2/19/2015
Ma/CS 6b
Class 19: Extremal Graph Theory
Paul Turán
By Adam Sheffer
Extremal Graph Theory

The subfield of extremal graph theory
deals with questions of the form:
◦ What is the maximum number of edges that a
graph with 𝑛 vertices can have without
containing a given subgraph 𝐻?
◦ Characterize the graphs that obtain the this
maximum number of edges.
A graph without 𝐾5
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𝑒𝑥 𝑛, 𝐻

We denote by 𝑒𝑥 𝑛, 𝐻 the maximum
number of edges in a graph with 𝑛
vertices and no subgraph 𝐻.

Example. Recall that 𝑃𝑖 is a simple path of
length 𝑖. What is 𝑒𝑥 4, 𝑃3 ? 3
Graphs Without Triangles

Problem. Find 𝑒𝑥 𝑛, 𝐾3 .
◦ What graph contains a large number of edges
but does not have 𝐾3 as a subgraph?
◦ Any complete bipartite graph.
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Maximizing the Number of Edges

What value of 𝑚 maximizes the number
of edges in 𝐾𝑚,𝑛−𝑚 ?
◦ The number of edges is 𝑚 𝑛 − 𝑚 .
◦ The maximum of 𝑛2 /4 is obtained for
𝐺𝑛/2,𝑛/2 .
◦ Can we find a graph without triangles and
with more edges?
Mantel’s Theorem


Theorem. 𝑒𝑥 𝑛, 𝐾3 = 𝑛2 /4 .
Proof. Let 𝐺 = 𝑉, 𝐸 be a triangle-free graph
with 𝑉 = 𝑛.
◦ 𝑑𝑖 = deg 𝑣𝑖 .
◦ If 𝑣𝑖 , 𝑣𝑗 ∈ 𝐸, then 𝑣𝑖 and 𝑣𝑗 do not have
common neighbors, so 𝑑𝑖 + 𝑑𝑗 ≤ 𝑛.
◦ We thus have
𝑣𝑖 ,𝑣𝑗 ∈𝐸
𝑑𝑖 + 𝑑𝑗 ≤ 𝑛 𝐸 .
◦ Since every 𝑑𝑖 appears in 𝑑𝑖 elements
𝑛𝐸 ≥
𝑑𝑖2 .
𝑑𝑖 + 𝑑𝑗 =
𝑑𝑖 ,𝑑𝑗 ∈𝐸
𝑖
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The Cauchy–Schwarz Inequality

The Cauchy–Schwarz inequality. For any
𝑎1 , 𝑎2 , … , 𝑎𝑛 , 𝑏1 , 𝑏2 , … , 𝑏𝑛 ∈ ℝ, we have
2
𝑛
𝑎𝑖 𝑏𝑖
𝑛
𝑎𝑖2
≤
𝑖=1

𝑛
𝑏𝑖2 .
𝑖=1
𝑖=1
Equality holds iff the vectors
𝑎1 , … , 𝑎𝑛 and 𝑏1 , … , 𝑏𝑛
are linearly dependent.
Available online
Completing the Proof

We proved that
𝑛𝐸 ≥
𝑖
𝑑𝑖 ,𝑑𝑗 ∈𝐸


𝑑𝑖2
𝑑𝑖 + 𝑑𝑗 =
Recall that 𝑖 𝑑𝑖 = 2|𝐸|.
By Cauchy-Schwarz with 𝑎𝑖 = 𝑑𝑖 and 𝑏𝑖 = 1:
2
𝑑𝑖
𝑖
◦
𝑑𝑖2
≤
12 .
𝑖
We thus have
𝑑𝑖2
𝑛𝐸 ≥
𝑖
4𝐸
≥
𝑛
𝑖
2
→
𝑛2
𝐸 ≤
4
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The Maximum Graph is Unique
Claim. The only graph that has the
maximum number of edges 𝑒𝑥 𝑛, 𝐾3
= 𝑛2 /4 is 𝐾 𝑛/2 , 𝑛/2 .
 Proof. We consider the case of an even 𝑛.

