1. 1 Zero-sum square matrices Paul Balister Yair Caro Cecil Rousseau Raphael Yuster

2. 2 Definitions Let A be a matrix over the integers, and let p be a positive integer. A submatrix B of A is called zero-sum mod p if the sum of each row and each column of B is a multiple of p. Let M(p,k) denote the least integer m for which every square matrix of order at least m over the integers, has a square submatrix of order k which is zero-sum mod p. As usual in zero- sum problems, we assume that k is a multiple of p, otherwise M(p,k) may be infinite. If k is a multiple of p, then M(p,k) is finite by a standard Ramsey-type argument: Consider the elements of our matrices as taken from Z p + : we have a coloring of an m*m grid with p colors, so, for sufficiently large m, there is a k*k sub-grid which is monochromatic. This translates to a (very specialized) zero-sum mod p submatrix of order k. 1 4 2 1 2 1 0 1 2 1 1 1

3. 3 Graph-theoretic interpretation (p=2) A zero-one matrix A=A m,n can be interpreted as the adjacency matrix of a bipartite graph G=(X  Y, E), where X=x 1,…,x m and Y=y 1,…,y n. In this interpretation, A(i,j)=1 if and only if x i is adjacent to y j. Hence, M(2,k) is the least integer m that guarantees that every bipartite graph with equal vertex classes having cardinality m, has an induced subgraph with equal vertex classes having cardinality k, such that all degrees are even. By the Ramsey-type argument mentioned above, it is obvious that M(2,k) ≤ B(k) where B(k) is the Bipartite Ramsey Number. The best upper bound and lower bound for B(k) are currently: √2e -1 k2 k/2 < B(k) ≤ 2 k (k-1)+1 In particular, we get M(2,k) ≤ 2 k(k-1) +1.

4. 4 Main result Unlike the exponential lower bound for B(k), we can show that M(2,k) is linear in k. We can also show that M(3,k) is linear in k for infinitely many values of k. Theorem: –limsup M(2,k)/k ≤ 4. –liminf M(3,k)/k ≤ 20. In particular, for every positive integer s, M(3,3 s ) ≤ 20 * 3 s (1+o(1)) Conjecture: For every positive integer p, there exists a constant c p such that limsup M(p,k)/k ≤ c p.

5. 5 The theorem implies the conjecture for p=2 with c 2 ≤ 4. It is not difficult to construct examples showing M(2,k) ≥2k+1. Thus, we cannot have c 2 < 2. The construction (example for k=4): 110 000 00 01100000 00110000 00011000 00001100 00000110 00000011 10000001

6. 6 It is extremely difficult to compute M(2,k) (moreover M(p,k) for p >2) even for relatively small values of k. It is an easy exercise to show that M(2,2)=5. A naive approach for computing M(2,4) with a computer program which generates all possible 0-1 matrices and testing them would fail. By using, once again, the fact that M(2,k) ≥ 2k+1, we know that M(2,4) ≥ 9. Even generating all 0-1 matrices of order 9 is not feasible as there are 2 81 such matrices (and even then, maybe the correct value is larger than 9). We used a sophisticated computer search which enables us to narrow down the checks considerably. With the aid of this search we were able to show: Proposition: –M(2,4)=10. A binary matrix of order 9 that demonstrates M(2,4) > 9 is the following (it is unique up to complement and row/column interchanges):

7. 7 J0IIJ00IJJ0IIJ00IJ M(2,6) is already an open problem. For general p, it is not difficult to show that M(p,k)/k grows exponentially with p. For fixed p the lower bound is linear in k M(p,k) ≥ (k/5)e p/2e. For p > 3, nothing but the trivial exponential upper bound that follows from bipartite Ramsey numbers with p colors is known. Thus, the conjecture is far from being proved except for p=2 (and p=3 with the liminf version).

8. 8 Proof of the main result First tool: A theorem of Enomoto, Frankl, Ito and Nomura. Recall that a linear binary code of length t contains the distance k if it contains as a codeword a vector with k ones and t-k zeroes. Their theorem: Let k be even, and let A be a binary matrix with k-1 rows and 2k columns. Then, A is a parity-check matrix of a linear code which contains the distance k. Example: k=4 and some arbitrary 3*8 matrix A 0 1 1 0 1 0 1 1 1 0 0 0 1 1 0 1 0 1 0 1 1 1 0 0 The theorem simply says that there exists a binary vector x of length 8 with 4 positions equaling 1 such that Ax=0 (in Z 2 ).

