NegativeWeight edges:

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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a
b
-4
h
i
-1
3
2
4
3
c
d
6
8
3
-8
5
11
5
g
0
s
-3
e
f
3
j
2
7
-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.
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 dv and v.
• RELAX(u,v,w)
• if dvgtduw(u,v)
• then dv ? duw(u,v) (based on
  Lemma 25.3)
• v ? u

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 dvgtduw(u,v)
prior to relaxation, the value of dv decreases.
(b)Here, dv duw(u,v) before the
relaxation step,so dv is unchanged by
relaxation.
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.

6
Continue

• BELLMAN-FORD(G,w,s)
• INITIALIZE-SINGLE-SOURCE(G,s)
• for i ? 1 to VG -1
• do for each edge (u,v) ? EG (The
  order here is arbitrary)
• do RELAX(u,v,w)
• for each edge (u,v)? EG
• do if dvgtduw(u,v)
• then return FALSE
• return TRUE
• Time complexity O(VE).

7
(a)
8
5
u
v
6
8
-2
6
-3
8
0
7
z
-4
2
7
7
8
9
x
y
(b)
9
5
u
v
6
4
-2
6
-3
8
0
7
z
-4
2
7
7
2
9
x
y
(c)
10
5
u
v
2
4
-2
6
-3
8
0
7
z
-4
2
7
2
7
9
x
y
(d)
11
5
u
v
2
4
-2
6
-3
8
0
7
z
-4
2
7
7
-2
9
x
y
(e)
12
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 i1.
• 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 ik. Then we
  are going to show that the theorem is true for
  ik1.
• Let P s-gt v1-gt v2 -gtvm-gt v be the shortest path
  from s to v containing at most k1 edges. Then
  P1 s-gt v1-gt v2 -gtvm 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 (k1)-th iteration, we tried
  to RELAX(vm, v, w) on node vm. Thus, path P is
  obtained after the (k1)-th iteration.

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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5
1
6
-2
8
7
7
9
2
2
5
-8
An example with negative-weight cycle
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5
6
11
1
6
-2
0
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12
7
7
9
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15
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-8
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1
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5
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11
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-2
0
8
12
7
7
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15
2
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-8
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0
x
22
2
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-3
1
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-2
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2
2
5
-6
Exercise use Bellman-Ford algorithm for the graph