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EMIS 8374

EMIS 8374. Optimal Trees updated 25 April 2006. Minimum Spanning Tree (MST). Input A (simple) graph G = (V,E) Edge cost c ij for each edge e  E Optimization Problem Find a minimum-cost spanning tree

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EMIS 8374

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  1. EMIS 8374 Optimal Trees updated 25 April 2006

  2. Minimum Spanning Tree (MST) • Input • A (simple) graph G = (V,E) • Edge cost cij for each edge e  E • Optimization Problem • Find a minimum-cost spanning tree • Spanning tree: a set of |V|-1 edges T such that each vertex is incident to at least one edge in T, and T contains no cycles.

  3. 1 3 2 5 4 6 3 5 4 7 1 2 MST Example: Input

  4. MST Example: Some Feasible Spanning Trees 6 6 3 3 2 2 3 3 5 cost = 17 1 1 7 cost = 16 1 2 5 5 4 4 6 3 3 2 2 cost = 14 cost = 16 4 1 1 7 7 1 1 2 2 5 5 4 4

  5. 1 3 2 5 4 MST Example: Optimal Solution 6 3 1 2 cost = 12

  6. 1 3 2 5 4 5 4 7 MST Optimality Conditions: Path Optimality c25 > c12 c25 > c15 6 c34 > c23 c34 > c12 c34 > c15 c34 > c45 3 1 2 c24 > c12 c24 > c15 c24 > c45

  7. Path Optimality Condition • A spanning tree T is a minimum spanning tree if and only for every out-of-tree edge (i,j), cij cuv for every in-tree edge (u,v) on the path from i to j in T . • This is clearly a necessary condition for a MST. If an out-of-tree edge (i, j) has a lower cost than any in-tree edge (u,v) on the path from i to j in T, then we can improve T by replacing edge (u,v) with edge (i, j).

  8. 1 3 2 5 4 Path Optimality: Necessity 6 3 5 1 2 Replacing in-tree edge (2,4) with out-of-tree edge (4,5) would decrease the cost of this tree. Therefore, it cannot be optimal.

  9. 1 3 2 5 4 c25 >c12 c24 > c12 c45 > c12 3 5 4 7 MST Optimality Conditions: Cut Optimality 6 1 2 Removing an in-tree edge (u,v) creates a cut in the tree. For any out-of-tree edge (i,j) crossing the cut cij cuv.

  10. Cut Optimality • Given a spanning tree T, let C[u,v] be the set of edges in the cut formed by removing in-tree edge (u, v) from T. • A spanning tree T is a MST if and only if every in-tree edge is a minimum cost edge in C[u,v].

  11. 1 3 2 5 4 3 Cut Optimality: Necessity 6 5 4 7 1 2 The in-tree edge (2, 4) is not a minimum cost edge in the cut formed by removing it from T. If we remove in-tree edge (2, 4), we can restore connectivity by adding out-of-tree edge (4, 5).

  12. 1 3 2 5 4 5 3 1 Cut Optimality: Necessity 6 4 If we remove in-tree edge (2, 4), we can restore connectivity by adding out-of-tree edge (4, 5). Since the resulting tree has a lower total cost, the first tree could not have been optimal.

  13. Sufficiency of Cut and Path Optimality • It is easy to see that any MST must satisfy the cut and path optimality conditions. • If a tree T happens to satisfy these conditions, does that imply that T is a MST?

  14. Sufficiency of Cut Optimality 4 4 2 2 G T* 1 1 3 3 5 5 • Let T* be a MST of G. • Suppose that TC satisfies the cut optimality condition. • Show that TC is also a MST. 4 2 TC 1 3 5

  15. Sufficiency of Cut Optimality 4 T* 2 Since T* is a MST, it must satisfy the path optimality condition. So, c25 c24 (and c25 c45). 1 3 5 Since TC satisfies the cut optimality condition, c25 c24 (and c25 c35 ). 4 2 TC This implies c25= c24. 1 Replacing (2,4) with (2,5) in T* creates a new MST, T**. 3 5

  16. Sufficiency of Cut Optimality 4 T** 2 Since T** is a MST, it must satisfy the path optimality condition. So, c23 c35 (and c23 c25). 1 3 5 Since TC satisfies the cut optimality condition, c23 c35 (and c25 c13 ). 4 2 TC This implies c23= c35. 1 Replacing (3,5) with (2,3) in T** creates a new MST, T***. 3 5

