Solve a Sudoku grid only using magic, recursion, and 140bytes of brute force.
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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 call 'a.toString()' 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*= // turn 'g' falsy if | |
a[j^i==j] // we are not on the cell 'i' | |
^m // and the cell 'j' is set to 'm' | |
|| // and we are in the same row|col|node as 'i' | |
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*=a[j^i==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*=a[j^i==j]^m||i/9^j/9&&i%9^j%9&&i/27^j/27|i%9/3^j%9/3}; | |
(function(f) | |
{ | |
var original=Array.prototype.toString; | |
Array.prototype.toString=function sudokuToString() | |
{ | |
if( (sudokuToString.caller||arguments.caller)==f ) | |
document.getElementById( "ret" ).textContent= original.call(this); | |
return original.call(this); | |
} | |
})( myFunction ); | |
myFunction( testGrid ); | |
</script> |
You people amaze me... +100 for the commented code!
Very nice.
ES5 indexOf could save us 1 byte:
function R(a,f,i,j,m,g){for(i=a.indexOf(0),~i||f(),m=10;g=a[i]=--m;g&&R(a,f))for(j in a)g*=a[j^i==j]^m||i/9^j/9&&i%9^j%9&&i/27^j/27|i%9/3^j%9/3}
@p01 Out of curiosity (related to a side project I'm working on) - what would happen if the grid provided had no solution?
ie: If it fell out-with of the 6,670,903,752,021,072,936,960 possible permutations
Would it be easy enough to return false, or something to that effect?
The number of actual calculated permutations is already reduced by the number of zeros in the input array; anyway, can a solution not be found, it would enter an infinite loop as at least one zero in the field cannot be cleared up.
I must say I find the closure thing wonky. Calling myFunction(a,f);
which in turn calls f
with no arguments just feel strange to me.
So I was thinking why not simply use the .toString
method of the array to output the grid, and ... Boom! We have a winner!
140 bytes
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*=a[j^i==j]^m||i/9^j/9&&i%9^j%9&&i/27^j/27|i%9/3^j%9/3}
Sleepy again? Why +a? Without checking further, either you meant ++a or the ||+a is completely useless...
@qfox +a is just invoking a.toString(), not intended as an increment.
the example doesn't work anymore, unfortunately.
@cjcliffe I understand, but it's in the third part of the for statement, who's return-value is discarded. Hence, useless, because the toString has no notable side effects. At least none in this code.
Please provide a working example.
@qfox ah, how I was reading I assumed it would just set a
to a.toString()
when i
reached 0 (when the || condition would trigger it)
Sorry for the broken example guys. Fixed now.
Yes +a
does invoke .toString()
here when i
is equal to 0 and the cell #0 was not empty.
hey @p01, can you figure out why it works in firefox and not in chrome? if indeed you golfed it down to a reliably working 140, that would be HUGE.
It does work in Opera and Firefox.
I'll try to figure what is going with the other browsers... if my baby girl "let" me, Priorities.
Meanwhile any help is appreciated.
Chrome/V8 is somewhat critical on having his prototypes overwritten. Perhaps we can do with an object looking like an array and having a toString property? like in
testSudoku=[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];
for(var sudoku={toString:function(){ document.getElementById( "ret" ).textContent=... }},s=0,t=testSudoku.length;s<t;s++)sudoku[s]=testSudoku[s];
Here's another (maybe a little less dirty) approach that raises an error in order to stop the recursion. Hence the invocation has to be wrapped inside a try/catch block: https://gist.github.com/1247640
Nice moves!
@fgnass: Clever (ab)use of a ReferenceError. Before even using a callback method I tried a similar but more expensive approach with throw a
Bravo sir, bravo.
I want to be like you when I grow up.
You make me happy.
This is awesome.
Just a notification: as I wrote on on Reddit the current version of the code is buggy. (See the linked post for details and a solution.)
Finally someone managed to reproduce this glitch. I tried the exact fix you suggested a few days ago: It works, unfortunately it is one byte bigger ( not smaller ), which bring us back to 141 bytes. One byte too many for 140byt.es
But I'm sure we can golf a valid version down to 140 bytes.
Oh, I only see now that you're adding sugar for the toString
"callback" in the test. I hadn't seen that before, just read the comments. That's why I was confused by the ||+a
part. If you modify toString
, the rules obviously change.
So I can see what you mean, but do you think this would be possible or acceptable in a js1k? I'd think the (temporary) toString
modification is boiler plate, but still required for the solution. Guess what I'm asking is, do you really think this is a proper 140b entry, or should the toString modification actually kind of be part of it. Especially when golfing, I think it matters. And I think it does. Just my two cents... :)
The toString
sugar would definitely count in a JS1K: Every thing counts in JS1K.
However things seem a more lax at 140byt.es. See the base64 entries for instance where the whole list of characters is passed as argument ( a whooping 64+ bytes for free ).
That being said, if people are OK with encapsulating the call in a try/catch
, then we can settle this at 140 bytes with a mix of fgnass ReferenceError trick and maksverver decomposition of the test i!=j && a[j]!=m
However things seem a more lax at 140byt.es. See the base64 entries for instance
Fair enough :)
I agree, though I'd rather find a way to shorten up the calculation for 1 bit...
My head just explo***BOOOOOOM***
this is just incredible.