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CSE 326: Data Structures Spanning Trees Ben Lerner Summer 2007.

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1 CSE 326: Data Structures Spanning Trees Ben Lerner Summer 2007

2 2 A Hidden Tree Start End

3 3 Spanning Tree in a Graph Vertex = router Edge = link between routers Spanning tree - Connects all the vertices - No cycles

4 4 Undirected Graph G = (V,E) –V is a set of vertices (or nodes) –E is a set of unordered pairs of vertices 1 2 3 4 5 6 7 V = {1,2,3,4,5,6,7} E = {{1,2},{1,6},{1,5},{2,7},{2,3}, {3,4},{4,7},{4,5},{5,6}} 2 and 3 are adjacent 2 is incident to edge {2,3}

5 5 Spanning Tree Problem Input: An undirected graph G = (V,E). G is connected. Output: T contained in E such that –(V,T) is a connected graph –(V,T) has no cycles

6 6 Spanning Tree Algorithm ST(i: vertex) mark i; for each j adjacent to i do if j is unmarked then Add {i,j} to T; ST(j); end{ST} Main T := empty set; ST(1); end{Main}

7 7 Example of Depth First Search 1 2 3 4 5 6 7 ST(1)

8 8 Example Step 2 1 2 3 4 5 6 7 ST(1) ST(2) {1,2}

9 9 Example Step 3 1 2 3 4 5 6 7 ST(1) ST(2) ST(7) {1,2} {2,7}

10 10 Example Step 4 1 2 3 4 5 6 7 ST(1) ST(2) ST(7) ST(5) {1,2} {2,7} {7,5}

11 11 Example Step 5 1 2 3 4 5 6 7 ST(1) ST(2) ST(7) ST(5) ST(4) {1,2} {2,7} {7,5} {5,4}

12 12 Example Step 6 1 2 3 4 5 6 7 ST(1) ST(2) ST(7) ST(5) ST(4) ST(3) {1,2} {2,7} {7,5} {5,4} {4,3}

13 13 Example Step 7 1 2 3 4 5 6 7 ST(1) ST(2) ST(7) ST(5) ST(4) ST(3) {1,2} {2,7} {7,5} {5,4} {4,3}

14 14 Example Step 8 1 2 3 4 5 6 7 ST(1) ST(2) ST(7) ST(5) ST(4) {1,2} {2,7} {7,5} {5,4} {4,3}

15 15 Example Step 9 1 2 3 4 5 6 7 ST(1) ST(2) ST(7) ST(5) ST(4) {1,2} {2,7} {7,5} {5,4} {4,3}

16 16 Example Step 10 1 2 3 4 5 6 7 ST(1) ST(2) ST(7) ST(5) {1,2} {2,7} {7,5} {5,4} {4,3}

17 17 Example Step 11 1 2 3 4 5 6 7 ST(1) ST(2) ST(7) ST(5) ST(6) {1,2} {2,7} {7,5} {5,4} {4,3} {5,6}

18 18 Example Step 12 1 2 3 4 5 6 7 ST(1) ST(2) ST(7) ST(5) ST(6) {1,2} {2,7} {7,5} {5,4} {4,3} {5,6}

19 19 Example Step 13 1 2 3 4 5 6 7 ST(1) ST(2) ST(7) ST(5) {1,2} {2,7} {7,5} {5,4} {4,3} {5,6}

20 20 Example Step 14 1 2 3 4 5 6 7 ST(1) ST(2) ST(7) {1,2} {2,7} {7,5} {5,4} {4,3} {5,6}

21 21 Example Step 15 1 2 3 4 5 6 7 ST(1) ST(2) {1,2} {2,7} {7,5} {5,4} {4,3} {5,6}

22 22 Example Step 16 1 2 3 4 5 6 7 ST(1) {1,2} {2,7} {7,5} {5,4} {4,3} {5,6}

23 23 Best Spanning Tree 1 2 3 4 5 6 7 Each edge has the probability that it won’t fail Find the spanning tree that is least likely to fail.80.75.95.50.95 1.0.85.84.80.89

24 24 Example of a Spanning Tree 1 2 3 4 5 6 7.80.75.95.50.95 1.0.85.84.80.89 Probability of success =.85 x.95 x.89 x.95 x 1.0 x.84 =.5735

25 25 Minimum Spanning Trees Given an undirected graph G=(V,E), find a graph G’=(V, E’) such that: –E’ is a subset of E –|E’| = |V| - 1 –G’ is connected – is minimal Applications: wiring a house, power grids, Internet connections G’ is a minimum spanning tree.

26 26 Minimum Spanning Tree Problem Input: Undirected Graph G = (V,E) and a cost function C from E to the reals. C(e) is the cost of edge e. Output: A spanning tree T with minimum total cost. That is: T that minimizes

27 27 Find the MST 4 7 1 5 9 2 Your Turn A C B D F H G E 1 7 6 5 11 4 12 13 2 3 9 10 4

28 28 Two Different Approaches Prim’s Algorithm Almost identical to Dijkstra’s Kruskals’s Algorithm Completely different!

