Graph Theory (Neal Bushaw)

Noted by Shule Yu

Fall 2015

2015.08.25

http://516.NealBushaw.com

• Graph.

– Vertex Set V (G). Needs to be non-empty.

– Edge Set E(G). E ⊆ V (G)(2)

– Simple Graphes. No loops (vi, vi).

– Multigraph. Can have multiple edges between two vertices, sometimes allow loops.

– Weighted graphs, directed graphs, etc.

• Def. The order of a graph G is the number of vertices, noted as v(G) or |V (G)|. We usually use n as

the number of vertices of G.

– Note: Maximum number of graphs for a given vertex set V (G) is ∼ O(2n2

), which is really a huge

number !

• Def. The size of a graph G is the number of edges in the graph, noted as e(G) or |E(G)|.

– Maxinum of e(G) isn(n− 1)

2∼ O(n2).

• Def. Empty graph with n vertices, En with e(En) = 0.

• Def. Complete graph with n vertices, Kn with e(Kn) =n(n− 1)

2.

• Def. The complement graph G: V (G) = V (G); E(G) = E(Kn)\E(G).

– En = Kn

• Def. Two graph is isomorphism if there exists a bijection between them. ψ : G⇔ H.

• Def. Path of length l, noted as Pl:

– V (Pl) = {x0, x1, x2, ..., xl}.

– E(Pl) = {x0x1, x1x2, x2x3, ..., xl−1xl}.

Note that:

– P visits every vertices just once, no cycles.

– Internal vertices of this path:{x1, x2, ..., xl−1}.

– Internal edges of this path:{x1x2, x2x3, ..., xl−2xl−1}.

• Def. Cycle of length l, noted as Cl:

– V (Cl) = {x1, x2, ..., xl}.

– E(Cl) = {x1x2, x2x3, ..., xl−1xl, xlx1}.

Note: For path and cycles, length is measured by the edge number.

• Def. For two vertex sets S, T ⊆ V (G), an S−T edge set, denoted as E(S, T ), is a subset of E(G) with

each elements joining one vertex in S and one vertex in T . Its size is denoted as e(S, T ).

• Def. A graph is called bi-partited, if there exist vertex sets U,W ⊆ V (G) such that:

– U ∩W = ∅.

– V (G) = U ∪W

– E(G) = E(U,W ).

Note: bi-partited graphs are small in number:

limn→∞

|{Gbi|V (G) = n}||{G|V (G) = n}|

= 0 (1)

• Theorem. G is bi-partied if and only if it doesn’t contain odd cycles.

– Sufficiency. Let G has no odd cycles. Pick a vertex v ∈ V (G), let A be the vertex set whose

elements has their shortest paths to v are odd in length. Let B be the vertex set whose elements

has their shortest paths to v are even in length. If ∃v1, v2 ∈ A are adjacent, then there exists an

odd closed walk (v, ..., v1, v2, ..., v) and therefore there exists an odd cycle. So no vertices in A is

adjacent to each other and the same situation in B. A,B becomes a way to bi-partite G.

– Necessity. Let G is bi-partited into A,B. Suppose we have an odd cycle (v1, v2, ..., v2k+1, v1). Let

v1 ∈ A, then v2 ∈ B, v3 ∈ A, ..., v2k ∈ B, v2k+1 ∈ A. Therefore there’s an edge (v2k+1, v1) ∈ E(A)

which contradicts to the bi-partition.

• Def. The degree of a vertex v ∈ V (G) is the number of incident edges to v. Noted as d(v).

d(v) =∑

e∈E(G)

1v∈e (2)

2015.08.27

Subgraph.

• Def. If G can be bi-partited into A,B with

– |V (A)| = m.

– |V (B)| = n.

– For ∀v1 ∈ A,∀v2 ∈ B, ∃e = (v1, v2).

Then G is a complete bi-partited graph, noted as Kmn.

– Naturally, Kmn ' Knm.

• Some Notations.

– Let Gn be a graph of order n.

– Let G(n,m) be a graph of order n and size m.

• Removing vertices. Let vertex set U ⊆ V (G). Note the remaining graph G′ = G− U :

– V (G′) = V (G)\U .

– E(G′) = {e ∈ E(G)|v 6∈ e, ∀v ∈ U}.

• Removing edges. Let edge set F ⊆ E(G). Note the remaining graph G′ = G− F :

– V (G′) = V (G).

– E(G′) = E(G)\F .

• Def. Let F,G be two graphs. F is a subgraph of G, noted as F ⊆ G, if and only if:

– V (F ) ⊆ V (G).

– E(F ) ⊆ E(G).

• Def. Let vertex set U ∈ V (G). G[U ] is a induced subgraph of G by vertex subset U :

– V (G[U ]) = U .

– E(G[U ]) = {e ∈ E(G)|V (e) ⊆ U}

Note: For ∀F ⊆ G, F ⊆ G[V (F )]. In other words, induced subgraph is edge-maximized subgraph for

a given vertex set.

• Def. The neighbourhood of vertex set S ⊆ V (G) is the set of vertices adjacent to some vertices in S,

Noted as NG(S). For open neighbourhood, NG(S) ∩ S = ∅. For closed neighbourhoold, S ⊆ NG(S).

We are discussing open neighbourhood.

– For ∀v ∈ V (G), d(v) = |NG(v)|.

• Lemma. The hand shake lemma.

∑v∈V (G)

d(v) = 2e(G) (3)

– Proof 1. Count the number of (vertex,edge) pairs. Loop through vertices for the first time, and

loop through edges for the second time.

– Proof 2.

∑v∈V (G)

d(v) =∑

v∈V (G)

∑u∈V (G)

1(u,v)∈E(G)

=∑

v∈V (G)

∑e∈E(G)

1v∈V (e)

=∑

e∈E(G)

∑v∈V (G)

1v∈V (e)

=∑

e∈E(G)

2

= 2e(G)

(4)

• Cor. The sum of all degrees is even.

• Prop. Every graph has two vertices of the same degree.

– Proof. For given G with v(G) = n, all possible degree is (0, 1, 2, 3, ..., n− 1). If every vertices gets

different degree, then 0 and n−1 must be in the degree sequence, which contradicts to each other.

• Def. A graph is connected if every pair of vertices x, y are joint by some path(s).

• Def. A connected component of a graph is a connection-maximized induced subgraph.

• Lemma. G(n,m) has at least n−m components.

– Proof by induction on the size of G. Base case. En has n components. For G(n, j), by adding the

j + 1 edge, number of components minus at most 1.

2015.09.01

Walk.

• Def. The maximum degree of G is noted as ∆(G).

• Def. The minimum degree of G is noted as δ(G).

• Two inequalities. Because

∑d(v) =2e(G)

nδ(G) ≤2e(G) ≤ n∆(G)

(5)

– δ(G) ≤⌊

2e(G)

n

⌋– ∆(G) ≥

⌈2e(G)

n

⌉• Def. A vertex set U ⊆ V (G) is independent if it induce an empty graph.

– namely. for ∀u1, u2 ∈ U ,(u1, u2) is not and edge in G.

– e.g. Vertex sets A,B of bi-partited G are both independent sets.

