself. Thus, given a subset X ⊆ Z, we see that equation L k (b) has a solution with all entries in X if and only if b ∈ k(X -X).
(See also Remark 3.5.) This will be used throughout the paper. For instance, using this formulation, here is a brief explanation for Fox and Kleitman's upper bound (1) stating that L k (b) is never 2k-regular. For integers a ≤ b, we shall denote [a, b] the integer interval consisting of all n ∈ Z such that a ≤ n ≤ b.

Lemma 2.1 ([FK06]). For all integers k, b ≥ 1, one has dor(L k (b)) ≤ 2k -1.
Proof. If b is not a multiple of k, then L k (b) is not even k-regular by the independent Lemma 2.4 below. If b = kr with r ∈ N + , we first 2k-color [1, 2b] as follows. The color class C 1 is given by

C 1 = [1, r]. For 2 ≤ i ≤ 2k, the color class C i is the translate of C 1 by (i -1)r, namely C i = C 1 + (i -1)r.
We then extend this coloring to the whole of N + by 2b-periodicity, namely where the color classes are

X i = C i + 2bZ = C i + 2krZ for all 1 ≤ i ≤ 2k. We claim that b / ∈ k(X i -X i ) for all 1 ≤ i ≤ 2k. It suffices to check it for i = 1, since X i -X i is independent of i. Now X 1 -X 1 = [-(r -1), r -1] + 2krZ, whence k(X 1 -X 1 ) = [-k(r -1), k(r -1)] + 2krZ. Therefore b / ∈ k(X 1 -X 1 ), as claimed. It follows that L k (b) is not 2k-regular.
We now show that dor(L k (b)) is monotone with respect to k.

Lemma 2.2. Let b, k 1 , k 2 be positive integers such that k 1 ≤ k 2 . Then

dor(L k 1 (b)) ≤ dor(L k 2 (b)). Proof. Let n = dor(L k 1 (b)). We claim that L k 2 (b) is n-regular. Let c be an n- coloring of N + . Then there is a c-monochromatic subset X ⊆ N + such that b ∈ k 1 (X -X). Now k 1 (X -X) ⊆ k 2 (X -X) since 0 ∈ X -X and k 1 ≤ k 2 . Hence b ∈ k 2 (X -X). Therefore L k 2 (b) is n-regular, as claimed.
Next, for fixed k, here are two basic lemmas about the behavior of dor(L k (b)) as a function of b. The first one shows that this function is monotone with respect to multiplication.

Lemma 2.3. Let b 1 , b 2 , k be positive integers such that b 1 divides b 2 . Then dor(L k (b 1 )) ≤ dor(L k (b 2 )).
Proof. By hypothesis, there is an integer t ∈ N + such that b 2 = tb 1 . Let r = dor(L k (b 1 )). We now show that L k (b 2 ) is r-regular. Let c be a r-coloring of N + . It suffices to establish the existence of a c-monochromatic solution of L k (b 2 ). Let c be the new r-coloring of N + defined by c (n) = c(tn) for all n ≥ 1. Since L k (b 1 ) is r-regular, there is a c -monochromatic solution a 1 , . . . , a k , d 1 , . . . ,

d k in N + of L k (b 1 ), i.e. satisfying k ∑ i=1 (a i -d i ) = b 1 .
Multiplying this equality by t, we get

k ∑ i=1 (ta i -td i ) = tb 1 = b 2 . ( 6
)
Since the a i 's, d i 's are c -monochromatic, it follows by construction that the ta i 's, td i 's are c-monochromatic. Moreover, by (6), they constitute a solution of L k (b 2 ). Hence r ≤ dor(L k (b 2 )), as desired.

On the other hand, here is an obvious upper bound on dor(L k (b)). Proof. It suffices to find an m-coloring of N + for which there is no monochromatic solution of L k (b). Consider the coloring π m given by the classes mod m. That is, the subset X i ⊆ N + of elements of color i, for 0 ≤ i ≤ m -1, is defined as

X i = i + m • N + . Then X i -X i ⊆ m • Z. Since b ≡ 0 mod m, it follows that b / ∈ k(X i -X i ). Hence equation L k (b) admits no π m -monochromatic solution, as desired. Proposition 2.5. Let b ≥ 1. Then dor(L k (b)) ≤ min(2k -1, ν(b) -1). Proof. We have dor(L k (b)) ≤ 2k -1 by Lemma 2.1, and dor(L k (b)) ≤ ν(b) -1 by Lemma 2.4 since ν(b) does not divide b.

Equation-regular sets

We now introduce a notion of regularity for sets which is closely linked to the usual notion of partition-regularity for Diophantine equations. This will turn out to be useful to determine, in some cases, the degree of regularity of the Fox-Kleitman equation L k (b).

Definition 3.1. Given a Diophantine equation L, or a system of such equations, we say that a set X ⊂ N + is regular with respect to L, or more shortly L-regular, if it contains a solution of L. We say that X is singular if X is not regular.

Here is an easy remark linking these notions of regularity.

Remark 3.2. Let L be a Diophantine equation. Then L is n-regular if and only if every partition of N + into n parts admits an L-regular part.

The notion of equation-regular sets moves the focus away from partitions and more towards those properties of a set which will force it to contain a solution of a given Diophantine system L.

For instance, given k, b ≥ 1, we shall see that any sufficiently dense subset of a sufficiently large integer interval is L k (b)-regular. That is, for sets, density alone implies regularity. This will allow us to determine the degree of regularity of L k (b) in some instances, and in particular to independently verify the Fox-Kleitman conjecture for k = 3, as well as to provide a shorter proof of it for k = 2.

Block sums

We first need some basic notions to be used throughout the paper. Let A = (a 1 , a 2 , . . . , a r ) be a sequence of positive integers.

• A block in A is any subsequence A of consecutive terms in A, i.e. of the form A = (a i , a i+1 , . . . , a j )

for some indices i, j such that 1 ≤ i ≤ j ≤ r.

• The empty subsequence of length 0 is also considered to be a block of A.

• We denote by σ(A) = ∑ r i=1 a i the sum of the elements of A, and by µ(A) the average of A, i.e. µ(A) = σ(A)/r.

• Finally, we denote by bs(A) the set of signed block sums in A, i.e.

bs(A)

= {±σ(A ) | A is a block in A}.
For instance, if A = (1, 10, 4), then bs(A) ∩ N = {0, 1, 4, 10, 11, 14, 15}. Note that 5 is not a block sum in A.

Note that if A is a block in A, then bs(A ) ⊆ bs(A). We will further discuss this property later on. Observe also that A ⊆ bs(A).

The discrete derivative

We shall need a variant of the discrete derivative, associating to a subset X ⊂ Z of cardinality r + 1 a sequence of length r, as follows.

Definition 3.3. Let X ⊂ Z be a finite subset of cardinality r + 1. Let the elements of X be x 0 < x 1 < • • • < x r . We associate to X the sequence

∆X = (x 1 -x 0 , x 2 -x 1 , . . . , x r -x r-1 )
of length r, and call ∆X the discrete derivative of X.

Of interest to us here is the fact that X -X can be read off from the signed block sums of ∆X.

Lemma 3.4. Let X ⊂ Z be a nonempty finite subset. Then X -X = bs(∆X).

