Some New Optimum Golomb Rectangles

We give some new optimumGolomb rectangles found by computer search. AMS Subject Classi cation. 05B99 In [2] Robinson de ned a Golomb rectangle as an N M array of ones and zeros such that the two-dimensional autocorrelation has three values: 0; 1 and K, where K is the number of ones in the array. This means that the positions of the ones in any nonzero integral translation of the rectangle will overlap with the positions of the ones in the original position of the rectangle in at most one place. Equivalently, the di erences between the positions of every pair of ones in the rectangle, considered as vectors, are distinct. See also [1]. Let G(N;M) be the maximum number of ones that can be present in an N M Golomb rectangle. For example G(2; 2) = 3. Robinson de ned an optimum Golomb rectangle to be one containing G(N;M) ones. We prefer to add the conditions G(N;M) > G(N 1;M) and G(N;M) > G(N;M 1). In table 1 we give a number of new optimumGolomb rectangles found by computer search. In most cases they are far from unique. Note there exists a 2 18 rectangle with 9 ones. The rectangle in Robinson's table V is 2 20 an apparent misprint. A brief description of the computer program used to nd these rectangles follows. Recall that a Golomb ruler is a set of integers a1 < a2 < < ak for which the k 2 ! di erences faj aij1 i < j kg are distinct. We will use the following easy lemma. Lemma 1: N M Golomb rectangles with K ones correspond 1 1 with K element Golomb rulers with elements chosen from the set fi+(2N 1)(j 1)j1 i N; 1 j Mg. Proof: Let f(bi; ci)j1 bi N; 1 ci M; 1 i Kg be a set of K positions in a N M rectangle. We claim this set consists of the positions of the ones in a N M Golomb rectangle with K ones i the set fai = bi+(2N 1)(ci 1)j1 i Kg, is a The Electronic Journal of Combinatorics 2 (1995) #R12 2 Golomb ruler with elements chosen from the set fi+ (2N 1)(j 1)j1 i N; 1 j Mg. For suppose (bi1; ci1) (bi2; ci2) = (bi3; ci3) (bi4; ci4). Then ai1 ai2 = (bi1 + (2N 1)(ci1 1)) (bi2 + (2N 1)(ci2 1)) = (bi1 bi2) + (2N 1)(ci1 ci2) = (bi3 bi4) + (2N 1)(ci3 ci4) = (bi3 + (2N 1)(ci3 1)) (bi4 + (2N 1)(ci4 1)) = ai3 ai4: Conversely suppose (ai1 ai2) = (ai3 ai4). Then (bi1 + (2N 1)(ci1 1)) (bi2 + (2N 1)(ci2 1)) = (bi3 + (2N 1)(ci3 1)) (bi4 + (2N 1)(ci4 1)):