9. 9 Second tool: A theorem of Olson Let G be a finite Abelian group. The Davenport Constant of G, denoted D(G), is the least positive integer t such that in any sequence of t elements of G there is a (nonempty) subsequence whose sum is zero. Let p be a prime, and let G be a p-group. Then, Olson proved the following:

10. 10 An even vector is a binary vector whose number of nonzero coordinates is even. More generally, a p-divisible vector is a vector from (Z p ) k whose sum of coordinates is divisible by p. We use the latter two tools to prove the following lemmas: The p=2 lemma: Let k be an even positive integer. Then, every sequence of 2k even vectors of length k each, contains a subsequence of exactly k vectors whose sum is the zero vector. (If k is a power of 2 we can improve 2k to 2k-1). Proof: Consider the matrix A whose columns are the elements of the sequence of 2k even vectors of length k. Ignoring the last row of A we have a parity check matrix of a linear code which contains the distance k. This means we can choose k columns whose sum is zero. Adding back the last coordinate to these columns still gives a zero sum, since the columns of A are even vectors.

11. 11 The p=3 lemma: Let k be a power of 3. Every sequence of 5k-2 vectors of length k each that are 3-divisible, contains a subsequence of exactly k vectors whose sum is the zero vector (in (Z 3 ) k ). Proof: Let a 1,…,a 5k-2 be a sequence of 3-divisible vectors where a i  (Z 3 ) k. Let b i be the same as a i, except that the last coordinate of b i is always one. Note that b i need not be 3- divisible anymore. We consider the b i as elements of G=(Z 3 ) k-1 ×Z 3k. Since k is a power of 3, G is a p-group for p=3. According to Olson’s Theorem, D(G)=1+2(k-1)+(3k-1)=5k-2. Hence, there is a nonempty subsequence of b 1,…,b 5k-2 whose sum is zero. Since the last coordinate of b i is the element 1 of Z 3k, such a subsequence must contain exactly 3k elements. Thus, w.l.o.g., b 1 +…+b 3k =0. Now let us consider these 3k vectors as elements of G’=(Z 3 ) k-1 ×Z k. Again, G’ is a p-group for p=3, so D(G’)=1+2(k-1)+(k-1)=3k-2. Hence, there is a

12. 12 nonempty subsequence of b 1,…,b 3k whose sum is zero. Since the last coordinate of b i is the element 1 of Z k, such a subsequence must contain either exactly k or exactly 2k elements. In the latter case, the sum of the remaining k vectors not in the subsequence is also zero. We proved there is always a subsequence of k elements whose sum is zero in G’. Now, since the a i are 3-divisible vectors we immediately get that after setting the last coordinate back to its original value the corresponding subsequence of the a i is also zero.

13. 13 Further definitions Let x be a vector. A k-subvector of x is taken by selecting k coordinates of x. Clearly, if x  (Z p ) t then every k-subvector is a member of (Z p ) k, and x has exactly t!/(k!(t-k)!) distinct k- subvectors. Let e k (x) denote the number of p-divisible k-subvectors of x. We shall always assume that k is divisible by p. If x=(1,…1) then trivially e k (x)= t!/(k!(t-k)!), but, in general, e k (x) may be a lot smaller. For example, if x=(1,1,1,0,0)  (Z 2 ) 5 then e 2 (x)=4<10. Now, put

14. 14 The counting lemma: –Let k be even, and let t satisfy Then, M(2,k) ≤ t. –Let k be a power of 2, and let t satisfy Then, M(2,k) ≤ t. –Let k be a power of 3, and let t satisfy Then, M(3,k) ≤ t. Proof: We prove the first part. Let A be a square 0-1 matrix of order t. We must show that A has a square submatrix A’ of order k, such that all rows and all columns of A’ are even vectors. The condition in the lemma implies that there exists a subset of k columns of A, such that the set of t line vectors they induce contains more than 2k-1 even vectors. Thus, there is also a subset of 2k rows such that in the induced 2k * k submatrix each row has zero sum. By the p=2 lemma there is a subset of k rows (out of the 2k rows) such that each column is also zero sum. The proofs of the other cases are identical using the p=2 lemma (Olson version) and the p=3 lemma.

15. 15 It is not difficult to compute explicit small values of f(p,k,t). Clearly, if x and y are two binary vectors having the same length, and the same number of zeroes then e k (x)=e k (y). More generally, if r denotes the number of zeroes in a binary vector of length t then, Similarly, for the case p=3, if r denotes the number of zeroes and s denotes the number of ones in a 3-ary vector of length t then, Example: f(2,4,13)=343. The minimum is obtained when r=9 (or r=4). We can use the second part of the counting lemma and see that 343 > (6/13)*13!/(4!9!). Thus, we immediately get M(2,4) ≤ 13.

16. 16 Computing lower bounds for f(p,k,t) in the general case is a more complicated task. e.g.: what would one guess is the minimum possible fraction of 3-divisible k-subvectors of a 20k-vector? The next two lemmas give a bound for every p ≥ 2, in case k=  (t) and t→∞. Before we state the lemmas, we need to define a particular constant: For a sequence of integers a 1,…,a m let w p (a 1,…,a m ) be the complex number Example: let z=e 2  i/3 w 3 (4,8)=w 3 (1,2)=0.25(4+(1+z)(1+z 2 )+(1+z 2 )(1+z))=1.5 This number is always real (in fact, rational). Define z p = inf w p (a 1,…,a m ) where the infimum is taken over all finite sequences a 1,…,a m of all lengths m ≥ 0.

