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1 Graphs: shortest paths & Minimum Spanning Tree(MST) 15-211 Fundamental Data Structures and Algorithms Ananda Guna April 8, 2003
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2 Announcements Homework #5 is due Tuesday April 15th. Quiz #3 feedback is enabled. Final Exam is Tuesday May 8 th at 8AM
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Recap
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4 Dijkstra’s algorithm S = {1} for i = 2 to n do D[i] = C[1,i] if there is an edge from 1 to i, infinity otherwise for i = 1 to n-1 { choose a vertex w in V-S such that D[w] is min add w to S (where S is the set of visited nodes) for each vertex v in V-S do D[v] = min(D[v], D[w]+c[w,v]) } Where |V| = n
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5 Features of Dijkstra’s Algorithm A greedy algorithm “Visits” every vertex only once, when it becomes the vertex with minimal distance amongst those still in the priority queue Distances may be revised multiple times: current values represent ‘best guess’ based on our observations so far Once a vertex is visited we are guaranteed to have found the shortest path to that vertex…. why?
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6 Correctness (via contradiction) D(x) must represent a shortest path to x, and D(x) D shortest (u). However, Dijkstra’s always visits the vertex with the smallest distance next, so we can’t possibly visit u before we visit x u x s visited unvisited Prove D(u) represent the shortest path to u (visited node) Assume u is the first vertex visited such that D(u) is not a shortest path (thus the true shortest path to u must pass through some unvisited vertex) Let x represent the first unvisited vertex on the true shortest path to u
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7 Quiz break Would it be better to use an adjacency list or an adjacency matrix for Dijkstra’s algorithm? What is the running time of Dijkstra’s algorithm, in terms of |V| and |E| in each case?
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8 Complexity of Dijkstra Adjacency matrix version Dijkstra finds shortest path from one vertex to all others in O(|V| 2 ) time If |E| is small compared to |V| 2, use a priority queue to organize the vertices in V-S, where V is the set of all vertices and S is the set that has already been explored So total of |E| updates each at a cost of O(log |V|) So total time is O(|E| log|V|)
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Negative Weighted Single-Source Shortest Path Algorithm (Bellman-Ford Algorithm)
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10 The Bellman-Ford algorithm (see Weiss, Section 14.4) Returns a boolean: TRUE if and only if there is no negative- weight cycle reachable from the source: a simple cycle, where v 0 =v k and FALSE otherwise If it returns TRUE, it also produces the shortest paths
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11 Example For each edge (u,v), let's denote its length by C(u,v)) Let d[i][v] = distance from start to v using the shortest path out of all those that use i or fewer edges, or infinity if you can't get there with <= i edges.
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12 Example ctd.. How can we fill out the rows? 012345 00 1050 15 2050801545 V i
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13 Example ctd.. Can we get i th row from i-1 th row? for v != start, d[v][i] = MIN d[x][i-1] + len(x,v) x->v We know minimum path to come to x using < i nodes.So for all x that can reach v, find the minimum such sum (in blue) among all x Assume d[start][i] = 0 for all i
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14 Completing the table 012345 00 1050 15 2050801545 3025801545 75 4025551545 75 5025551545 65 d[v][i] = MIN d[x][i-1] + len(x,v) x->v
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15 Key features If the graph contains no negative-weight cycles reachable from the source vertex, after |V| - 1 iterations all distance estimates represent shortest paths…why?
