Printing Longest Common Subsequence

Question

the problem i am facing in the longest subsequence: For Ex:

“ABCDGH” and “AEDFHR” is “ADH” of length 3

Code:

void lcs( char *X, char *Y, int m, int n )
{
   int L[m+1][n+1];

   /* Following steps build L[m+1][n+1] in bottom up fashion. Note
      that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */
   for (int i=0; i<=m; i++)
   {
     for (int j=0; j<=n; j++)
     {
       if (i == 0 || j == 0)
         L[i][j] = 0;
       else if (X[i-1] == Y[j-1])
         L[i][j] = L[i-1][j-1] + 1;
       else
         L[i][j] = max(L[i-1][j], L[i][j-1]);
     }
   }

I don't understand why this line of code:

L[i][j] = max(L[i-1][j], L[i][j-1]);

What if i want to print the sequence ie ADH how can i do that?

Answer 1

I don't understand why this line of code:

 L[i][j] = max(L[i-1][j], L[i][j-1]); 

If the last character of X[0..i-1] does not match the last character of Y[0..j-1] , then for every common subsequence at least one of these characters does not belong. The answer thus is given by the maximum length for X[0..i-2] and Y[0..j-1] , or the maximum length for X[0..i-1] and Y[0..j-2] .

To recover the actual subsequence, we have to trace back these decisions like so.

char lcs[min(m, n) + 1];
char *end = lcs;
int i = m;
int j = n;
while (i > 0 && j > 0) {
    if (X[i - 1] == Y[j - 1]) {
        *end = X[i - 1];
        end++;
        i--;
        j--;
    } else if (L[i - 1][j] >= L[i][j - 1]) {
        i--;
    } else {
        j--;
    }
}
reverse(lcs, end);
*end = '\0';
puts(lcs);