Comments on: Java Sudoku Solver in 6 lines

By: Paul King

Paul King — Sat, 28 Oct 2006 20:55:54 +0000

3 line groovy version:
def r(a){def m=a=~’0′; if(m.find()){def i=m.start();((49..57)-(0..80).collect{
j->j/9==i/9||j%9==i%9||j/27==i/27&&j%9/3==i%9/3?a[j]:0}).collect{
r(a[0..i-1]+(char)it+a[i+1..-1])}}else print a}

By: Jeff

Jeff — Sun, 06 Aug 2006 16:47:59 +0000

Using Kofa’s idea of a boolean array I got it down to under 5 lines.

public class S{static char[] A;static void R(){int i=0,j;for(;i

By: Daniel

Daniel — Tue, 01 Aug 2006 10:31:47 +0000

Wow, still not sure how it does it. Need to get my head around it’s recursiveness (R() is called 1107(!) times for that example problem).

By: tiny but slow

tiny but slow — Wed, 26 Jul 2006 15:42:37 +0000

import java.util.*;public class S{static char[]A;static void R(){int i=0,j;for(;i

By: Kofa

Kofa — Wed, 26 Jul 2006 10:04:01 +0000

You can get smaller and faster by replacing the hash with a boolean array (‘x’ is for eXcluded chars). You need to (ab)use the fact that booleans are false by default.
public class S{static char[] A;static void R(){int i,j;for(i=0;i

By: Chris

Chris — Mon, 24 Jul 2006 23:43:59 +0000

… you can save four more bytes by using ‘Map h=new HashMap()’

By: Richard Rodger

Richard Rodger — Mon, 24 Jul 2006 12:11:05 +0000

And to think I managed to write solver in only 100 or so line of Perl. And it wasn’t even complete!

Bah.