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

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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.

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.

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}

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}.

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.

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.

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}.

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)

Comparison of two strings

- Longest common subsequence
- Shortest common supersequence
- Edit distance between two sequences

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

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

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.

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.

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 xmyn, then zkxm implies that Z is an LCS of X[1..m-1] and Y.
- 2. If xmyn, then zkyn implies that Z is an LCS of X and Y[1..n-1].

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 in a specific order.

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. 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 }

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).

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.

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.

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.

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.

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.

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 complexity

Both 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-tracking). Each will take O(n2).

Linear space algorithm

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

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).

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).

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.

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).

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.

The algorithm

1. 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.

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).

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