Design and Analysis of Computer Algorithm September 10, Design and Analysis of Computer Algorithm Lecture 5-2 Pradondet Nilagupta Department of Computer Engineering This lecture note has been modified from lecture note for by Prof. Francis Chin, CS332 by David Luekbe
Design and Analysis of Computer Algorithm September 10, Greedy Method (Cont.)
Design and Analysis of Computer Algorithm September 10, TREE ● A tree is an undirected graph which is connected and acyclic. It is easy to show that if graph G ■ G (V,E) is connected. ■ G (V,E) is acyclic. ■ |E| =|V| -1
Design and Analysis of Computer Algorithm September 10, Spanning Trees ● A spanning tree in an undirected graph G(V,E) is a subset of edges that are acyclic and connect all the vertices in V. ● It follows from the above conditions that a spanning tree must consist of exactly n-1 edges. ● Now suppose that each edge has a weight associated with it: Say that the weight of a tree T is the sum of the weights of its edges;
Design and Analysis of Computer Algorithm September 10, Example: Spanning Tree connected graph spanning tree cost = 129 spanning tree cost = 110
Design and Analysis of Computer Algorithm September 10, Definition ● The minimum spanning tree in a weighted graph G(V,E) is one which has the smallest weight among all spanning trees in G(V,E).
Design and Analysis of Computer Algorithm September 10, A Greedy Strategy Generic-MST(G, w) A = {} // Invariant: A is a subset of some MST while A does not form a spanning tree do find an edge (u, v) that is safe for A A = A + {(u, v)} // A remains a subset of some MST return A How to find a safe edge?
Design and Analysis of Computer Algorithm September 10, Example ● Example: spanning tree T {(a,b), (a,c), (b,d), (d,e), (d,f)} ● w(T) = = 30 ● T * = {(a,b), (b,e), (c,e), (c,d), (d,f)} ● w(T * ) = = 17 (minimum ?, why ? see note 1)
Design and Analysis of Computer Algorithm September 10, Method 1: Brute force ● Enumerate all spanning trees, evaluate their costs and find the minimum. ● Drawback - the number of spanning trees must be very large, how large? ● Let N s be the number of spanning trees for an n-node graph. ● lower bound, (the least value of N s ): n!/2 for a chain, why? ● upper bound, (the largest value of N s ): C e,n-1, where e = n(n-1)/2, i.e. O(n (n-1) )
Design and Analysis of Computer Algorithm September 10, Method 2: Removing redundant edges ● Edge of the largest weight is removed if the resultant graph remains connected. Repeat the edge removal until not possible. ● Correctness: If the largest-weight edge was not removed, it can always be replaced by an edge of lower weight because of the redundancy. ● Time complexity: ● sorting the edges - O(n 2 logn) or O(elogn) ● testing connectivity - O(n 2 ) or O(e) ● connectivity test for all edges - O(n 4 ) or O(e 2 )
Design and Analysis of Computer Algorithm September 10, Minimum Spanning Trees ● Theorem Given a spanning forest which is a subset of the MST, the minimum weight edge (u, v) emerging from a tree in the forest must be "safe", i.e., an edge in the MST.
Design and Analysis of Computer Algorithm September 10, Proof: ● Assume that the minimum weight edge (u, v) is not in MST. Let us add edge(u, v) to the MST to result a unique cycle which must contain another edge (u', v'), where u and u' belong to the some tree. As edge (u, v) is shorter than edge (u', v'), the weight of the new MST containing (u, v) and not (u', v') should be lower than the weight of the original MST. This contradicts our assumption that (u, v) is not in the MST.
Design and Analysis of Computer Algorithm September 10, Method 3: Ordered edge insertion (Krushal) ● The edge with minimum weight is added to the "forest" as long as it is not redundant, i.e., it connects two trees in the forest. Repeat the edge insertion until not possible. ● Correctness: The proof follows directly from the theorem and can be proved by induction. Initially, the forest has n trees each of which contains a single node of G. The edge with minimum weight edge which is not redundant, i.e., emerging from some trees in the forest, must be a minimum weight edge in the MST.