• Def. A set of paths is vertex independent or vertex internal disjoint when the only vertices

which belongs to more than one paths are within their endpoint set.

• Def. A walk is a sequence of vertices / edges.

– Walks allow repeating vertices and edges.

– Walks includes cycles and paths.

• Theorem If there is a walk from u to v, then there is a path joining u and v.

• Prop. Every graph G with δ(G) ≥ 2 contains one path of length at least δ and one cycle of length at

least δ + 1.

– Proof. Take a longest path P in the graph, name its starting vertex v0. Since P is the longest

path, all neighbours of v0 should be visited by P , otherwise we can extend P by including that

neighbour and choosing that neighbour as the starting vertex. Since v0 has at least δ neighbours,

the length of P must be greater or equal to δ.

– The edge connecting v0 and farthest neighbour on P create the cycle of length at least δ + 1.

• Cor. Let G be a graph with minimum degree of 2, then it contains non-trivial cycle.

2015.09.03

Tree.

• Homework Problem. How to find all such graph: in its degree sequence, there’s a unique pair of same

degree.

– First notice that if G with V (G) = n is such graph, then G is also such graph.

– Then we claim if G with V (G) = n, n ≥ 3 is a connected graph with such property, then

Kn+1 − E(G) is a graph with n+ 1 vertices which has such property.

– Is that all such graphs?

• Theorem If G is disconnected graph, then G is connected.

– Proof. Let G be disconnected. For ∀x, y from the same component of G, let vertex u from another

component. Then {x, u, y} is a path in G; For ∀x, y from different component of G, (x, y) is an

edge in G. Therefore G is connected.

• Def. A tree is a connected acycle graph.

• Def. A leaf is a vertex of degree 1 in a tree.

• Def. A leaf which is not in a tree is a pendant.

• Lemma Every tree of order at least 2 has at least 2 leaves.

– Take the longest path. Then at the ends you can find these two leaves.

• Theorem The following are equal. (TFAE)

– (i). T is a tree.

– (ii). T is a minimal connected graph.

– (iii). T is a maximal acycle graph.

• Def. An edge of a connected graph is a cut edge if when you delete it, the graph becomes disconnected.

• Lemma No edge of a cycle is a cut edge.

• Cor. Every edge in a tree is a cut edge.

• Proof of the TFAE.

– (i) → (ii). Let (u, v) be a redundent edge of tree G to be a mininal connected graph. Then by

removing (u, v), G is still connected, therefore there exists a u − v path in G − (u, v). And by

adding (u, v) back, we create a cycle in G. Therefore no such (u, v) exists in a tree.

– (ii) → (i). Let’s suppose G is a minimum connected graph but not a tree. Then there exists a

cycle. Since no edge of a cycle is a cut edge, we can remove one edge from the cycle while keep

the graph connected. Then it means G is not minimum connected. Therefore no such cycle and

G is a tree.

– (i) → (iii). Let’s suppose tree G is not a maximum acycle graph, we can add at least one edge(s)

while keeping it acycle. Since the result is still a tree, it’s a minimum connected graph. Then by

removing the added edge(s) will result in a disconnected graph G. Therefore no such edge(s) can

be added.

– (iii) → (i). Let’s suppose G is maximum acycle graph but not a tree. Then G is disconnected. By

add an edge to connect two components of G we at the same time create a cycle containing the

edge just added. Then removing it will left these two components connected, which contradicts to

where we started.

• Theorem TFAE II.

– (i). G is a tree.

– (ii). G has n− 1 edges, with either acycle or connected.

– (iii). For every pair of vertices x and y there exists a unique x− y path.

• Proof of TFAE II.

– (i) → (ii). Induction basic case: V (G) = 1. Induction on number of vertices of the graph: Let’s

suppose for any tree G, if V (G′) = n−1, then e(G′) = n−2. Let’s consider tree G with V (G) = n.

Since it’s a tree, it has at least two leaves u, v. Let G′ = G− v, then G′ is still a connected acycle

graph. So G′ is a tree with n− 1 vertex. So G′ has n− 2 edges. Therefore G has n− 1 edges.

– other relations are quite easy to prove.

• Cor. Adding an edge to a tree yields exactly one cycle.

2015.09.08

Tree.

• Def. A spanning tree of G (T ⊆ G) is a tree that use all the vertices in V (G).

– Q1: How many spanning tree (don’t consider isomorphism) exists for n vertices?

– A1: Find a way to naming each of them and the number of names is the number of spanning trees.

– Q2: How many spanning tree (consider isomorphism as the same tree) exists for n vertices?

– A2: open question. No body knows the answer.

• Cor. Every connected graph on n vertices has at least n−1 edges and have a spanning tree as subgraph.

• Cayley’s formula. There are nn−2 spanning trees for a n-vertex graph G. (No isomorphism.)

• Prufer code. The algorithm of naming a spanning tree.

– (i). Number all vertices in G in order.

– (ii). For the current spanning tree, find the leaf with the smallest number, let it be v.

– (iii). Remove v from T , write down the number of its neighbour.

– (iv). Repeat (ii) and (iii) until what remains is an edge.

– e.g.

– Claim that every Prufer code can be reconstructed to a spanning tree.

∗ Proof. If we can find a way to reconstruct the n − 1 edges, we can reconstruct the spanning

tree. Algorithm of re-construct:

· (a). Let V = {1, 2, 3, ..., n}, Prufer code P = {p1, p2, ..., pn−2}

· (b). Find the smallest label in V but not in P as k, let (k, p1) be an edge.

· (c). Remove k from V , remove p1 from P .

· (d). Repeat (b) and (c) until there’s not element left in P . Now there’s 2 element left in

V . Let them be an edge.

– Conclusion. For n-vertex spanning tree, there are nn−2 Prufer code. Therefore the number of

spanning tree is nn−2.

2015.09.10

Conncetivity. Mader’s Theorem.

• Def. A vertex cut of a connected graph G is a vertex set X ⊆ V (G) such that G[V (G) − X] is

disconnected.

– Note: the interesting topic will be the minimum size of a vertex cut for G.

• Def. A graph is called k-connected if |V (G)| > k and for ∀X ⊆ V (G) with |X| < k, X can’t be a

vertex cut.

• Def. The connnectivity of G, noted as κ(G), is the largest k such that G is k-connected. In other

words, the miminum size of any vertex cut of G.

– Facts.

∗ κ(G) ≤ δ(G).

∗ If a graph is k-conncected, then it’s also (k − l)-connected, k ≥ l > 0.

∗ 0-connected: every graph.

∗ 1-connected: a connected graph.

∗ 2-connected: a connected graph with no cut vertex.

∗ κ(Kn) = n− 1

• Theorem, Mader 72’ Every graph which has an averaged degree at least 4k contains a k-connected

subgraph.

– ∀G(n,m) with d(G) ≥ 4k satisfy:

∗ n ≥ 2k − 1.

∗ e(G) = m ≥ (2k − 3)(n− k + 1) + 1.

– Proof.

∗ Since n− 1 ≥ ∆(G) ≥ d(G) ≥ 4k, n ≥ 4k + 1 ≥ 2k − 1.

∗ We have:

∑d(vi) = 2e(G)

nd(G) = 2m

m =n

2d(G) ≥ 2nk.

