Dynamic Planning of LCS

Overview

Dynamic Planning is effective when finding the optimal solution with many overlapping sub-problems. It combines the problem into a subproblem. In order to avoid multiple solutions to these subproblems, their results are gradually calculated and saved, from a simple problem until the entire problem is solved. Therefore, dynamic planning saves recursive results, so it does not take time to solve the same problem.

Dynamic Planning can only be applied to problems with optimal sub-structures. The optimal sub-structure means that the local optimal solution can determine the global optimal solution (for some problems, this requirement cannot be fully met, so some approximation is sometimes required ). Simply put, the problem can be resolved into sub-problems.

Longest Common subsequence (LCS ).

Question: If all the characters in string 1 appear in the order of the strings in another string,
Then, string 1 is called a substring of string 2.

Note that the character of a substring (string 1) must appear in string 2 consecutively.
Compile a function, enter two strings, calculate their longest public substrings, and print the longest public substrings.
For example, if two strings bdcaba and abcbdab are input, and both bcba and bdab are their longest common subsequences, the output length is 4 and any subsequence is printed. 1. describe an LCS given a sequence X = {x1, x2, X3 ,..., xm} for I = 0, 1, 2 ,..., m, which defines the I-th prefix of X as xi = {x1, x2, X3 ,..., XI }. For example, if X = {A, B, C, B, D, a, B}, X4 = {A, B, C, B}, and x0 is an empty sequence. LCS optimal sub-structure x = {x1, x2 ,..., xm} and Y = {y1, Y2 ,..., yn} is two sequences with Z = {Z1, Z2 ,..., ZK} is any LCS of X and Y. 1). If XM = YN, then zk = XM = YN and Zk-1 is the LCS of Xm-1 and Yn-1.2). If XM! = YN, then if ZK! = XM => Z is the LCS of Xm-1 and y.3). If XM! = YN, then if ZK! = YN => Z is the LCS of Yn-1 and x.2. a recursive solution for C [I, j] is the length of Xi and yjthen | ---- 0 if I = 0 or J = 0C [I, j] = | ---- C [I-1, J-1] + 1, if I, j> 0 & xi = YJ | ---- max {C [I, J-1], C [I-1, J]}, if I, j> 0 & Xi! = Yj3. calculate the LCS Length

LCS-LENGTH(X, Y)m <- X.lengthn <- Y.lengthfor(i <- 1 to m)    do c[i, 0] <- 0for(j <- 0 to n)    do c[0, j] <- 0for i <- 1 to m    do for j <- 1 to n       if xi==yj          c[i,j]=c[i-1, j-1]+1       else if c[i-1, j] >= c[i, j-1]          c[i,j]=c[i-1, j]       else           c[i, j]=c[i, j-1]return c