I. Reiman. Uber ein Problem von K. Zarankiewicz. Acta. Math. Acad. Sci. Hungar., 9:269--273, 1958.  Home/Search   Document Not in Database   Summary   Related Articles   Check

....during the algorithm is ) 2 u u t 1000 4 100 400n Therefore, for the resulting graph, On the other hand, each component of G has relatively few vertices: j ) 2=3) Claim C. [R58] Let ex(n; K 2;2 ) denote the maximum number of edges that a K 2;2 free graph with n vertices can have. Then ex(n; K 2;2 ) 1 4n 3 n Applying the Claim to each G ; we obtain j ) n j ) n(G therefore, j ) 4n(G the desired contradiction. ....

I. Reiman,  Uber ein Problem von K. Zarankiewicz, Acta Math. Acad. Sci. Hungar. 9 (1958), 269-279.

....# # # j ) 100 4 1 # 8 25 e # 100n . Therefore, for the resulting graph, On the other hand, each component of G has relatively few vertices: j ) 2 3) n(G # ) j = 1, 2, M k ) Claim C. [R58] Let ex(n, K 2,2 ) denote the maximum number of edges that a K 2,2 free graph with n vertices can have. Then ex(n, K 2,2 ) 1 # 4n # 4 # . Applying the Claim to each G j ) n(G # therefore, j ) the desired contradiction. The tightness of Theorem ....

I. Reiman, Uber ein Problem von K. Zarankiewicz, Acta Math. Acad. Sci. Hungar. 9 (1958), 269--279.

....bound and its proof can be viewed as an adaptation of the main argument of [1] to the case of bipartite graphs. Previous relevant bounds for bipartite graphs were known for girth 4; 6; 8. For g = 4, 1) reduces to the trivial statement n L dR . For g = 6 the best lower bound is due to Reiman ([6], see [3] p.312 Theorem 2.6) His bound is the same as the bound given by (1) n L 1 (d R 1) d R 1) d L 1) For the case of g = 8 the best bound is due to Neuwirth [5] following work of de Caen and Sz ekely [4] This bound can be shown to be the same as (1) n L 1 (d R 1) d R ....

I. Reiman.  Uber ein Problem von K. Zarankiewicz. Acta. Math. Acad. Sci. Hungar., 9:269-273, 1958. 6

....100n 3 2 4e # n # e 4 100n e 2 . Therefore, for the resulting graph, e(G k ) # e 2 . On the other hand, each component of G k has relatively few vertices: n(G k j ) 2 3) k n(G # ) e 2 16n 2 (G # ) e 2 16n 2 (G k ) j = 1, 2, M k ) Claim C. [R58] Let ex(n, K 2,2 ) denote the maximum number of edges that a K 2,2 free graph with n vertices can have. Then ex(n, K 2,2 ) # n # 1 # 4n 3 # 4 # n 3 2 . Applying the Claim to each G j k , we obtain e(G k j ) # n 3 2 (G k j ) n(G k j ) # e 2 16n 2 (G k ) ....

I. Reiman, Uber ein Problem von K. Zarankiewicz, Acta Math. Acad. Sci. Hungar. 9 (1958), 269--279.

....1000n 3 2 4e # n e 10 400n e 2 . Therefore, for the resulting graph, e(G k ) # e 2 . On the other hand, each component of G k has relatively few vertices: n(G k j ) 2 3) k n(G # ) e 2 16n 2 (G # ) e 2 16n 2 (G k ) j = 1, 2, M k ) Claim C. [R58] Let ex(n, K 2,2 ) denote the maximum number of edges that a K 2,2 free graph with n vertices can have. Then ex(n, K 2,2 ) # n 1 # 4n 3 4 # n 3 2 . Applying the Claim to each G j k , we obtain e(G k j ) # n 3 2 (G k j ) n(G k j ) v u u t e 2 16n 2 (G k ) ....

I. Reiman, Uber ein Problem von K. Zarankiewicz, Acta Math. Acad. Sci. Hungar. 9 (1958), 269--279.

....the following results. Lemma 2 For n 4, fi(K n ) 1 2 (n Gamma 2) n Gamma 3) 2 Note that Lemmas 1 and 2 imply Theorem 1. Lemma 3 For p; q 1, fi(K p;q ) p Gamma 1) q Gamma 1) 2 As to complete bipartite subgraphs, the following results are known. Lemma 4 [2] Chapter 6, Theorem 2. 6) [11] Any bipartite graph G = V 1 ; V 2 ; E) with p = jV 1 j q = jV 2 j and m = jEj contains a subgraph K r;2 with r (2m Gammap) 2 Gammap 2 4p(p Gamma1)q . 2 Lemma 5 [4] For any constant 0 ffi 1, a bipartite graph G = V 1 ; V 2 ; E) with p = jV 1 j = jV 2 j and m = jEj contains a ....

I. Reiman, Uber ein Problem von K. Zarankiewicz, Acta. Math. Acad. Sci. Hunger. 9 (1958) 269-279.

....p; q 1 [7] Thus, we have the following results. Lemma 2 For n 4, K n ) 1 2 (n 2) n 3) 2 Note that Lemmas 1 and 2 imply Theorem 1. Lemma 3 For p; q 1, K p;q ) p 1) q 1) 2 As to complete bipartite subgraphs, the following results are known. Lemma 4 [2] Chapter 6, Theorem 2. 6) [11] Any bipartite graph G = V 1 ; V 2 ; E) with p = jV 1 j q = jV 2 j and m = jEj contains a subgraph K r;2 with r (2m p) 2 p 2 4p(p 1)q . 2 Lemma 5 [4] For any constant 0 1, a bipartite graph G = V 1 ; V 2 ; E) with p = jV 1 j = jV 2 j and m = jEj contains a subgraph K a;b with a = ....

I. Reiman,  Uber ein Problem von K. Zarankiewicz, Acta. Math. Acad. Sci. Hunger. 9 (1958) 269-279.

....existence of a single K 2;2 in an equibipartite graph. It would be of interest to narrow this gap of p 2 in the multiplicative constant. If in the statement of Corollary 2. 3 we do not insist that G be equibipartite, then, as is found by many constructions, for example, see, 12] Problem 10.36, [14], or [3] for more references) n 3=2 (1 Gamma o(1) 2 are necessary for the appearance of a K 2;2 . 2.2 Hypergraphs to integers Now we demonstrate a bijection between edges in a complete hypergraph G 2 G(d; a) recall Definition 2.1) and the elements of an initial interval of positive ....

I. Reiman, Uber ein Problem von K. Zarankiewicz, Acta Math. Acad. Sci. Hung. 9 (1959), 269--279.

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I. Reiman. Uber ein Problem von K. Zarankiewicz. Acta. Math. Acad. Sci. Hungar., 9:269--273, 1958.

No context found.

I. Reiman. Uber ein Problem von K. Zarankiewicz. Acta. Math. Acad. Sci. Hungar., 9:269--273, 1958.

No context found.

I. Reiman. Uber ein Problem von K. Zarankiewicz. Acta. Math. Acad. Sci. Hungar., 9:269--273, 1958.

No context found.

Reiman, I.,  Uber ein Problem von K. Zarankiewicz, Acta. Math. Acad. Sci. Hungar. 9, (1958), 269-279.

No context found.

Reiman, I., Uber ein Problem von K. Zarankiewicz, Acta. Math. Acad. Sci. Hungar. 9, (1958), 269--279.

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