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

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

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

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  1. 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.

  2. Floyd-Warshall algorithm for shortest path: • Use a different dynamic-programming formulation 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.

  3. 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}

  4. 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}.

  5. 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.

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

  7. A recursive solution to the all-pairs shortest paths problem: • Let dij(k) be the weight of a shortest path from vertex i to vertex 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}.

  8. Computing the shortest-path weights bottom up: • FLOYD-WARSHALL(W) • n rows[W] • D(0) W • for k1 to n • dofor 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)

  9. Example: • Figure 3 2 4 3 1 3 8 1 -5 -4 2 7 5 4 6

  10. D(0)= (0)= (1)= D(1)=

  11. D(2)= (2)= D(3)= (3)=

  12. D(4)= (4)= (5)= D(5)=

  13. 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

  14. 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

  15. 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.

  16. 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.

  17. 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].

  18. 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 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.

  19. 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. elseif 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 }

  20. Example 3: X=BDCABA and Y=ABCBDAB.

  21. 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).

  22. 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.

  23. 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.

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