17. 17 If all a i =1 and  ≠1 then |(1+  a j )/2| ≤ cos(  /p)=  <1, and so w p ≤ 1+(p-1)  m. Letting m→∞ we see that z p ≤ 1. Also note that, trivially, z 2 =1. The z p lemma: z p =p2 1-p. A combinatorial proof: The sum of  j taken over all roots is p if p | j and 0 otherwise. Hence w p (a 1,…,a m ) is just p2 -m times the number of subsets X of {1,…,m} such that ∑ i  x a i = 0 mod p. If m=p-1 and all a i =1 then the only such subset X is . Hence z p ≤ p2 1-p and the result will follow if we can show the number of subsets X is always at least 2 m+1-p. We prove this by induction on m. Let S m be the set of residue classes mod p that can be written in the form ∑ i  x a i for some X  {1,…,m} and assume that for each element of S m there are at least 2 s m such subsets X. Now S m =S m-1  (S m-1 +a m ). We consider two cases:

18. 18 1.S m ≠ S m-1. Then |S m |>|S m-1 |. 2.S m =S m-1. Then S m-1 +a m =S m-1 and so if x  S m then both x and x-a m lie in S m-1. Thus x can be written as a sum of the a i 's in at least 2 s m- 1 ways which do not involve a m and 2 s m- 1 ways that do involve a m. Thus s m ≥ s m-1 +1. In all cases |S m |+s m ≥ |S m-1 |+s m-1 +1. Now S 0 ={0} and s 0 =0, so by induction |S m |+s m ≥ m+1. Since |S m | ≤ p we have s m ≥ m+1-p. Thus 0  S m can be written as a sum of a i mod p in at least 2 m+1-p ways.

19. 19 The f(p,k,t) lower bound lemma: Let k ≤ t/2. Then for any fixed prime p: Proof: Let T={1,…,t} and let x=(x 1,…,x t )  (Z p ) t be a fixed vector for which e k (x)=f(p,k,t). Assume p | k and k ≤ t/2. We wish to estimate the number f(p,k,t) of sets K  T, |K|=k with ∑ i  K x i = 0 mod p. Fix the k element subset K={1,…,k} of T and instead count the number p k of permutations   S t such that ∑ i=1..k x  (i) =0 mod p. It is clear that p k =f(p,k,t) k!(t-k)! Define Then  (1)=1, |  (  )| ≤ 1, p k /t!=p -1 ∑   (  ). Let  be a partition of T into k pairs S i ={a i,b i }, i=1,…,k, and a remaining set S 0 of t-2k numbers. We also regard  as the set of permutations  such that {  (i),  (i+k)}=S i for all i=1,…,k.

20. 20 For each  there are exactly 2 k (t-2k)! choices for   . where and the expectation is over a uniform random choice of  (there are exactly t!/(2 k (t-2k)!) choices for  ). Assume now  ≠1. If a ≠ b mod p then | (  a +  b )/2| ≤ cos(  /p)=  <1. Thus |  ( ,  )| ≤  n  where n  is the number of i=1,…,k such that x a i ≠ x b i mod p. Hence |  (  )| ≤ E  [  n  ]. Let r s be the number of coordinates with x i = s mod p. Then E  [n  ]=  =k ∑ i<j (2r i r j )/(t(t-1)). It is not hard to show that Var  [n  ]=O(  ): Indeed, note that n  = ∑ j=1..k n  (j) where n  (j) are indicator random variables that are equal to 1 in case x a j ≠ x b j mod p. Since Cov[n  (j 1 ), n  (j 2 )] = O(  2 /(k 2 t)) and since  ≤ k < t we have: Var  [n  ] ≤  + k 2 O (  2 /(k 2 t)) = O (  ).

21. 21 By Tchebychev's Inequality, if  →∞ then n  is almost surely large as well and  (  )=o(1). In this case Now assume  is bounded and k →∞. Let r=t- max r i be the number of coordinates of x not equal to the most common value. Then  ≥ 2kr(t/p)/t 2, so r=O(t/k) and kr 2 /t 2 =o k (1). By adding a constant vector to x and using p | k we can assume the most common coefficient in x is 0. Let m  be the number of S i with both x a i and x b i non-zero. Then, trivially, Pr[m  >0] ≤ E  [m  ]=O(kr 2 /t 2 )=o(1). If m  =0 then  ( ,  ) is a product of n  terms of the form (1+  a i )/2. Hence, the sum over all roots of unity of  ( ,  ) is at least z p. It follows that

22. 22 Completing the proof of the main theorem: Let  >0 and let 0<  <  /(1+  ). Let k be an even integer sufficiently large such that Such a k exists by the lower bound lemma. Now, Since 0.5(1-  )>(2k-1)/(4k(1+  )) we get by the counting lemma that M(2,k) ≤ 4k(1+  ). Similarly, by using the second part of the counting lemma we get that for k sufficiently large which is a power of 3, and so M(3,k) ≤ 20k(1+  ).