(6)

Therefore

(2k − 3)(n− k + 1) + 1 = 2nk − (3n+ 2k2 − 5k)− 2

≥ 2nk − (3(4k + 1) + 2k2 − 5k)− 2

= 2nk − 2− (2k2 + 7k + 3)

≤ 2nk ≤ m

(7)

– G(n,m) contains a k-connected (k ≥ 2) subgraph if

∗ n ≥ 2k − 1.

∗ e(G) = m ≥ (2k − 3)(n− k + 1) + 1.

– Proof.

∗ n = 2k − 1. then by the second requirements

e(G) =m ≥ (2k − 3)(n− k + 1) + 1

=(n− 2)1

2(n+ 1) + 1

=1

2n(n− 1)

(8)

In this case, G is a complete graph with connectivity n−1 = 2k−2 ≥ k. Choose the subgraph

to be G.

∗ n ≥ 2k. Applying induction on n. Let’s suppose ∀|G′| = n − 1 satisfy the claim. That

is, for every |G′| = n − 1 with n − 1 ≥ 2k − 1, e(G′) ≥ (2k − 3)(n − 1 − k + 1) + 1,

then G′ has a k-connected subgraph. And now for any |G| = n satisfying n ≥ 2k − 1,

e(G) ≥ (2k − 3)(n− k + 1) + 1 we are finding G a k-connected subgraph.

· Case 1. There is some vertex v with d(v) ≤ 2k − 3. Then by removing vertex v, we have

a G′ satisfying:

n− 1 ≥ 2k − 1.(because n ≥ 2k)

e(G′) ≥ e(G)− d(v) = (2k − 3)(n− 1− k + 1) + 1.

(9)

Therefore G′ = G\v has a k-connected subgraph, therefore G also has a k-connected

subgraph.

· Case 2. All vertices have degree ≥ 2k − 2. Assuming G is not k-connected, then we can

find a vertex cut X with v(X) < k such that G−X is a disconnected graph with at least

two components C1, C2. Let G1 = G[V (C1)∪X] and G2 = G[V (C2)∪X]. Then we have:

1. e(G) ≤ e(G1) + e(G2).

2. Assume G1, G2 are both not k-connected (otherwise we find some k-connected subgraph

of G).

3. In graphs G1, G2, some vertices still have degree ≥ 2k − 2. Namely, v(Gi) ≥ 2k − 1.

Using the G′ condition on Gi: since Gi are not k-connected, then

e(Gi) < (2k − 3)(|Gi| − k + 1) + 1 (10)

Therefore

e(G) ≤ e(G1) + e(G2)

< (2k − 3)(|G1| − k + 1) + 1 + (2k − 3)(|G2| − k + 1) + 1

≤ (2k − 3)(n− k + 1) + 1

(11)

contradicting our assumptions. Therefore G is k-connected and we can choose the sub-

graph to be G.

2015.09.15

Block.

• Def. A disconnecting set of edges is a subset F ⊆ E(G) such that G− F is disconnected.

• Def. For ∀S ⊂ V (G), E(S, S) is an edge cut.

• Cor. An edge cut is a disconnecting set. A minimum disconnecting set of G is an edge cut.

• Def. A graph is k-edge connected if every disconnecting set of G has at least k elements. (in another

words, ∀|F | ≤ k − 1, G− F is connected.)

• Def. The edge connectivity κ′(G) is the maximum k such that G is k-edge connected. (in another

words, the minimum size of any disconnecting set of G)

• κ(Kn) = κ′(Kn) = n− 1.

– But κ(G) and κ′(G) are not the same in general. The graph bellow have κ(G) = 2, κ′(G) = 3:

• Theorem κ(G) ≤ κ′(G) ≤ δ(G).

– Their relation to δ(G) is obvious.

– Proof of κ(G) ≤ κ′(G). Let S be the vertex set such that E(S, S) is a minimum edge cut of G.

Therefore |E(S, S)| = κ′(G).

∗ Assume every vertex in S is adjacent to every vertex in S and vice versa, then |E(S, S)| =

|v(S)||v(S)|. Since v(S) + v(S) = n, we have |E(S, S)| = κ′(G) ≥ 1 · (n− 1) = n− 1 ≥ κ(G).

∗ Assume ∃x ∈ S, y ∈ S such that (x, y) 6∈ E(S, T ). Let set up a vertex set T contiaining:

· Every neighbours of x in S.

· Every vertices in S (except x) that has neighbour(s) in S.

· schematic figure:

Then T is a vertex cut of G (by deleting T we guarantee x and y remains and non-adjacent),

and v(T ) ≤ e(S, S). Therefore κ(G) ≤ v(T ) ≤ e(S, S) = κ′(G).

• Def. A block of a graph G is a connection-maximized induced subgraph contains no cut vertex (of

the subgraph).

– A block is either 2-connected, or a bridge of G plus its endpoints.

– Every block with at least 3 vertices is 2-connected.

– If a block contains only a single edge, then that edge is a bridge in G.

– Every edge is in a block.

– Isolated vertex is a connected subgraph.

– One vertex can belong to different blocks.

• Lemma Two blocks B1, B2 ⊆ G share at most 1 vertex.

– If two blocks shares an edge, then B1 ∪B2 should be a larger block.

• Def. The block graph of G is a bipartited graph with one partite set containing all the cut vertices

vi of G and the other partite set containing one vertex bi for each block Bi of G. And (vi, bj) is an edge

if vi ∈ Bj .

• Theroem For any connected graph G, its block graph is a tree.

– Connection. Because G is connected.

– Acycle. Any cycle is 2-connected. Therefore the cycle must be within certain block. So in a block

graph, there won’t be any cycles.

2015.09.17

Whitney’s Theorem.

• Def. The distance between two vertices u, v, noted as d(u, v), is the length of the shortest path joining

them.

• Theorem, Whitney 1932 A graph G with |V (G)| ≥ 3 is 2-connected if and only if every vertex pair

u, v are joint by at least two internal disjoint paths.

– Sufficiency. If there are two internal disjoint paths joining any pair of vertices u, v, then by

removing any vertex (except u, v), there’s still at least one path joining u, v. Therefore G is

2-connected.

– Necessity. By induction on d(u, v).

∗ Basic case. d(u, v) = 1. Since κ(G) = 2 ≤ κ′(G). Therefore by removing (u, v), the graph is

still connected, meaning there’s some path(s) joining u, v. So originally there are at least two

internal disjoint paths joining u and v.

∗ Let us suppose that for d(u, v) = k − 1, given G is 2-connected means there’s at least two

internal disjoint paths connecting u, v, and now consider the case where d(u, v) = k in a 2-

connected G. Since k > 1, we can name a vertex w, which is adjacent to v on the shortest

u−v path P . Since d(u,w) = k−1, then there exists another internal disjoint path Q between

u,w other than P − v. Now, since G is 2-connected, then by removing w, there exists another

path R connecting u, v. Notice in this way, w 6∈ R.

· If R doesn’t intersects with neither P nor Q. Then P and R are the two internal disjoint

paths between u, v.

· If R intersects with P and Q. Call the common vertex nearest to v on Q as x.

