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By James Pommersheim and Irena SwansonMarko Petkovsek, James Pommersheim and Irena Swanson


The well-known Zarankiewicz problem [Za] asks to determine the least positive integer Z(m � n � r�s)suchthat each m n zero-one matrix containing Z(m � n � r�s) ones has an r s submatrix consisting entirely of ones. In graph-theoretic language, this is equivalent to nding the least positive integer Z(m � n � r�s) such that each bipartite graph on m black vertices and n white vertices with Z(m � n � r�s) edges has a complete bipartite subgraph on r black vertices and s white vertices. A complete solution of the Zarankiewicz problem has not been given. While exact values of Z(m � n � r�s) are known for certain in nite subsets of m � n � r and s, only asymptotic bounds are known in the general case � see, for example, Cul k [C], Furedi [F], Guy [G], Hartmann, Mycielski and Ryll-Nardzewski [HMR], Hylten-Cavallius [HC], Irving [I], Kovari, Sos and Turan [KST], Mors [M], Reiman [Re], Roman [Ro], Znam [Zn]. Even the case r = s = 2 has not been solved in general. Here we quote some known facts about this case: Hartmann, Mycielski and Ryll-Nardzewski [HMR] proved the asymptotic bounds c1n 4=3 <Z(n � n � 2 � 2) <c2n 3=2 for some constants c1 3 4 and c2 2. Kovari, Sos and Turan [KST] proved that Z(n � n � 2 � 2) 2n + n 3=2 � lim n!1 n;3=2 Z(m � n � 2 � 2) = 1: Moreover, when p is a prime number, [KST] proved that Z(p2 + p � p2 � 2 � 2) = p2 (p +1)+1. n Hylten-Cavallius [HC] proved that Z(m � n � 2 � 2) 2

Topics: q
Year: 2000
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