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 = - . 2019/5/23 chapter25

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. 2019/5/23 chapter25

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 2019/5/23 chapter25

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. 2019/5/23 chapter25

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. 2019/5/23 chapter25

Continue: BELLMAN-FORD(G,w,s): INITIALIZE-SINGLE-SOURCE(G,s) for i  1 to |V[G]| -1 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|). 2019/5/23 chapter25

6 7 9 2 5 -2 8 -3 -4 z u v x y (a) 2019/5/23 chapter25

u 5 v 6 8 -2 6 -3 8 7 z -4 2 7 7 8 9 x y (b) 2019/5/23 chapter25

u 5 v 6 4 -2 6 -3 8 7 z -4 2 7 7 2 9 x y (c) 2019/5/23 chapter25

u 5 v 2 4 -2 6 -3 8 7 z -4 2 7 7 2 9 x y (d) 2019/5/23 chapter25

u 5 v 2 4 -2 6 -3 8 7 z -4 2 7 7 -2 9 x y (e) 2019/5/23 chapter25

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. 2019/5/23 chapter25

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. 2019/5/23 chapter25

An example with negative-weight cycle 5  1 6 -2 8 7 7 9 2 2 5 -8 An example with negative-weight cycle 2019/5/23 chapter25

7 6  8 5 -2 1 2 9 -8 2019/5/23 chapter25

7 6 16  11 9 8 5 -2 1 2 -8 2019/5/23 chapter25

7 6 16 12 11 1 9 8 5 -2 2 -8 2019/5/23 chapter25

6 16 12 11 1 9 7 8 5 -2 2 -8 2019/5/23 chapter25

5 6 11 1 6 -2 8 7 12 7 9 2 6 15 2 5 -8 8 1 2019/5/23 chapter25

6 15 12 11 8 7 5 -2 1 2 9 -8 2019/5/23 chapter25

5 6 11 1 6 -2 8 7 12 7 9 2 5 15 2 5 -8 8 x 2019/5/23 chapter25

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 2019/5/23 chapter25

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 2019/5/23 chapter25

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. 2019/5/23 chapter25

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. 2019/5/23 chapter25

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]. 2019/5/23 chapter25

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: 0 if i=0 or j=0, c[i,j]= c[i-1,j-1]+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. 2019/5/23 chapter25

The algorithm to compute an LCS 1. for i=1 to m do 2. c[i,0]=0; 3. for j=0 to n do 4. c[0,j]=0; 5. for i=1 to m do 6. for j=1 to n do 7. { 8. if x[I] ==y[j] then 9. c[i,j]=c[i-1,j-1]+1; 10 b[i,j]=1; 11. else if c[i-1,j]>=c[i,j-1] then 12. c[i,j]=c[i-1,j] 13. b[i,j]=2; 14. else c[i,j]=c[i,j-1] 15. b[i,j]=3; 14 } b[i,j] stores the directions. 1—diagnal, 2-up, 3-forward. 2019/5/23 chapter25

Example 3: X=BDCABA and Y=ABCBDAB. 2019/5/23 chapter25

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]==2 i=i-1 6. if b[i,j]==3 j=j-1 7. Goto Step 3. The time complexity: O(nm). 2019/5/23 chapter25

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. 2019/5/23 chapter25

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]+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. 2019/5/23 chapter25

2019/5/23 chapter25