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Dynamic programming algorithms for all-pairs shortest path and longest common subsequences

• We will study a new technique—dynamic programming algorithms (typically for optimization problems)

• Ideas:

– Characterize the structure of an optimal solution

– Recursively define the value of an optimal solution

– Compute the value of an optimal solution in a bottom-up fashion (using matrix to compute)

– Backtracking to construct an optimal solution from computed information.

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Floyd-Warshall algorithm for shortest path:• Use a different dynamic-programming formulatio

n to solve the all-pairs shortest-paths problem on a directed graph G=(V,E).

• The resulting algorithm, known as the Floyd-Warshall algorithm, runs in O (V3) time. – negative-weight edges may be present, – but we shall assume that there are no negative-

weight cycles.

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The structure of a shortest path:

• We use a different characterization of the structure of a shortest path than we used in the matrix-multiplication-based all-pairs algorithms.

• The algorithm considers the “intermediate” vertices of a shortest path, where an intermediate vertex of a simple path p=<v1,v2,…,vl> is any vertex in p other than v1 or vl, that is, any vertex in the set {v2,v3,…,vl-1}

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Continue:

• Let the vertices of G be V={1,2,…,n}, and consider a subset {1,2,…,k} of vertices for some k.

• For any pair of vertices i,j V, consider all paths from i to j whose intermediate vertices are all drawn from {1,2,…,k},and let p be a minimum-weight path from among them.

• The Floyd-Warshall algorithm exploits a relationship between path p and shortest paths from i to j with all intermediate vertices in the set {1,2,…,k-1}.

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Relationship:

• The relationship depends on whether or not k is an intermediate vertex of path p.

• If k is not an intermediate vertex of path p, then all intermediate vertices of path p are in the set {1,2,…,k-1}. Thus, a shortest path from vertex i to vertex j with all intermediate vertices in the set {1,2,…,k-1} is also a shortest path from i to j with all intermediate vertices in the set {1,2,…,k}.

• If k is an intermediate vertex of path p,then we break p down into i k j as shown Figure 2.p1 is a shortest path from i to k with all intermediate vertices in the set {1,2,…,k-1}, so as p2.

1p 2p

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i j

kp1 p2

All intermediate vertices in {1,2,…,k-1}

P:all intermediate vertices in {1,2,…,k}

Figure 2. Path p is a shortest path from vertex i to vertex j,andk is the highest-numbered intermediate vertex of p. Path p1, the portion of path p from vertex i to vertex k,has all intermediatevertices in the set {1,2,…,k-1}.The same holds for path p2 fromvertex k to vertex j.

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A recursive solution to the all-pairs shortest paths problem:

• Let dij(k) be the weight of a shortest path from vertex i to ve

rtex j with all intermediate vertices in the set {1,2,…,k}. A recursive definition is given by

• dij(k)= wij if k=0,

• min(dij(k-1),dik

(k-1)+dkj(k-1)) if k 1.

• The matrix D(n)=(dij(n)) gives the final answer-dij

(n)= for all i,j V-because all intermediate vertices are in the set {1,2,…,n}.

),( ji

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Computing the shortest-path weights bottom up:

• FLOYD-WARSHALL(W)

• n rows[W]

• D(0) W

• for k 1 to n

• do for i 1 to n

• do for j 1 to n

• dij(k) min(dij

(k-1),dik(k-1)+dkj

(k-1))

• return D(n)

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Example:

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5 4

3

2

3 4

7-4

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1 -52

• Figure 3

6

10

06

052

04

710

4830

D(0)=

NILNILNILNIL

NILNILNIL

NILNILNILNIL

NILNILNIL

NILNIL

5

44

3

22

111

(0)=

D(1)= (1)=

06

20552

04

710

4830

NILNILNILNIL

NIL

NILNILNILNIL

NILNILNIL

NILNIL

5

1414

3

22

111

11

06

20552

11504

710

44830

D(2)=

NILNILNILNIL

NIL

NILNIL

NILNILNIL

NIL

5

1414

223

22

1211

(2)=

D(3)= (3)=

06

20512

11504

710

44830

NILNILNILNIL

NIL

NILNIL

NILNILNIL

NIL

5

1434

223

22

1211

12

06158

20512

35047

11403

44130

D(4)=

NIL

NIL

NIL

NIL

NIL

5434

1434

1234

1244

1241

(4)=

D(5)= (5)=

06158

20512

35047

11403

42310

NIL

NIL

NIL

NIL

NIL

5434

1434

1234

1244

1543

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Comparison of two strings

• Longest common subsequence

• Shortest common supersequence

• Edit distance between two sequences

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1. 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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• 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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• 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.

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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 approachSuppose that mn. 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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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 zkxm implies that Z is an LCS

of X[1..m-1] and Y.• 2. If xm yn, then zkyn implies that Z is an LCS o

f X and Y[1..n-1].

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The 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: 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 xiyj.

