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

Introduction to Graphs

Historical Motivation

Seven Bridges of K¨ onigsberg K¨ onigsberg (now Kaliningrad, Russia) around 1735 Problem: Find a walk through the city that would cross each bridge once and only once. 3 / 22

Definitions

Graph Graph A graph G = ( V , E ) is a set V of vertices connected by an edge set E . 5 / 22

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 . 6 / 22

Adjacency Adjacency Two vertices u and v are adjacent if there is an edge connecting them. This is sometimes writen as u ∼ v . v a c b v is adjacent with b and c but not with a . 7 / 22

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 a c b N ( v ) = { b , c } N [ v ] = { v , b , c } 8 / 22

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 a c b deg ( v ) = 3 deg ( a ) = 1 deg ( b ) = 5 deg ( c ) = 1 9 / 22

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 a c b deg + ( v ) = 2 deg + ( a ) = 0 deg + ( b ) = 3 deg + ( c ) = 1 deg − ( v ) = 1 deg − ( a ) = 1 deg − ( b ) = 2 deg − ( c ) = 0 10 / 22

Degree Let G = ( V , E ) be a graph. Then Lemma � deg ( v ) = 2 | E | v ∈ V If G is directed Lemma � � deg + ( v ) = deg − ( v ) = | E | v ∈ V v ∈ V It follows that the number of odd degree vertices is even. 11 / 22

Path Path In a graph a path p of length l is sequence of l + 1 adjacent vertices, i. e. p = ( v 0 , . . . , v l ) with v i v i + 1 ∈ E . A path is simple if all vertices are different, i. e. v i � = v j ⇔ i � = j . v v a a c c b b u u path simple path ( b , a , v , c , u ) ( a , v , b , u ) 12 / 22

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 v a a c c b b u u cycle simple cycle ( a , v , b , c , u , b , a ) ( a , v , c , u , b , a ) 13 / 22

Eulerian Cycles

Recall: Seven Bridges of K¨ onigsberg K¨ onigsberg (now Kaliningrad, Russia) around 1735 Problem: Find a walk through the city that would cross each bridge once and only once. 15 / 22

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 once? 16 / 22

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 v 1 , e 1 , . . . , v m , e m , v 1 such that e i = v i v i + 1 and e i � = e j ⇔ i � = j . Eulerian Path An Eulerian path in a graph G is a sequence of vertices and edges v 1 , e 1 , . . . , v m , e m , v m + 1 such that e i = v i v i + 1 and e i � = e j ⇔ i � = j . In an Eulerian cycle the start and end vertex hast to be equal. In an Eulerian path they can be different. 17 / 22

Euler’s Solution Theorem A connected graph has an Eulerian cycle if and only if the degree of all vertices is even, i. e. ∀ v ∈ V ∃ k ∈ N : deg ( v ) = 2 k . Theorem A graph has an Eulerian path if and only if it is connected and has at most two vertices with an odd degree. 18 / 22

Seven bridges of K¨ onigsberg Degree of all vertices is odd. 3 5 3 3 Thus, there is no walk through the city that would cross each bridge once and only once. 19 / 22

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 21 / 22

Algorithm 1. Determine the degree of all vertices. 3 4 5 3 3 4 21 / 22

Algorithm 2. If There are more than two vertices with an odd degree T hen Stop , G has no Eulerian path. 3 4 5 3 3 4 21 / 22

Algorithm 3. Choose any starting vertex v . Select a vertex with odd degree if possible. 3 4 4 2 3 4 21 / 22

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 21 / 22

Algorithm 5. W hile 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 21 / 22

Algorithm 5.1 Start another trail T ′ from v , following unused edges until returning to v . a b c d e f T ′ = b , f , c , a , b T = a , e , f , d , b , c , e 21 / 22

Algorithm 5.1 Start another trail T ′ from v , following unused edges until returning to v . a b c d e f T ′ = b , f , c , a , b T = a , e , f , d , b , c , e 21 / 22

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 21 / 22

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 T hen 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. 22 / 22