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16 Correctness By induction: D(s) = 0 after initialization Assume D(v i-1 ) is a shortest path after iteration (i-1) Since edge (v i-1,v i ) is updated on the i th pass, D(v i ) must then reflect the shortest path to v i. Since we perform |V| - 1 iterations, D(v i ) for all reachable vertices v i must now represent shortest paths The algorithm will return true because on the |V| th iteration, no distances will change Case 1: Graph G=(V,E) doesn’t contain any negative- weight cycles reachable from the source vertex s Consider a shortest path p =, which must have k |V| - 1 edges
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17 Correctness Proof by contradiction: Assume the algorithm returns TRUE Thus, D(v i-1 ) + weight(v i-1, v i ) D(v i ) for i = 1,…,k Summing the inequalities for the cycle: leads to a contradiction since the first sums on each side are equal (each vertex appears exactly once) and the sum of weights must be less than 0. Case 2: Graph G=(V,E) contains a negative-weight cycle reachable from the source vertex s
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18 Performance Initialization: O(|V|) Path update and cycle check: |V| calls checking |E| edges, O(|VE|) Overall cost: O(| VE |)
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The All Pairs Shortest Path Algorithm (Floyd’s Algorithm)
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20 Finding all pairs shortest paths Assume G=(V,E) is a graph such that c[v,w] 0, where C is the matrix of edge costs. Find for each pair (v,w), the shortest path from v to w. That is, find the matrix of shortest paths Certainly this is a generalization of Dijkstra’s. Note: For later discussions assume |V| = n and |E| = m
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21 Floyd’s Algorithm A[i][j] = C(i,j) if there is an edge (i,j) A[i][j] = infinity(inf) if there is no edge (i,j) Graph “adjacency” matrix A is the shortest path matrix that uses 1 or fewer edges
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22 Floyd ctd.. To find shortest paths that uses 2 or fewer edges find A 2, where multiplication defined as min of sums instead sum of products That is (A 2 ) ij = min{ A ik + A kj | k =1..n} This operation is O(n 3 ) Using A 2 you can find A 4 and then A 8 and so on Therefore to find A n we need log n operations Therefore this algorithm is O(log n* n 3 ) We will consider another algorithm next
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23 Floyd-Warshall Algorithm This algorithm uses nxn matrix A to compute the lengths of the shortest paths using a dynamic programming technique. Let A[i,j] = c[i,j] for all i,j & ij If (i,j) is not an edge, set A[i,j]=infinity and A[i,i]=0 A k [i,j] = min (A k-1 [i,j], A k-1 [i,k]+ A k-1 [k,j]) Where A k is the matrix after k-th iteration and path from i to j does not pass through a vertex higher than k
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24 Example – Floyd-Warshall Algorithm 1 23 8 2 3 5 Find the all pairs shortest path matrix A k [i,j] = min (A k-1 [i,j], A k-1 [i,k]+ A k-1 [k,j]) Where A k is the matrix after k-th iteration and path from i to j does not pass through a vertex higher than k
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25 Floyd-Warshall Implementation initialize A[i,j] = C[i,j] initialize all A[i,i] = 0 for k from 1 to n for i from 1 to n for j from 1 to n if (A[i,j] > A[i,k]+A[k,j]) A[i,j] = A[i,k]+A[k,j]; The complexity of this algorithm is O(n 3 )
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26 Questions Question: What is the asymptotic run time of Dijkstra (adjacency matrix version)? O(n 2 ) Question: What is the asymptotic running time of Floyd-Warshall?
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27 Minimum Spanning Trees (some material adapted from slides by Peter Lee)
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28 Problem: Laying Telephone Wire Central office
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29 Wiring: Naïve Approach Central office Expensive!
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30 Wiring: Better Approach Central office Minimize the total length of wire connecting the customers
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31 Minimum Spanning Tree (MST) (see Weiss, Section 24.2.2) it is a tree (i.e., it is acyclic) it covers all the vertices V contains |V| - 1 edges the total cost associated with tree edges is the minimum among all possible spanning trees not necessarily unique A minimum spanning tree is a subgraph of an undirected weighted graph G, such that
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32 Applications of MST Any time you want to visit all vertices in a graph at minimum cost (e.g., wire routing on printed circuit boards, sewer pipe layout, road planning…) Internet content distribution $$$, also a hot research topic Idea: publisher produces web pages, content distribution network replicates web pages to many locations so consumers can access at higher speed MST may not be good enough! content distribution on minimum cost tree may take a long time!
