# Shortest Common Supersequence via LCS in C

In my Shortest Common Supersequence post I gave a dynamic programming algorithm that computes the SCS in a manner very similar to the Longest Common Subsequence algorithm. Another way of computing the SCS is to find the LCS first, and then insert into it those characters from the two strings that are not in the LCS.

As an example, given the two strings “pAqBrCs”, and “wAxByCz”, their LCS is “ABC”, and we can form their SCS by inserting the remaining characters from each string into it in the correct order. One such SCS would be “pwAqxBryCsz”, but there are several possibilities because it doesn’t matter whether we insert the characters from the first string first or the second string at any point, as long as we add them in the order they occur in that string.

The algorithm just needs to scan the two strings, adding the characters not in the LCS while keeping track of where the LCS characters are, and making sure they are only added once. This is how it looks in C:

#include <string.h>
#include <stdlib.h>

char *shortest_common_supersequence_by_lcs(const char *str1, const char *str2)
{
char *lcs = longest_common_subsequence(str1, str2);
const size_t lcs_len = strlen(lcs);
const size_t scs_len = lcs_len + (strlen(str1) - lcs_len) + (strlen(str2) - lcs_len);
char *scs = malloc(scs_len + 1);
unsigned int lcs_pos = 0, str1_pos = 0, str2_pos = 0, scs_pos = 0;

/* Add the characters up to the last LCS character */
while (lcs_pos < lcs_len) {
while (str1[str1_pos] != lcs[lcs_pos]) {
scs[scs_pos++] = str1[str1_pos++];
}
while (str2[str2_pos] != lcs[lcs_pos]) {
scs[scs_pos++] = str2[str2_pos++];
}
scs[scs_pos++] = lcs[lcs_pos++];
str1_pos++;
str2_pos++;
}
/* Add the characters after the LCS */
while (str1[str1_pos]) {
scs[scs_pos++] = str1[str1_pos++];
}
while (str2[str2_pos]) {
scs[scs_pos++] = str2[str2_pos++];
}
scs[scs_pos] = '\0';
free(lcs);
return scs;
}


An example program:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
char a[] = "pAqBrCs";
char b[] = "wAxByCz";
char *scs = shortest_common_supersequence_by_lcs(a, b);
printf("Shortest Common Supersequence is %s\n", scs);
free(scs);
return 0;
}


Output:

Shortest Common Supersequence is pwAqxBryCsz


# Shortest Common Supersequence using dynamic programming in C

The Shortest Common Supersequence (SCS) problem is to find the shortest sequence that is a supersequence of two or more sequences. It is closely related to the Longest Common Subsequence problem, in that the SCS of two strings is precisely their LCS, with those letters from each that are not in the LCS spliced into the string in their proper place relative to the SCS letters. For example, given the strings “pAqBrCs” and “wAxByCz”, their LCS is “ABC”, and an SCS is “wpAxqByrCzs”. Note that the order in which the letters occurring between the LCS letters is chosen from the two strings doesn’t matter, as long as the letters from each string are in the same order as they occur in it, so “pwAqxByrCsz” is also an SCS, and in general the SCS is not unique.

One way of computing an SCS is to use a dynamic programming algorithm similar to the Longest Common Subsequence one, but in which the letters occurring between the LCS letters are added to the solution. This is what I have implemented in C here. Rather than using pointers in the table to retrieve the solution as I have done in other DP algorithms, this time I have used enumerated values that indicate in which direction the next cell in the sequence is to be found – DOWN, RIGHT, or DIAG (diagonally down and to the right).

typedef enum  {
DOWN,
RIGHT,
DIAG,
STOP
} direction;

typedef struct {
size_t len;
char letter;
direction dir;
} cell;

char *shortest_common_supersequence(const char *str1, const char *str2)
{
size_t lstr1 = strlen(str1);
size_t lstr2 = strlen(str2);
cell **matrix;
int i;
int j;
char *result;
unsigned int r;

matrix = malloc((lstr2 + 1) * sizeof(cell*));
for (i = 0; i <= lstr2; i++) {
matrix[i] = malloc((lstr1 + 1) * sizeof(cell));
}

for (i = 0; i < lstr2; i++) {
matrix[i][lstr1].len = lstr2 - i;
matrix[i][lstr1].letter = str2[i];
matrix[i][lstr1].dir = DOWN;
}

for (j = 0; j < lstr1; j++) {
matrix[lstr2][j].len = lstr1 - j;
matrix[lstr2][j].letter = str1[j];
matrix[lstr2][j].dir = RIGHT;
}

matrix[lstr2][lstr1].len = 0;
matrix[lstr2][lstr1].letter = 0;
matrix[lstr2][lstr1].dir = STOP;

for (i = lstr2 - 1; i >= 0; i--) {
for (j = lstr1 - 1; j >= 0; j--) {
cell *pcell = &matrix[i][j];
if (str2[i] == str1[j]) {
pcell->dir = DIAG;
pcell->letter = str1[j];
pcell->len = matrix[i + 1][j + 1].len + 1;
}
else if (matrix[i][j + 1].len < matrix[i + 1][j].len) {
pcell->dir = RIGHT;
pcell->letter = str1[j];
pcell->len = matrix[i][j + 1].len + 1;
}
else {
pcell->dir = DOWN;
pcell->letter = str2[i];
pcell->len = matrix[i + 1][j].len + 1;
}
}
}

result = malloc(matrix[0][0].len + 1);
i = 0;
j = 0;
r = 0;
while (i <= lstr2 && j <= lstr1) {
result[r] = matrix[i][j].letter;
r++;
if (matrix[i][j].dir == DOWN) {
i += 1;
}
else if (matrix[i][j].dir == RIGHT) {
j += 1;
}
else {
i += 1;
j += 1;
}
}

for (i = 0; i <= lstr2; i++) {
free(matrix[i]);
}
free(matrix);

return result;
}


Example program:

int main(void)
{
char a[] = "pAqBrCs";
char b[] = "wAxByCz";
char *scs = shortest_common_supersequence(a, b);
printf("%s\n", scs);
free(scs);
return 0;
}


Output:

wpAxqByrCzs