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Longest Common Subsequence(LCS)
研究生 鍾聖彥 指導老師 許慶昇
Dynamic Programming
12014/05/07
最長共同子序列
Dynamic Programming
Optimal substructure(當一個問題存在著最佳解，則表示其所有子問題也必存在著最佳解)
Overlapping subproblems(子問題重複出現)
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Longest Common Subsequence???
Biological applications often need to compare the DNA of tow(or more) different organisms.
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Subsequence
A subsequence of a given sequence is just the given sequence with zero or more elements left out.
Ex: app、le、ple and so on are subsequences of “apple”.
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Common Subsequence
X = (A, B, C, B, D, A, B)Y = (B, D, C, A, B, A)
Two sequences:
Sequence Z is a common subsequence of X and Y if Z is a subsequence of both X and Y
Z = (B, C, A) — length 3 Z = (B, C, A, B) - length 4 Z = (B, D, A, B) — length 4
Z= — length 5 ???
longest
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What is longest Common Subsequence problem?
X = (x1, x2,……., xm) Y = (y1, y2,……., yn)
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Find a maximum-length common subsequence of X and Y
How to do?Dynamic Programming!!!
Brute Force!!!
Step 1: Characterize optimality
Sequence X = (x1, x2,……., xm)
Define the ith prefix of X, for i = 0, 1,…, m as Xi = (x1, x2, ..., xi)
with X0 representing the empty sequence.
EX: if X = (A, B, C, A, D, A, B) then X4 = (A, B, C, A)X0 = ( ) empty sequence
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Theorem (Optimal substructure of LCS)
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1. If Xm = Yn, then Zk = Xm = Yn and Zk-1 is a LCS of Xm-1 and Yn-1
2. If Xm ≠ Yn, then Zk ≠ Xm implies that Z is a LCS of Xm-1 and Y
3. If Xm ≠ Yn, then Zk ≠ Yn implies that Z is a LCS of X and Yn-1
X = (X1, X2,…, Xm) and Y = (Y1, Y2,…, Yn)
Sequences
Z = (Z1, Z2,…, Zk) be any LCS of X and Y
We assume:
Optimal substructure problem
The LCS of the original two sequences contains a LCS of prefixes of the two sequences.
(當一個問題存在著最佳解，則表示其所有子問題也必存在著最佳解)
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Step 2: A recursive solutionXi and Yj end with xi=yj
Zk is Zk -1 followed by Zk = Xi = Yj where Zk-1 is an LCS of Xi-1 and Yj -1
LenLCS(i, j) = LenLCS(i-1, j-1)+1
Xi x1 x2 … xi-1 xi
Yj y1 y2 … yj-1 yj=xi
Zk z1 z2…zk-1 zk =yj=xi
Case 1:
Step 2: A recursive solutionCase 2,3: Xi and Yj end with xi ≠ yj
Xi x1 x2 … xi-1 xi
Yj y1 y2 … yj-1 yj
Zk z1 z2…zk-1 zk ≠yj
Xi x1 x2 … xi-1 x i
Yj yj y1 y2 …yj-1 yj
Zk z1 z2…zk-1 zk ≠ xi
Zk is an LCS of Xi and Yj -1 Zk is an LCS of Xi-1 and Yj
LenLCS(i, j)=max{LenLCS(i, j-1), LenLCS(i-1, j)}
Step 2:A recursive solution
Let c[i,j] be the length of a LCS for Xi and Yj the recursion described by the above cases as
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Case 1 Reduces to the single subproblem of finding a LCS of
Xm-1, Yn-1 and adding Xm = Yn to the end of Z.
Cases 2 and 3 Reduces to two subproblems of finding a LCS of Xm-1, Y and X, Yn-1 and selecting the longer of the two.
Step 3: Compute the length of the LCS
LCS problem has only ɵ(mn) distinct subproblems.
So? Use Dynamic programming!!!
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Step 3: Compute the length of the LCSProcedure 1
LCS-length takes two Sequences X = (x1, x2,…, xm) and Y = (y1, y2,…, yn) as input.
Procedure 2It stores the c[i, j] values in a table c[0..m, 0..n] and
it computes the entries in row-major order.Procedure 3
Table b[1..m, 1..n] to construct an optimal solution. b[i, j] points to the table entry corresponding to the
optimal solution chosen when computing c[i, j]Procedure 4
Return the b and c tables; c[m, n] contains the length of an LCS X and Y
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LCS-Length(X, Y)1 m = X.length 2 n = Y.length 3 let b[1..m, 1..n] and c[0..m, 0..n] be new tables. 4 for i 1 to m do 5 c[i, 0] = 0 6 for j 1 to n do 7 c[0, j] = 0 8 for i 1 to m do 9 for j 1 to n do 10 if xi ==yj11 c[i, j] = c[i-1, j-1]+112 b[i, j] = “ ” 13 else if c[i-1, j] ≥ c[i, j-1]14 c[i, j] = c[i-1, j] 15 b[i, j] = “ ”16 else 17 c[i, j] = c[i, j-1]18 b[i, j] = “ ” 19 return c and b 15
The table produced by LCS-Length on the sequences X = (A, B, C, B, D, A, B) and Y = (B, D, C, A, B, A).
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The running time of the procedure is O(mn), since each table entry table O(1) time to compute
Step 4: Construct an optimal LCS
PRINT-LCS(b, X, i, j)PRINT-LCS(b, X, X.length, Y.length)
1 if i == 0 or j == 0 2 return 3 if b[i, j] == “ ” 4 PRINT-LCS(b,X,i-1, j-1) 5 print Xi 6 else if b[i, j] == “ ” 7 PRINT-LCS(b,X,i-1, j) 8 else PRINT-LCS(b,X,i, j-1)
This procedure prints BCBA. The procedure takes time O(m+n)
ExampleX = <A, B, C, B, A> Y = <B, D, C, A, B>
We will fill in the table in row-major order starting in the upper left corner using the following formulas:
ExampleX = <A, B, C, B, A> Y = <B, D, C, A, B>
We will fill in the table in row-major order starting in the upper left corner using the following formulas:
Answer
Thus the optimal LCS length is c[m,n] = 3.Optimal LCS starting at c[5,5] we get Z = <B, C, B>
Alternatively start at c[5,4] we would produce Z = <B, C, A>.
*Note that the LCS is not unique but the optimal length of the LCS is.
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ReferenceLecture 13: Dynamic Programming - Longest Common Subsequence http://faculty.ycp.edu/~dbabcock/cs360/lectures/lecture13.html
http://www.csie.ntnu.edu.tw/~u91029/LongestCommonSubsequence.html
Longest common subsequence (Cormen et al., Sec. 15.4)
https://www.youtube.com/watch?v=Wv1y45iqsbk
https://www.youtube.com/watch?v=wJ-rP9hJXO0