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 obtained is at most the cost of a shortest path from s to any node v containing at most i edges.
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. Longest common subsequence
Definition 1: Given a sequence X=x1x2...xm, another sequence Z=z1z2...zk is a subsequence of X if there exists a strictly increasing sequence i1i2...ik of indices of X such that for all j=1,2,...k, we have xij=zj.
Example 1: If X=abcdefg, Z=abdg is a subsequence of X. X=abcdefg, Z=ab d g
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23. Definition 2: Given two sequences X and Y, a sequence Z is a common subsequence of X and Y if Z is a subsequence of both X and Y.
Example 2: X=abcdefg and Y=aaadgfd. Z=adf is a common subsequence of X and Y.
X=abc defg
Y=aaaadgfd
Z=a d f
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24. Longest common subsequence may not be unique. Example: abcd acbd
Definition 3: A longest common subsequence of X and Y is a common subsequence of X and Y with the longest length. (The length of a sequence is the number of letters in the seuqence.)
Longest common subsequence may not be unique.
Example: abcd
acbd
Both acd and abd are LCS.
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25. Longest common subsequence problem
Input: Two sequences X=x1x2...xm, and
Y=y1y2...yn.
Output: a longest common subsequence of X and Y.
A brute-force approach
Suppose that mn. Try all subsequence of X (There are 2m subsequence of X), test if such a subsequence is also a subsequence of Y, and select the one with the longest length.
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26. Charactering a longest common subsequence
Theorem (Optimal substructure of an LCS)
Let X=x1x2...xm, and Y=y1y2...yn be two sequences, and
Z=z1z2...zk be any LCS of X and Y.
1. If xm=yn, then zk=xm=yn and Z[1..k-1] is an LCS of X[1..m-1] and Y[1..n-1].
2. If xm yn, then zkxm implies that Z is an LCS of X[1..m-1] and Y.
2. If xm yn, then zkyn implies that Z is an LCS of X and Y[1..n-1].
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27. The recursive equation
Let c[i,j] be the length of an LCS of X[1...i] and Y[1...j].
c[i,j] can be computed as follows:
if i=0 or j=0,
c[i,j]= c[i-1,j-1] if i,j>0 and xi=yj,
max{c[i,j-1],c[i-1,j]} if i,j>0 and xiyj.
Computing the length of an LCS
There are nm c[i,j]’s. So we can compute them in a specific order.
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28. The algorithm to compute an LCS
1. for i=1 to m do
c[i,0]=0;
3. for j=0 to n do
c[0,j]=0;
5. for i=1 to m do
for j=1 to n do {
if x[I] ==y[j] then
c[i,j]=c[i-1,j-1]+1;
b[i,j]=1;
else if c[i-1,j]>=c[i,j-1] then
c[i,j]=c[i-1,j]
b[i,j]=2;
else c[i,j]=c[i,j-1]
b[i,j]=3; }
b[i,j] stores the directions. 1—diagnal, 2-up, 3-forward.
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29. Example 3: X=BDCABA and Y=ABCBDAB.
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30. Constructing an LCS (back-tracking)
We can find an LCS using b[i,j]’s.
We start with b[n,m] and track back to some cell b[0,i] or b[i,0].
The algorithm to construct an LCS
1. i=m
2. j=n;
3. if i==0 or j==0 then exit;
4. if b[i,j]==1 then {
i=i-1;
j=j-1;
print “xi”; }
5. if b[i,j]== i=i-1
6. if b[i,j]== j=j-1
7. Goto Step 3.
The time complexity: O(nm).
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31. Shortest common supersequence
Definition: Let X and Y be two sequences. A sequence Z is a supersequence of X and Y if both X and Y are subsequence of Z.
Shortest common supersequence problem:
Input: Two sequences X and Y.
Output: a shortest common supersequence of X and Y.
Example: X=abc and Y=abb. Both abbc and abcb are the shortest common supersequences for X and Y.
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32. Let c[i,j] be the length of an LCS of X[1...i] and X[1...j].
Recursive Equation:
Let c[i,j] be the length of an LCS of X[1...i] and X[1...j].
c[i,j] can be computed as follows:
j if i=0
i if j=0,
c[i,j]= c[i-1,j-1] if i,j>0 and xi=yj,
min{c[i,j-1]+1,c[i-1,j]+1} if i,j>0 and xiyj.
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