Proof. Let the elements of X be x

0 < x 1 < • • • < x r . Then X -X = {x t -x s | 0 ≤ s,t ≤ r}.
Let now A = ∆X = (a 1 , . . . , a r ), where a i = x i -x i-1 for 1 ≤ i ≤ r. Then for indices s,t such that 0 ≤ s ≤ t ≤ r, we have

x t -x s = t ∑ i=s+1 (x i -x i-1 ) = t ∑ i=s+1 a i .
Thus x tx s = σ(A ), where A is the block (a s+1 , . . . , a t ) of A. Hence x tx s ∈ bs(A) ∩ N for s ≤ t, and x tx s ∈ -(bs(A) ∩ N) if s ≥ t. The claim follows.

L k (b)-regular sets

We now turn to equation L k (b) for given integers k, b ≥ 1. We first recall an informal observation made at the beginning of Section 2.

Remark 3.5. Let L denote equation L k (b) for some integers k, b ≥ 1. Let X ⊂ N + . Then X is regular with respect to L if and only if b ∈ k(X -X).
Indeed, we have b ∈ k(X -X) if and only if there exist x 1 , . . . , x k , y 1 , . . . , y k ∈ X such that b = ∑ k i=1 (x iy i ), which therefore constitute a solution of L in X. This allows us to express the L k (b)-regularity of a set X in terms of the discrete derivative ∆X. The resulting lemma will be tacitly used in the sequel. Lemma 3.6. Let X ⊂ Z be a nonempty finite subset. Then X is regular with respect to equation L k (b) if and only if b ∈ k bs(∆X).

Proof. Directly follows from the equality X -X = bs(∆X) of Lemma 3.4 together with Remark 3.5.

Forbidden sequences and subsets

In the sequel, we will try to construct finite subsets X ⊂ Z which are singular with respect to L k (b), with the purpose of showing that it is hard to achieve if X is sufficiently dense in a suitable integer interval. Moreover, working with ∆X rather than X is more convenient. This justifies the following terminology. Definition 3.7. Let L denote equation L k (b) for some integers k, b ≥ 1. Let A = (a 1 , . . . , a r ) be a sequence of positive integers. We say that A is admissible (with respect to L) if b / ∈ k bs(A). We say that A is forbidden if it is not admissible, i.e. if b ∈ k bs(A).

Remark 3.8. With respect to equation L k (b), a subset X ⊂ N is regular if and only if its discrete derivative A = ∆X is forbidden. Equivalently, X is singular if and only if ∆X is admissible.

The following result is the heart of our approach to evaluate the degree of regularity of the Fox-Kleitman equation L k (b). Recall that µ(A) denotes the average of the sequence A = (a 1 , . . . , a r ).

X ⊂ [1, dN + 1] of cardinality |X| = N + 1. Let A = ∆X. Then A is of length N. Let x 0 = min X, x N = max X. Then x N -x 0 = σ(A). Since X ⊂ [1, dN + 1], it follows that x N -x 0 ≤ dN, whence µ(A) ≤ d.
It follows from the hypothesis that A is forbidden with respect to L, and hence that b ∈ k bs(A). Therefore b ∈ k(X -X), i.e. X is L-regular. Since the d-coloring was arbitrary, it follows that L is d-regular, as claimed.

We end this section by looking at ways to construct new forbidden sequences from a given one. Definition 3.10. Let A = (a 1 , . . . , a r ) be a sequence of positive integers.

• An elementary contraction of A is any sequence A obtained by replacing a block A in A by its sum σ(A ). That is, if A = (a i , . . . , a j ) for some 1 ≤ i ≤ j ≤ r, then A = (a 1 , . . . , a i-1 , σ(A ), a j+1 , . . . , a r ).

• A contraction of A is any sequence obtained from A by successive elementary contractions.

For instance, let A = (1, 2, 3, 4). Then (3, 3, 4), (6, 4) and (3, 7) are contractions of A, the first two ones being elementary. Definition 3.11. Let A = (a 1 , . . . , a r ) be a sequence of positive integers. A minor of A is either a block A in A or a contraction A of A.

We now show that if a sequence A has a forbidden minor, then A itself is forbidden.

Proposition 3.12. Let L denote equation L k (b) for some integers k, b ≥ 1. Let A be a finite sequence of positive integers, and let B be a minor of A. If B is forbidden with respect to L, then A also is.

Proof. We have b ∈ k bs(B) by hypothesis. Therefore, to prove that A is forbidden, it suffices to show that bs(B) ⊆ bs(A). This inclusion clearly holds if B is a block in A, since any block sum in B is a block sum in A. If now B is an elementary contraction of A, then again, any block sum in B is a block sum in A. Therefore, the same holds if B is obtained from A by successive elementary contractions.

Here is another condition forcing a sequence A to be forbidden. Observe finally that if A = (a 1 , . . . , a r ) is forbidden with respect to L k (b), then so is the reverse sequence A = (a r , . . . , a 1 ). Indeed, A and A have identical block sums, i.e. bs(A) = bs(A ). Hence, when looking for admissible or forbidden sequences, we only need do it up to reversal.

The case k = 2

As an illustration of the method, we provide here a proof of the Fox-Kleitman conjecture for k = 2, and more generally of Theorem 1.2, which is shorter than the ones given in [START_REF] Adhikari | On a conjecture of Fox and Kleitman on the degree of regularity of a certain linear equation[END_REF][START_REF] Schoen | The degree of regularity of the equation ∑ n i=1 x i = ∑ n i=1 y i + b[END_REF]. The key tool is Proposition 3.9. Proposition 4.1. Let L denote equation L 2 (2). Every positive integer sequence A of length 1 and average µ(A) ≤ 2 is forbidden with respect to L.

A 2 = A 1 by the same argument as for A 1 . Hence A = (2, 5, 2, 2, 5, 2). But this sequence cannot be admissible, since it contains the forbidden minor (2, 2). This contradiction establishes the claim, and the proof is complete.

Corollary 4.4. We have dor(L 2 (6)) = 3.

Proof. Propositions 3.9 and 4.3 imply dor(L 2 (6)) ≥ 3. The reverse inequality follows from Proposition 2.5 using ν(6) = 4.

Note that this equality alone settles the Fox-Kleitman conjecture for k = 2. We are now ready to prove Theorem 1.2 in the following reformulation.

The case k = 3

Our purpose here is to determine dor(L 3 (b)) for all b ≥ 1 and, in the process, independently settle the Fox-Kleitman conjecture for k = 3. The development is self-contained, except for b = 120 where we need to rely on some computer calculations.

The determination of dor(L 3 (b)) is achieved in Theorem 5.11. We start with the more challenging case b ≡ 0 mod 4, treated in the next three sections.

On equation L 3 (b) when b ≡ 4 mod 8

Let L denote equation L 3 (b) for some b ≡ 4 mod 8. Our present purpose is to prove that L is not 4-regular. We shall work with the group G = Z/8Z. Quite naturally, we shall say that a subset X ⊆ G is regular if b ∈ 3 (X -X), singular otherwise. We start by partitioning G into four singular subsets of cardinality 2.

Lemma 5.1. The four subsets {0, 1}, {2, 3}, {4, 5}, {6, 7} of G = Z/8Z constitute a partition of G into singular subsets. and δ 1 , δ 2 ∈ {4, 8, 12, 16, 20, 24, 28} since S is constant mod 4. We must show that δ 1 = δ 2 = 16. Since S is singular, we have 24 / ∈ 3 bs(A). Note that bs(A) = ±{0, δ 1 , δ 2 , δ 1 + δ 2 }.