29 29 Prim’s algorithm Idea: Grow a tree by adding an edge from the “known” vertices to the “unknown” vertices. Pick the edge with the smallest weight. G v known

30 30 Prim’s Algorithm for MST A node-based greedy algorithm Builds MST by greedily adding nodes 1.Select a node to be the “root” mark it as known Update cost of all its neighbors 2.While there are unknown nodes left in the graph a.Select an unknown node b with the smallest cost from some known node a b.Mark b as known c.Add (a, b) to MST d.Update cost of all nodes adjacent to b

31 31 Find MST using Prim’s v4v4 v7v7 v2v2 v3v3 v5v5 v6v6 v1v1 Start with V 1 2 2 5 4 7 110 46 3 8 1 VKwnDistancepath v1 v2 v3 v4 v5 v6 v7 Your Turn Order Declared Known: V1V1

32 32 Prim’s Algorithm Analysis Running time: Same as Dijkstra’s:O(|E| log |V|) Correctness: Proof is similar to Dijkstra’s

33 33 Kruskal’s MST Algorithm Idea: Grow a forest out of edges that do not create a cycle. Pick an edge with the smallest weight. G=(V,E) v

34 34 Kruskal’s Algorithm for MST An edge-based greedy algorithm Builds MST by greedily adding edges 1.Initialize with empty MST all vertices marked unconnected all edges unmarked 2.While there are still unmarked edges a.Pick the lowest cost edge (u,v) and mark it b.If u and v are not already connected, add (u,v) to the MST and mark u and v as connected to each other Doesn’t it sound familiar?

35 35 Example of Kruskal 1 1 6 5 4 7 2 3 3 3 4 0 22 1 3 {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4 1 3

36 36 Example of Kruskal 2 1 6 5 4 7 2 3 3 3 4 0 22 1 3 {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4 1 3

37 37 Example of Kruskal 2 1 6 5 4 7 2 3 3 3 4 0 22 1 3 {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4 1 3

38 38 Example of Kruskal 3 1 6 5 4 7 2 3 3 3 4 0 22 1 3 {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4 1 3

39 39 Example of Kruskal 4 1 6 5 4 7 2 3 3 3 4 0 22 1 3 {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4 1 3

40 40 Example of Kruskal 5 1 6 5 4 7 2 3 3 3 4 0 22 1 3 {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4 1 3

41 41 Example of Kruskal 6 1 6 5 4 7 2 3 3 3 4 0 22 1 3 {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4 1 3

42 42 Example of Kruskal 7 1 6 5 4 7 2 3 3 3 4 0 22 1 3 {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4 1 3

43 43 Example of Kruskal 7 1 6 5 4 7 2 3 3 3 4 0 22 1 3 {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4 1 3

44 44 Example of Kruskal 8,9 1 6 5 4 7 2 3 3 3 4 0 22 1 3 {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4 1 3

45 45 Data Structures for Kruskal Sorted edge list Disjoint Union / Find –Union(a,b) - union the disjoint sets named by a and b –Find(a) returns the name of the set containing a {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4

46 46 Example of DU/F 1 1 6 5 4 7 2 3 3 3 4 0 22 1 3 {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4 1 3 7 1 3 Find(5) = 7 Find(4) = 7

47 47 Example of DU/F 2 1 6 5 4 7 2 3 3 3 4 0 22 1 3 {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4 1 3 7 1 3 Find(1) = 1 Find(6) = 7

48 48 Example of DU/F 3 1 6 5 4 7 2 3 3 3 4 0 22 1 3 {7,4} {2,1} {7,5} {5,6} {5,4} {1,6} {2,7} {2,3} {3,4} {1,5} 0 1 1 2 2 3 3 3 3 4 1 3 7 3 Union(1,7)

49 49 Kruskal’s Algorithm with DU / F Sort the edges by increasing cost; Initialize A to be empty; for each edge {i,j} chosen in increasing order do u := Find(i); v := Find(j); if not(u = v) then add {i,j} to A; Union(u,v);

50 50 Kruskal code void Graph::kruskal(){ int edgesAccepted = 0; DisjSet s(NUM_VERTICES); while (edgesAccepted < NUM_VERTICES – 1){ e = smallest weight edge not deleted yet; // edge e = (u, v) uset = s.find(u); vset = s.find(v); if (uset != vset){ edgesAccepted++; s.unionSets(uset, vset); } 2|E| finds |V| unions |E| heap ops

51 51 Find MST using Kruskal’s A C B D FH G E 2 2 3 2 1 4 10 8 1 9 4 2 7 Your Turn Total Cost: Now find the MST using Prim’s method. Under what conditions will these methods give the same result?

52 52 Kruskal’s Algorithm: Correctness It clearly generates a spanning tree. Call it T K. Suppose T K is not minimum: Pick another spanning tree T min with lower cost than T K Pick the smallest edge e 1 =(u,v) in T K that is not in T min T min already has a path p in T min from u to v  Adding e 1 to T min will create a cycle in T min Pick an edge e 2 in p that Kruskal’s algorithm considered after adding e 1 (must exist: u and v unconnected when e 1 considered)  cost(e 2 )  cost(e 1 )  can replace e 2 with e 1 in T min without increasing cost! Keep doing this until T min is identical to T K  T K must also be minimal – contradiction!

53 53 Reducing Best to Minimum Let P(e) be the probability that an edge doesn’t fail. Define: Minimizing is equivalent to maximizing because

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