Negative-Weight 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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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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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

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RELAX(u,v)
RELAX(u,v)
u
v
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(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.

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Continue

• BELLMAN-FORD(G,w,s)
• INITIALIZE-SINGLE-SOURCE(G,s)
• for i ? 1 to VG -1
• do for each edge (u,v) EG
• 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).

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(a)
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(b)
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(c)
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(e)
12
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 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.

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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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An example with negative-weight cycle
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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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Solve the problem by Single-Source shortest paths
algorithm

• If all edge weights are nonnegative, we can use
  Dijkstras 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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Preliminary knowledge

• We use an adjacency-matrix representation of
  graphs
• wij
  if ij
• 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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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 ij,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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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 m0, dij(m) 0 ij and otherwise ? .
• For mgt1, 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 kj.

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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(n1)...

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Computing the shortest-path weights bottom up

• We compute the shortest-path weights by extending
  shortest paths edge by edge.Letting AB denote
  the matrix product returned by
  EXTEND-SHORTEST-PATH(A,B) .We compute the
  sequence of n-1 matrices
• D(1)D(0)WW,
• D(2)D(1)WW2,
• D(n-1)D(n-2)WWn-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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Continue

• SLOW-ALL-PAIRS-SHORTEST-PATH(W)
• n rowsW
• D(1) W
• for m 2 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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Continue

• EXTEND-SHORTEST-PATH(D,W)
• n rowsD
• for i 1 to n
• do for j 1 to n
• for k 1 to n
• do dij
  min(dij,dikwkj)
• return D
• Time O(n3).

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Example

• Figure 1

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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 mgtn-1.So, we can compute D(n-1) with
  only matrix products by
  computing the sequence
• D(1)W,
• D(2)W2WW
• D(4) W2 W2

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Continue

• FASTER-ALL-PAIRS-SHORTEST-PATHS(W)
• n rowsW
• D(1) W
• while n-1gtm
• do D(2m) EXTEND-SHORTEST-PATHS(D(m
  ),D(m))
• m 2m
• return D(m)
• Time complexity O(n3 log n).

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The End
36
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 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.