1. If x ∈ P , then the first path is Q,w, v, the second path is u, x,R, v.

2. If x ∈ Q, then the first path is Q, x,R; the second path is P .

• Cor. For a 2-connected G, every pair of vertices are in a cycle.

• Def. Let vertex sets A,B ⊆ V (G), then an A,B path is any path which starts in A and ends in B,

and all its vertices live inside A ∪B.

2015.09.22

Menger’s theorem.

• Def. Let the line graph of G, L(G), is the graph whose vertex set is the edge set of G, and for

e, f ∈ E(G), (e, f) is and edge if e, f shares a vertex in G.

• Menger’s Theorem, 1927

– Vertex form. Let s and t be distinct nonadjacent vertices of a graph G. Then the minimal

number of vertices separating s from t is equal to the maximal number of vertex disjoint s − t

paths.

– Edge form. Let s and t be distinct vertices of G. Then the minimal number of edges separating

s from t is equal to the maximal number of edge-disjoint s− t paths.

– S − T version. Let vertex set S, T ⊆ V (G). Then the maximum number of vertex disjoint

S − T paths (none of the paths have common vertices) is equal to the minimum size of an S − T

separating set.

– Notice: this theorem tells us connectivity can be measured both in invulnerabiliity to deletions

and in the number of alternative paths.

– Proof of the vertex form. Let’s suppose k is the size of minimum separting set between two non-

adjacent vertices s and t. For k ≤ 1, we have a disconnected graph or a tree, and therefore k vertex

disjoint paths. Let’s assume there are counterexamples to the theorem for k ≥ 2, and consider

the conter example G with the minimum number of k and minimum number of edges. Then

in G, s and t can’t share any common vertices: suppose x is their common vertex, then G − x,

satisfying the theorem, has a s− t separating set of size k − 1, and therefore exists k − 1 vertex-

disjoint paths, which results in G has k vertex-disjoint paths and therefore G is not a counter

example.

Let W be a s− t separating set of size k. Let’s assume neither of s nor t is connected to all vertices

in W . By collapsing the component of G −W containing s to a single vertex s′, we find s′ need

to be adjacent to all the vertices in W and |s′| > 1. We have a new graph G′ = s′ −W − t. Since

|s′| > 1, we have e(G′) < e(G). Therefore G′ satisfies the theorem. Since we still need k vertices

to disconnect s′ and t, there exists k vertex disjoint paths between s′ and t. So in the W −t part of

G, there exists k paths connecting every vertices in W and t, with their only common vertex at t.

Similarly, if we do the collapsing on the component of G−W containing t, we can prove there exists

k paths connecting every vertices in W and s, with s be their only common vertex. Therefore by

combining this two sets of paths, we found k vertex disjoint path connecting s and t, which makes

G not a counter example. Therefore either s or t is connected to all the vertices in W .

Let P be the shortest path connecting s and t in G. Since s and t have no common vertex, length

of P ≥ 2, and we can write P = sx1x2...t. Now consider G−(x1, x2), it needs k or k−1 vertices to

disconnect s and t. If it is k, since G−(x1, x2) is smaller than the minimum couter example, so the

theorem applies, which means there are k vertex disjoint path connecting s and t in G− (x1, x2),

and G can’t be a counter example. Therefore we can find a s− t separating set W0 of G− (x1, x2)

with |W0| = k−1. Let W1 = W0∪x1 and W2 = W0∪x2. Then W1 and W2 are two s−t separating

set of size k. Notice that either of s or t must be adjacent to all vertices in any s − t separating

set W . Since s is not adjacent to x2, t must be adjacent to all vertices in W2. Similarly, s must

adjacent to all vertices in W1. Therefore s and t have common vertices: the vertices in W0. Since

|W0| = k − 1 ≥ 1, we find it contraditing to the statement that s and t doesn’t have any common

vertex. Therefore there’s no such counter example G.

– By adding a virtual vertex v1 connecting to every vertices in S, adding another virtual vertex v2

connecting to every vertices in T , we can prove the S − T version.

∗ Note, we can just adding one virtual vertex and get a v − T version of the theorem.

– By applying the S − T version on L(G), we get the edge form.

• Cor. By removing an edge, the connectivity of the graph decrease at most 1.

– Apply vertex form of menger’s theorem on a k-connected graph.

• Global Version.

– A graph G is k-connected if and only if it contains k internally vertex disjoint paths between any

pair of vertices.

– A graph G is k-connected if and only if it contains k edge disjoint paths between any pair of

vertices.

– Sufficiency. If there exist k internally disjoint paths between any pair of vertices u, v, then we

need to remove at least one vertex from each of the paths to disconnect u, v. Therefore G is

k-connected.

– Necessity. Let’s assume G is k-connected but with less than k internally disjoint paths between

some pair of vertices u, v. Then by vertex form of Menger’s theorem, u, v must be adjacent. And

G − (u, v) have at most k − 2 internally disjoint paths between u and v. By Menger’s again

(G − (u, v) satisfies the theorem), there exist a u − v separating set with size of k − 2. However,

by removing one edge the connectivity of G decrease at most 1. Therefore such separating set

doesn’t exist.

2015.09.24

Eulerian.

• Def. A multi-graph is a graph which may have more than one edge between a given pair of vertices.

• Def. A Trail is a walk with no repeated edges. (can repeat vertices)

• Def. A Eulerian walk/path is a trail visits all vertices.

• Def. A Eulerian circuit is a closed Eulerian walk/path.

• Theorem If G has a Eulerian circuit, then d(v) is even for every vertices.

• Lemma If every d(v) is even, then the maixmal trail is closed.

• Theorem If G is connected, then G has an Eulerian circuit if and only if all degrees of G are even.

– Sufficiency. Suppose G is a graph and all its vertices have even degree. Apply induction on the

number of edges.

∗ Basic case: δ(G) = ∆(G) = 2. G is a big cycle, check.

∗ Induction: Since δ(G) ≥ 2, therefore there exists at least one non-trivial cycle in G. Let the

maximum cycle be C and pick a v ∈ C with d(v) > 2. Consider G′ = G − E(C). Let H be

the component of G′ containing v. Since H has less number of edges than G, and all vertices

of H still has even degree, H has an Eulerian circuit D. Since D and C intersects (at least at

v) and edge disjoint, there exists a larger cycle in G than C.

– Sufficiency2. Split vertex v with d(v) = 2k into k sub-vertices, so that every sub-vertices has

degree 2. Do this for all vertices in G, then G becomes a big cycle, which is the Eulerian circuit

we want.

• Theorem A connected graph G has an Eulerian trail/walk/path if and only if it has at most two

vertices of odd degree.

– 2 odd degrees: connecting these two vertices and there’s an Eulerian circuit.

– more than 2 odd degrees: removing more than 2 non-adjacent edges from an Eulerian circuit will

create segaments, can’t guarantee an Eulerian trial.

2015.09.29

Hamiltonian. Dirac’s Theorem.

• Def. A Hamiltonian path/cycle is a path/cycle which visits every vertex exactly once. Graphs contain

hamiltonian cycle are Hamiltonian graphs.

• Prop. If G is Hamiltonian, then for any vertex set S ⊆ V (G), the graph G − S has at most |S|

component.

