Proof technique (pigeonhole principle) Principle n+1 to n boxes ⇒ ≥2 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

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

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)

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 2 cycles such that any two cycles meet only in x. Proof Consider maximal (longest) path 𝑣 1 𝑣 2 … 𝑣 𝑘 . Construct 𝑑 𝑥 2 cycles for 𝑣 𝑘 … x

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 ≥ 1 2 𝑑 𝐺 𝑣 for any 𝑣∈𝑉. Proof Take bipartite F with largest |𝐸(𝐹)|.

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

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 ?!