1. Proof technique (pigeonhole principle)
Principle n to n boxes ⇒ ≥ in the same box
More general Let 𝑆=( 𝑎 1 , 𝑎 2 ,…, 𝑎 𝑛 ) be a multiset
(repetitions are allowed). Let 𝑎= ∑ 𝑎 𝑖 𝑛 be the average.
min 𝑖 𝑎 𝑖 ≤𝑎≤ max 𝑖 𝑎 𝑖 .
If 𝑎 𝑖 are integers min 𝑖 𝑎 𝑖 ≤ 𝑎 and 𝑎 ≤ max 𝑖 𝑎 𝑖

2. n choose 2 - # of different pairs among n vertices
Examples
Fact Simple graph 𝐺 with 𝑛≥2 vertices has 𝑢≠𝑣: 𝑑 𝑢 = 𝑑 𝑣 .
Proof Pigeonhole principle: 𝑑 𝑣 ∈{0,1,…,𝑛−1} + ∄ 𝑑 𝑣 1 =0, 𝑑 𝑣 2 =n-1
Problem Show that simple graph 𝐺=(𝑉,𝐸) with
contains a quadrilateral subgraph
Proof For each 𝑣∈𝑉 count pairs : 𝑣
n choose 2 - # of different pairs among n vertices

3. Proof technique (Minimal/maximal subgraph)
Principle Choose a property X. Take minimal subgraph with property X
Longest (maximal) simple path
Fact Every acyclic graph (no cycle subgraphs) with ≥2 vertices has
≥2 vertices 𝑑 𝑣 ≤1, i.e., v∈𝑉 d v ≤1 ≥2
Proof Consider the longest simple path 𝑃= 𝑣 1 … 𝑣 𝑘−1 𝑣 𝑘 . Then
𝑣 𝑘 𝑣 𝑖 ∉𝐸 for 𝑖<𝑘−1. (otherwise we found a cycle)
𝑣 𝑘 𝑣∉𝐸 for 𝑣∉𝑃 (otherwise, we can increase P by adding 𝑣)

4. Proof technique (Minimal/maximal subgraph)
Principle Choose a property X. Take minimal subgraph with property X
Longest (maximal) simple path
Problem every simple graph has a vertex x and a family of 𝑑 𝑥 cycles
such that any two cycles meet only in x.
Proof Consider maximal (longest) path 𝑣 1 𝑣 2 … 𝑣 𝑘 . Construct 𝑑 𝑥 cycles for 𝑣 𝑘 … x

5. Proof technique (Minimal/maximal subgraph)
Principle Choose a property X. Take minimal subgraph with property X
2. Largest bipartite subgraph
Theorem[Erdesh 1965] Every loopless multi-graph 𝐺 has a bipartite subgraph F such that 𝑑 𝐹 𝑣 ≥ 𝑑 𝐺 𝑣 for any 𝑣∈𝑉.
Proof Take bipartite F with largest |𝐸(𝐹)|.

6. Trees Fact T- is a tree (minimal connected graph). If 𝑉 𝑇 ≥2, then
There is at least one vertex 𝑑 𝑣 =1 (called leaf)
There are ≥2 vertices 𝑑 𝑣 =1
Fact (Recall) T is a tree ⇒ T is acyclic (no cycle subgraphs).
If G is acyclic ⇒ G is a tree.
Fact a) T – e is not connected b) if 𝑑 𝑣 =1, then T-v is a tree.
Fact
T with n vertices has 𝑚=𝑛−1 edges
G-connected graph, then ∃ 𝑆⊂𝐸:𝐺−𝑆 – is a tree.
Every connected graph has ≥𝑛−1 edges

7. Bipartite graphs
Fact Any tree is a bipartite graph and parts A and B are unique.
Proof E.g. by induction on the number of vertices.
Fact G- bipartite graph ⇔ all cycles in G have even length.
Proof “⇒” is easy (every cycle alternates between parts A and B)
“⇐” [for connected G] Take spanning tree 𝑇⊂𝐺. Define parts A and B for T. Check for each 𝑒∈𝐸 𝐺 ∖𝐸(𝑇): e is between A and B.
Proof2 “⇐”[for connected G]. By induction on 𝑛=|𝑉 𝐺 |. There is
𝑣∈𝑉:𝐺−𝑣 is connected ( 𝑑 𝑣 =1 in a spanning tree). Use induction hypothesis to partition 𝐺−𝑣. Add 𝑣: if 𝑣 𝑢 1 ,𝑣 𝑢 2 ∈𝐸, 𝑢 1 ∈𝐴, 𝑢 2 ∈𝐵, then
𝑣 𝑢 1 ,Path( 𝑢 1 − 𝑢 2 ), u 2 𝑣 – is an odd cycle ?!