– Proof. Let C1, C2, ..., Cl be the l component of G − S, let vi be the first vertex in S to be visit

after the Hamiltonian cycle leaves component Ci. Then |S| ≥ l.

• Cor. If G is Hamiltonian and bi-partited into A,B. Then |A| = |B|.

• Dirac’s Theorem If G is a simple graph with n ≥ 3 vertices, and if δ(G) ≥ n

2, then G is Hamiltonian.

– Proof. Consider the edeg-maximized counter example G with δ(G) ≥ n

2. Since G iss edge-

maximized, we can find (u, v) 6∈ E(G), and G + (u, v) is Hamilton. Therefore G originally has a

Hamilton path from u to v.

From the figure we can see for ∀xi ∈ NG(u), xi−1 6∈ NG(v). Because if xi−1 ∈ NG(v), then we

can find a Hamiltonian cycle in G (red order in the figure above). Therefore to keep G non-

Hamiltonian, we need max{d(u) + d(v)} ≤ n− 1 (critical case, xi ∈ NG(u), xi ∈ NG(v)), which is

contradicting to δ(G) ≥ n

2.

• Ore’s Theorem Let G be a simple graph with n ≥ 3 vertices. If for ∀u, v ∈ G, we have d(u)+d(v) ≥ n,

then G is Hamiltonian.

2015.10.01

Matching.

• Def. An edge set M is called a matching if no two edges share an endpoint.

– Notice: max size of matching for a graph of order n is

⌊n

2

⌋.

• Def. A matching is perfect if it containsn

2edges (v(G) = n = 2k). In other words, every vertices of

G belongs to one edge.

• Def. The matching number of a graph G, called ν(G) is the max size of any matching in G.

– ν(Kn) =

⌊n

2

⌋– ν(Kmn) = min{m,n}

• SideTrack. Travelling salesman problem.

– What’s the shortest (lowest weight) collected when a walk go through all vertices?

∗ It’s a NP complete problem.

– Method: solve it on a complete graph, then set the non-existed edges to a super high weight.

– Thoughts: Can Kn be decomposed into edge disjoint Hamiltonian cycles?

∗ E(Kn) =1

2n(n − 1). Therefore the number of Hamiltonian cycles (if such decomposition

exists) is E(Kn)/n =1

2(n − 1). Therefore for n is an even number, such decomposition

doesn’t exists.

∗ Now consider odd n. A way to decompose is rotating the zig-zag path and finally connecting

the start/end vertex with the temperorily removed vertex (shown here is n = 7):

∗ Summary.

· If n ≥ 3 and n is odd, then Kn can be decomposed into1

2(n−1) edge disjoint Hamiltonian

cycles.

· If n ≥ 3 and n is even, then Kn can be decomposed inton

2edge disjoint Hamiltonian

paths.

• Application of matching.

– The Scheduling Problem. Certain workers can operate certain machines. Therefore G is a bi-

partited graph with A = {workers},B = {machines}. Target is find ν(G).

– Medical Residency Problem. Each graduates rank the hospitals they chose; each hospitals rank

every graduates. Find a stable matching.

• Def. Given a vertex set A ⊆ V (G), the matching M of G convers A if every vertices in A is one

endpoint of some edge in M .

• Def. A vertex set T ⊆ V (G) is a cover of G if every edges in E(G) has at least an endpoint in T .

• Def. The cover number of G, denoted as τ(G), is the minimum size of any cover of G.

• Prop. ν(G) ≤ τ(G) ≤ 2ν(G).

– First half is obvious. Let’s prove τ(G) ≤ 2ν(G). Consider the worst case, in the maximum

matching, there’s an extra edge hanging over one vertex of each pair. In this case τ(G) = 2ν(G).

2015.10.06

Hall’s Theorem.

• Hall’s Theorem If G is a bi-partited graph with bi-partition A,B, then G has a matching convering

A if and only if for ∀S ⊆ A, |NG(S)| ≥ |S|.

– Necessity. Obvious.

– Sufficiency. Let’s suppose for ∀S ⊆ A we have |NG(S)| ≥ |S|. Assume there’s no matching

convering A. Take the maximum matching M in terms of covering most vertices in A. Let’s name

the vertex set in A not covered by M as S′. Since M is maximized, we have NG(S′) ⊆ V (M).

Moreover, M is a matching, we have: NG(A − S′) = |M |. Pick one vertex v ∈ S′, and let

R = {A− S′} ∪ {v}, then NG(R) = |M |. On the other hand, since |R| = |A− S′|+ 1 = |M |+ 1.

Therefore, for R ⊆ A, |NG(R)| = |M | < |R|+ 1 = |R|.

– Defected Hall’s Theorem. If G is a bi-partited graph with bi-partition A,B, and for every S ⊆ A,

|NG(S)| ≥ |S| − d for some fixed integer d. Then G has a matching covering all but at most d

vertices in A.

∗ Add a virtual vertex set H of size d to B and set very vertices in H adjacent to every vertices

in A. Apply Hall’s theorem and then remove set H.

• Def. If every vertices in G is k degree, then we say G is k-regular.

• Cor. If G is bi-partited graph with A,B and G is k-regular for some k ≥ 1, then G has a perfect

matching.

– Proof. If a graph is k-regular and it’s bi-partited, then the size of two parts are equal: |A| = |B|.

Because the number of incident edges into this two sets are equal (bi-partited) and each vertex

has the same number of incident edges. Secondly, for ∀S ⊆ A, the minimum size of NG(S) is |S|.

Because there are k|S| edges incident on NG(S), and each vertex of NG(S) can receive at most k

edges. Therefore G has a matching covering A. Since |A| = |B|, this matching also covers B and

thus a perfect macthing.

• Konig’s Theorem For bi-partited graph G = (A ∪B), ν(G) = τ(G).

– Proof. Recall τ(G) is the minimum size of vertex cover of G, while ν(G) is the maximum size of

matching of G. Take a minimum vertex cover U , let R = U ∩ A, T = U ∩ B, H = R ∪ {B\T},

H ′ = T ∪{A\R}. Take S ⊆ R, then since U is minimized, |NH(S)| ≥ |S| (otherwise we can replace

S by NH(S) in U and get a smaller vertex cover), therefore there’s matching covering R. In the

same way, there’s matching covering T . By combining these two matchings, we got a matching

the same size as U . Notice that U is a vertex cover guaranteed there’s no edges between {A\R}

and {B\T}. Therefore the matching we got is a maximized matching.

2015.10.08

Tutte’s Theorem.

• Def. Let q(G) be the number of odd size component in graph G.

• Def. A vertex is called universal if it is adjacent to all the other vertices in G.

• Tuttle’s Theorem G has a perfect matching if and only if q(G− S) ≤ |S| for ∀S ⊆ V (G).

– Necessity. Given G with a perfect matching. For ∀S ⊆ V (G), let Ci be the components of G− S.

For each odd component, at least one of the vertices need to be adjacent to S to form a matching.

And for each vertices within S, it at most respond to one odd component in terms of matching.

Therefore, |S| ≥ q(G− S).

– Sufficiency. Let’s consider the edge maximized counter example of the theorem. That is, a G

without perfect matching, but for ∀e 6∈ E(G), G + e has a perfect matching (edge-maximized).