Computing the length of an LCS• There are nm c[i,j]’s. So we can compute them i

n a specific order.

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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 xi ==yj 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 }

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Example 3: X=BDCABA and Y=ABCBDAB.

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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=m2. 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-16. if b[i,j]==3 j=j-17. Goto Step 3.

• The time complexity: O(nm).

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2. 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.

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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 xiyj.

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3. Edit distance between two sequences

Three operations:

• insertion: inserting an x into abc (between a and b), we get axbc.

• deletion: deleting b from abc, we get ac.

• replacement: Given a sequence abc, replacing a with x, we get xbc.

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Definition: Suppose that we can use three edit operations (insertion, deletion, and replacement) to edit a sequence into another. The edit distance between two sequences is the minimum number of operations required to edit one sequence into another.

• Note: each operation is counted as 1.

Weighted edit distance:• There is a weight on each operation. • For example: s(a,b)=1, s(a, _)=1.5, s(b,a)=1, s(b,_)=1.5.• Where the weight comes from:• For DNA and protein sequences, it is from statistics.

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Alignment of sequences -- an alternative

• An alignment of two sequences is obtained by inserting spaces into or at either end of X and Ysuch that the two resulting sequences X’ and Y’ are of the same length. That is, every letter in X’ is opposite to a unique letter in Y’.

• The alignment value is defined as

• where X’[i] and Y’[i] denote the two letters in column i of the alignment and s(X’[i], Y’[i]) is the score (weight) of these opposing letters.

• There are several popular socre schemes for DNA and protein sequences.

n

i

iYiXs1

])['],['(

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• Facts: The edit distance between two sequences is the same as the alignment value of two sequences if we use the same score scheme.

• Recursive equation:c[i,j]=min{ c[i-1, j-1]+s(X[i], Y[j]), c[i, j-1]+s(_,Y[j]),

c[i-1, j)+s(X[i],_)}.

• Time and space complexityBoth are O(nm) or O(n2) if both sequences have equal

length n.

• Why?We have to compute c[i,j] (the cost) and b[i,j] (for back-tr

acking). Each will take O(n2).

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Linear space algorithm

• Hints: Computing c[i,j] needs linear space whereas back-tracking needs O(nm) time.

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• To compute c[i,j], we need c[i-1,j-1], c[i,j-1], & c[i-1,j].

• So, to get c[n,m], we only have to keep dark cells.

• However, if we do not have all the b[i,j]’s, we can not get the alignment (nor the edit process, the subsequence, the supersequence).

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• Discussion: Each time we only keep a few b[i,j]’s and we can re-compute the b[i,j]’s again. In this way, we can get a linear space algorithm. However, the time complexity is increased to O(n3).

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• A Better Idea: find a cuting point.

• For the problems of smaller size, we do the same thing until one of the segment contains 1 letter.

• Key: each time, we fix the middle point (n/2) of X.

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• Example: X=abcdefgh and Y=aacdefhh.• Score scheme: match -- 0 and mismatch

-- 1.The alignment abcdefgh abcd efgh aacdefhh aacd efhh /|\ cutting point (4,4).

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Finding the cutting point:

Let X=x1x2x3...xn and Y=y1y2y3...ym.

Define XT=xnxn-1...x1 and YT=ymym-1 ...y1.Let c[i,j] be the cost of optimal alignment for X[1...i]

and Y[1...j] and cc[k,l] be the cost of optimal alignment for XT[1...k] and YT[1...l].

for (i=1, i<=n i++) if( (c[[n/2], i]+cc[n-[n/2], m-i]) ==c[n,n]) point = i;We need to check two rows, c[[n/2],1], c[[n/2],2], ...

c[[n/2],m] and cc[n-[n/2], 1], cc[n-[n/2],2], ... cc[n-[n/2],m]. O(m) space.

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The algorithm1. compute c[n,n], the [n/2]-th row and the ([n/2]+1)-th row

of c.2. find the cutting point ([n/2], i) as shown above. 3. if i-[n/2] == 1 then compute the alignment of X[1...[n/2])

and Y[1...i]. 4. if n-[n/2]+1 == 1 then compute the alignment of

X[[n/2]+1...n] and Y[i+1...n].5. if i-[n/2] != 1 and n-[n/2]+1 !=1 then recursive on step 1-4 for the two pairs of sequences X[1...

[n/2]) and Y[1...i], and X[[n/2]+1...n] and Y[i+1...n]; finally combine the two alignments for the two pairs of sequences.

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Time complexity analysis:

• The first round needs T’ time, where T’ is the time for the normal algorithm. (O(n2).)

• 2nd round needs 1/2 T’. (0.5 n i +0.5 n (n-i)=0.5n2.)

• 3rd round need 1/4 T’.

• i-th round needs 1/2i-1 T’.

• Total time T=(1/2+1/4+1/8+ ... )T’ =2T’ =O(n2).