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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 new characterization of the structure of a shortest path

• Different from that for 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,…,vk> is any vertex in p other than v1 or vk, that is, any vertex in the set {v2,v3,…,vk-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.

• Case 1: k is not an intermediate vertex of path p– 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}.

• Case 2: k is an intermediate vertex of path p– 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 is 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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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.

• Example: abcd acbd Both acd and abd are LCS.

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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 approach

Suppose 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 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 }

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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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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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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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