While for ∀S ⊆ V (G), |S| ≥ q(G − S) (counter example). We will choose a particular vertex set

S and then find a perfect mathcing for G to reach a contradiction. The particular vertex set is

S = {x|d(x) = V (G) − 1}, namely, the set of universal vertices (it doesn’t matter if S = ∅). Let

Ci be the components of G− S.

∗ Case 1. All Ci themselves are complete graphs. Since |S| ≥ q(G − S), we can easily find a

perfect matching in this case.

∗ Case 2. Some Ci have at least two non-adjacent vertices (x, z).

· Claim: there exist a y in the same component that is adjacent to both x and z. Let P be the

shortest path joining x and z, if there’s no such y, then l(P ) ≥ 3. Let P = (x, v1, v2, ..., z),

since it’s shortest, (x, v2) is not an edge. Then we can redifine z = v2, y = v1.

Since y is not in S, then there is w 6= x, z (since (x, y), (y, z) are known edges) that (y, w)

is not an edge. Because G is edge-maximized, there’s perfect matching M1 for G + xz with

(x, z) ∈M1 and perfect matching M2 for G+ yw with (y, w) ∈M2. We can construct a new

graph F :

V (F ) = V (G)

E(F ) = {e|M1 ⊕M2}(12)

Then for ∀v ∈ V (F ), d(v) = 0, 2. Therefore each component of F are even cycles or inde-

pendent vertex. Moreover, Each cycles in F have edges sequence M1,M2,M1, ...,M2,M1 in

terms of their original matching. Let’s now consider the cycle C which contains (x, z).

· If (y, w) isn’t in C. Then the combination of all M1 outside of C and all M2 inside of C

is a matching M using only the original edges in G.

· If (y, w) is also in C. We still can find a matching M for V (C) by using original edges in

G (that is, don’t use yw ∈ M2 and xz ∈ M1), by the method described in the following

graph:

(the position of x and z can be interchanged.)

• Cor. If G is 3-regular and has no cut edge, then it contains a perfect matching.

– Proof. By the restriction, let S ⊆ V (G), and Ci be the components of G− S. Because there’s no

cut edge, each component should joint S with more than 1 edge. Let’s focus on the components

with odd nubmer of vertices, choose C be one of them. If there’s only 2 edges between S and C,

then within C, the sum of the degree is 3|V (C)| − 2, which is an odd number. Therefore for each

odd component, there’s at least 3 edges between them and S, while for S the total number of

edges it can provide is 3|S|, therefore the number of odd component is less equal to |S|. Therefore

Tutte’s requirement holds for G.

2015.10.15

Vertex coloring.

• Motivation.

– Schedualing.

∗ No student takes two finals at the same time.

∗ G(V,E), V = {finals}.

∗ (vi, vj) ∈ E(G), if ∃ a student in both vi and vj .

∗ We need to group V (G) into independent groups of vertices. (coloring)

∗ What’s the minimum number of groups?

– Zoo planning.

∗ vertices: animals.

∗ edges: connecting don’t-get-along animals. (edges connecting incompatible elements)

∗ What’s the minimum number of groups?

• Def. A k-coloring of G is a function c : V (G)→ {1, 2, 3, ..., k}.

• Def. A proper k-coloring is the function c such that if (v1, v2) ∈ E(G), then c(v1) 6= c(v2).

• Def. G is k-colorable if ∃ a proper k-coloring c of G.

• Def. Chromatic number, noted as ψ(G), is the minimum k such that G is k-colorable.

– ψ(En) = 1− δn,0.

– ψ(Pn) = ψ(bi− Partited) = ψ(C2k) = 2.

– ψ(C2k+1) = 3, k > 0.

– ψ(Kn) = n.

• Notations.

– α(G) is the size of the largest independent set.

– ω(G) is the size of the larget complete sub-graph of G.

• Properties.

– Color classes are independent sets themselves. G can be viewed as C1 − C2 − C3 − ... − Cψ(G).

Namely, G is k-colorable ⇒ G is k-partited.

– ψ(G) is also the minimum number of independent vertex sets required to cover V (G).

– α(G)× ψ(G) ≥ |G|

– If H ⊆ G, then ψ(H) ≤ ψ(G).

• Cor. ψ(G) ≥ ω(G). (Let H in the properties be the largest complete subgraph of G).

• Prop. Let vertex set U ⊆ V (G), then ψ(G) ≤ ψ(G[U ]) + ψ(G[V (G)\U ]).

• Prop. Let subgraphs G1, G2 ⊆ G such that V (G1) = V (G2) (share the same vertices), then ψ(G1 ∪

G2) ≤ ψ(G1) · ψ(G2).

– Proof. Let c1,c2 be the k1-coloring and k2-coloring of G1 and G2. Define

f : v ∈ (G1 ∪G2)→ (c1(v), c2(v)) (13)

then f is a proper coloring ofG1∪G2, which have a value set whose size is at most the multiplication

of k1 and k2.

• Cor. |G| ≤ ψ(G)ψ(G).

• Prop. ψ(G) + ψ(G) ≤ |G|+ 1.

– Proof. Induction on v(G) = n. Basic Case: n = 1. check. Induction step: suppose for ∀G′ with

v(G′) = n − 1, we have: ψ(G′) + ψ(G′) ≤ |G′| + 1 = n. Let’s consider |G| = n. Pick any vertex

v ∈ V (G) and let G′ = G− v. Let c be the k-coloring of G′ and c′ be the k′-coloring of G′, then

k + k′ ≤ n.

∗ k + k′ ≤ n − 1. We can just select two new colors to color v in G and v in G. Then

ψ(G) + ψ(G) ≤ n+ 1 = |G|+ 1.

∗ k + k′ = n. Then k = n− k′.

· dG(v) < k. Then we can find one proper color in c to color v in G. Then just select one

new color to color v in G: ψ(G) + ψ(G) ≤ k + (k′ + 1) = n+ 1 = |G|+ 1.

· dG(v) ≥ k. Then since d(G)(v) = (n − 1) − dG(v) ≤ (n − 1) − k = (n − 1) − (n −

k′) = k′ − 1 < k′, this becomes counter version for the previous case. We can find one

proper color in c′ to color v in (G). Then just select one new color to color v in G:

ψ(G) + ψ(G) ≤ (k + 1) + k′ = n+ 1 = |G|+ 1

Therefore ψ(G) + ψ(G) ≤ |G|+ 1.

• Prop. ψ(G) ≤ ∆(G) + 1.

– Proof. Greedy coloring: When coloring the present vertex, choose the lowest proper color code.

The coloring order is in random. Therefore by preparing ∆(G)+1 colors, we can guarantee there’s

a proper coloring.

– This will give the upper bound of ψ(G).

– What about the lower bound, or a tighter upper bound of ψ(G)? Can we find it by following a

certain order in greedy coloring?

2015.10.20

Brooks’ Theorem.

• SideNote. 3-regular graph has a 3-coloring.

• Def. A graph is k-degenerate if its any subgraph H ⊆ G has one vertex with degree (within the

subgraph) at most k.

• Prop. G is k-degenerate if and only if there is an ordering of V (G) such that each vi has at most k

neighbours among {v1, v2, ..., vi−1}.