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33 How Can We Generate a MST? a c e d b 2 45 9 6 4 5 5 a c e d b 2 45 9 6 4 5 5
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34 Prim’s Algorithm Let V ={1,2,..,n} and U be the set of vertices that makes the MST and T be the MST Initially : U = {1} and T = while (U V) let (u,v) be the lowest cost edge such that u U and v V-U T = T {(u,v)} U = U {v}
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35 Prim’s Algorithm implementation Initialization a. Pick a vertex r to be the root b. Set D(r) = 0, parent(r) = null c. For all vertices v V, v r, set D(v) = d. Insert all vertices into priority queue P, using distances as the keys a c e d b 2 45 9 6 4 5 5 eabcd 0 VertexParent e -
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36 Prim’s Algorithm While P is not empty: 1. Select the next vertex u to add to the tree u = P.deleteMin() 2. Update the weight of each vertex w adjacent to u which is not in the tree (i.e., w P) If weight(u,w) < D(w), a. parent(w) = u b. D(w) = weight(u,w) c. Update the priority queue to reflect new distance for w
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37 Prim’s algorithm a c e d b 2 45 9 6 4 5 5 dbca 455 VertexParent e- be ce de The MST initially consists of the vertex e, and we update the distances and parent for its adjacent vertices
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38 Prim’s algorithm a c e d b 2 45 9 6 4 5 5 acb 245 VertexParent e- be c d de a d
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39 Prim’s algorithm a c e d b 2 45 9 6 4 5 5 cb 45 VertexParent e- be c d de a d
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40 Prim’s algorithm a c e d b 2 45 9 6 4 5 5 b 5 VertexParent e- be c d de a d
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41 Prim’s algorithm VertexParent e- be c d de a d a c e d b 2 45 9 6 4 5 5 The final minimum spanning tree
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42 Prim’s Algorithm Invariant At each step, we add the edge (u,v) s.t. the weight of (u,v) is minimum among all edges where u is in the tree and v is not in the tree Each step maintains a minimum spanning tree of the vertices that have been included thus far When all vertices have been included, we have a MST for the graph!
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43 Running time of Prim’s algorithm Initialization of priority queue (array): O(| V |) Update loop: | V | calls Choosing vertex with minimum cost edge: O(| V |) Updating distance values of unconnected vertices: each edge is considered only once during entire execution, for a total of O(| E |) updates Overall cost: O(| E | + | V | 2 )
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44 Another Approach – Kruskal’s a c e d b 2 45 9 6 4 5 5 Create a forest of trees from the vertices Repeatedly merge trees by adding “safe edges” until only one tree remains A “safe edge” is an edge of minimum weight which does not create a cycle forest: {a}, {b}, {c}, {d}, {e}
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45 Kruskal’s algorithm Initialization a. Create a set for each vertex v V b. Initialize the set of “safe edges” A comprising the MST to the empty set c. Sort edges by increasing weight a c e d b 2 45 9 6 4 5 5 {a}, {b}, {c}, {d}, {e} A = E = {(a,d), (c,d), (d,e), (a,c), (b,e), (c,e), (b,d), (a,b)}
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46 Kruskal’s algorithm For each edge (u,v) E in increasing order while more than one set remains: If u and v, belong to different sets a. A = A {(u,v)} b. merge the sets containing u and v Return A Use Union-Find algorithm to efficiently determine if u and v belong to different sets
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47 Kruskal’s algorithm E = {(a,d), (c,d), (d,e), (a,c), (b,e), (c,e), (b,d), (a,b)} Forest {a}, {b}, {c}, {d}, {e} {a,d}, {b}, {c}, {e} {a,d,c}, {b}, {e} {a,d,c,e}, {b} {a,d,c,e,b} A {(a,d)} {(a,d), (c,d)} {(a,d), (c,d), (d,e)} {(a,d), (c,d), (d,e), (b,e)} a c e d b 2 45 9 6 4 5 5
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48 After each iteration, every tree in the forest is a MST of the vertices it connects Algorithm terminates when all vertices are connected into one tree Kruskal’s Algorithm Invariant
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49 Greedy Approach Like Dijkstra’s algorithm, both Prim’s and Kruskal’s algorithms are greedy algorithms The greedy approach works for the MST problem; however, it does not work for many other problems!
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50 Thursday P vs NP Models of Hard Problems Work on Homework 5
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