Let j ∈ {1, 2}. Since 3 bs(A) contains ±{3δ j , 2δ j , δ j } and does not contain 24, it follows that δ j / ∈ {8, 12, 24}. Further, we may assume δ 1 ≤ δ 2 . This is achieved by replacing S by S = 32 -S if necessary, and noting that S is singular if and only if S is, since S -S = S -S. It follows that δ 1 ≤ 16.

Let us now show that δ 1 = 4. For assume, on the contrary, that δ 1 = 4. Each possible value of δ 2 ∈ {4, 16, 20, 28} is then excluded by the following table, which in each case would explicitly write 24 as an element of 3 bs(A), in contradiction with the hypothesis. For that, it suffices to note that 3 bs(A) contains 3(δ 1 + δ 2 ), 2δ 1 + δ 2 , δ 1 + δ 2 + 0 and δ 2δ 1 + 0. This lemma will be used below in conjunction with the following obvious remark.

Remark 5.5. The only subset B ⊂ [0, 32] of cardinality at least 3 which is constant mod 16 is B = {0, 16, 32} of cardinality 3.

We are now ready to prove the main result of this section.

Theorem 5.6. Let L denote equation L 3 (24). Every subset X ⊂ [0, 32] of cardinality |X| = 9 is regular with respect to L.

Proof. Let X ⊂ [0, 32] be such that |X| = 9. Assume for a contradiction that X is singular. Let us partition X according to the class mod 4:

X = X 0 ∪ X 1 ∪ X 2 ∪ X 3 ,
where X i = X ∩ (i + 4N) for all 0 ≤ i ≤ 3. There are several steps.

Step 1. Since subsets of singular sets are singular, and since X is singular, it follows that X i is singular for all 0 ≤ i ≤ 3.

Step 2. We claim that X 0 = {0, 16, 32} and that

|X j | = 2 for 1 ≤ j ≤ 3. Indeed, since 9 = |X| = |X 0 | + |X 1 | + |X 2 | + |X 3 |, (7)
there is some index 0 ≤ i ≤ 3 for which |X i | ≥ 3. As X i is singular and constant mod 4, it follows from Lemma 5.4 that X i = B = {0, 16, 32}. Thus i = 0 and X 0 = {0, 16, 32} as claimed. Further, it now follows from (7) that

|X 1 | = |X 2 | = |X 3 | = 2.
Step 3. Taking the discrete derivative of X 1 , X 2 , X 3 , let ∆X j = (δ j ) for j = 1, 2, 3. Since the X j 's are positive and constant mod 4, it follows that δ j ∈ 4N + . Moreover, since 0 < δ j < 32 for 1 ≤ j ≤ 3, we have

δ j ∈ {4, 8, 12, 16, 20, 24, 28}.
Step 4. We claim that δ 1 = δ 2 = δ 3 = 16. Indeed, let j ∈ {1, 2, 3}. Since X is singular, we have 24 / ∈ 3 (X -X). On the other hand, since X 0 and X ⊃ X j , we have 3 (X -X) ⊇ {3δ j , 2δ j , δ j }. Hence δ j / ∈ {8, 12, 24}. Further, since 24 = 4 + 4 + 16 = 20 + 20 -16 = 28 + 28 -32 and since X -X ⊇ {±16, -32, δ j }, it follows that δ j / ∈ {4, 20, 28}. Hence δ j = 16, as claimed.

Step 5. For j ∈ {1, 2, 3} and z ∈ N, let us denote

Y j (z) = {4z + j, 4z + j + 16}.
Since X j is of class j mod 4 and since δ j = 16, we have X j = Y j (a j ) for some a j ∈ N. Further, since X j ⊂ [0, 32], we have a j ∈ {0, 1, 2, 3}.

Step 6. Thus X depends on the three parameters a 1 , a 2 , a 3 . We shall then write

X = X(a 1 , a 2 , a 3 ) = X 0 ∪Y 1 (a 1 ) ∪Y 2 (a 2 ) ∪Y 3 (a 3 ).
Since 0 ≤ a j ≤ 3 for j ∈ {1, 2, 3}, there are 64 cases to consider. Our task is to show that X(a 1 , a 2 , a 3 ) is regular in each one, thereby leading to a contradiction and concluding the proof of the theorem. Fortunately, it turns out that only 8 cases need to be considered.

Step 7. To start with, we may assume a 2 ∈ {0, 2}. Indeed, for any subset Z ⊆ [0, 32], denote Z = 32 -Z as in the proof of the preceding lemma. Then Z ⊆ [0, 32] and Z is singular if and only if Z is. Now note that for 0 ≤ a ≤ 3, we have

Y 2 (a) = Y 2 (3 -a)
as easily verified. In particular, we have Y 2 (1) = Y 2 (2) and Y 2 (3) = Y 2 (0). The claim follows by replacing X by X if a 2 is odd. This reduces the number of cases to consider from 64 to 32.

Step 8. We now see that it suffices to consider the 8 cases given by (a 2 , a 3 ) ∈ {0, 2} × {0, 1, 2, 3}, the value of the parameter a 1 being irrelevant. Indeed, for the 8 listed cases, the following table shows that X = X(a 1 , a 2 , a 3 ) is regular by explicitly writing 24 as an element of 3 (X -X). These contradictions conclude the proof of the theorem.

(a 2 , a 3 ) X 2 ∪ X 3 = Y 2 (a 2 ) ∪Y 3 (a 3 ) 24 = (0, 0) {2, 18, 3, 19} (18 -0) + (3 -0) + (3 -0) (0, 1) {2, 18, 7, 23} (23 -2) + (7 -2) + (0 -2) (0, 2) {2, 18, 11, 27} (11 -0) + (11 -0) + (2 -0) (0, 3) {2, 18, 15, 31} (15 -2) + (15 -2) + (0 -2) (2, 0) {10, 26, 3, 19} (
Corollary 5.7. We have dor(L 3 (24)) = 4.

Proof. Consider an arbitrary 4-coloring of the integer interval [0, 32]. Since that interval has cardinality 33, one of the color classes contains a subset X of cardinality 9. By the theorem, X is regular, and hence contains a solution to L, which is monochromatic by construction. This implies dor(L 3 (24)) ≥ 4. The reverse inequality follows from Lemma 2.4 and the fact that 5 does not divide 24.

On equation L 3 (120)

We establish here that equation L 3 (120) is 5-regular, thereby independently settling the Fox-Kleitman conjecture for k = 3. We give two different proofs. They both require some computer calculations, but of a very different nature. Here is the first approach. Then |X| = 111 and, for every 5-coloring of X, there is a monochromatic solution of equation L 3 (120) in X.

Proof. By translating this coloring problem as a Boolean satisfiability problem and then feed it to a SAT solver. The solver march reached the conclusion in about 20 seconds on a standard desktop computer. The set X itself was discovered through a patient and delicate computer-aided purification process using march.

Corollary 5.9. We have dor(L 3 (120)) = 5.

Proof. Proposition 5.8 implies that L 3 (120) is 5-regular, i.e. dor(L 3 (120)) ≥ 5. The reverse inequality follows from Lemma 2.1.

