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min {dist(u, i 1) + (u, v)}
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What if we want to nd the shortest path not just between s and t but between all pairs of vertices One approach would be to execute our general shortest-path algorithm from Section 461 (since there may be negative edges) |V | times, once for each starting node The total running time would then be O(|V | 2 |E|) We ll now see a better alternative, the O(|V | 3 ) dynamic programming-based Floyd-Warshall algorithm Is there is a good subproblem for computing distances between all pairs of vertices in a graph Simply solving the problem for more and more pairs or starting points is unhelpful, 172
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because it leads right back to the O(|V | 2 |E|) algorithm One idea comes to mind: the shortest path u w 1 wl v between u and v uses some number of intermediate nodes possibly none Suppose we disallow intermediate nodes altogether Then we can solve all-pairs shortest paths at once: the shortest path from u to v is simply the direct edge (u, v), if it exists What if we now gradually expand the set of permissible intermediate nodes We can do this one node at a time, updating the shortest path lengths at each stage Eventually this set grows to all of V , at which point all vertices are allowed to be on all paths, and we have found the true shortest paths between vertices of the graph! More concretely, number the vertices in V as {1, 2, , n}, and let dist(i, j, k) denote the length of the shortest path from i to j in which only nodes {1, 2, , k} can be used as intermediates Initially, dist(i, j, 0) is the length of the direct edge between i and j, if it exists, and is otherwise What happens when we expand the intermediate set to include an extra node k We must reexamine all pairs i, j and check whether using k as an intermediate point gives us a shorter path from i to j But this is easy: a shortest path from i to j that uses k along with possibly other lower-numbered intermediate nodes goes through k just once (why because we assume that there are no negative cycles) And we have already calculated the length of the shortest path from i to k and from k to j using only lower-numbered vertices:
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Thus, using k gives us a shorter path from i to j if and only if dist(i, k, k 1) + dist(k, j, k 1) < dist(i, j, k 1), in which case dist(i, j, k) should be updated accordingly Here is the Floyd-Warshall algorithm and as you can see, it takes O(|V | 3 ) time for i = 1 to n: for j = 1 to n: dist(i, j, 0) = for all (i, j) E: dist(i, j, 0) = (i, j) for k = 1 to n: for i = 1 to n: for j = 1 to n: dist(i, j, k) = min{dist(i, k, k 1) + dist(k, j, k 1), dist(i, j, k 1)}
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A traveling salesman is getting ready for a big sales tour Starting at his hometown, suitcase in hand, he will conduct a journey in which each of his target cities is visited exactly once 173
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Figure 69 The optimal traveling salesman tour has length 10
2 1 3
before he returns home Given the pairwise distances between cities, what is the best order in which to visit them, so as to minimize the overall distance traveled Denote the cities by 1, , n, the salesman s hometown being 1, and let D = (d ij ) be the matrix of intercity distances The goal is to design a tour that starts and ends at 1, includes all other cities exactly once, and has minimum total length Figure 69 shows an example involving ve cities Can you spot the optimal tour Even in this tiny example, it is tricky for a human to nd the solution; imagine what happens when hundreds of cities are involved It turns out this problem is also dif cult for computers In fact, the traveling salesman problem (TSP) is one of the most notorious computational tasks There is a long history of attempts at solving it, a long saga of failures and partial successes, and along the way, major advances in algorithms and complexity theory The most basic piece of bad news about the TSP, which we will better understand in 8, is that it is highly unlikely to be solvable in polynomial time How long does it take, then Well, the brute-force approach is to evaluate every possible tour and return the best one Since there are (n 1)! possibilities, this strategy takes O(n!) time We will now see that dynamic programming yields a much faster solution, though not a polynomial one What is the appropriate subproblem for the TSP Subproblems refer to partial solutions, and in this case the most obvious partial solution is the initial portion of a tour Suppose we have started at city 1 as required, have visited a few cities, and are now in city j What information do we need in order to extend this partial tour We certainly need to know j, since this will determine which cities are most convenient to visit next And we also need to know all the cities visited so far, so that we don t repeat any of them Here, then, is an appropriate subproblem For a subset of cities S {1, 2, , n} that includes 1, and j S, let C(S, j) be the length of the shortest path visiting each node in S exactly once, starting at 1 and ending at j When |S| > 1, we de ne C(S, 1) = since the path cannot both start and end at 1 Now, let s express C(S, j) in terms of smaller subproblems We need to start at 1 and end at j; what should we pick as the second-to-last city It has to be some i S, so the overall path length is the distance from 1 to i, namely, C(S {j}, i), plus the length of the nal edge, 174
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