◦ In our proof for 𝑒𝑥 𝑛, 𝐾3 = 𝑛2 /4, we used
Cauchy-Schwarz to obtain
2
2
≤ 𝑖 𝑑𝑖2
𝑖 𝑑𝑖
𝑖 1 , which implied
𝑛𝐸 ≥
2
𝑖 𝑑𝑖 ≥
4𝐸2
𝑛
.
◦ Equality holds if and only if for every
1 ≤ 𝑖 ≤ 𝑛, we have 𝑑𝑖 = 𝑛/2.
Inequality of Arithmetic and
Geometric Means

AM-GM Inequality. For any 𝑎1 , 𝑎2 , … , 𝑎𝑛
∈ ℝ, we have
𝑎1 + ⋯ + 𝑎𝑛
≥ 𝑎1 ⋯ 𝑎𝑛
𝑛
1/𝑛
.
No Shirt available.
Marketing opportunity?
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Recall: Independent Sets
Consider a graph 𝐺 = 𝑉, 𝐸 . An
independent set in 𝐺 is a subset 𝑉 ′ ⊂ 𝑉
such that there is no edge between any
two vertices of 𝑉′.
 Finding a maximum independent set in a
graph is a major problem in theoretical
computer science.

◦ No polynomial-time algorithm
is known.
Mantel’s Theorem: Second Proof
𝐺 = 𝑉, 𝐸 – a triangle-free graph with 𝑉 = 𝑛.
 𝛼 – the size of the largest independent set 𝐴 in
𝐺.
 The set of neighbors of any 𝑣 ∈ 𝑉 is an
independent set, so 𝛼 ≥ deg 𝑣.
 𝐵 = 𝑉 ∖ 𝐴, so |𝐵| = 𝑛 − 𝛼. Also, 𝐵 is a vertex
cover.
 By the AM-GM inequality, we have
2
𝛼+ 𝑛−𝛼
𝑛2
𝐸 ≤
deg 𝑣 ≤ 𝛼 𝐵 ≤
=
.
2
4

𝑣∈𝐵
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Generalizing the Problem

We move from 𝑒𝑥 𝑛, 𝐾3 to 𝑒𝑥 𝑛, 𝐾𝑟 .
◦ For some 𝑟 > 3, what graph contains many
edges but no 𝐾𝑟 ?
◦ An 𝑟-partite graph is a graph with 𝑟 parts, and
no edge between two vertices of the same
part (generalizing bipartite graphs).
◦ Example. The figure presents the complete
4-partite graph 𝐾3,3,3,4 .
Turán Graphs

The Turán graph 𝑇 𝑛, 𝑟 is a complete
𝑟-partite graph with 𝑛 vertices, such that
each part consists of either 𝑛/𝑟 or
𝑛/𝑟 vertices.
◦ The graph in the figure is 𝑇 13,4 .
◦ 𝑇 𝑛, 𝑟 − 1 contains no copy of 𝐾𝑟 and has
about 𝑛 𝑛 −
𝑛
𝑟−1
/2 edges.
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Turán’s Theorem

Theorem. If 𝐺 = 𝑉, 𝐸 is a graph that
contains no copy of 𝐾𝑟 and 𝑉 = 𝑛, then
𝐸 ≤

𝑛2
2
1−
1
𝑟−1
.
Proof. By induction on 𝑛.
◦ Induction basis. Easily holds when 𝑛 < 𝑟.
◦ Induction step. 𝐺 = 𝑉, 𝐸 – a graph that
contains no copy of 𝐾𝑟 , with 𝑉 = 𝑛 and
𝐸 = 𝑒𝑥 𝑛, 𝐾𝑟 .
◦ 𝐺 contains a copy of 𝐾𝑟−1 . Otherwise, adding
an edge to 𝐺 would not yield a copy of 𝐾𝑟 ,
contradicting the maximality of 𝐺.




𝐺 = 𝑉, 𝐸 – a graph that contains no copy of
𝐾𝑟 , with 𝑉 = 𝑛 and 𝐸 = 𝑒𝑥 𝑛, 𝐾𝑟 edges.
𝐴 – an induced subgraph of 𝐺 that is a 𝐾𝑟−1 .
𝑟−1
𝐴 contains
edges.
2
By the hypothesis, 𝐺 ∖ 𝐴 has at most
(𝑛−𝑟+1)2
2


1−
1
𝑟−1
edges.
Since each vertex of 𝑉 ∖ 𝐴 is adjacent to at most
𝑟 − 2 vertices of 𝐴, there are
𝑛 − 𝑟 + 1 𝑟 − 2 edges between 𝑉\𝐴 and 𝐴.
Combining the above, we have
𝐸 ≤
𝑟−1 𝑟−2
𝑛−𝑟+1 2
+
1 − 1/ 𝑟 − 1
2
2
𝑛2
+ 𝑛−𝑟+1 𝑟−2 ≤
1 − 1/ 𝑟 − 1 .
2
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Which Actor from the Big Bang
Theory is a Real Scientist?
Proof #2: Balancing Weights

𝐺 = 𝑉, 𝐸 – a graph that contains no copy of
𝐾𝑟 , with 𝑉 = 𝑣1 , … , 𝑣𝑛 .