Our second proof of the 5-regularity of L 3 (120) uses Proposition 3.9. Here is the precise statement.

Theorem 5.10. Every positive integer sequence A of length 80 and average µ(A) ≤ 5 is forbidden with respect to L 3 (120). Moreover, 80 is minimal with respect to that property.

Proof. The first part of this result has been obtained by exhaustive computer search. Combined with Proposition 3.9, it directly implies Corollary 5.9.

The fact that length 80 is minimal for the stated property is witnessed by the following instances. First, for 1 ≤ r ≤ 39, the sequence of length 2r + 1. Then A(r) is admissible for all r ≤ 39, and it is of average µ(A(r)) ≤ 5 for all r ≥ 20. Since any block of an admissible sequence is admissible, we get admissible sequences of average at most 5 and of any length 41 ≤ l ≤ 79. Finally, an admissible sequence of length 40 and average exactly 5 is provided by the sequence A(20) with the last 1 removed.

where the last inclusion derives from 0 ∈ bs(A). Thus (a) is indeed forbidden with respect to L 4 (12).

By Proposition 3.12, any sequence admitting either (3), (4), ( 6) or (12) as a minor is forbidden. Let us now look at forbidden minors of length 2.

• The sequences A = (1, a) with 2 ≤ a ≤ 12 are forbidden. This is clear if 2 ≤ a ≤ 6 by the above. If a = 7, then 14 ∈ 2 bs(A), whence 12 = -1 -1 + 14 ∈ 4 bs(A). If 8 ≤ a ≤ 9, then 9 ∈ bs(A), whence 12 = 1 + 1 + 1 + 9 ∈ 4 bs(A). If a = 10, then 12 = 1 + 1 + 10 ∈ 4 bs(A). And finally, if 11 ≤ a ≤ 12, then (12) is a forbidden minor of A.

• The sequences A = (2, a) with 2 ≤ a ≤ 8 are forbidden. Indeed, if 2 ≤ a ≤ 4, then either (3) or (4) is a forbidden minor. If a = 5, then 12 = 2 + 5 + 5 ∈ 4 bs(A). If a = 6, then (6) is a forbidden minor. If a = 7, then 12 = -2 + 7 + 7 ∈ 4 bs(A). And finally, if a = 8, then 12 = 2 + 2 + 8 ∈ 4 bs(A).

Let now A = (a 1 , a 2 , a 3 ), and assume µ(A) ≤ 4, i.e. σ(A) ≤ 12. Let x = min A. Then x ≤ 4. If x = 3 or 4, then A is forbidden. Assume now x ≤ 2.

Let y, z be the other two members of A. We may assume that y is a neighbor of x in A, so that either (x, y) or (y, x) is a block in A. We have x ≤ y, z by hypothesis.

If x = 2, then y ≤ 8 since σ(A) ≤ 12, whence A is forbidden since (2, y) or (y, 2) is a forbidden minor as seen above.

Finally, if x = 1, then y ≤ 10 since σ(A) ≤ 12, whence A is forbidden since (1, y) or (y, 1) is a forbidden minor.

In all cases, we conclude that A is forbidden, as claimed.

Corollary 6.2. We have dor(L 4 (12)) = 4.

Proof. Follows from Lemma 2.4 and from Proposition 3.9 applied to the above statement.

On equation L 4 (6!)

Our second instance of discrepancy between dor(L 4 (b)) and min(7, f (b)) is for b = 6! = 720. The equality f (6!) = 6 is obvious.

Theorem 6.3. We have dor(L 4 (6!)) = 5.

Proof. Let L denote equation L 4 (6!). On the one hand, by successively applying Corollary 5.9, Lemma 2.2 and Lemma 2.3, we have 5 = dor(L 3 (5!)) ≤ dor(L 4 (5!)) ≤ dor(L 4 (6!)).

It remains to show dor(L 4 (6!)) < 6. For that, it suffices to exhibit a partition of Z into 6 subsets which are singular with respect to L. Denote X 0 = {0, 1} + 11Z, and X i = X 0 + 2i for 1 ≤ i ≤ 4, and finally X 5 = {10} + 11Z. The sets X 0 , X 1 , X 2 , X 3 , X 4 , X 5 constitute a partition of Z. We claim that they are all Lsingular. To this end, it suffices to show that X 0 is L-singular, since the other five sets are either translates of X 0 or subsets thereof. Thus, it remains to show that 720 / ∈ 4(X 0 -X 0 ). To do this, let us apply the canonical morphism from Z to the quotient group G = Z/11Z. Let X = {0, 1} ⊂ G. Then X 0 is mapped to X, and 720 is mapped to 5 since 720 = 11 • 65 + 5. It remains to show that 5 / ∈ 4(X -X) in G. But this is obvious, since X -X = {-1, 0, 1} and hence 4(X -X) = {-4, -3, . . . , 3, 4}.

Is dor(L 4 (b)) = 7 realizable?

That the answer to this question is positive is precisely the Fox-Kleitman conjecture for k = 4. We show here, quite surprisingly, that any integer b which would satisfy dor(L 4 (b)) = 7 must be much bigger than 7! = 5040. This is our third instance of discrepancy between dor(L 4 (b)) and min(7, f (b)). For b = 18! for instance, we have f (18!) = 18 and so min(7, f (18!)) = 7, whereas the following result implies dor(L 4 (18!)) < 7. Proof. To start with, if b ≡ 0 mod 3 × 4 × 5 × 7 then ν(b) ≤ 7, and Proposition 2.5 implies dor(L 4 (b)) < 7. In order to treat the remaining prime factors 11, 13 and 19, we proceed as in the proof of Theorem 6.3.

First, let p ∈ {11, 13}. Let X 0 = {0, 1}+ pZ, and consider its image X = {0, 1} in the quotient group G = Z/pZ. Then 4(X -X) = {-4, -3, . . . , 3, 4} = G. For instance, 5 / ∈ 4(X -X). Let b ∈ N + such that b ≡ 0 mod p. Since b is invertible mod p, there exists c ∈ N + such that bc ≡ 5 mod p. Then bc / ∈ 4(X 0 -X 0 ), since 5 / ∈ 4(X -X) in G. Hence X 0 is singular with respect to L 4 (bc). The same is true for its translates t + X 0 with t ∈ {2, 4, 6, 8, 10, 12}. It follows that equation L 4 (bc) is not 7-regular. By Lemma 2.3, we get dor(L 4 (b)) ≤ dor(L 4 (bc)) < 7.

Assume now p = 19. The proof proceeds in an analogous way, except that here we need consider X 0 = {0, 1, 2} + pZ and its image X = {0, 1, 2} in the quotient group G = Z/pZ. Then 4(X -X) = {-8, -7, . . . , 7, 8} = G. For instance, 20 9 / ∈ 4(X -X). Let b ∈ N + such that b ≡ 0 mod p. Since b is invertible mod p, there exists c ∈ N + such that bc ≡ 9 mod p. Hence X 0 is singular with respect to L 4 (bc), and the same is true for its translates t + X 0 with t ∈ {3, 6, 9, 12, 15, 18}. It follows that equation L 4 (bc) is not 7-regular, and by Lemma 2.3 again we conclude dor(L 4 (b)) < 7.

The arguments in the above proof cannot be extended to prime factors greater than 19. We may thus ask whether the equality dor(L 4 (19!)) = 7 might hold. For the time being, we can neither prove nor disprove it.

