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### Re^4: Longest Common SubSequence Not Working Correctly

 on Nov 14, 2007 at 14:58 UTC ( #650763=note: print w/replies, xml ) Need Help??

The lcsbruteforce algorithm maintains this big table of solutions to subproblems. In this example, it's maintaining just the length of the LCS. Just change it to maintain the actual substring itself:
```sub lcsbruteforce {
my(\$x, \$y) = @_;
my(@v, \$cx, \$cy, \$left, \$above);
for my \$xi (0 .. length(\$x) - 1) {
\$cx = substr \$x, \$xi, 1;
for my \$yi (0 .. length(\$y) - 1) {
\$cy = substr \$y, \$yi, 1;
if (\$cx eq \$cy) {
#        \$v[\$xi][\$yi] = 1 + ((\$xi && \$yi) ? \$v[\$xi - 1][\$yi - 1] : 0);
\$v[\$xi][\$yi] = (\$xi && \$yi ? \$v[\$xi-1][\$yi-1] : "") . \$cx;
} else {
#        \$left = (\$xi && \$v[\$xi - 1][\$yi]) || 0;
#        \$above = (\$xi && \$v[\$xi][\$yi - 1]) || 0;
#        \$v[\$xi][\$yi] = (\$left > \$above) ? \$left : \$above;
\$left = (\$xi && \$v[\$xi - 1][\$yi]) || "";
\$above = (\$xi && \$v[\$xi][\$yi - 1]) || "";
\$v[\$xi][\$yi] = length(\$left) > length(\$above) ? \$left : \$above
+;

}
}
}
return \$v[length(\$x) - 1][length(\$y) - 1];
}
To change it to an actual brute force algorithm? That would be pretty strange. The brute force algorithm is:
```\$best = "";

for every subsequence \$s of \$x:
if \$s is also a subsequence of \$y:
\$best = \$s if length(\$s) > length(\$best);

return \$best;
Of course, the part where you get all subsequences and check for subsequence-ness is a pain. You can probably generate all subsequences using Algorithm::Loops, and perhaps use some regex stuff to check whether a string was a subsequence of another.

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