We assign a weight 𝑤𝑖 = to every 𝑣𝑖 . We set
𝑛
𝑊 = 𝑣 ,𝑣 ∈𝐸 𝑤𝑖 𝑤𝑗 .
1
𝑖 𝑗

Initially we have 𝑊 = 𝐸
weights to increase 𝑊.
1/3
1/3
1/3
1 1 1 1 2
⋅ + ⋅ =
3 3 3 3 9
1
𝑛2
. We wish to move
1/2
0
1/2
1 1 1
1
⋅ + ⋅0=
2 2 2
4
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Balancing Weights (cont.)

𝑣𝑖 , 𝑣𝑗 – two vertices of 𝑉 with positive weights,
such that 𝑣𝑖 , 𝑣𝑗 ∉ 𝐸.

𝑁𝑖 , 𝑁𝑗 – the sum of the weights of the neighbors
of 𝑣𝑖 and 𝑣𝑗 , respectively.

WLOG, assume that 𝑁𝑖 ≤ 𝑁𝑗 .
By moving the weight of 𝑣𝑖 to 𝑣𝑗 , we change 𝑊
by 𝑤𝑖 𝑁𝑗 − 𝑤𝑖 𝑁𝑖 ≥ 0.
𝑣𝑖
𝑣𝑖
0.1
0.1

0.1
0.1
0.1
0.1
𝑣𝑗
0
0.1
0.2
0.2
0.1
𝑣𝑗
0.2
More Balancing of Weights.



𝜀
If there exist 𝑣𝑖 , 𝑣𝑗 with positive weights and
𝑣𝑖 , 𝑣𝑗 ∉ 𝐸, we can move the weight vertex of
one to the other without decreasing 𝑊.
We repeatedly do this until the entire weight is
on the vertices of a complete subgraph 𝐾𝑚 .
Consider 𝑣𝑖 , 𝑣𝑗 ∈ 𝐾𝑚 with 𝑤𝑗 − 𝑤𝑖 ≥ 2𝜀 > 0.
Then moving 𝜀 from 𝑤𝑗 to 𝑤𝑖 changes 𝑊 by
1 − 𝑤𝑖 − 𝜀 − 1 − 𝑤𝑗
= 𝜀 𝑤𝑗 − 𝑤𝑖 − 𝜀 ≥ 𝜀 2 .
0.2
0.25
0.25
0.25
0.3
0.25
0.25
0.25
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Concluding the Proof
We conclude that we can move the entire
weight to a subgraph 𝐾𝑚 where each vertex of
1
the subgraph has a weight of .

𝑚
By the assumption, 𝑚 ≤ 𝑟 − 1.
1
𝑟−2
𝑟−1
That is, we obtain 𝑊 ≤
=
.
2
2𝑟−2
2 (𝑟−1)
1
Recall that we started with 𝑊 = 𝐸 2 .




𝑛
Since we only increased 𝑊, we have
1
𝑟−2
𝑛2
1
𝐸 2≤
→ 𝐸 ≤
1−
.
𝑛
2𝑟 − 2
2
𝑟−1
Cliques
Given a graph 𝐺 = 𝑉, 𝐸 , a clique of 𝐺 is
any subgraph that is a complete graph.
 No polynomial-time algorithm for finding
a maximum clique in a graph is known.

◦ The problem is equivalent to finding a
maximum independent set in a graph.
◦ Define the graph 𝐺 𝑐 = (𝑉, 𝐸 𝑐 ) such that
𝑣, 𝑢 ∈ 𝐸 𝑐 iff 𝑣, 𝑢 ∉ 𝐸. Finding
a maximum clique in 𝐺 is
equivalent to finding a maximum
independent set in 𝐺 𝑐 .
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A Lower Bound on the Clique Size