  Theorem 1.2 ([AE15]). For all positive integers b, we havedor(L 2 (b)) =    1 if b ≡ 1 mod 2, 2 if b ≡ 2, 4 mod 6, 3 if b ≡ 0 mod 6.

  dor(L 2 (b)) = min(3, f (b)) = min(3, ν(b) -1) (2) for all b ≥ 1. Our purpose in this paper is to similarly determine dor(L 3 (b)) for all b ≥ 1. Indeed, with the upper bound (1) in mind, we shall establish the equality dor(L 3 (b)) = min(5, f (b)) (3) for all b ≥ 1. This implies dor(L 3 (5!)) = 5, thereby verifying the Fox-Kleitman conjecture for k = 3. Unfortunately, in contrast with (2), there is no equality in general between dor(L 3 (b)) and min(5, ν(b) -1). For b = 60 for instance, we shall see that dor(L 3 (60)) = 3, whereas min(5, ν(60) -1) = 5. However, the inequality dor(L k (b)) ≤ min(2k -1, ν(b) -1) (4) always holds, as will be shown further down. Moreover, for fixed b ≥ 1, we shall prove that dor(L k (b)) = min(2k -1, ν(b) -1) = ν(b) -1 (5) for all sufficiently large k, in fact for all k ≥ b. As for the function f , can one expect, based on (2), (3) and (4), an equality or at least an inequality between dor(L k (b)) and min(2k -1, f (b)) for k ≥ 4? Here again, the answer turns out to be negative. Indeed, for k = 4, we shall establish the following values. b dor(L 4 (b)) min(7, f (b)) the behavior of dor(L k (b)) as a function of b, for fixed k ≥ 4, is much more tricky than what formulas (2) and (3) for k ≤ 3 might lead one to expect. In the same vein, we shall show that if b is a positive integer satisfying the Fox-Kleitman conjecture for k = 4, i.e. such that dor(L 4 (b)) = 7, then necessarily b must be divisible by 2 2 • 3 • 5 • 7 • 11 • 13 • 19 = 1141140. At the time of writing, we do not know whether dor(L 4 (b)) reaches 7 for b = 1141140.

Lemma 2. 4 .

 4 Let b, m be positive integers such that m does not divide b. Then dor(L k (b)) ≤ m -1.

  Proposition 3.9. Let L denote equation L k (b) for some integers k, b ≥ 1. Let d ≥ 1. Assume that there exists an integer N ≥ 1 such that all positive integer sequences A of length N and average µ(A) ≤ d are forbidden with respect to L. Then L is d-regular. Proof. Consider an arbitrary d-coloring of the integer interval [1, dN + 1]. By the pigeonhole principle, there exists a monochromatic subset

  Proposition 3.13. Let L denote equation L k (b) for some integers k, b ≥ 1, and let A be a finite sequence of positive integers. If bs(A) contains a subset Z such that b ∈ k (Z ∪ -Z), then A is forbidden with respect to L. Proof. If Z ⊆ bs(A), then (Z ∪ -Z) ⊆ bs(A) since bs(A) = -bs(A). Therefore k (Z ∪ -Z) ⊆ k bs(A). The hypothesis on b then implies b ∈ k bs(A), and we are done.

  Theorem 4.5. We have dor(L 2 (b)) = min(3, ν(b) -1) for all b ≥ 1. Proof. If ν(b) = 2, then b is odd, whence dor(L 2 (b)) = 1 by Lemma 2.4. If ν(b) = 3, then b is even but not divisible by 3. Since dor(L 2 (2)) = 2 and since 2 divides b, it follows from Lemma 2.3 that dor(L 2 (b)) ≥ 2. The reverse inequality follows from Lemma 2.4. Finally, if ν(b) ≥ 4, then 6 divides b. Again, since dor(L 2 (6)) = 3, it follows that dor(L 2 (b)) = 3 by Lemmas 2.1 and 2.3.

  4 + 16 20 4 + 20 + 0 28 28 -4 + 0 It follows that δ 1 = 16, whence δ 2 = 16 as well, since δ 1 ≤ δ 2 and δ 1 + δ 2 ≤ 32.

  10 -3) + (10 -3) + (10 -0) (2, 1) {10, 26, 7, 23} (10 -0) + (7 -0) + (7 -0) (2, 2) {10, 26, 11, 27} (27 -10) + (27 -10) + (0 -10) (2, 3) {10, 26, 15, 31} (15 -0) + (15 -0) + (10 -16)

Proposition 5. 8 .

 8 Let X = -Y ∪ {0} ∪Y , where Y = [1, 30] ∪ 5 • [7, 20] ∪ (110 + 10 • [0, 5]) ∪ (220 + 60 • [0, 4]) = {1, . . . , 30} ∪ {35, 40, . . . , 100} ∪ {110, 120, . . . , 160} ∪ {220, 280, . . . , 460}.

( 1 ,

 1 . . . , 1 r ) is admissible and of average 1. Next, let A(r) = (1, . . . , 1 r , 121 + 2r, 1, . . . , 1 r )

Theorem 6. 4 .

 4 If b ≡ 0 mod 3 • 4 • 5 • 7 • 11 • 13 • 19, then dor(L 4 (b)) < 7.

7

  The case k ≥ b While for k ≥ 1 fixed, it looks hard to determine dor(L k (b)) as a function of b, we now show that, for b ≥ 1 fixed, it is easy to determine dor(L k (b)) for all sufficiently large k, in fact for all k ≥ b. The answer again involves the function ν. Proposition 7.1. Let b ≥ 1. Then, for all k ≥ b, we have dor(L k (b)) = ν(b) -1. Proof. Let m = ν(b). We have dor(L k (b)) ≤ m -1 by Corollary 2.5. For the reverse inequality, we shall invoke Proposition 3.9 at N = 1. So let A = (a) be any positive integer sequence of length 1 and average µ(A) ≤ m -1. Thus a ≤ m -1, whence a divides b by definition of m = ν(b). We then have b = (b/a)a ∈ (b/a) bs(A) ⊆ k bs(A) since b/a ≤ k, whence A is forbidden with respect to L k (b). It then follows from Proposition 3.9 that L k (b) is (m -1)-regular, i.e. dor(L k (b)) ≥ m -1.

Acknowledgement. The authors wish to thank the referees for their very careful reading of a preliminary version of this paper, their sharp comments correcting of a few dubious points, and for pointing out [ST].

Proof. The only sequences to consider are A = (a) with a ∈ {1, 2}. Then bs(A) = {0, a, -a} and 2 bs(A) = {0, a, 2a, -a, -2a}. Hence 2 ∈ 2 bs(A) in either case, showing that A is forbidden.

Corollary 4.2. We have dor(L 2 (2)) = 2.

Proof. Propositions 3.9 and 4.1 imply that L 2 (2) is 2-regular, i.e. dor(L 2 (2)) ≥ 2. Since ν(2) = 3, the reverse inequality follows from the bound dor(L k (b)) ≤ min(2k -1, ν(b) -1) of Proposition 2.5. Proposition 4.3. Let L denote equation L 2 (6). Every positive integer sequence A of length 6 and average µ(A) ≤ 3 is forbidden with respect to L 2 (6).

Proof. In length 1, the sequences (3) and (6) are clearly forbidden.

