1 / 25

Minimum spanning trees (MST)

Minimum spanning trees (MST). Def: A spanning tree of a graph G is an acyclic subset of edges of G connecting all vertices in G. A sub- forest of G is an acyclic subset of edges of G. Minimum spanning trees (MST). 1. Def:

wherbert
Download Presentation

Minimum spanning trees (MST)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Minimum spanning trees (MST) Def: A spanning tree of a graph G is an acyclic subset of edges of G connecting all vertices in G. A sub-forest of G is an acyclic subset of edges of G.

  2. Minimum spanning trees (MST) 1 Def: A spanning tree of a graph G is an acyclic subset of edges of G connecting all vertices in G. A sub-forest of G is an acyclic subset of edges of G. 15 7 8 4 3 5 7 2 1 1 6 2 Def: Given is a weighted (undirected) graph G=(V,E,w) where w:E->Reals defines a weight of every edge in E. A minimum spanning tree of G is a spanning tree with the minimum total weight of edges.

  3. Minimum spanning trees (MST) - Kruskal 1 15 7 8 4 3 5 7 2 1 1 6 2 • Kruskal ( G=(V,E,w) ) • Let T=; • Sort the edges in increasing order of weight • 3. For edge e do • 4. If T [ e does not contain a cycle then • Add e to T • Return T

  4. Minimum spanning trees (MST) - Kruskal 1 15 7 8 4 3 5 7 2 1 1 6 2 • Kruskal ( G=(V,E,w) ) • Let T=; • Sort the edges in increasing order of weight • 3. For edge e do • 4. If T [ e does not contain a cycle then • Add e to T • Return T

  5. Minimum spanning trees (MST) - Kruskal 1 Lemma: Algo is correct. 15 7 8 4 3 5 7 2 1 1 6 2 • Kruskal ( G=(V,E,w) ) • Let T=; • Sort the edges in increasing order of weight • 3. For edge e do • 4. If T [ e does not contain a cycle then • Add e to T • Return T

  6. Minimum spanning trees (MST) - Kruskal 1 Implementation? 15 7 8 4 3 5 7 2 1 1 6 2 • Kruskal ( G=(V,E,w) ) • Let T=; • Sort the edges in increasing order of weight • 3. For edge e do • 4. If T [ e does not contain a cycle then • Add e to T • Return T

  7. Minimum spanning trees (MST) - Kruskal 1 Implementation? - Union-Find datastructure 15 7 8 4 • Init (V) • for every vertex v do • boss[v]=v • size[v]=1 • set[v]={v} • Union (u,v) • if size[boss[u]]>size[boss[v]] then • set[boss[u]]=set[boss[u]] union set[boss[v]] • size[boss[u]]+=size[boss[v]] • for every z in set[boss[v]] do • boss[z]=boss[u] • else do steps 2.-5. with u,v switched 3 5 7 2 1 1 6 2

  8. Minimum spanning trees (MST) - Kruskal 1 Analysis of Union-Find 15 7 8 Lemma: n-1 Unions take O(n log n) time 4 3 5 7 2 1 1 6 2 • Union (u,v) • if size[boss[u]]>size[boss[v]] then • set[boss[u]]=set[boss[u]] union set[boss[v]] • size[boss[u]]+=size[boss[v]] • for every z in set[boss[v]] do • boss[z]=boss[u] • else do steps 2.-5. with u,v switched

  9. Minimum spanning trees (MST) - Kruskal 1 Analysis of Union-Find 15 7 8 Lemma: k Unions take O(k log k) time 4 3 5 7 2 1 1 6 2 Corollary: The running time of Kruskal is: O(|E| log |E|) + O(|V| log |V|)

  10. Minimum spanning trees (MST) - Prim 1 15 7 8 4 3 5 7 2 1 1 6 2

  11. Minimum spanning trees (MST) - Prim 1 • Prim ( G=(V,E,w) ) • Let T=;, H=; • For every vertex v do • cost[v]=1, parent[v]=null • Let u be a vertex • Update (u) • 6. For i=1 to n-1 do • 7. u=vertex from H of • smallest cost (remove) • Add (u,parent[u]) to T • Update(u) • Return T 15 7 8 4 3 5 7 2 1 1 6 2 • Update (u) • For every neighbor v of u • If cost[v]>w(u,v) then • cost[v]=w(u,v), parent[v]=u • If v not in H then • Add v to H

