Introduction to Graphs Historical Motivation Seven Bridges of K - - PowerPoint PPT Presentation

Introduction to Graphs

Historical Motivation

Seven Bridges of K¨

• nigsberg

K¨

• nigsberg (now Kaliningrad, Russia) around 1735

Problem: Find a walk through the city that would cross each bridge

• nce and only once.

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Definitions

Graph

Graph A graph G = (V, E) is a set V of vertices connected by an edge set E.

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Variations

Multi-Graph: Multiple edges between two vertices. Directed: Edges have a direction. Weighted: Vertices and/or edges have weights. Simple: No multiple edges, no loops. Simple Undirected Graph A simple undirected graph G = (V, E) is a set V of vertices con- nected by an edge set E ⊆ {{u, v} | u, v ∈ V, u = v}. An edge {u, v} is also writen as uv.

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Adjacency

Adjacency Two vertices u and v are adjacent if there is an edge connecting

• them. This is sometimes writen as u ∼ v.

v b a c

v is adjacent with b and c but not with a.

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Neighbourhood

Neighbourhood The open neighbourhood N(v) = {u ∈ V | u = v, u ∼ v} of a vertex v is the set of vertices adjacent to v (not including v). The closed neighbourhood N[v] = N(v) ∪ {v} includes v.

v b a c

N(v) = {b, c} N[v] = {v, b, c}

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Degree

Degree The degree deg(v) of a vertex v is the number of incident edges. Note that the degree is not necessarily equal to the cardinality of neighbours.

v b a c

deg(v) = 3 deg(a) = 1 deg(b) = 5 deg(c) = 1

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Degree (directed)

Indegree, Outdegree For a vertex v, the indegree deg−(v) is the number of incoming edges, and the outdegree deg+(v) the number of incoming edges.

v b a c

deg+(v) = 2 deg+(a) = 0 deg+(b) = 3 deg+(c) = 1 deg−(v) = 1 deg−(a) = 1 deg−(b) = 2 deg−(c) = 0

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Degree

Let G = (V, E) be a graph. Then Lemma

• v∈V

deg(v) = 2|E| If G is directed Lemma

• v∈V

deg+(v) =

• v∈V

deg−(v) = |E| It follows that the number of odd degree vertices is even.

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Path

Path In a graph a path p of length l is sequence of l +1 adjacent vertices,

• i. e. p = (v0, . . . , vl) with vivi+1 ∈ E. A path is simple if all vertices

are different, i. e. vi = vj ⇔ i = j.

v b a c u

path (b, a, v, c, u)

v b a c u

simple path (a, v, b, u)

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Cycle

Cycle A cycle is a path that begins and ends with same vertex. A cycle is simple, if it doesn’t cross itself.

v b a c u

cycle (a, v, b, c, u, b, a)

v b a c u

simple cycle (a, v, c, u, b, a)

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Eulerian Cycles

Recall: Seven Bridges of K¨

• nigsberg

K¨

• nigsberg (now Kaliningrad, Russia) around 1735

Problem: Find a walk through the city that would cross each bridge

• nce and only once.

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Euler’s Solution

Represent problem as graph: Original problem: Find a walk through the city that would cross each bridge once and only once. Graph problem: Does G contain a path using each edge exactly

• nce?

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Euler’s Solution

Given a graph G = (V, E) with |E| = m and index arithmetic modulo m. Eulerian Cycle An Eulerian cycle in a graph G is a sequence of vertices and edges v1, e1, . . . , vm, em, v1 such that ei = vivi+1 and ei = ej ⇔ i = j. Eulerian Path An Eulerian path in a graph G is a sequence of vertices and edges v1, e1, . . . , vm, em, vm+1 such that ei = vivi+1 and ei = ej ⇔ i = j. In an Eulerian cycle the start and end vertex hast to be equal. In an Eulerian path they can be different.

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Euler’s Solution

Theorem A connected graph has an Eulerian cycle if and only if the degree

• f all vertices is even, i. e. ∀v ∈ V ∃k ∈ N: deg(v) = 2k.

Theorem A graph has an Eulerian path if and only if it is connected and has at most two vertices with an odd degree.

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Seven bridges of K¨

• nigsberg

Degree of all vertices is odd.

5 3 3 3

Thus, there is no walk through the city that would cross each bridge

• nce and only once.

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Algorithm

Algorithm

Given a graph G. Does G have an Eulerian path? Algorithm based on Hierholzer’s algorithm from 1837.

a b c d e f

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Algorithm

• 1. Determine the degree of all vertices.

3 4 5 3 3 4

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Algorithm

• 2. If There are more than two vertices with an odd degree Then

Stop, G has no Eulerian path.

3 4 5 3 3 4

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Algorithm

• 3. Choose any starting vertex v. Select a vertex with odd degree if

possible.

3 4 4 2 3 4

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Algorithm

• 4. Follow a trail T of edges from v until geting stuck.

a b c d e f

T = a, e, f , d, b, c, e

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Algorithm

• 5. While There is a vertex v belonging to the current trail T but

has adjacent edges not part of T.

a b c d e f

T = a, e, f , d, b, c, e

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Algorithm

5.1 Start another trail T ′ from v, following unused edges until returning to v.

a b c d e f

T = a, e, f , d, b, c, e T ′ = b, f , c, a, b

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Algorithm

5.1 Start another trail T ′ from v, following unused edges until returning to v.

a b c d e f

T = a, e, f , d, b, c, e T ′ = b, f , c, a, b

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Algorithm

5.2 Join the tour T ′ formed in this way to the previous trail T.

a b c d e f

T = a, e, f , d, b, f , c, a, b, c, e

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Complete Algorithm

Given a graph G = (V, E).

• 1. Determine the degree of all vertices.
• 2. If There are more than two vertices with an odd degree Then

Stop, G has no Eulerian path.

• 3. Choose any starting vertex v. Select a vertex with odd degree if

possible.

• 4. Follow a trail T of edges from v until geting stuck.
• 5. While There is a vertex v belonging to the current trail T but

has adjacent edges not part of t.

5.1 Start another trail T ′ from v, following unused edges until returning to v. 5.2 Join the tour T ′ formed in this way to the previous trail T.

This algorithm can be implemented in O(|E|) time.

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