Design and Analysis of Computer Algorithm September 10, Time Complexcity ● Time Complexity: sorting the edges - O(n 2 logn) or O(elogn)
Design and Analysis of Computer Algorithm September 10, Minimum Spanning Tree (MST) Connect all the vertices Minimize the total edge weight // must be a tree Problem Select edges in a connected graph to f d a b c e g
Design and Analysis of Computer Algorithm September 10, Cut A cut of G = (V, E) is a partition (S, V-S) of V. Ex. S = {a, b, c, f}, V-S = {e, d, g} Edges {b, d}, {a, d}, {b, e}, {c, e} all cross the cut. The remaining edges do not cross the cut. f d a b c e g
Design and Analysis of Computer Algorithm September 10, Respected by a Cut Ex. S = {a, b, c, d} A1 = {(a, b), (d, g), (f, b), (a, f)} A2 = A1 + {(b, d)} The cut (S, V-S) respects A1 but not A2. f d a b c e g
Design and Analysis of Computer Algorithm September 10, Light Edge Ex. S = {a, b, c, d} Edge {b, e} has weight 3, “lighter” than the other edges {a, d}, {b, d}, and {b, e} that cross the cut. The edge that crosses a cut with the minimum weight. f d a b c e g light edge
Design and Analysis of Computer Algorithm September 10, Light Edge Is Safe! Theorem Let (S, V-S) be any cut of G(V, E) that respects a subset A of E, which is included in some MST for G. Let (u, v) be a light edge crossing (S, V-S). Then (u, v) is safe for A; that is, A + {(u, v)} is also included in some MST. S V- S u v A includes all red edges.
Design and Analysis of Computer Algorithm September 10, Corollary Let A be a subset of E that is included in some MST of G, and C be a connected component in the forest G =. If (u, v) is a light edge connecting C to some other component in G, then (u, v) is safe for A. A A u v C 7 4
Design and Analysis of Computer Algorithm September 10, Minimum Spanning Tree ● Problem: given a connected, undirected, weighted graph:
Design and Analysis of Computer Algorithm September 10, Minimum Spanning Tree ● Problem: given a connected, undirected, weighted graph, find a spanning tree using edges that minimize the total weight
Design and Analysis of Computer Algorithm September 10, Minimum Spanning Tree ● Which edges form the minimum spanning tree (MST) of the below graph? HBC GED F A
Design and Analysis of Computer Algorithm September 10, Minimum Spanning Tree ● Answer: HBC GED F A
Design and Analysis of Computer Algorithm September 10, Minimum Spanning Tree ● MSTs satisfy the optimal substructure property: an optimal tree is composed of optimal subtrees ■ Let T be an MST of G with an edge (u,v) in the middle ■ Removing (u,v) partitions T into two trees T 1 and T 2 ■ Claim: T 1 is an MST of G 1 = (V 1,E 1 ), and T 2 is an MST of G 2 = (V 2,E 2 ) (Do V 1 and V 2 share vertices? Why?) ■ Proof: w(T) = w(u,v) + w(T 1 ) + w(T 2 ) (There can’t be a better tree than T 1 or T 2, or T would be suboptimal)
Design and Analysis of Computer Algorithm September 10, Minimum Spanning Tree ● Thm: ■ Let T be MST of G, and let A T be subtree of T ■ Let (u,v) be min-weight edge connecting A to V-A ■ Then (u,v) T
Design and Analysis of Computer Algorithm September 10, Minimum Spanning Tree ● Thm: ■ Let T be MST of G, and let A T be subtree of T ■ Let (u,v) be min-weight edge connecting A to V-A ■ Then (u,v) T ● Proof: in book (see Thm 24.1)
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v);
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); Run on example graph
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); Run on example graph
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); 0 Pick a start vertex r r
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); 0 Red vertices have been removed from Q u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); 0 3 Red arrows indicate parent pointers u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); 14 0 3 u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); 14 0 3 u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); 14 08 3 u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); 10 08 3 u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); 10 08 3 u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); 102 08 3 u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); 102 u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); 102 u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); u
Design and Analysis of Computer Algorithm September 10, Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); u
Design and Analysis of Computer Algorithm September 10, Review: Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); What is the hidden cost in this code?
Design and Analysis of Computer Algorithm September 10, Review: Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; DecreaseKey(v, w(u,v));
Design and Analysis of Computer Algorithm September 10, Review: Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; DecreaseKey(v, w(u,v)); How often is ExtractMin() called? How often is DecreaseKey() called?
Design and Analysis of Computer Algorithm September 10, Review: Prim’s Algorithm MST-Prim(G, w, r) Q = V[G]; for each u Q key[u] = ; key[r] = 0; p[r] = NULL; while (Q not empty) u = ExtractMin(Q); for each v Adj[u] if (v Q and w(u,v) < key[v]) p[v] = u; key[v] = w(u,v); What will be the running time? A: Depends on queue binary heap: O(E lg V) Fibonacci heap: O(V lg V + E)