  12. Minimum spanning trees (MST) - Prim 1 Lemma: Prim is correct. 15 7 8 4 3 5 7 2 1 1 6 2 Running time:

  13. Single source shortest paths - Dijkstra 1 Input: Output: G=(V,E,w) and a vertex s (w non-negative) shortest paths from s to every other vertex 15 7 8 4 3 5 7 2 1 1 6 Can use similar idea to Prim? 2

  14. Single source shortest paths - Dijkstra 1 • Dijkstra ( G=(V,E,w), s ) • Let H=; • For every vertex v do • dist[v]=1 • dist[s]=0 • Update (s) • For i=1 to n-1 do • 7. u=extract vertex from H • of smallest cost • 8. Update(u) • Return dist[] 15 7 8 4 3 5 7 2 1 1 6 2 • Update (u) • For every neighbor v of u • If dist[v]>dist[u]+w(u,v) then • dist[v]=dist[u]+w(u,v) • If v not in H then • Add v to H

  15. Single source shortest paths - Dijkstra 1 Lemma: Dijkstra is correct. 15 7 8 4 3 5 7 2 1 1 6 2 Running time:

  16. All pairs shortest paths – Floyd-Warshall 1 Input: Output: G=(V,E,w), w non-negative shortest paths between all pairs of vertices 15 7 8 4 3 5 7 2 1 1 6 2

  17. All pairs shortest paths – Floyd-Warshall 1 Input: Output: G=(V,E,w), w non-negative shortest paths between all pairs of vertices 15 7 8 4 3 5 7 2 1 1 6 • Idea 1: • Use Dijkstra from every vertex 2

  18. All pairs shortest paths – Floyd-Warshall 1 Input: Output: G=(V,E,w), w non-negative shortest paths between all pairs of vertices 15 7 8 4 3 5 7 2 1 1 6 • Idea 1: • Use Dijkstra from every vertex • Idea 2: • How about dynamic programming? 2

  19. All pairs shortest paths – Floyd-Warshall 1 Heart of the algorithm: 15 7 8 the length of the shortest path from i to j using only vertices · k 4 3 S[i,j,k] = 5 7 2 1 1 6 2

  20. All pairs shortest paths – Floyd-Warshall 1 Heart of the algorithm: 15 7 8 the length of the shortest path from i to j using only vertices · k 4 3 S[i,j,k] = 5 7 2 1 1 6 2 How to compute S[i,j,k] ? S[i,j,k] =

  21. All pairs shortest paths – Floyd-Warshall 1 • Floyd-Warshall ( G=(V,E,w) ) • For i=1 to |V| do • For j=1 to |V| do • S[i,j,0] = w(i,j) • For k=1 to |V| do • For i=1 to |V| do • For j=1 to |V| do • S[i,j,k] = min { • S[i,j,k-1], • S[i,k,k-1]+S[k,j,k-1] } • Return ? 15 7 8 4 3 5 7 2 1 1 6 2 w(i,j) if k = 0 min { S[i,j,k-1], S[i,k,k-1] + S[k,j,k-1] } if k > 0 S[i,j,k] =

  22. Single source shortest paths – Bellman-Ford -1 • Input: • directed G=(V,E,w) and a vertex s • Output: • FALSE if exists reachable negative-weight cycle, • distance to every vertex, otherwise. 4 7 -7 4 3 5 -3 2 1 1 6 2

  23. Single source shortest paths – Bellman-Ford -1 • Input: • directed G=(V,E,w) and a vertex s • Output: • FALSE if exists reachable negative-weight cycle, • distance to every vertex, otherwise. 4 7 -1 4 3 5 -3 2 1 1 6 2

  24. Single source shortest paths – Bellman-Ford -1 4 7 -7 8 3 5 -3 • Bellman-Ford ( G=(V,E,w), s ) • For every vertex v • d[v] = 1 • d[s]=0 • For i=1 to |V|-1 do • For every edge (u,v) in E do • If d[v]>d[u]+w(u,v) then • d[v]=d[u]+w(u,v) • For every edge (u,v) in E do • If d[v]>d[u]+w(u,v) then • Return NEGATIVE CYCLE • Return d[] 2 1 1 6 2

  25. Single source shortest paths – Bellman-Ford -1 Lemma: Bellman-Ford is correct. Running time: 4 7 -7 8 3 5 -3 • Bellman-Ford ( G=(V,E,w), s ) • For every vertex v • d[v] = 1 • d[s]=0 • For i=1 to |V|-1 do • For every edge (u,v) in E do • If d[v]>d[u]+w(u,v) then • d[v]=d[u]+w(u,v) • For every edge (u,v) in E do • If d[v]>d[u]+w(u,v) then • Return NEGATIVE CYCLE • Return d[] 2 1 1 6 2

More Related