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Solutions to Problem Sheet 4 Extremal Graph Theory summer term 2016 Solutions to problem sheet 4 Problem 11. A graph is triangle-formed if each edge is contained in exactly one triangle. Prove that for every λ > 0 there is n0 such that each triangle-formed graph on n vertices, 2 with n n0, has at most λn edges. ≥ Solution. Let t(G) denote the number of triangles in a graph G. If G is triangles-formed, then E(G) = 3 t(G), since each edge is contained in exactly one triangle. Hence one needs j j jE(G)j λ to remove t(G) = 3 edges from G to make G triangle-free. Let = 3 . Due to the triangle removal lemma, there is δ > 0 such that every graph on n vertices with at most δn3 triangles can be made triangle-free by removing at most n2 edges. 1 2 Let n0 = 6δ . Assume that G is triangle-formed with n n0 vertices and more than λn λn2 2 ≥ edges. Then more than 3 = n edges are needed to make G triangle-free, as argued above. Due to the triangle removal lemma, G has more than 3 2 1 1 2 1 n 1 δn δn0n = n E(G) = t(G) ≥ 3 2 ≥ 3 2 ≥ 3 j j triangles, a contradiction. Problem 12. Let t be a positive integer and δ > 0. Prove that there is n0 such that for any n n0 2 ≥ and any bipartite graph with parts u1; : : : ; un and v1; : : : ; vn and at least δn edges f g f g there are arithmetic progressions x1; : : : ; xt and y1; : : : ; yt in [n] of length t and of common difference such that ux ; : : : ; ux vy ; : : : ; vy induces a complete bipartite graph. f 1 t g [ f 1 t g Solution. Let P = (i; j) 1 i; j t . Due to the 2-dimensional Szemer´edi'stheorem there is an f j ≤ ≤ g 2 2 integer n0 such that for any n n0 each set Q [n] with at least δn elements contains ≥ ⊆ 2 a subset of the form kP + ` for some k Z and ` Z . 2 2 Let n n0 and consider a bipartite graph G with parts u1; : : : ; un and v1; : : : ; vn and at≥ least δn2 edges. Define a subset Q of [n]2 by f g f g 2 Q = (i; j) [n] uivj E(G) : f 2 j 2 g Then Q has at least δn2 elements. Hence for some k Z and ` = (x; y) Z2 there is a subset in Q of the form 2 2 kP + ` = (ki + x; kj + y) 1 i; j t : f j ≤ ≤ g Therefore ki + x 1 i t and kj + y 1 j t form arithmetic progressions of f j ≤ ≤ g f j ≤ ≤ g length t and of common difference. Moreover uki+x 1 i t vkj+y 1 j t f j ≤ ≤ g [ f j ≤ ≤ g induces a complete bipartite subgraph of G. Problem 13. Let G be a bipartite graph with parts A and B and density d. Jonathan Rollin http://www.math.kit.edu/iag6/edu/extremalgt2016s Extremal Graph Theory summer term 2016 (a) Prove that there are A0 A, B0 B, with A0 d A , B0 2−|Aj B such that A0 B0 induces a complete⊆ bipartite⊆ graph inj Gj. ≥ j j j j ≥ j j [ (b) Prove that there is a constant c that depends on d only such that, if A = B = n, j j j j then there is a copy of Ka;b in G with a c log(n) and b pn. ≥ ≥ Hint: Count pairs like in the proof of Theorem 1 in Lecture 8. Solution. (a) Remark: We can guarantee A0 d A only. ≥ b j jc Let d(v) denote the degree of a vertex v in G. We consider pairs (A0; v) with A0 A, A0 = d A , and v B, v adjacent to each vertex in A0. The number of such pairs⊆ isj givenj b byj jc 2 1 P d(v) X d(v) jBj B v2B d A ≥ j j v2B d A b j jc b j jc 1 E(G) d A = B jBj j j = B j j B j j d A j j d A ≥ j j b j jc b j jc A 2−|Aj B j j : ≥ j j d A d j je x Here the second inequality is a variant of Jensen's inequality for the function djAj . Therefore, by pigeonhole principle, there is a set A0 A of size d A contained in at least 2−|Aj B such pairs. Hence there is a set B0 of⊆ size at leastd 2j−|Ajej B such that each v B isj adjacentj to each vertex in A0. j j 2 Remark: Here is an example of a graph that we can not ensure A0 d A . ≥ d j je Consider a bipartite graph G with parts A, A = 3, and B, B = 24, that is j j constructed from a vertex disjoint union of a star K1;22 and a K2;2 by adding four edges like in the figure. 22 } B A This graph has density 30 = 5 . Moreover d A = 5 , 5 = 2, and 2−|Aj B = 3. 3·24 12 j j 4 d 4 e j j But G contains no copy of K2;3. (b) Let c = d 1 and s = c log(n) . Then 2 1−log(d) d e c log(n) 1 dn (1) d e ≤ 2 d c log(n) 1+c log(d=2) 1−c(1−log(d)) 1− d 1 n = n = n = n 2 n 2 : (2) 2 ≥ Similarly as in part (a) consider pairs (A0; v) with A0 A, A0 = s, and v B, v adjacent to each vertex in A0. The number of such pairs⊆ is givenj j by 2 1 c log(n) X d(v) dn (1) ( dn) c log(n) n (2) n n n 2 n d pn : s ≥ s ≥ c log(n) ! ≥ 2 s ≥ s v2B d e Jonathan Rollin http://www.math.kit.edu/iag6/edu/extremalgt2016s Extremal Graph Theory summer term 2016 Therefore, by pigeonhole principle, there is a set A0 A of size s contained in at least pn such pairs. Hence there is a set B0 B of size⊆ at least pn such that each v B is adjacent to each vertex in A0. ⊆ 2 Jonathan Rollin http://www.math.kit.edu/iag6/edu/extremalgt2016s.
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