Matrix approximation and Tusnády s problem

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1 European Journal of Combinatorics 28 (2007) Matrix approximation and Tusnády s problem Benjamin Doerr Max-Planck-Institut für Informatik, Stuhlsatzenhausweg 85, Saarbrücken, Germany Received 26 October 2004; accepted 29 September 2005 Available online 18 January 2006 Abstract We consider the problem of approximating a given matrix by an integer one such that in all geometric submatrices the sum of the entries does not change by much. We show that for all integers m, n 2 and real matrices A R m n there is an integer matrix B Z m n such that (a i j b i j ) i I j J < 4 log 2 (min{m, n}) holds for all intervals I [m], J [n]. Such a matrix can be computed in time O(mn log(min{m, n})). The result remains true if we add the requirement a i j b i j < 2 for all i [m], j [n]. This is surprising, as the slightly stronger requirement a i j b i j < 1 makes the problem equivalent to Tusnády s problem. c 2005 Elsevier Ltd. All rights reserved. 1. Introduction and results 1.1. Matrix rounding problems We consider the problem of approximating a given matrix by an integer one such that a particular error measure is small. Such matrix rounding and approximation problems have a rich history and found several applications. Baranyai [3] showed that any matrix can be rounded to an integer one in such a way that the errors in all rows, all columns and the whole matrix are less than one. He used this integer making lemma to prove several deep partitioning theorems for complete uniform hypergraphs, among them the precise estimate of its chromatic index. Independently, Causey et al. [9] proved the same rounding result and applied it to several problems in statistics. Further applications and results in this direction can be found in [24,4,14,10] /$ - see front matter c 2005 Elsevier Ltd. All rights reserved. doi: /j.ejc

2 B. Doerr / European Journal of Combinatorics 28 (2007) Raghavan [20], Karp et al. [15] and other authors use matrix rounding algorithms to solve VLSI routing problems. Asano et al. [1] (see also [2]) relate the digital half-toning problem to a matrix rounding problem. They suggest that if the average error in suitable regions is small, then the image represented by the rounded matrix is a high quality half-toning of the greyscale image represented by the input matrix. In this context, the problem of rounding with small average error in all 2 2 geometric (contiguous) submatrices is investigated. A recent improvement of these rounding results can be found in [13]. In this paper, we are concerned with the errors in all geometric submatrices. Throughout the paper let m, n 2 be integers. For m n matrices A and B let d(a, B) = max I,J (a i j b i j ), i I j J where I [m] := {1,..., m} and J [n] are intervals. We say that B is a rounding of A if b i j { a i j, ai j } for all i [m], j [n]. In other words, B is integral and ai j b i j < 1 holds for all i [m], j [n]. The matrix rounding problem is, given A R m n, to find a rounding B of A that minimizes d(a, B). Denote by r m,n the smallest r R such that for any A R m n a rounding B exists with d(a, B) r. This matrix rounding problem is equally hard as a famous problem from computational geometry known as Tusnády s problem. We briefly state the problem and the connection between the two Tusnády s problem Tusnády s problem is coloring a finite collection of points in the plane with two colors in such a way that in each rectangle the number of (say) red points deviates not too much from that of (say) blue points. More precisely, let P R 2 be finite and χ : P { 1, 1}. For a rectangle R R 2 let χ(r) = p P R χ(p) denote the difference in number of (+1)-colored points and ( 1)-colored points in R. The discrepancy of χ is T (P, χ) = max R χ(r), where the maximum is taken over all rectangles R in the plane. The discrepancy T (P) is the minimum T (P, χ) among all colorings χ : P { 1, 1}. By t n we denote the maximum discrepancy among all n-point sets in the plane. In other words, t n is the smallest number such that any n points can be two-colored in such a way that in no rectangle does the number of points in one color exceed that of points in the other color by more than t n. Note that this combinatorial discrepancy is to be distinguished from geometric discrepancies of point sets and their application in numerical integration (see e.g. [18] or [19]). Tusnády s problem has a rich history. With the focus on extending results of Komlós et al. [16], Gábor Tusnády asked whether t n can be bounded by a constant independent of n. Beck [5] was the first to disprove this by showing a non-constant lower bound of Ω(log n) for t n via a reduction to geometric discrepancies and Schmidt s lower bound [21]. In the same paper, he gave an O(log 4 n) upper bound, which he improved to O(log 3.5+ε ) in [7]. Bohus [8] derived an O(log 3 n) bound from his bound on the k permutation problem. Matoušek [18] proved O(log 2.5 n log log n) by a clever application of Beck s partial coloring method. The current best upper bound of O(log 2.5 n) due to Srinivasan [22] is obtained via the so-called entropy method (cf. [18]). Note that only Beck s first-mentioned and Bohus results are constructive. The following connection between the matrix rounding problem and Tusnády s problem was used without proof in [1], so for reasons of completeness we give a short proof.

