Extremal Graph Theory
summer term 2016
Solutions to problem sheet 5
Problem 14.
For a prime number p consider the construction of a K2,2 -free graph G(p) in Lecture 8.
(a) Draw the graph G(5).
Hint: Compare the automorphisms with the symmetries of a square in the plane.
(b) Prove that for distinct (x, y), (x0 , y 0 ) ∈ Zp × Zp \ {(0, 0)} the system of equations
ux + vy = 1, ux0 + vy 0 = 1 has at most one solution (u, v) ∈ Zp × Zp . Conclude that
G(p) is K2,2 -free.
Solution.
(a) The graph G(5) is given in the following picture.
The picture is drawn in such a way that reflection at a horizontal, vertical or diagonal
line through the middle point or rotation about π2 , π, or 3π
around the middle point
2
corresponds to an automorphism of the graph. This works as the automorphism
group of G(5) is isomorphic to the dihedral group D4 , i.e, the symmetry group of a
square in the plane. To see this one can check that the automorphisms

(0, j) 7→ (0, j)





(1, j) 7→ (4, j)
φ1 : (i, j) 7→ (j, i),
φ2 : (2, j) 7→ (3, j)



(3, j) 7→ (2, j)



(4, j) 7→ (1, j)
generate the automorphism group of G(5) and satisfy the relation φ21 = φ22 =
(φ1 φ2 )4 = 1.
Jonathan Rollin
www.math.kit.edu/iag6/edu/extremalgt2016s
(1)
Extremal Graph Theory
summer term 2016
(b) Do the linear algebra here.
Now assume that u = (u1 , u2 ), v = (v1 , v2 ), x = (x1 , x2 ) and y = (y1 , y2 ) form a K2,2
in G(p) with edges ux, uy, vx and vy. Then (u1 , u2 ) as well as (v1 , v2 ) is a solution
(a1 , a2 ) of the system a1 x1 + a2 x2 = 1, a1 y1 + a2 y2 = 1, a contradiction.
Problem 15.
Prove that for each bipartite graph H with t vertices and m edges and for each n ≥ 2
ex(n, H) ≥
1
64
t−2
n2− m−1 .
Deduce Theorem 1 from Lecture 10.
Hint: Consider a random graph.
Solution.
Of course, we need to assume m ≥ 2 here.
t−2
We will construct an H-free graph on n vertices with cn2− m−1 edges randomly. Let
t−2
p = 81 n− m−1 . Consider a graph G = Gn,p with vertex set [n] where each edge is contained
in G with probability p. Let X be the random variable that is given by the number of
copies of H in G. Further let H be the set of possible copies of H on [n]. By linearity of
expectation
X
E(|E(G)|) =
P (uv ∈ E(G)) = n2 p ≥ 14 p n2 ,
{u,v}∈([n]
2 )
X
E(X) =
P (E(H 0 ) ∈ E(G)) = |H| pm ≤ nt pm = pnt pm−1 ≤ 81 p n2 .
H 0 ∈H
Let G0 be a (random) graph obtained from G by removing one edge from each copy of H
in G. Then G0 is H-free and
E(|E(G0 )|) ≥ E(|E(G)| − X) = E(|E(G)|) − E(X) ≥ 81 p n2 =
1
64
t−2
n2− m−1 .
Hence there is a graph G0 on n vertices that is H-free and contains at least
edges.
1
64
t−2
n2− m−1
For a cycle C2k this yields Theorem 1 from Lecture 10 as
ex(n, C2k ) ≥
1
64
2k−2
n2− 2k−1 =
1
64
1
n1+ 2k−1 .
Problem 16.
A 1-subdivision of a graph is obtained by subdividing each edge exactly once. Let ≤ 12 .
2
Prove that each graph on n ≥ 16− 3 vertices and n2 edges contains a 1-subdivision of
3
1
Kt with t ≥ 2 n 4 .
Hint: Dependent Random Choice
Solution.
Like in the proof of Theorem 1 in Lecture 9 there is a bipartite subgraph G0 of G with
Jonathan Rollin
www.math.kit.edu/iag6/edu/extremalgt2016s
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Extremal Graph Theory
summer term 2016
parts A and B that contains at least 12 n2 edges. We apply the dependent random choice
lemma with r = 2 and density α = 21 . This yields a subset A0 ⊆ A with
3
1
1
1
3
1
|A0 | ≥ 12 α2 n = 18 2 n = 18 2 n 4 n 4 ≥ 18 2 8− 2 n 4 = 2 n 4
such that each pair of vertices in A0 has at least
1
1
αn 2
4
=
1
1
n 2
8
≥
1 3 12
n
2
3 1
2 n4
>
2
common neighbors in B. Therefore we can embed a copy of a 1-subdivision of Kt with
3
1
t = d 2 n 4 e in G0 by choosing t arbitrary vertices from A0 and the subdividing vertices
from B.
Jonathan Rollin
www.math.kit.edu/iag6/edu/extremalgt2016s
(3)