– Sufficiency. Let {v1, v2, ..., vn} be such an order of given G. Consider any subgraph H ⊆ G and

pick vj ∈ H to be the last H vertices appearing in the order. Then all vj ’s neighbours in H

are among {v1, v2, ..., vj−1}, so |NH(vj)| ≤ k. Therefore we find the vertex in H with at most k

neighbours.

– Necessity. Given a k-degenerate graph G, we need to find such a vertex ordering. Applying

induction on |V (G)|. Let’s suppose for ∀G′ with |V (G′)| = n− 1, we have such a vertex ordering.

Now consider |V (G)| = n. Let H = G, then since G is k-degenerate, there’s a vertex, call it vn ∈ H,

has at most k neighbours. Let G′ = G − vn, then any subgraph of G′ is still a subgraph of G.

Therefore G′ is k-degenerate. So there exists such a vertex ordering for V (G′): {v1, v2, ..., vn−1}.

By putting vn at the end of this order we find a vertex order for G.

• Cor. If G is k-degenerate, then there’s a way to order the vertices such that the greedy coloring uses

at most k + 1 colors.

• Def. The degeneracy of G, denoted as dg(G) is the smallest k such that G is k-degenerate.

• Lemma. δ(G) ≤ dg(G) ≤ ∆(G).

• Cor. ψ(G) ≤ dg(G) + 1 ≤ ∆(G) + 1.

– For Kn and C2k+1, k > 0, dg(G) = ∆(G).

• Brooks’ Theorem If G is a connected graph and it is neither a complete graph nor an odd cycle, then

ψ(G) ≤ ∆(G).

– Proof. For notation convenience, we let ∆(G) = k.

– k = 0, 1. Impossible case for a connected graph.

– k = 2. Even cycles. ψ(G) = 2 ≤ k.

– k ≥ 3.

∗ G is not k-regular. Then we can pick a vertex v with d(v) < k. Start from v as the root, we

can build a spanning tree. If we label the root as n and label each vertices within the tree

layer by layer, then we find a vertex order such that each vertices has at least one neighbour

whose label is greater than its label. Therefore we find a ordering {v1, v2, ..., vn}, (v = vn)

such that for any vertex vj , it has at most k − 1 neighbours among {v1, v2, ..., vj−1}. So we

have a k-coloring of G.

∗ G is k-regular and has a cut vertex. Let the cut vertex be v and the components of G − x

be Ci. Therefore in G[V (Ci) ∪ {x}] d(x) < k, and by the result of the previous case, we can

color each G[V (Ci)∪{x}] with k colors. By permutating the colors, we can let x has the same

color acrossing every G[V (Ci) ∪ {x}]. Therefore we can find a k-coloring for G.

∗ G is k-regular and 2-connected. Same spanning tree idea, but we need such three vertices

v1, v2, v:

· v1 is not adjacent to v2.

· v is adjacent to both v1, v2.

· G− v1, v2 is a connected graph.

If we have these three vertices, then we can build a spanning tree of G − v1, v2 (because it’s

connected) starting from v and get a vertex ordering {v3, v4, ..., vn}, v = vn. Adding {v1, v2}

to the front, we get {v1, v2, v3, v4, ..., vn−1, vn}, v = vn such that for every vertex except v, it

has at least one neighbour whose label is greater than its label. Therefore we can k-coloring

{v1, v2, v3, v4, ..., vn−1}. Since v1 and v2 are not adjacent, we can choose the same color for

them. Thus vn can get a color in the k-coloring because two of its k neighbours get the same

color.

So we need to show that we can find such three vertices in a k-regular k ≥ 3, 2-connected

graph G. Pick an arbitrary vertex x ∈ V (G).

· κ(G − x) ≥ 2. If ∀y ∈ NG(x), d(x, y) = 1, then G is a complete graph. So we can find a

vertex y ∈ V (G) that d(x, y) = 2. Choose v1 = x, v2 = y and v be the vertex between x

and y. In this way, we guarantee G− v1, v2 is still a connected graph.

· κ(G− x) = 1. Let Ca, Cb be the two leaf block in G− x’s block graph. Let vertex cut of

G− x c1, c2 be the cut vertex in Ca, Cb. Since κ(G) = 2 and κ(G− x) = 1, x need to be

connected to some vertex in C1, C2 other than c1, c2. Choose v1 to be the vertex adjacent

to x in Ca, v2 to be the vertex adjacent to x in Cb and v = x. Because k ≥ 3, x has at

least one more edge connecting to the components of G− x. Therefore G− v1, v2 is still

a connected graph.

2015.10.22

Planar.

• Def. A polygonal path is a connected series of line segments with no self-intersecting points.

• Def. A drawing of a graph G is a function that maps v ∈ V (G) to a point f(v) in the plane and maps

every edge (u, v) ∈ E(G) to a polygonal path from f(u) to f(v). Require f(u) 6= f(v) for any u 6= v.

• Def. A graph is planar if it has a drawing that has no intersecting paths. We call this particular

drawing a planar embedding.

• Def. A planar graph is a graph along with one of its planar embedding.

• Def. An open set is a measurable set such that for any of the element within the set, we can define

a small circle around the element so that every element within the small circle is also within the set.

(informal definition.)

• Def. A region is an open set that for ∀u, v ∈ U , there’s a path connecting u, v in U and the path ∈ U .

• Def. If G is a planar graph, then any plane drawing of G divides the plane into regions, called faces.

– Any forest planar embedding has 1 face.

• Jordon curve Theorem A closed polygonal curve C ∈ R2 divides the plane into exactly two faces,

the interior and the exterior. Both of them has C as their boundary.

• Def. The length of a face f , denoted as l(f), in a planar embedding is the length of the minimal closed

walk which is its boundary. (Notice, need to cound the hanging edges twice)

• Lemma.∑i l(fi) = 2e(G).

• Euler’s Formular If a connected graph G has a planar embedding with e edges, n vertices and f

faces, then n− e+ f = 2.

– Proof. Induction on the number of edges. Basic case: e(G) = n − 1, a tree. Then n − e + f =

n − (n − 1) + 1 = 2. checked. Induction: Suppose for any given graph G′ with |V (G′)| = n,

e(G′) = e′ ≥ n, f(G′) = f ′, we have n′ − e′ + f ′ = 2. Consider G with e(G) = e′ + 1. Since G has

a cycle, we can pick the cycle C and name an edge (c1, c2) ∈ C. Let G′ = G− (c1, c2), then since

we break a cycle, f ′ = f − 1. Therefore n− e+ f = n′ − (e′ + 1) + (f ′ + 1) = 2.

2015.10.27

• Mid-Term. How to prove everything.

– Look for specific/special/outstanding cases where you can prove the result (2-connected: cycle,

longest paths, etc.).

– Draw it out.

– Search for counter examples (usually they are edge-maximized, connection-maximized, etc.)

– Try induction on number of edges, vertices, matchings, degrees, etc.

• Def. A polytope is a 3D solid with flat faces, line segments as edges and sharp points as vertices.

• Def. A polytope is convex if the line connecting any two points within any line segments lives inside

the convex.

• Lemma. Every convex polytope can be associated to a planar graph.

