Download presentation

Presentation is loading. Please wait.

Published byAshlynn Dabney Modified over 2 years ago

2
§5 Minimum Spanning Tree 【 Definition 】 A spanning tree of a graph G is a tree which consists of V( G ) and a subset of E( G ) 〖 Example 〗 A complete graph and three of its spanning trees Note: The minimum spanning tree is a tree since it is acyclic -- the number of edges is |V| – 1. It is minimum for the total cost of edges is minimized. It is spanning because it covers every vertex. A minimum spanning tree exists iff G is connected. Adding a non-tree edge to a spanning tree, we obtain a cycle. Note: The minimum spanning tree is a tree since it is acyclic -- the number of edges is |V| – 1. It is minimum for the total cost of edges is minimized. It is spanning because it covers every vertex. A minimum spanning tree exists iff G is connected. Adding a non-tree edge to a spanning tree, we obtain a cycle. 1/9

3
§5 Minimum Spanning Tree Greedy Method Make the best decision for each stage, under the following constrains : (1) we must use only edges within the graph; (2) we must use exactly |V| 1 edges; (3) we may not use edges that would produce a cycle. 1. Prim’s Algorithm – grow a tree /* very similar to Dijkstra’s algorithm */ v1v1 v2v2 v6v6 v7v7 v3v3 v4v4 v5v5 2 4 2 1310 7 5 8 4 6 1 2/9

4
§5 Minimum Spanning Tree 2. Kruskal’s Algorithm – maintain a forest void Kruskal ( Graph G ) { T = { } ; while ( T contains less than |V| 1 edges && E is not empty ) { choose a least cost edge (v, w) from E ; delete (v, w) from E ; if ( (v, w) does not create a cycle in T ) add (v, w) to T ; else discard (v, w) ; } if ( T contains fewer than |V| 1 edges ) Error ( “No spanning tree” ) ; } /* DeleteMin */ /* Union / Find */ A more detailed pseudocode is given by Figure 9.58 on p.321 T = O( |E| log |E| ) Home work: p.341 9.15 A test case and a discussion on uniqueness. 3/9

5
§6 Applications of Depth-First Search /* a generalization of preorder traversal */ void DFS ( Vertex V ) /* this is only a template */ { visited[ V ] = true; /* mark this vertex to avoid cycles */ for ( each W adjacent to V ) if ( !visited[ W ] ) DFS( W ); } /* T = O( |E| + |V| ) as long as adjacency lists are used */ 0 123 45 6 DFS ( 0 ) 1. Undirected Graphs void ListComponents ( Graph G ) { for ( each V in G ) if ( !visited[ V ] ) { DFS( V ); printf(“\n“); } 7 8 0 1 4 6 5 2 3 7 8 4/9

6
§6 Applications of Depth-First Search 2. Biconnectivity Articulation point Biconnected graph v is an articulation point if G’ = DeleteVertex( G, v ) has at least 2 connected components. G is a biconnected graph if G is connected and has no articulation points. A biconnected component is a maximal biconnected subgraph. 〖 Example 〗 0 1 23 4 8 7 5 6 9 Connected graph 0 1 1 23 4 35 8 7 7 5 6 7 9 Biconnected components Note: No edges can be shared by two or more biconnected components. Hence E(G) is partitioned by the biconnected components of G. 5/9

7
§6 Applications of Depth-First Search Finding the biconnected components of a connected undirected G 0 1 23 4 8 7 5 6 9 Use depth first search to obtain a spanning tree of G DFS ( 3 ) 0 1 2 3 4 5 6 7 89 Depth first spanning tree 3 45 2 1 0 6 7 98 0 1 2 3 4 5 6 7 89 Depth first number (Num) Find the articulation points in G The root is an articulation point iff it has at least 2 children Any other vertex u is an articulation point iff u has at least 1 child, and it is impossible to move down at least 1 step and then jump up to u’s ancestor. Back edges ::= (u, v) tree and u (or v) is an ancestor of v (or u). Note: If u is an ancestor of v, then Num( u ) < Num( v ). 6/9

8
§6 Applications of Depth-First Search 3 45 2 1 0 6 7 98 0 1 2 3 4 5 6 7 89 vertex Num Low 0123456789 4320156798 0054005598 Therefore, u is an articulation point iff (1) u is the root and has at least 2 children; or (2) u is not the root, and has at least 1 child such that Low( child ) Num( u ). Please read the pseudocodes on p.327 and p.329 for more details. 7/9

9
§6 Applications of Depth-First Search 3. Euler Circuits Draw each line exactly once without lifting your pen from the paper – Euler tour Draw each line exactly once without lifting your pen from the paper, AND finish at the starting point – Euler curcuit 〖 Proposition 〗 An Euler circuit is possible only if the graph is connected and each vertex has an even degree. 〖 Proposition 〗 An Euler tour is possible if there are exactly two vertices having odd degree. One must start at one of the odd-degree vertices. 8/9

10
§6 Applications of Depth-First Search 2 6 9 8 3 7 1 12 4 10 5 11 Note: The path should be maintained as a linked list. For each adjacency list, maintain a pointer to the last edge scanned. T = O( |E| + |V| ) Note: The path should be maintained as a linked list. For each adjacency list, maintain a pointer to the last edge scanned. T = O( |E| + |V| ) Find a simple cycle in an undirected graph that visits every vertex – Hamilton cycle Home work: Self-study the digraphs and strong components on p. 331-334 and do p.342 9.27 DFS 9/9

Similar presentations

OK

Chapter 20: Graphs. Objectives In this chapter, you will: – Learn about graphs – Become familiar with the basic terminology of graph theory – Discover.

Chapter 20: Graphs. Objectives In this chapter, you will: – Learn about graphs – Become familiar with the basic terminology of graph theory – Discover.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google