There are probably tons of these already available, but here is a quick suduko solver (in Perl):

#! /usr/bin/perl # Example: # # echo $' 2 \n 1 9 4 \n 2 1 5 9\n3 6 \n 68 41 5\n 427 8 \n 51 \n 7 3 \n79 5 '| perl suduko-solve.pl # use strict; use warnings; my $s = read_soduku(); my $res= solve($s); if($res) { print_soduku($res); print "Got it!\n"; } else { print "Failed :-(\n"; } exit 0; sub solve { my ($s)= @_; my $res= try_solve($s); return $s if $res eq 'SOLVED'; return undef if $res eq 'FAIL'; # Make a guess, backtracking if we were wrong. # Try to find some field where there are only two possibilities. my ($a, $b); OUTER: for my $i (0..8) { INNER: for my $j (0..8) { next INNER if keys(%{$s->[$i][$j]}) == 1; if(keys(%{$s->[$i][$j]}) == 2) { ($a,$b)= ($i,$j); last OUTER; } elsif(!defined($a)) { ($a,$b)= ($i,$j); } } } die "Internal?!?" unless defined($a); for my $choice (keys %{$s->[$a][$b]}) { my $s_copy = [ map { [ map { { %$_ } } @$_ ] } @$s ]; $s_copy->[$a][$b] = { $choice => 1 }; my $res= solve($s_copy); return $res if defined($res); # Got it! } return undef; # Failed. } sub read_soduku { my $s= [ ]; for(my $i = 0; $i < 9; $i++) { my $x = <STDIN>; chomp($x); if(length($x) < 9) { print STDERR "Short line: '$x'\n"; redo; } for(my $j = 0; $j < 9; $j++) { my $entry= substr($x, $j, 1); $s->[$i][$j] = { map(($_ => 1), ($entry eq ' ' ? (1..9) : ($entry))) }; } } return $s; } sub print_soduku { my ($s) = @_; for(my $i = 0; $i < 9; $i++) { print "---------------------\n" unless $i % 3; for(my $j= 0; $j < 9; $j++) { print "|" unless $j % 3; print((keys(%{$s->[$i][$j]}) > 1 ? ' ' : keys(%{$s->[$i][$j]})), ($j == 8 ? "|\n" : " ")); } } print "---------------------\n"; } sub try_solve { my ($s)= @_; my $done; my $progress; do { $done = 1; # Set false when non-determined field found $progress= undef; for(my $i = 0; $i < 9; $i++) { for(my $j = 0; $j < 9; $j++) { my $x = $s->[$i][$j]; return 'FAIL' if keys(%$x) == 0; $done = undef if keys(%$x) > 1; my $h1= { %$x }; my $h2= { %$x }; my $h3= { %$x }; for(my $a = 0; $a < 9; $a++) { if($a != $i) { my $y = $s->[$a][$j]; delete $h1->{$_} for keys(%$y); if(keys %$y == 1) { $progress = 1 if delete $x->{(keys(%$y))[0]}; } } if($a != $j) { my $y = $s->[$i][$a]; delete $h2->{$_} for keys(%$y); if(keys %$y == 1) { $progress = 1 if delete $x->{(keys(%$y))[0]}; } } my $b = 3*int($i/3) + int($a / 3); my $c = 3*int($j/3) + $a % 3; if($b != $i || $c != $j) { my $y = $s->[$b][$c]; delete $h3->{$_} for keys(%$y); if(keys %$y == 1) { $progress = 1 if delete $x->{(keys(%$y))[0]}; } } } return 'FAIL' if keys(%$h1) > 1 || keys(%$h2) > 1 || keys(%$h3) > 1; if(keys(%$h1) == 1) { delete($x->{$_}) for grep(!$h1->{$_}, keys %$x); } elsif(keys(%$h2) == 1) { delete($x->{$_}) for grep(!$h2->{$_}, keys %$x); } elsif(keys(%$h3) == 1) { delete($x->{$_}) for grep(!$h3->{$_}, keys %$x); } } } } while(!$done && $progress); return $done ? 'SOLVED' : 'UNSOLVED'; }

I wanted to use strategy rather than brute force (not sure what the complexity of that would be; the above seems to solve puzzles in a split second). Strategies are

- If a digit is already in a row/column/square, it cannot occur a second time.
- A digit is known to occur in a field if it cannot occur in any of the 8 other fields of a row/column/square.

These are not sufficient to solve (all) puzzles. Here is another example strategy: If we know in some 3×3 square that some particular digit can only occur in a particular row/column, then that digit can not occur in that digit/column in any other square.

However, at this point I just added a quick backtracking step. It seems the implemented strategies are sufficiently close to a full solution that only very little backtracking is needed, and a solution is usually arrived at instantly.

This example took perhaps an hour to put together. It is a nice example of the usefulness of learning Perl. Because Perl is well suited for a very wide range of programming tasks, the investment needed to learn the language pays off very well. Perl can be used for 2-minute one-liners like this to massage a .h file:

perl -nle 'print "#define prefix_$1 $1" if /^\s*[a-zA-z0-9_ ]+\s+([_a-zA-Z0-9]+)\s*\(/ ' *.h

It can be used for quick hacks like this suduko solver. And it can be used for full-scale application development.

If I were to use different languages for these different tasks, I would have less time to spend on learning each one, and thus would work less efficiently at *all* of them.

hmm interesting, I just finished my sudoku solver site sudoku solver using an open source, It does work and solves sudoku, but it’s not very visually appealing.

Would you mind if I played around with your sudoku solver code?

Feel free! I hereby release the source code for the sudoku solver under the GPL, either version 2 or, at your option, any later version of the GPL license.