Lemma. Let 𝐺 = 𝑉, 𝐸 be a graph with
𝑉 = 𝑣1 , … , 𝑣𝑛 , and let 𝑑𝑖 = deg 𝑣𝑖 .
Then 𝐺 contains a clique of size at least
1
𝑛
.
𝑖=1
𝑛−𝑑𝑖
1
1
1
1
1
1
+
+
+
+
=3+
5−4 5−4 5−3 5−3 5−2
3
Probabilistic Proof
We uniformly choose a permutation 𝜋 of
1, … , 𝑛 .
 We build a subset 𝐶𝜋 ⊂ 𝑉: for every
1 ≤ 𝑖 ≤ 𝑛, we add 𝑣𝜋 𝑖 to 𝐶𝜋 if 𝑣𝜋 𝑖 is
adjacent to 𝑣𝜋 1 , 𝑣𝜋 2 , … , 𝑣𝜋 𝑖−1 .
 Notice that 𝐶𝜋 is a clique.
 𝑋𝑖 – an indicator variable for whether
𝑣𝜋 𝑖 ∈ 𝐶𝜋 .


𝐸 𝑋𝑖 = Pr 𝑋𝑖 = 1 =
1
𝑛−𝑑𝜋 𝑖
.
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Proof (cont.)

𝑋𝑖 – an indicator variable for whether
𝑣𝜋 𝑖 ∈ 𝐶𝜋 .

𝐸 𝑋𝑖 = Pr 𝑋𝑖 = 1 =
1
𝑛−𝑑𝜋 𝑖
.
Set 𝑋 = 𝑛𝑖=1 𝑋𝑖 .
 𝑋 is the size of the clique 𝐶𝜋 .
 By linearity of expectation

𝑛
𝐸𝑋 =
𝑛
𝐸 𝑋𝑖 =
𝑖=1
𝑖=1
1
𝑛 − 𝑑𝜋
𝑖
𝑛
=
𝑖=1
1
.
𝑛 − 𝑑𝑖
Completing the Proof

𝑋 is the size of the clique 𝐶𝜋 .
𝑛
𝐸𝑋 =
𝑖=1

1
.
𝑛 − 𝑑𝑖
Thus, there exists a permutation that leads to a
clique of at least this size.
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Proof #3 of Turán’s Theorem
𝐺 = 𝑉, 𝐸 – a graph that contains no copy of 𝐾𝑟 ,
with 𝑉 = 𝑣1 , … , 𝑣𝑛 .
1
𝑑𝑖 = deg 𝑣𝑖 ,
𝑎𝑖 = 𝑛 − 𝑑𝑖 ,
𝑏𝑖 =
.
𝑛 − 𝑑𝑖
 By Cauchy-Schwarz

2
𝑛
𝑎𝑖 𝑏𝑖
𝑛
≤
𝑖=1

𝑛
𝑎𝑖2
𝑖=1
𝑖=1
Since 𝑎𝑖 𝑏𝑖 = 1, we have
𝑛
𝑛
𝑎𝑖2
𝑛2 ≤
𝑏𝑖2 .
𝑖=1
𝑑𝑖 = deg 𝑣𝑖 ,

We have
𝑛
𝑛2
𝑖=1

𝑛 − 𝑑𝑖 ,
𝑛
𝑛
𝑏𝑖2
𝑏𝑖 =
𝑛
=
𝑖=1
𝑛 − 𝑑𝑖
𝑖=1
𝑖=1
1
𝑛 − 𝑑𝑖
.
1
.
𝑛 − 𝑑𝑖
By the previous lemma, 𝐺 has a clique of size at
1
least 𝑛𝑖=1
.
𝑛−𝑑𝑖

𝑖=1
𝑎𝑖 =
𝑎𝑖2
≤
𝑏𝑖2 .
By the assumption,
𝑛
𝑛2 ≤ 𝑟 − 1
𝑛2
𝑟−1
1
𝑛
𝑖=1 𝑛−𝑑
𝑖
𝑛 − 𝑑𝑖
𝑖=1
≤
𝑛2
−2 𝐸
→
≤ 𝑟 − 1. Thus,
= 𝑟 − 1 𝑛2 − 2 𝐸 .
𝑛2
1
𝐸 ≤
1−
2
𝑟−1
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Recap

We saw three proofs of Turán’s Theorem:
◦ A simple proof by induction.
◦ A proof by assigning a weight to every vertex
and manipulating these weights.
◦ A proof by combining the probabilistic
method and the Cauchy-Schwarz inequality.
The End: Mayim Bialik

Just like the character Amy Farrah Fowler
that she plays in the show, Mayim Bialik
has a Ph.D. in neuroscience.
◦ From UCLA.
15