Therefore, by Proposition 3.12, any sequence admitting either (3) or (6) as a minor is forbidden. Let us now look at forbidden minors of length 2.

• The sequences A = (1, a) with 2 ≤ a ≤ 7 are forbidden. This is clear if a ∈ {2, 3, 6} by the above. If 4 ≤ a ≤ 5, then 1, 5 ∈ bs(A), hence 6 ∈ 2 bs(A). Finally, if a = 7, then 6 = -1 + 7 ∈ 2 bs(A).

• The sequences A = (2, a) with a ∈ {2, 3, 4, 6, 8} are forbidden. This is clear if a ∈ {3, 6} by the above. If a = 2 or 4, then 2, 4 ∈ bs(A), hence 6 ∈ 2 bs(A). Finally, if a = 8, then 6 = -2 + 8 ∈ bs(A). Claim. If A = (a 1 , a 2 , a 3 ) is admissible and if µ(A) ≤ 3, then A = (2, 5, 2). Indeed, we have σ(A) ≤ 9 by hypothesis. Let x = min A. Then x ≤ 3. As A is admissible, it cannot contain 3 for otherwise (3) would be a forbidden minor. Hence x ≤ 2. Assume x = 1. Since (1, 1, 1) is forbidden, because (3) is a minor of it, there can be at most two 1's in A. Let a ∈ A with a = 1. We may assume that (1, a) or (a, 1) is a block in A. As A is admissible, we must have a ≥ 8 by the length 2 case above, whence σ(A) ≥ 10. But this is impossible since σ(A) ≤ 9 by assumption.

Hence x = 2. Let (2, a) or (a, 2) be a block in A. Since A is admissible, then a ≥ 5 by the length 2 case, whence σ(A) ≥ 2 + 2 + a ≥ 9. But since σ(A) ≤ 9, we must have a = 5 and A = (2, 5, 2). This settles the claim.

Indeed, assume for a contradiction that A is admissible. Then every block in A is admissible, and in particular so are its two halves A 1 = (a 1 , a 2 , a 3 ), A 2 = (a 4 , a 5 , a 6 ). Since µ(A) ≤ 3, we have µ(A i ) ≤ 3 for some i ∈ {1, 2}, say µ(A 1 ) ≤ 3 up to renumbering. But since A 1 is admissible, we must have A 1 = (2, 5, 2) by the above claim, and in particular µ(A 1 ) = 3. Therefore µ(A 2 ) ≤ 3 as well, whence

Thus X is L-singular, as claimed. Since the property of being L-singular is stable under translation, the three translates X +t with t ∈ {2, 4, 6} are also singular. ∈ 3(X -X), since in the quotient group G = Z/8Z, we have b = 4 / ∈ 3(X -X) as seen in the above lemma. Said otherwise, the subset X ⊂ Z is singular with respect to equation L 3 (b). This property remaining true under translation, the subsets X +t with t ∈ {0, 2, 4, 6} constitute a partition of Z into four L 3 (b)-singular sets. This implies dor(L 3 (b)) ≤ 3 as stated.

On equation L 3 (24)

Let L denote equation L 3 (24). Our aim here is to prove that L is 4-regular. During research on this paper, our first proof of this followed the same line of reasoning as above, using Proposition 3.9 as the key ingredient. Indeed, the 4-regularity of L directly follows from that tool applied to the following statement.

Proposition 5.3. Let L denote equation L 3 (24). Every positive integer sequence A of length 8 and average µ(A) ≤ 4 is forbidden with respect to L. Now, our detailed proof of this proposition is several pages long. In its place, we shall present here an alternate shorter proof of the 4-regularity of L with a slightly different and more ad-hoc approach. We start with a lemma on L-singular subsets of [0, 32] which are constant mod 4, i.e. contained in a single class a + 4N for some integer a.

Lemma 5.4. Let S ⊂ [0, 32] be an L-singular subset of cardinality at least 3 which is constant mod 4. Then S is constant mod 16.

Proof. We may assume |S| = 3, since if the statement is valid in that particular case, its validity automatically extends to the general case |S| ≥ 3. Let A = ∆S = (δ 1 , δ 2 ) be the discrete derivative of S. We have δ 1 + δ 2 ≤ 32 since S ⊂ [0, 32], , where again f (b) denotes the largest integer r such that r! divides b. However, it turns out that this is far from being the case, as shown in the next three sections.

We start with the case b = 12, for which f (12) = 3 but where dor(L 4 (12)) turns out to be equal to 4.

On equation L 4 (12)

We show here that equation L 4 (12) is 4-regular. Proposition 6.1. Let L denote equation L 4 (12). Every positive integer sequence A of length 3 and average µ(A) ≤ 4 is forbidden with respect to L 4 (12).

Proof. In length 1, the sequences (3), (4), (6) and (12) are all forbidden. Indeed, let A = (a) with a ∈ {3, 4, 6, 12}, and let t = 12/a ≤ 4. Since 12 = ta, we have 12 ∈ t A ⊆ t bs(A) ⊆ 4 bs(A),


Vasiliki Myrodia

Aurélie Calabrèse

Ambre Denis-Noel

Frédéric Matonti

Pierre Kornprobst

Eric Castet

Pointing at static targets in a virtual reality environment: performance of visually impaired vs. normally-sighted persons

de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d'enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Pointing at static targets in a virtual reality environment: performance of visually impaired vs. normally-sighted persons

Introduction

The pointing task mobilizes visuo-attentional and oculomotor functions. Our hypothesis is that it could be a good rehabilitation tool. We have developed this pointing task as an exploratory experiment in which we try to know if the patients manage to do it.

Question

How can a visually impaired person with central vision impairment perform a pointing task with the head in a virtual environment?

Method

Virtual Reality & Microperimetry

Subjects must point at the target using a reticle that moves with the head. This study is developed with PTVR, an opensource VR toolbox (https://ptvr.inria.fr/).

Discussion

The virtual reality was well received by the visually impaired subjects. Results show a significant difference in performance between patients and normally-sighted persons. An effect of the activation radius on the reaction time is observed as well.

Results

Virtual Reality

Perspectives

Analyze the performance of patients as the function of their scotoma characteristics. Monocular vision experiment.

FigureFigure 2 .

 2 Figure 1. Trial procedure

Figure 3 .

 3 Figure 3. The effect of the activation radius on the reaction time

FigureFigure 6 .

 6 Figure 5. Average reaction time for a patient

Aix-Marseille University, CNRS

Centre Monticelli Paradis d'Ophtalmologie

Université Côte d'Azur, Inria

Vision Sciences Society May 19-24, 2023 vasiliki.myrodia@univ-amu.fr


Florian Faucher
email: florian.faucher@inria.fr

Otmar Scherzer

Synthetic dataset for visco-acoustic imaging

Keywords:

a b s t r a c t

We provide computationally generated dataset simulating propagation of ultrasonic waves in viscous tissues in two and three dimensional domains. The dataset contains physical parameters of a human breast with a high-contrast inclusion, the acquisition setup with positions of sources and receivers, and the associated pressure-wave data at ultrasonic frequencies. We simulated the wave propagation based on seven different viscous models using the physical parameters of the breast. Furthermore, different choices of conditions for the medium's boundaries are given, namely absorbing and reflecting boundaries. The dataset allows to evaluate the performance of reconstruction methods for ultrasound imaging under attenuation model uncertainty, that is, when the precise attenuation law that characterizes the medium is unknown. In addition, the dataset enables to evaluate the robustness of inverse scheme in the context of reflecting boundary conditions where multiple reflections illuminate the sample, and/or the performance of data-processing to suppress these multiple reflections.