– Put the polytop into a sphere, make sure the north pole is not a vertex and use stereographic

projection on vertices.

• Def. A platonic solid is a convex polytope whose every faces are congruent (in exact same shape) and

every vertex is adjacent to the same number of faces.

• Lemma. The number of different type of platonic solids is 5.

– For any platonic solid, let number edges of each face be n, number of faces adjacent to any vertex

be m.

– Since we have Euler’s Formular: v − e+ f = 2. We can write:

2

me− e+

2

ne = 2

1

m+

1

n=

1

2+

1

e>

1

2

(14)

Therefore min{n,m} = 3.

– Consider the angle restriction:360

m>

180(n− 2)

n, we have: m <

2n

n− 2.

– In summary, we have 5 results for (n,m):

∗ (3, 3), Tetrahedron, 4 triangles.

∗ (3, 4), Octahedron, 8 triangles

∗ (3, 5), Icosahedron, 20 triangles.

∗ (4, 3), Hexahedron, 6 squares.

∗ (5, 3), Dodecahedron, 12 pentagons.

2015.11.03

• Theorem If G is a planar graph, then e(G) ≤ 3|v(G)| − 6; If G is tiangle free (no C3), then e(G) ≤

2|v(G)| − 4.

– Proof.

∗ Assume G is connected. If G is disconnected, we can prove the theorem for each its compo-

nents.

∗ Assume G has no leaf vertices. If G has leaves, we can prove the subgraph of G with leaf

vertices removed still satisfies the theorem. By adding leaves back, no vialation to the theorem.

∗ Assume e(G) ≥ 1. e(G) = 0 always satisfies the theorem provided v(G) ≥ 2

∗ Then we can take an edge e ∈ E(G). Since G has no leaves, this edge belongs to a cycle.

Therefore we have: 2e(G) =∑i fi. While fi represent each faces in the planar graph G.

Because each face is bounded by at least 3 edges, we have:∑i fi ≥ 3f . While f is the

number of faces. Using Euler’s Formular, we have:

n− e+ f = 2 ≤ n− e+2

3e

n− e

3≥ 2

3n− 6 ≥ e

(15)

∗ Triangle free means each face is bounded by at least 4 edges.

2015.11.10

• Hajos 1961 Theorem Every graph with ψ(G) ≥ k > 2 contains a k-constructible subgraph.

– Proof. Take an edge-maximized counter example G with ψ(G) ≥ k > 2 which has no k-

constructible subgraph. Let G is r-partited. Because ψ(G) ≥ k, we have r ≥ k, otherwise we

can color G with r < k colors.

∗ Claim. G is not a complete r-partited graph.

· If G is complete r-partited. We can find a Kr ⊆ G which is r-constructible. Because

r ≥ k, there’s also a Kk ⊆ Kr which is k-constructible, which means G is not a counter

example.

Since G is not a complete r-partited graph, there exists three vertices {y1, x, y2} such that

xy1, xy2 6∈ E(G) and y1y2 ∈ E(G). Because G is edge-maximized, G+xy1 and G+xy2 both have

k-constructible subgraph. Let’s call their k-constructible subgraph H1 and H2, respectively. Then

xy1 ∈ E(H1) and xy2 ∈ E(H2). By the third construction rule, (H1 ∪H2)− xy1 − xy2 + y1y2 is a

k-constructible subgraph of G, contradicting to the assumption that G is a counter example.

∗ Note: If H1 and H2 overlaps at vertices other than x, we can split each of these vertex into

two sub-vertices until H1 and H2 only overlaps at x. After applying the third construction

rule in the proof above, we combine these sub-vertice pairs using the second construction rule,

which maintain the subgraph is k-constructible.

• Def. Edge-coloring. A proper k-edge coloring of a graph G is a function c : E(G)→ {1, 2, 3, ..., k} such

that if e1, e2 ∈ E(G) share one vertex, then c(e1) 6= c(e2).

• Def. Let ψ′(G) be the edge-chromatic number of G (also called chromatic index of G) is the smallest

k such that G has a proper k-edge coloring.

– Notice: ∆(G) ≤ ψ′(G). This gives a lower bound of ψ′(G).

• Def. A color is missing at vertex x if no edge incidence to x uses that color.

• konig 1916 Theorem If G is bi-partited, then ∆(G) = ψ′(G). The lower bound can be reached.

– Proof. Trivial case: e(G) = 0. Let’s apply induction on graphs with e(G) ≥ 1. Assume for any

given bi-partitable G with e(G) = k, for ∀xy ∈ E(G), G′ = G − xy has a proper ∆(G′)-edge

coloring.

∗ Case 1. ∆(G′) = ∆(G) − 1. We can use a brand new color on xy so that G has a proper

∆(G)-edge coloring.

∗ Case 2. ∆(G′) = ∆(G). Then vertices x, y both have at least one missing color in G′. Call the

missing color α at x and β at y. Trivial case: α = β (we can color xy using this color while

coloring G). Furthermore, if x is not incident to a β edge, we can color xy by β. Similarly,

if y is not incident to a α edge, we can color xy by α. Therefore the only non-trivial case

is when x is incident to a β edge and y is incident to a α edge with α 6= β. Let’s take the

edge-maximum walk in G′ starting at x with an edge-coloring sequence of alternating colors:

β, α, β, α, .... Then

· The walk is a path, otherwise it’s not a proper edge-coloring of G′.

· y is not an internal vertex of the path. Because color β should be missing at y.

· y is not the end point of the path. Assume y is the end point of the path, then the last

edge must be color α. Therefore the length of the path is even. In G, xy is an edge,

therefore we find an odd cycle in G, which can not be the case for a bi-partible G.

Therefore when we interchanging α and β on this path

· the proper coloring of G′ is maintained.

· y is not affected.

· x has β as its missing color.

Then we can color xy with β in G and obtain a proper ∆(G)-edge coloring of G.

• Vizing 1964 Theorem For any G, ψ′(G) ≤ ∆(G) + 1. This gives the upper bound of ψ′(G).

– Proof. Take the edge-minimum counter example G. Which means there exists one certain edge

xy1 ∈ E(G) such that we can (∆ + 1)-edge coloring G′ = G− xy1, but we can’t (∆ + 1)-coloring

G. In G′, since we have ∆(G) + 1 colors, we can find at least one missing color for each vertex.

Let color s is missing at x, color t1 missing at y1. Let’s construct a sequency of edges incident at

x: {xy1, xy2, xy3, ...xyh} in the following algorithem:

∗ choose a color ti that is missing at yi.

∗ find the edge xyi+1 not yet in the sequence, with c(xyi+1) = ti.

· Notice there exits at most one edge incident to x with a give color in the proper coloring

of G′.

Then the sequnce stops at h when

∗ Case 1. No such edge xyh+1 with color th. (Notice this includes the case h = ∆(G′), and the

case th = s.) Then by shifting color of each edge to its precedent edge, and coloring xyh by

th, we obtain an proper coloring for G, contradicting to the assumption that G doesn’t have

a (∆ + 1)-coloring.

∗ Case 2. Such edge xyh+1 with color th is already in the sequence: Let edge c(xyj) = th, 1 <

j < h. To be continued ...