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Re^3: Search for identical substrings

by BrowserUk (Pope)
on Aug 21, 2005 at 08:12 UTC ( #485504=note: print w/replies, xml ) Need Help??


in reply to Re^2: Search for identical substrings
in thread Search for identical substrings

Here's my Inline C implementation.

#! perl -slw use strict; #use Inline 'INFO'; use Inline C => 'DATA', NAME => 'LCS', CLEAN_AFTER_BUILD => 1; my( $len, $offset0, $offset1 ) = LCS( @ARGV ); $ARGV[ 0 ] =~ s[(.{$offset0})(.{$len})][$1<$2>]; $ARGV[ 1 ] =~ s[(.{$offset1})(.{$len})][$1<$2>]; print for @ARGV; __END__ [ 9:10:28.57] P:\test>DynLCS-C hello aloha hel<lo> a<lo>ha __C__ #define IDX( x, y ) (((y) * an)+(x)) /* LONGEST COMMON SUBSTRING(A,m,B,n) for i := 0 to m do Li,0 := 0 for j := 0 to n do L0,j := 0 len := 0 answer := <0,0> for i := 1 to m do for j := 1 to n do if Ai ? Bj then Li,j := 0 else Li,j := 1 + Li-1,j-1 if Li,j > len then len := Li,j answer = <i,j> */ void LCS ( char* a, char*b ) { Inline_Stack_Vars; int an = strlen( a ); int bn = strlen( b ); int*L; int len = 0; int answer[2] = { 0,0 }; int i, j; Newz( 42, L, an * bn, int ); for( i = 1; i < an; i++ ) { for( j = 1; j < bn; j++ ) { if( a[ i ] != b[ j ] ) { L[ IDX(i,j) ] = 0; } else { L[ IDX(i,j) ] = 1 + L[ IDX(i-1, j-1) ]; if( L[ IDX(i,j) ] > len ) { // xs(70) len = L[ IDX(i,j) ]; answer[ 0 ] = i; answer[ 1 ] = j; } } } } Safefree( L ); Inline_Stack_Reset; Inline_Stack_Push(sv_2mortal(newSViv( len ))); Inline_Stack_Push(sv_2mortal(newSViv( answer[ 0 ] - len + 1 ))); Inline_Stack_Push(sv_2mortal(newSViv( answer[ 1 ] - len + 1 ))); Inline_Stack_Done; }

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
Lingua non convalesco, consenesco et abolesco. -- Rule 1 has a caveat! -- Who broke the cabal?
"Science is about questioning the status quo. Questioning authority".
The "good enough" maybe good enough for the now, and perfection maybe unobtainable, but that should not preclude us from striving for perfection, when time, circumstance or desire allow.

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