1. Algorithms LECTURE 14 Shortest Paths II Bellman-Ford algorithm
Floyd-Warshal algorithm 4
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

2. Negative-weight cycles
Recall: If a graph G = (V, E) contains a negative-
weight cycle, then some shortest paths may not exist. …
Example:
< 0 uu vv
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

3. Negative-weight cycles
Recall: If a graph G = (V, E) contains a negative-
weight cycle, then some shortest paths may not exist. …
Example:
< 0
uu vv
Bellman-Ford algorithm: Finds all shortest-path lengths from a source s ∈ V to all v ∈ V or determines that a negative-weight cycle exists.
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

4. Bellman-Ford algorithm
d[s] ← 0
for each v ∈ V – {s}
do d[v] ← ∞
for i ← 1 to |V| – 1
initialization
do for each edge (u, v) ∈ E
do if d[v] > d[u] + w(u, v)
then d[v] ← d[u] + w(u, v)
for each edge (u, v) ∈ E
relaxation step
then report that a negative-weight cycle exists At the end, d[v] = δ(s, v), if no negative-weight cycles.
Time = O(V E).
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

5. Example of Bellman-FordBB 2 –1 2 AA 3 EE 1 –3 4 CC DD 5
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

6. Example of Bellman-Ford∞ BB 2 –1 ∞ EE 2 AA 3 1 –3 4 CC ∞ DD ∞ 5
Initialization.
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

7. Example of Bellman-Ford∞ BB 1 2 –1 ∞ 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC ∞ DD ∞ 5
Order of edge relaxation.
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

8. Example of Bellman-Ford∞ BB 1 2 –1 ∞ 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC ∞ DD ∞ 5
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

9. Example of Bellman-Ford∞ BB 1 2 –1 ∞ 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC ∞ DD ∞ 5
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

10. Example of Bellman-Ford∞ BB 1 2 –1 ∞ 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC ∞ DD ∞ 5
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

11. Example of Bellman-Ford∞ −1 BB 1 2 –1 ∞ 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC ∞ DD ∞ 5
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

12. Example of Bellman-Ford−1 BB 1 2 –1 ∞ 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 4 DD ∞ 5 ∞
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

13. Example of Bellman-Ford−1 BB 1 2 –1 ∞ 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 4 DD ∞ 5
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

14. Example of Bellman-Ford−1 BB 1 2 –1 ∞ 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 2 DD ∞ 5 4
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

15. Example of Bellman-Ford−1 BB 1 2 –1 ∞ 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 2 DD ∞ 5
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

16. Example of Bellman-Ford−1 BB 1 2 –1 ∞ 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 2 DD ∞ 5
End of pass 1.
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

17. Example of Bellman-Ford−1 BB 1 2 –1 ∞ 1 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 2 DD ∞ 5
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

18. Example of Bellman-Ford−1 BB 1 2 –1 1 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 2 DD ∞ 5
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

19. Example of Bellman-Ford−1 BB 1 2 –1 1 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 2 DD 1 5 ∞
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

20. Example of Bellman-Ford−1 BB 1 2 –1 1 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 2 DD 1 5
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

21. Example of Bellman-Ford−1 BB 1 2 –1 1 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 2 DD 1 5
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

22. Example of Bellman-Ford−1 BB 1 2 –1 1 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 2 DD 1 5
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

23. Example of Bellman-Ford−1 BB 1 2 –1 1 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 2 DD 1 5
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

24. Example of Bellman-Ford−1 BB 1 2 –1 1 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 2 DD −2 5 1
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

25. Example of Bellman-Ford−1 BB 1 2 –1 1 7 3 4 2 AA 3 EE 1 5 4 8 –3 2 6 CC 2 DD −2 5
End of pass 2 (and 3 and 4).
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

26. Correctness
Theorem. If G = (V, E) contains no negative- weight cycles, then after the Bellman-Ford algorithm executes, d[v] = δ(s, v) for all v ∈ V.
November 16, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

27. Shortest paths Single-source shortest paths Nonnegative edge weights
Dijkstra’s algorithm: O(E + V lg V)
General
Bellman-Ford algorithm: O(VE)
DAG
One pass of Bellman-Ford: O(V + E)
November 21, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

28. Shortest paths Single-source shortest paths Nonnegative edge weights
Dijkstra’s algorithm: O(E + V lg V)
General
Bellman-Ford: O(VE)
DAG
One pass of Bellman-Ford: O(V + E)
All-pairs shortest paths
Dijkstra’s algorithm |V| times: O(VE + V 2 lg V)
Three algorithms today.
November 21, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

29. All-pairs shortest paths
Input: Digraph G = (V, E), where V = {1, 2,
…, n}, with edge-weight function w : E → R.
Output: n × n matrix of shortest-path lengths
δ(i, j) for all i, j ∈ V.
November 21, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

30. All-pairs shortest paths
Input: Digraph G = (V, E), where V = {1, 2,
…, n}, with edge-weight function w : E → R.
Output: n × n matrix of shortest-path lengths
δ(i, j) for all i, j ∈ V.
IDEA:
Run Bellman-Ford once from each vertex.
Time = O(V 2E).
Dense graph (n2 edges) ⇒ Θ(n 4) time in the worst case.
Good first try!
November 21, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson

31. Pseudocode for Floyd- Warshall
for k ← 1 to n
do for i ← 1 to n
do for j ← 1 to n
do if cij > cik + ckj
then cij ← cik + ckj
relaxation
Notes:
Okay to omit superscripts, since extra relaxations can’t hurt.
Runs in Θ(n3) time.
Simple to code.
Efficient in practice.
November 21, 2005
Copyright © by Erik D. Demaine and Charles E. Leiserson