Extremal 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 | | |E(G)| 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 | |

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 with parts u1, . . . , un and v1, . . . , vn and at least δn edges { } { } 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 . { 1 t } ∪ { 1 t } Solution. Let P = (i, j) 1 i, j t . Due to the 2-dimensional Szemer´edi’stheorem there is an { | ≤ ≤ } 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 . ∈ ∈ 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 { } { }

2 Q = (i, j) [n] uivj E(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 ∈ ∈ kP + ` = (ki + x, kj + y) 1 i, j t . { | ≤ ≤ } Therefore ki + x 1 i t and kj + y 1 j t form arithmetic progressions of { | ≤ ≤ } { | ≤ ≤ } length t and of common difference. Moreover uki+x 1 i t vkj+y 1 j t { | ≤ ≤ } ∪ { | ≤ ≤ } 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−|A| B such that A0 B0 induces a complete⊆ bipartite⊆ graph in| G|. ≥ | | | | ≥ | | ∪ (b) Prove that there is a constant c that depends on d only such that, if A = B = n, | | | | then there is a copy of Ka,b in G with a c log(n) and b √n. ≥ ≥ Hint: Count pairs like in the proof of Theorem 1 in Lecture 8. Solution.

(a) Remark: We can guarantee A0 d A only. ≥ b | |c 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⊆ is| given| b by| |c ∈    1 P d(v) X d(v) |B| B v∈B d A ≥ | | v∈B d A b | |c b | |c  1 E(G)   d A  = B |B| | | = B | | B | | d A | | d A ≥ | | b | |c b | |c  A  2−|A| B | | . ≥ | | d A d | |e x  Here the second inequality is a variant of Jensen’s inequality for the function d|A| . Therefore, by pigeonhole principle, there is a set A0 A of size d A contained in at least 2−|A| B such pairs. Hence there is a set B0 of⊆ size at leastd 2|−|A|e| B such that each v B is| adjacent| to each vertex in A0. | | ∈ Remark: Here is an example of a graph that we can not ensure A0 d A . ≥ d | |e Consider a bipartite graph G with parts A, A = 3, and B, B = 24, that is | | 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−|A| B = 3. 3·24 12 | | 4 d 4 e | | 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 given| | by ∈

    1 c log(n)     X d(v) dn (1) ( dn) c log(n) n (2) n n n 2 n d  √n . s ≥ s ≥ c log(n) ! ≥ 2 s ≥ s v∈B 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 √n such pairs. Hence there is a set B0 B of size⊆ at least √n such that each v B is adjacent to each vertex in A0. ⊆ ∈ 

Jonathan Rollin http://www.math.kit.edu/iag6/edu/extremalgt2016s