Tutorial 7

Yuen-Man Pun

Department of Systems Engineering and Engineering ManagementThe Chinese University of Hong Kong

23 November 2017

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Definition

(i) A graph G is a pair (V ,E ) of sets, where V is the vertex set, and E isthe edge set, which contains unordered pair of vertices.

(ii) The degree of a vertex v , denoted by deg(v), is the number of edgesincident to it.

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Definition

(i) A path in G is either a single vertex or an ordered list of distinctverticesv1 − v2 − ...− vk , where (v1, v2),(v2, v3),...,(vk − 1, vk) are edges.

(ii) A cycle in G is a path in which the starting and ending vertices arethe same.

(iii) G is connected if for every pair of vertices u, v in G , there is a pathbetween them.

(iv) G is an acyclic graph if it has no cycle. A connected acyclic graph iscalled a tree.

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Examples

G1 and G2 are trees, but G3 and G4 are not.

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Handshaking Lemma

Theorem (Handshaking lemma)

Let v1, v2, ..., vn be the vertices of a graph G, and E be the edge set. Then

n∑i=1

deg(vi ) = 2|E |

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Example 1

Problem 1 (Q7 in 15/16 Exam)Let T be a tree whose average degree is a. Show that T has 2/(2− a)vertices.

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Example 1

Problem 1 (Q7 in 15/16 Exam)Let T be a tree whose average degree is a. Show that T has 2/(2− a)vertices.

Solution: Let n be the number of vertices in T .

T is a tree implies it has n − 1 edges. By handshaking lemma,

n · a = 2(n − 1) =⇒ n = 2/(2− a)

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Example 2

Problem 2Let G be a graph with n vertices and assume that each vertex of G hasdegree dn−12 e. Show that G must be connected.

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Example 2

Problem 2Let G be a graph with n vertices and assume that each vertex of G hasdegree dn−12 e. Show that G must be connected.

Solution:

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Example 2

Problem 2Let G be a graph with n vertices and assume that each vertex of G hasdegree dn−12 e. Show that G must be connected.

Solution:

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Example 2

Problem 2Let G be a graph with n vertices and assume that each vertex of G hasdegree dn−12 e. Show that G must be connected.

Solution:Pick any u ∈ G .

Since deg(u) = dn−12 e, u is connected to dn−12 e distinct vertices,say {u1, u2, · · · , ud(n−1)/2e}.

Let G1 = {u, u1, u2, · · · , ud(n−1)/2e}, which is connected.

|G1| = dn − 1

2e+ 1

Let G2 = G \ G1. Pick any v ∈ G2.

Note that |G2| = n − (dn−12 e+ 1) = bn−12 c and deg(v) = dn−12 e.

By pigeonhole principle, there must exist some vertex w ∈ G1 connectedto v by an edge. Therefore, G must be connected.

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Example 3

Problem 3 (Q7 in 14/15 Exam)Show that every simple graph G = (V ,E ) with |V | ≥ 2 has at least 2vertices of the same degree.

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Example 3

Problem 3 (Q7 in 14/15 Exam)Show that every simple graph G = (V ,E ) with |V | ≥ 2 has at least 2vertices of the same degree.

Solution: Let

Set of possible degree of vertices = S = {0, 1, 2, · · · , |V | − 1}

Case 1: There exists a vertex with degree 0.Observe that a vertex u with deg(u) = 0 implies none of the vertices in Vforms an edge with u.

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Example 3

Problem 3 (Q7 in 14/15 Exam)Show that every simple graph G = (V ,E ) with |V | ≥ 2 has at least 2vertices of the same degree.

Solution(cont’d):Then maximum degree of the remaining vertices is at most |V | − 2.

Case 2: There does not exist a vertex with degree 0. Therefore, in anycase,

|S | = |V | − 1

By pigeonhole principle, G has at least 2 vertices of the same degree.

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