Specifications Table

Subject

Mathematical modeling -Computational Mathematics Specific subject area Viscous materials -ultrasonic wave imaging -inverse problems -visco-acoustic wave propagation Type of data Models of physical properties of breast sample -Pressure wave data -2D and 3D synthetic data How the data were acquired Synthetic data of visco-acoustic wave propagation are generated using software hawen, [START_REF] Faucher | hawen: time-harmonic wave modeling and inversion using hybridizable discontinuous Galerkin discretization[END_REF] ( https://ffaucher.gitlab.io/hawen-website ). The geometry of the structures for the samples are obtained from a cross-section of the OABreast Phantom dataset [START_REF] Lou | Generation of anatomically realistic numerical phantoms for photoacoustic and ultrasonic breast imaging[END_REF] ( https://anastasiolab.wustl.edu/downloadable-content/oa-breast-database ). The values of the physical properties in each layer of the sample is selected according to the IT'IS Database ( https://itis.swiss/virtual-population/tissue-properties ).

Value of the Data

• The dataset provides simulations of the pressure fields showing the propagation of ultrasonic waves in a breast sample for seven models of attenuation. This is the first collection of such simulations. It is therefore useful for visualization effects of different viscosity models on the propagation of waves. In addition, these data can be used for evaluation of computational ultrasound imaging (inversion) techniques. In particular, two medium configurations for the data are provided: An implementation with absorbing conditions that approximate the freespace propagation which is very often used in the physics literature, and a configuration with reflecting (wall) conditions. • The dataset offers two and three dimensional benchmark data for visco-acoustic imaging that can be used by researchers and engineers working on inverse wave problems. • The data can be used by experimentalists and computational scientists as a basis for testing computational ultrasound inversion techniques under several emitter and receiver configurations, as well as for different attenuation models characterizing the propagation within the sample.

Objective

The dataset made available for the community can be used for the following purposes:

1. For the comparison of computationally simulated wave propagation in viscous media: We provide data for different viscous models and different variants of implementations, in particular free-space and implementation on compact sets. 2. For ultrasound inversion algorithms based on (partial) data and for investigating reconstructions with different models of attenuation. 3. For the investigation of the scalability of simulations (forward model) and ultrasound inversion algorithms as we provide simulations of wave propagation in viscous media in space dimensions two and three.

In the related research paper [START_REF] Faucher | Quantitative inverse problem in visco-acoustic media under attenuation model uncertainty[END_REF] , the dataset allows to develop and validate an efficient methodology for the reconstruction of visco-acoustic media with attenuation model uncertainty, and with different choices of boundary conditions surrounding the sample.

Data Description

The folder hierarchy separates the two and three-dimensional datasets:

• Folder 2D/ contains simulated data of two-dimensional wave propagation and the associated material parameters of the 2D sample. • Folder 3D/ contains simulated data of three-dimensional wave propagation and the associated material parameters of the 3D sample.

Each folder further contains three subfolders which are detailed in the below table for the 2D numerical experiment, and the 3D experiment uses the same organization and structure of files, however with less choices of attenuation models. We further refer to Table 1 for more details.

Table 1

Directories for the 2D dataset.

2D dataset folder

2D/acquisition

It contains two text files: sources.txt and receivers.txt that correspond to the list of the 2D positions of the sources and of the receivers. The first line of each file indicates the number of sources (36) and of receivers (360), then each subsequent line gives the index of the source/receiver and the two coordinates of its position, illustrated in Figs. 1 and2 .

2D/models 2D/models/main 2D/models/attenuation

It contains the stored parameters of the breast phantom. In the other subfolders are the material parameters depending on the choice of attenuation models, with seven options, cf. [START_REF] Faucher | Quantitative inverse problem in visco-acoustic media under attenuation model uncertainty[END_REF] . For instance, folder kelvin-voigt/ contains the physical parameters to simulate the Kelvin-Voigt attenuation model within the medium.

2D/wave-dataset/ It contains simulated pressure fields for various visco-acoustic models. In the first subdirectory, there are two subfolders, each contains simulated pressure fields for a pair of attenuation models and boundary conditions. The data are saved in text files, with one file per source and per frequency: For instance, file data-record_p_frequency_0.0 0 0 0 0E + 0 0_2.0 0 0 0 0E + 05Hz_src0 0 0 010.txt corresponds to simulation at 200kHz (indicated by 2.0 0 0 0 0E + 05Hz ), for the source number 10 (indicated by src0 0 0 010 ).

Moreover, the first number corresponds to the imaginary part of the frequency, see Eq. ( 1) : This is only non-zero in the case of wall boundary conditions, and we specify below the computed frequencies.

Each file contains 360 complex values that correspond to the pressure value at the positions of the receivers. Note that the positions of the sources and receivers can be found in the acquisition folder described above.

In Fig. 3 , we illustrate graphically one of this file with the real and imaginary parts of the provided pressure field.

Model description. For the two-dimensional experiment, the physical parameters are given on a Cartesian grid (folder 2D/models/ ) of size 1001 × 1001 with a discretization step of 0.18mm. That is, we use models of size 18 × 18cm 2 . The three-dimensional physical parameters are given on a Cartesian grid of size 121 × 191 × 201 with discretization steps of 0.833mm along xaxis, 0.842mm along y-axis and 0.800mm along z-axis. Therefore, the 3D models are of size 10 × 16 × 16cm 3 .

Frequency content of the data. We consider complex frequencies according to Eq. ( 1) . For the two-dimensional experiment, the frequencies simulated are the following:

• With absorbing boundary conditions, we provide the pressure waves for real frequencies 20 0, 30 0, 40 0, 50 0 and 60 0 kHz. • With wall boundary conditions, the pressure field are provided for the real frequencies 20 0, 30 0, 40 0, 50 0 and 600 kHz, and the same ones including imaginary part 10 × 10 3 , 15 × 10 3 and 20 × 10 3 .

The frequencies simulated for the three-dimensional experiment are the following:

• With absorbing boundary conditions, we provide frequencies 100, 200, 300 and 400 kHz.

• With wall boundary conditions, we provide the ordinary frequencies 10 0, 20 0, 30 0 and 400 kHz, and the same frequencies including imaginary part: 10 × 10 3 , 15 × 10 3 and 50 × 10 3 .

Fig. 1. Configuration of the two-dimensional experiment which is of size 18 × 18cm 2 , with the wave speed (left subfigure) that corresponds to binary file 2D/models/main/cp_1001 ×1001.bin and density (right subfigure) to file 2D/models/main/rho_1001 ×1001.bin. The models of viscosity have the same structures, however with different scales adapted to each attenuation models ( Table 2 ); these scales are chosen such that all attenuation models coincide at frequency 300 kHz, [START_REF] Faucher | Quantitative inverse problem in visco-acoustic media under attenuation model uncertainty[END_REF] . The positions of the sources are indicated with • and the receivers are along the dashed line. These positions are listed in files 2D/acquisition/sources.txt and 2D/acquisition/receivers.txt respectively.

