1. Negative-Weight edges:
Edge weight may be negative.
negative-weight cycles– the total weight in the cycle (circuit) is negative.
If no negative-weight cycles reachable from the source s, then for all v V, the shortest-path weight remains well defined,even if it has a negative value.
If there is a negative-weight cycle on some path from s to v, we define =
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2. a b -4 h i 3 -1 2 4 3 c d 6 8 5 -8 3 5 11 g s -3 e 3 f 2 7 j -6
Figure1 Negative edge weights in a directed graph.Shown within each vertex is its
shortest-path weight from source s.Because vertices e and f form a negative-weight
cycle reachable from s,they have shortest-path weights of Because vertex g is
reachable from a vertex whose shortest path is ,it,too,has a shortest-path weight
of Vertices such as h, i ,and j are not reachable from s,and so their shortest-path
weights are , even though they lie on a negative-weight cycle.
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3. Relaxation:
The process of relaxing an edge (u,v) consists of testing whether we can improve the shortest path to v found so far by going through u and,if so,updating d[v] and [v].
RELAX(u,v,w)
if d[v]>d[u]+w(u,v)
then d[v]  d[u]+w(u,v) (based on Lemma 25.3)
[v]  u
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4. u v u v 2 2 5 9 5 6 RELAX(u,v) RELAX(u,v) u v u v 2 2 5 7 5 6 (a) (b)
Figure2 Relaxation of an edge (u,v).The shortest-path estimate of each vertex
is shown within the vertex. (a)Because d[v]>d[u]+w(u,v) prior to relaxation,
the value of d[v] decreases. (b)Here, d[v] d[u]+w(u,v) before the relaxation
step,so d[v] is unchanged by relaxation.
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5. Bellman-Ford algorithm:
The Bellman-Ford algorithm solves the single-source shortest-paths problem in the more general case in which edge weights can be negative.
It report FALSE if a negative-weight circuit exists.
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6. Continue: BELLMAN-FORD(G,w,s): INITIALIZE-SINGLE-SOURCE(G,s)
for i  1 to |V[G]|
do for each edge (u,v)  E[G] (The order here is arbitrary)
do RELAX(u,v,w)
for each edge (u,v) E[G]
do if d[v]>d[u]+w(u,v)
then return FALSE
return TRUE
Time complexity: O(|V||E|).
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7. 6 7 9 2 5 -2 8 -3 -4 z u v x y
(a)
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8. u 5 v 6 8 -2 6 -3 8 7 z -4 2 7 7 8 9 x y
(b)
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9. u 5 v 6 4 -2 6 -3 8 7 z -4 2 7 7 2 9 x y
(c)
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10. u 5 v 2 4 -2 6 -3 8 7 z -4 2 7 7 2 9 x y
(d)
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11. u 5 v 2 4 -2 6 -3 8 7 z -4 2 7 7 -2 9 x y
(e)
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12. Proof: We prove it by induction on i.
Theorem: (hard) after the i-th iteration, the cost of a shortest path from s to any node v containing i edges is obtained.
Proof: We prove it by induction on i.
The theorem is true for i=1.
The shortest path from s to v containing at most one edge is the the edge (s, v) (if exists).
Assume that the theorem is true for i=k. Then we are going to show that the theorem is true for i=k+1.
Let P: s-> v1-> v2 …->vm-> v be the shortest path from s to v containing at most k+1 edges. Then P1: s-> v1-> v2 …->vm must be the shortest path from s to vm containing at most k edges. By assumption, P1 has been obtained after (k)-th iteration. In the (k+1)-th iteration, we tried to RELAX(vm, v, w) on node vm. Thus, path P is obtained after the (k+1)-th iteration.
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13. Corollary: If negative-weight circuit exists in the given graph, in the n-th iteration, the cost of a shortest path from s to some node v will be further reduced.
Demonstrated by the following example.
Note: For each iteration, the order of edges used does not effect the Theorem. However, the intermediate results may vary. The final results will be the same.
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14. An example with negative-weight cycle5  1 6 -2 8 7 7 9 2 2 5 -8
An example with negative-weight cycle
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15. 7 6  8 5 -2 1 2 9 -8
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16. 7 6 16  11 9 8 5 -2 1 2 -8
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17. 7 6 16 12 11 1 9 8 5 -2 2 -8
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18. 6 16 12 11 1 9 7 8 5 -2 2 -8
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19. 5 6 11 1 6 -2 8 7 12 7 9 2 6 15 2 5 -8 8 1
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20. 6 15 12 11 8 7 5 -2 1 2 9 -8
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21. 5 6 11 1 6 -2 8 7 12 7 9 2 5 15 2 5 -8 8 x
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22. Exercise: use Bellman-Ford algorithm for the graph２ 5 -3  1 6 -2 8 7 7 9 2 2 5 -６
Exercise: use Bellman-Ford algorithm for the graph
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