3 992 B. Doerr / European Journal of Combinatorics 28 (2007) Theorem 1. For all m, n N, m n, we have 1 2 t n r m,n t mn. Proof. Let P R 2, P = n. Note that T (P) is invariant under strictly monotone transformations in the coordinates: Let σ, τ : R R be strictly monotone mappings and π : R 2 R 2 ; (x, y) (σ (x), τ(y)). Then T (P) = T (π(p)), as T (P, χ) = T (π(p), χ π 1 ) for all χ : P { 1, 1}. Thus we may assume that P [n] [n]. Let A {0, 2 1 }m n such that a i j = 1 2 if and only if (i, j) P. Let B be an optimal rounding of A. Let χ : P { 1, 1} such that χ((i, j)) = 1, if b i j = 0, and χ((i, j)) = 1, if b i j = 1, for all (i, j) P. Now for any rectangle R, we have χ(p) = 2 (b i j a i j ) = 2 (b i j a i j ). p P R (i, j) P R (i, j) R [m] [n] Hence T (P) d(a, B) r m,n. To prove the second inequality, we use the result of Lovász et al. [17] stating that the hereditary linear discrepancy of a hypergraph is at most twice its hereditary discrepancy (the factor of 2 can be replaced by 2(1 2m 1 ), cf. [11], but we shall not need this). Since r mn is nothing else than the hereditary linear discrepancy of the hypergraph of rectangles in [m] [n], we have r mn 2 max min d(a, B), A {0, 2 1 }m n B where B runs over all roundings of A. Let A {0, 1 2 }m n. Put P = {(i, j) [m] [n] a i j = 1 2 }. Let χ : P { 1, 1} such that T (P, χ) t P t mn. Define B {0, 1} m n through b i j = 1 if and only if (i, j) P and χ((i, j)) = 1. Then for all intervals I [m] and J [n], we compute (a i j b i j ) = 1 χ((i, j)) 2 1 T (P, χ). 2 i I j J (i, j) P (I J) Hence d(a, B) 1 2 t mn. We conclude r m,n t mn. Corollary 2. For m n, r m,n = Ω(log n), r m,n = O(log 2.5 m) Our results The result presented in this paper is that we do get an upper bound of O(log n) if we do not require B to be a rounding of A. Theorem 3. Let m, n 2 be integers. Then for all real matrices A R m n there is an integer matrix B Z m n such that d(a, B) < 4 log 2 (min{m, n}). Such a matrix can be computed in time O(mn log(min{m, n})). The result remains true if we add the requirement a i j b i j < 2 for all i [m], j [n].