Experimental Design, Materials and Methods

The objective of inversion is to reconstruct the structures of the medium from wave measurements: these configurations arise in applications such as medical imaging [5 , 10] , seismic imaging [14], and non-destructive testing. The dataset we provide consists in simulated pressure waves that propagate through a breast sample and that are recorded at receivers positioned outside the sample, as illustrated in Fig. 1 . One difficulty of visco-acoustic media is that several mathematical models of attenuation exist, [2 , 13] , and the precise model that characterizes the medium of interest is usually unknown a priori. This increases the difficulty of the reconstruction.

Attenuation is frequency-dependent such that time-harmonic formulation of wave equation is more appropriate [1 , 12] for discretization. Indeed, time-domain equations with attenuation can result in integro-differentiable equations while the frequency domain can be described by partial differential equations and complex-valued parameters, [13 , 3] . In our work, we further allow for complex frequency [8 , 9] such that ω = 2 πν + i ω i , [START_REF] Bland | The Theory of Linear Viscoelasticity[END_REF] where ν is the ordinary frequency and ω i is the imaginary part of the frequency. The propagation is given by the pressure field p and particle velocity v that solve the system of equations:

-i ω ρ( x ) v ( x, ω ) + ∇ p ( x, ω ) = 0 in , (2a) - iω κ † ( x, ω ) p ( x, ω ) + ∇ • v ( x, ω ) = δ( x s ) in . ( 2b
)
Two alternatives are considered for boundary conditions defined on ∂ . Denoting n the normal direction, we have:

absorbing boundary conditions : -ρ κ † -1 / 2 p + v • n = 0 reflecting boundary conditions : ∇ p • n = 0
Here x s is the position of the source which is taken as a Dirac function in x . The density is ρ and the complex bulk modulus κ † depends on the choice of attenuation model and is expressed in terms of a real-valued bulk modulus κ 0 (x ) = c 0 (x ) 2 ρ(x ) where c 0 is the wave speed ( Fig. 1 ).

In order to solve the wave Equation ( 2) in the different configurations to generate the synthetic data, we use the software hawen, cf. [START_REF] Faucher | hawen: time-harmonic wave modeling and inversion using hybridizable discontinuous Galerkin discretization[END_REF] . It relies on the hybridizable discontinuous Galerkin discretization method [START_REF] Cockburn | Unified hybridization of discontinuous Galerkin, mixed, and continuous Galerkin methods for second order elliptic problems[END_REF] , with precise implementation described in [START_REF] Faucher | Adjoint-state method for Hybridizable Discontinuous Galerkin discretization, application to the inverse acoustic wave problem[END_REF] . Consequently, the data generation follows those steps:

1. Using software hawen, find ( p, v ) that solve the wave Equation ( 2 As indicated above, the 2D domain is of size 18 × 18cm 2 and in 3D, the domain is of size 10 × 16 × 16cm 3 . For the numerical resolution, the input parameters are given on a Cartesian grid (see the model description above). However, the computations are performed on an unstructured mesh. The parameters (the bulk modulus κ † and density ρ) are first represented with respect to the mesh via Lagrange basis functions to ensure the accurate representation (see Fig. 7 in [START_REF] Faucher | Adjoint-state method for Hybridizable Discontinuous Galerkin discretization, application to the inverse acoustic wave problem[END_REF] ). The solutions are represented using a piecewise-polynomial representation such that the values of the pressure field are saved at the exact position of the receivers (without need for interpolation).

The formulation of κ † depending on the choice of attenuation model is given in the below table. The different models are extracted from [13 , 2 , 6] see also [START_REF] Faucher | Quantitative inverse problem in visco-acoustic media under attenuation model uncertainty[END_REF] and the references therein.

In Table 3 , we summarize the main characteristics of the dataset for the two and threedimensional cases.

CRediT Author Statement

Florian Faucher: Conceptualization, Methodology, Software, Validation, Formal analysis, Investigation, Resources, Data curation, Writing -original draft, Writing -review & editing, Visualization, Project administration, Funding acquisition; Otmar Scherzer: Conceptualization, Formal analysis, Investigation, Resources, Writing -review & editing, Project administration, Supervision, Funding acquisition.

  Each parameter is described by two files: a header (extension .H ) that indicates the discretization step and sizes of the associated binary file (extension .bin ) that contains the values of the physical parameter. Each binary consists of a 1001 × 1001 array of real numbers which correspond to the pixel representation of the parameter. It contains the wave speed ( cp_1001 × 1001 ), density ( rho_1001 × 1001 ) and quality factor at 300kHz ( Q_30 0kHz_10 01 × 1001 ). The wave speed and density are further pictured Fig. 1 .

Fig. 2 .

 2 Fig.2. Three-dimensional wave speed model of size 10 × 16 × 16cm 3 that corresponds to binary file 3D/models/main/cp_121 × 191 × 201.bin. We picture the cross-sections corresponding to y-z plane (top right), x-z plane (bottom right) and x-y plane (bottom left). In the numerical simulations with wall conditions, an absorbing boundary condition is imposed on the y-z plane for x = x max = 10cm, to mimic the body.

Fig. 3 .

 3 Fig. 3. Pressure field saved for the 2D case at frequency 200kHz for the source numbered 10, in the configuration of absorbing boundary conditions with Kelvin-Voigt model of attenuation. It corresponds to file 2D/wave-dataset /data_absorbing-boundary-conditions/kelvin-voigt/receivers/data-record_p_frequency_0.0 0 0 0 0E + 0 0_2.0 0 0 0 0E + 05Hz_src 0 0 0 010.txt.

  ) for given a. domain (2D or 3D); b. boundary conditions for the domain (absorbing or reflecting); c. parameters ρ and κ † . d. source position; e. frequency ω. 2. Extract the value of the pressure field p solution to the wave equation at the positions of the receivers (illustrated in Fig. 1 for 2D).

Acknowledgments

FF is funded by the Austrian Science Fund (FWF) under the Lise Meitner fellowship M 2791-N. OS is supported by the Austrian Science Fund (FWF) with SFB F6 8, project F6 807-N36 (Tomography with Uncertainties). The financial support by the Austrian Federal Ministry for Digital and Economic Affairs, the National Foundation for Research, Technology and Development and the Christian Doppler Research Association is gratefully acknowledged.

Synthetic dataset for visco-acoustic imaging (Original data) Dataset link: Synthetic dataset for visco-acoustic imaging (Original data)

Data Availability

Synthetic dataset for visco-acoustic imaging (Original data) (phaidra). Synthetic dataset for visco-acoustic imaging (Original data) (Mendeley Data).

Table 2

List of complex bulk modulus depending on the attenuation model.

Attenuation model

Complex bulk modulus formula Simplified Kolsky-Futterman Cartesian grid with step 0.833mm along x-axis, 0.842mm along y-axis and 0.800mm along z-axis Frequency content in the data • with absorbing boundary conditions.

Ordinary frequency: 100, 200, 300 and 400 kHz, Imaginary part: 0.

• with wall boundary conditions. Ordinary frequency: 100, 200, 300 and 400 kHz, Imaginary part: 0, 10 × 10 3 , 15 × 10 3 and 50 × 10 3 .

Ethics Statements

This work meets the ethical requirements of the Journal Data in Brief. Furthermore, this work does not involve experiments on humans or animals.

Declaration of Competing Interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.