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Bellman-Ford Algorithm

Bellman-Ford Algorithm. Shortest path alg , even for negative weights. 2. 2. 9. 6. 5. 5. Relax. Relax. 2. 2. 7. 6. 5. 5. Relaxation. A key technique in shortest path algorithms is relaxation Idea: for all v , maintain upper bound on distance Relax(u,v,w ) {

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Bellman-Ford Algorithm

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  1. Bellman-Ford Algorithm Shortest path alg, even for negative weights

  2. 2 2 9 6 5 5 Relax Relax 2 2 7 6 5 5 Relaxation • A key technique in shortest path algorithms is relaxation • Idea: for all v, maintain upper bound on distance Relax(u,v,w) { if (d[v] > d[u]+w) then d[v]=d[u]+w; }

  3. Initialize d[], which will converge to shortest-path value  Relaxation: Make |V|-1 passes, relaxing each edge Test for solutionUnder what condition do we get a solution? Bellman-Ford Algorithm BellmanFord() for each v V d[v] = ; d[s] = 0; for i=1 to |V|-1 for each edge (u,v)  E Relax(u,v, w(u,v)); for each edge (u,v)  E if (d[v] > d[u] + w(u,v)) return “no solution”; Relax(u,v,w): if (d[v] > d[u]+w) then d[v]=d[u]+w

  4. Bellman-Ford Algorithm BellmanFord() for each v  V d[v] = ; d[s] = 0; for i=1 to |V|-1 for each edge (u,v)  E Relax(u,v, w(u,v)); for each edge (u,v)  E if (d[v] > d[u] + w(u,v)) return “no solution”; Relax(u,v,w): if (d[v] > d[u]+w) then d[v]=d[u]+w What will be the running time?

  5. Bellman-Ford Algorithm BellmanFord() for each v  V d[v] = ; d[s] = 0; for i=1 to |V|-1 for each edge (u,v)  E Relax(u,v, w(u,v)); for each edge (u,v)  E if (d[v] > d[u] + w(u,v)) return “no solution”; Relax(u,v,w): if (d[v] > d[u]+w) then d[v]=d[u]+w What will be the running time? A: O(VE)

  6. 5 t x -2 ¥ ¥ 6 -3 8 s 0 7 -4 2 7 ¥ ¥ 9 y z Bellman-Ford Example

  7. 5 5 t t x x -2 -2 6 ¥ ¥ ¥ 6 6 -3 -3 8 8 s 0 s 0 7 7 -4 -4 2 2 7 7 7 ¥ ¥ ¥ 9 9 y z y z Bellman-Ford Example

  8. 5 5 t t x x -2 -2 6 ¥ ¥ ¥ 6 6 -3 -3 8 8 s 0 s 0 7 7 -4 -4 2 2 7 7 7 ¥ ¥ ¥ 9 9 y z y z 5 t x -2 6 4 6 -3 8 s 0 7 -4 2 7 7 2 9 y z Bellman-Ford Example

  9. 5 5 t t x x -2 -2 6 ¥ ¥ ¥ 6 6 -3 -3 8 8 s 0 s 0 7 7 -4 -4 2 2 7 7 7 ¥ ¥ ¥ 9 9 y z y z 5 5 t t x x -2 -2 6 4 2 4 6 6 -3 -3 8 8 s 0 s 0 7 7 -4 -4 2 2 7 7 7 2 7 2 9 9 y z y z Bellman-Ford Example

  10. t x -2 2 4 6 -3 8 s 0 7 -4 2 7 7 -2 9 y z Bellman-Ford Example • Bellman-Ford running time: • (|V|-1)|E| + |E| = Q(VE) 5

  11. Bellman-Ford • Prove: after |V|-1 passes, all d values correct • Consider shortest path from s to v:s  v1  v2  v3  v4  v • Initially, d[s] = 0 is correct, and doesn’t change (Why?) • After 1 pass through edges, d[v1] is correct (Why?) and doesn’t change • After 2 passes, d[v2] is correct and doesn’t change • … • Terminates in |V| - 1 passes: (Why?) • What if it doesn’t?

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