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Created January 8, 2012 09:03
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Sudoku Solver in 140bytes

Sudoku Solver in 140 bytes

Solve a Sudoku grid only using magic, recursion, and 140bytes of brute force.

This entry was inspired and built on top of the itsy bitsy shoulders of the smallest sudoku solvers in Perl (120b), Ruby (122b), Python (178b), ...

Credits go this way:

166b @p01 .......... initial implementation
147b @p01 .......... initial golf
146b @qfox ......... loop optimization
145b @qfox ......... output with closured callback
140b @p01 .......... output with hijacked Array.prototype.toString()
141b @maksverver ... fixed the glitchy j^i==j test
140b @fgnass ....... ReferenceError exit trick + cross browser fix

Thanks to everyone who helped golf and fix this puppy.

// /!\ WARNING this function assumes that there is no variable 'A' in the global scope
function R
(
a, // the array representing the sudoko grid
// placeholder arguments
i, // index of the last empty cell
j, // index to check the candidates for the cell 'i'
m, // candidate number for the the cell 'i'
g // flag whether 'm' is a already used in the same row|col|node as 'i'
){
// phase 1: look for an empty cell
for
(
i=80;
a[i]; // keep going if the cell isn't empty
i--||A // decrease the index and throw a ReferenceError exception if we went through the whole grid
);
// phase 2: check all candidate numbers for the cell 'i'
for
(
m=10;
g=a[i]=--m; // put the candidate in the cell 'i' already and set 'g' to something truthy
// at the end of phase 2, the cell 'i' is reset to 0 for "higher" branches of the recursion
g&&R(a) // recurse if 'm' isn't already used in the same row|col|node as 'i'
)
// phase 3: check if the candidate number is used already
for(j in a) // loop through the whole grid again
g*= //
j==i // keep 'g' as is if we are on the cell 'i'
|| // otherwise, turn 'g' falsy if
a[j]^m // the cell 'j' is set to 'm'
|| // and 'i' and 'j' are in the same row|col|node
i%9^j%9&&i/9^j/9&&i/27^j/27|i%9/3^j%9/3
}
function R(a,i,j,m,g){for(i=80;a[i];i--||A);for(m=10;g=a[i]=--m;g&&R(a))for(j in a)g*=j==i||a[j]^m||i%9^j%9&&i/9^j/9&&i/27^j/27|i%9/3^j%9/3}
DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE
Version 2, December 2004
Copyright (C) 2011 Mathieu 'p01' Henri - http://www.p01.org/releases/
Everyone is permitted to copy and distribute verbatim or modified
copies of this license document, and changing it is allowed as long
as the name is changed.
DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE
TERMS AND CONDITIONS FOR COPYING, DISTRIBUTION AND MODIFICATION
0. You just DO WHAT THE FUCK YOU WANT TO.
{
"name": "sudokuSolver",
"description": "Brute force sudoku solver.",
"keywords": [
"sudoku",
"solver",
"recursive",
"brute force"
]
}
<!doctype html>
<title>Sudoku Solver in 140bytes</title>
<div>Expected value: <b>4,2,8,1,5,9,6,7,3,1,9,6,3,7,4,8,2,5,3,7,5,8,6,2,9,4,1,9,8,1,4,2,3,5,6,7,5,6,4,7,1,8,3,9,2,7,3,2,5,9,6,1,8,4,2,4,3,6,8,1,7,5,9,6,1,7,9,4,5,2,3,8,8,5,9,2,3,7,4,1,6</b></div>
<div>Actual value: <b id="ret"></b></div>
<script>
var testGrid = [0,0,0,1,5,0,0,7,0,1,0,6,0,0,0,8,2,0,3,0,0,8,6,0,0,4,0,9,0,0,4,0,0,5,6,7,0,0,4,7,0,8,3,0,0,7,3,2,0,0,6,0,0,4,0,4,0,0,8,1,0,0,9,0,1,7,0,0,0,2,0,8,0,5,0,0,3,7,0,0,0];
var myFunction = function R(a,i,j,m,g){for(i=80;a[i];i--||A);for(m=10;g=a[i]=--m;g&&R(a))for(j in a)g*=j==i||a[j]^m||i%9^j%9&&i/9^j/9&&i/27^j/27|i%9/3^j%9/3};
var solutionFound = false;
try
{
myFunction( testGrid );
}
catch(e)
{
solutionFound = true;
}
document.getElementById('ret').innerHTML = testGrid;
</script>
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