  17. Sufficiency of Cut Optimality 4 T*** 2 Since TC is identical to T***, it is optimal. 1 This argument can be formalized to apply to the general case and show that any tree that satisfies the cut optimality condition must be optimal. 3 5 4 2 TC 1 Thus, cut optimality is a sufficient condition for a tree to be a MST. 3 5

  18. Kruskal’s Algorithm (Path Optimality) • F := E • T := {} • Repeat Until |T| = |V| - 1 • Begin • Select (i,j)  F such that cij = min(cuv: (u,v)  F) • F := F \ {(i,j)} • If T  {(i,j)} does not contain a cycle then • T := T  {(i,j)} • End Can be implemented in O(|E|+|V| log |V}) time plus the time for sorting the edges.

  19. 1 4 2 3 5 Kruskal’s Algorithm: Example 1 6 3 5 4 7 1 2

  20. Testing for Cycles • Let GT be the subgraph of G induced by the set of edges in T. • As edges are added to T, the algorithm creates a forest (i.e., a collection of trees). • Each tree in the forest forms a connected component of GT. • By keeping track of which component each node is in, we can quickly, and easily determine whether or not adding a new edge to T will create a cycle.

  21. Testing for Cycles Initialize component[i] = 0 for all i  V. When edge (i,j) is inspected, there are 5 cases to • component[i] = component[j] = 0 Add (i,j) to T; (i,j) becomes a new component of GT. • component[i] = 0, component[j] > 0. Add (i,j) to T; vertex i will go into the same component as j. • component[i] > 0, component[j] = 0. Add (i,j) to T; vertex j will go into the same component as i. • component[i] > 0, component[j] > 0, component[i]  component[j] Add (i,j) to T; merge components. • component[i] = component[j] > 0 Adding (i,j) would create a cycle.

  22. 1 4 2 3 5 Kruskal’s Algorithm: Example 2 4 2 5 3 1 6 2

  23. 1 4 2 3 5 Kruskal’s Algorithm: Example 2 2 4 1 2 2 5 3 1 6 2 1 2 (2, 3) creates a cycle because vertices 2 and 3 are in the same connected component.

  24. 1 4 2 3 5 Kruskal’s Algorithm: Example 2 2 4 1 2 2 5 3 1 6 2 1 2 (2, 4) does not create a cycle because vertices 2 and 4 are in different connected components.

  25. 1 4 2 3 5 Kruskal’s Algorithm: Example 2 2 4 2 2 2 5 3 1 6 2 2 2 Merge components 1 and 2.

  26. Prim’s Algorithm (Cut Optimality) • Choose any node v in V. • S1 := {v} • T := {} • k := 1 • Repeat Until |T| = |V| - 1 • Begin • Select a minimum cost edge (i,j) in the cut [Sk,V\{Sk}] • T := T  {(i,j)} • If i  Sk then Sk+1 := Sk {j} • Else If j  Sk then Sk+1 := Sk {i} • k := k+1 • End Can be implemented in O(|E|+|V| log |V}) time.

  27. 1 4 2 3 5 Prim’s Algorithm 4 2 5 3 1 6 2 S1 = {3} S2 = {1,3}

  28. 1 4 2 3 5 Prim’s Algorithm 4 2 5 3 1 6 2 S2 = {1,3} S3 = {1, 2, 3}

  29. 1 4 2 3 5 Prim’s Algorithm 4 2 5 3 1 6 2 S3 = {1, 2, 3} S4 = {1, 2, 3, 4}

  30. 1 4 2 3 5 Prim’s Algorithm 4 2 5 3 1 6 2 S4 = {1, 2, 3, 4} S5 = {1, 2, 3, 4, 5}

  31. Directed Spanning Tree (DST) • Input • A network (directed graph) G = (N,A) • Arc cost cij for each edge arc (i,j)  A • A designated root node s  N • Optimization Problem • Find a minimum-cost directed-out tree T = (N, A*) rooted at node s. • Directed-Out Tree: T = (N, A*) contains a unique, directed path from s to every other node in N (every other node in the tree has in-degree = 1).

  32. Is DST a (Pure) Network Flow Problem? • Suggestion: • Let b(s) = |N| - 1 • Let b(i) = -1 for i in N\{s} • Let ij = 0 • Let uij = |N| • Each node gets 1 unit of flow • Total cost of flow is minimized • Arcs carrying flow form a directed spanning tree

  33. 1 3 2 5 4 DST Example 10 10 1 s 5 5

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