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]
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 of at most 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.
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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. All-Pairs Shortest Paths
Problem definition:
Given a weighted,directed graph G=(V,E), for every pair of vertices u, v  V, find a shortest (least weight) path from u to v, where the weight of a path is the sum of the weights of its constituent edges.
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23. Solve the problem by Single-Source shortest paths algorithm:
If all edge weights are nonnegative, we can use Dijkstra’s algorithm.
If negative-weight edges are allowed, then we use Bellman-Ford algorithm instead.
Can you analysis the complexity of the two solutions ? Is it possible for us to find a more efficient algorithm?
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24. Preliminary knowledge:
We use an adjacency-matrix representation of graphs:
wij= if i=j
the weight of directed edge (i,j) if i j and (i,j) E
if i j and (i,j) E
The output of the algorithm presented in this chapter is an nn matrix D=(dij),where entry dij contains the weight of a shortest path from vertex i to vertex j. That is , if we let denotes the shortest-path weight from vertex i to vertex j, then dij= at termination.
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25. The structure of a shortest path:
Consider a shortest path p from vertex i to vertex j, and suppose that p contains at most m edges.Assuming that there are no negative-weight cycles,m is at most |V|-1. If i=j,then p has weight 0 and no edges.If vertices i and j are distinct, then we decompose path p into
i k j,
where path p’ now contains at most m-1 edges.
Moreover, p’ is a shortest path from i to k. Thus, we have
+wkj.
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26. A recursive solution to the all-pairs shortest-paths problem:
dij(m) ---minimum weight of any path from vertex i to vertex j that contains at most m edges.
When m=0, dij(m) = 0 i=j and otherwise  .
For m>=1, we compute dij(m) as follows:
dij(m)=min ( dij(m-1), {dik(m-1)+wkj})
= {dik(m-1)+wkj}.
Note that dij(m-1)= dik(m-1)+wkj for k=j.
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27. Continue:
If the graph contains no negative-weight cycles,then all shortest paths are simple and thus contain at most n-1 edges.A path from vertex i to vertex j with more than n-1 edges cannot be shorter than a shortest path from i to j.The actual shortest-path weights are therefore given by
=dij(n-1)=dij(n)=dij(n+1)=...
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28. Computing the shortest-path weights bottom up:
We compute the shortest-path weights by extending shortest paths edge by edge.Letting A*B denote the matrix “product” returned by EXTEND-SHORTEST-PATH(A,B) .We compute the sequence of n-1 matrices:
D(1)=D(0)*W=W,
D(2)=D(1)*W=W2, …
D(n-1)=D(n-2)*W=Wn-1.
As we argued above, the matrix D(n-1)=Wn-1 contains the shortest-path weights.So we get the algorithm.
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29. Continue: SLOW-ALL-PAIRS-SHORTEST-PATH(W) n rows[W] D(1) W
for m to n-1
do D(m) EXTEND-SHORTEST-PATHS(D(m-1),W)
return D(n-1)
Time O(nESP), where ESP is the time for EXTEND-SHORTEST-PATHS(), which is O(n3). Thus, the total time is O(n4).
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30. Continue: EXTEND-SHORTEST-PATH(D,W) n rows[D] for i 1 to n
do for j to n
for k to n
do dij’ min(dij’,dik+wkj)
return D’
Time: O(n3).
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31. Example:
Figure 1 2 4 3 1 3 8 1 -5 -4 2 7 5 4 6
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32. D(1)=
D(2)=
D(3)=
D(4)=
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33. Improving the running time:
Our goal,however, is not to compute all the D(m) matrices: we are interested in matrix D(n-1).Recall that in the absence of negative-weight cycles, D(m)=D(n-1),for all integers m>=n-1.So, we can compute D(n-1) with only matrix products by computing the sequence:
D(1)=W,
D(2)=W2=W*W
D(4)= W2* W2 …
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34. Continue: FASTER-ALL-PAIRS-SHORTEST-PATHS(W) n rows[W] D(1) W
while n-1>m
do D(2m) EXTEND-SHORTEST-PATHS(D(m),D(m))
m m
return D(m)
Time complexity O(n3 log n).
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35. The End:
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36. Proof: We prove it by induction on i.
Theorem: (hard) after the i-th iteration, the cost obtained is at most the cost of a shortest path from s to any node v containing at most i edges (New slide 12 of lecture 5)
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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