4 B. Doerr / European Journal of Combinatorics 28 (2007) Our result has an interesting implication: Either the rounding restriction a i j b i j < 1 makes the matrix approximation problem significantly harder, or both Tusnády s problem and the matrix rounding problem can be solved with error O(log n). We believe the latter to be more likely. Since we do not even have a non-constant lower bound, we did not try to optimize the constant in Theorem 3. Indeed, there is room for improvement, e.g., by writing arbitrary intervals not only as disjoint unions of canonical intervals, but also allowing differences in Lemma 5. Note that the one-dimensional version of our problem is elementary. A simple greedy algorithm, sometimes called 1D error diffusion, yields the following result. Lemma 4. Let a 1,..., a m R. Then there are b 1,..., b m Z such that i 2 i=i1 (a i b i ) < 1 holds for all 1 i 1 i 2 m. They can be computed in time O(m). In fact, much more about this problem is known, e.g., the number of such roundings and how to compute an optimal one in time O(m log m), see [23,12]. The novel contribution of this paper is combining Lemma 4 successfully with classical ideas like the decomposition into canonical intervals. In fact, our approach works for any set of regions of type I J, where J is an interval and I is a hyperedge of a totally unimodular hypergraph (note that Lemma 4 apart from the computational statement holds for arbitrary totally unimodular hypergraphs instead of the hypergraph of intervals). 2. Proof of the main result We denote by N the positive integers, by N 0 the nonnegative ones. For arbitrary numbers p, q R we set [p] = {i N i p} and [p... q] = {i N 0 p i q}. Let n, m N. By Int([n]) = {[i... j] 1 i j n} we denote the set of intervals in [n]. A set of form [λ2 k (λ + 1)2 k ] for some λ Z is called a canonical interval of length 2 k. We write CInt k ([n]) for the set of all canonical intervals of length 2 k that are contained in [n]. Like in [6], we have the following decomposition lemma. Lemma 5. For n 2, any interval in [n] is the disjoint union of at most 2 log 2 n canonical intervals. Let A and B be m n matrices. For any S [m] [n] we use A(S) to denote the sum (i, j) S a i j of the entries of A in S. The distance of A and B with respect to S is d(a, B, S) = A(S) B(S). Lemma 6. Let l N 0, n = 2 l and A R m n. Then there is an à R m n that has integral row sums j [n] ãi j, i [m], and d(a, Ã, I J) < 2 l+k for all I Int([m]), k [0... l] and J CInt k ([n]). It can be computed in time O(mn). Proof. Let s i = n j=1 a i j for all i [m]. By Lemma 4, there are t i Z, i [m], such that D(i 1, i 2 ) := i 2 i=i1 (t i s i ) has absolute value less than one for all i 1, i 2 [m]. Put ã i j = a i j + n 1 (t i s i ) for all i [m], j [n]. Since j [n] ãi j = t i for all i [m], à has integral row sums. Let I Int([m]), k [0... l] and J CInt k ([n]). Then d(a, Ã, I J) = J n max I i=min I (t i s i ) = 1 n 2k D(min I, max I ) < 2 l+k.

5 994 B. Doerr / European Journal of Combinatorics 28 (2007) The proof of Theorem 3 depends heavily on the following lemma. Lemma 7. Let l N 0, n = 2 l and A R m n such that all row sums j [n] a i j, i [m], are integral. Then there is a B Z m n such that for all I Int([m]), k [0... l] and J CInt k ([n]) we have l 1 k 2 i if k < l d(a, B, I J) i=0 0 if k = l. It can be computed in time O(mn log n). Proof. We use induction on l. For l = 0, we have n = 1 and thus A Z m n by assumption. Hence B = A proves the claim. Let l > 0 and assume the claim to be true for l 1. Let A L, A R R m (n/2) such that A = (A L A R ). Obtain à L from A L as in Lemma 6. Set à R = A R (à L A L ). Then both à L and à R have integral row sums. By induction, there are B L, B R Z m (n/2) such that the pairs (à L, B L ) and (à R, B R ) in the roles of (A, B) both fulfill the assertion of this lemma. Put B = (B L B R ). Let I Int([m]), k [0... l] and J CInt k ([n]). If J = [n], then ( [ n ]) ( [ n ]) B(I J) = B L I + B R I n ( [ n ]) ( [ n ]) = à L I + à R I n = A(I J). Hence d(a, B, I J) = 0 as claimed. If J [n], i.e., k < l, then either J [ n 2 ] or J [ n n]. In the first case, we compute d(a, B, I J) = d(a L, B L, I J) d(a L, à L, I J) + d(ã L, B L, I J) < 2 l+1+k l 2 k + 2 i = l 1 k i=0 2 i. An analogous argument shows the claim for J [ n n]. i=0 Proof of Theorem 3. Let A R m n. Without loss of generality, we may assume m n. Let n 0 = 2 l be the smallest power of two not less than n. By appending columns of zeros, we obtain A R m n 0 from A. By Lemma 6, we compute a matrix à having integral row sums such that for all I Int([m]), k [0... l] and J CInt k ([n 0 ]) we have d(a, Ã, I J) < 2 l+k. Applying Lemma 7 to Ã, we obtain a matrix B that fulfills d(a, B, I J) < l k i=0 2 i < 2 for all I Int([m]), k [0... l] and J CInt k ([n 0 ]). By Lemma 5, we infer d(a, B, I J) = d(a, B, I J) < 4 log 2 n for all I Int([m]), J Int([n]).

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