**definition of the problem:**Sub sequence

X= (A, B, C, B, D, b) z= (b, C, D, b) is a subsequence example of x w= (b, D, A) is not a subsequence of X's subroutine common subsequence

Z is a common subsequence of sequence x and y if z is a subsequence of X and also a sub sequence of Y.

**Maximum common child sequence (LCS) problem**
Input: X = (x1, x2, ..., xn), Y = (y1, y2, ..., ym)

Output: Z = Longest common child sequence for x and y

**Brute Force method:** enumerates each of the subsequence Z of x, checks whether Z is a subsequence of y, T (n) =o (mx2n).

**Dynamic Planning:**

To optimize a child structure:

Set x = (x1, x2, ..., xn), y = (y1, y2, ..., ym) are two sequences, z= (z1, Z2, ..., Zn) are the X and Y LCS, we have: if xm = yn, then Zk = XM = yn, Zk-1 is the Xm-1 and Yn-1 LCS , that is, Lcsxy = lcsxm-1yn-1+ (xm = yn). If Xm≠yn, and ZK≠XM, then Z is the LCS for Xm-1 and y, that is, lcsxy= lcsxm-1y if Xm≠yn, and Zk≠yn, then Z is the LCS for X and Yn-1, that is lcsxy= LCSXYn-1

Recursive equation: c[i, J] = XI and YJ of LCS length of LCS

C[i, J] = 0, if i=0 or j=0 c[i, j] = C[i-1, j-1] + 1, if I, j>0 and XII = YJJ c[i, j] = Max (C[i, j-1), C[i-1, J]), if I, j>0 and XIIYJJ

Basic idea:

**in the case of known c[i-1, J-1], C[i, j-1], C[i-1, J], the C[i, J, is computed according to the recursive equation. **

Data structure: C[0:m,0:n]: c[i,j] is the length of the LCS of Xi and YJ B[1:m,1:n]: b[i,j] is a pointer to the table entry for the C table corresponding to the optimal solution of the selected child problem when calculating c[i,j]

Pseudo code:

Lcs-length (x, Y)
m = length (x);
n = Length (Y); for
i = 1 to M do
c[i,0] = 0;
For j = 1 to n do
c[0,j] = 0;
For i = 1 to M doing for
j = 1 to n do
If XI = YJ Then
c[i,j] = c[i-1,j-1] + 1;
b[i,j] = "↖";
Else If c[i-1,j]>=c[i,j-1] Then
c[i,j] = c[i-1,j];
B[I,J] = "↑";
Else C[i,j] = c[i,j-1];
B[I,J] = "←";
Return C and B.

**C + + code:**
#include <iostream> #include <vector> #include <string> using namespace std;
Vector<vector<int>> B, C;
void LCS (String str1, string str2, int len1, int len2) {b.resize (len1 + 1);
C.resize (len1 + 1);
for (int i = 0; i < len1 + 1; i++) {b[i].resize (len2 + 1, 0);
C[i].resize (len2 + 1, 0); for (int i = 1; I <= len1. i++) {for (int j = 1; J <= Len2; j +) {if str1[
I-1] = = Str2[j-1]) {c[i][j] = c[i-1][j-1] + 1;
B[I][J] = 0;
else if (C[i-1][j] >= c[i][j-1]) {c[i][j] = c[i-1][j];
B[I][J] = 1;
else {C[i][j] = c[i][j-1];
B[I][J] =-1; }} void Printlcs (Vector<vector<int>> B, string str1, int i, int j) {if (i = = 0 | | j
= = 0) Return
if (b[i][j] = = 0) {Printlcs (b, str1, i-1, j-1);
cout << str1[i-1] << "";
else if (b[i][j] = = 1) {Printlcs (b, str1, I-1, J);
else {Printlcs (b, str1, I, j-1);
int main () {string str1 = ' abcdef ';
String str2 = "ACEF";
int len1 = Str1.size ();
int len2 = Str2.size ();
B.resize (len1 + 1);
C.resize (len2 + 1);
LCS (STR1, STR2, Len1, len2);
cout << C[len1][len2] << Endl;
Printlcs (b, str1, LEN1, len2);
return 0; }