Thursday, July 13, 2006

shortest sudoku solver

bearophile sent me a link to this really cool program:

def r(a):i=a.find('0');~i or exit(a);[m
in[(i-j)%9*(i/9^j/9)*(i/27^j/27|i%9/3^j%9/3)or a[j]for
j in range(81)]or r(a[:i]+m+a[i+1:])for m in'%d'%5**18]
from sys import*;r(argv[1])

apparently it's a sudoku solver (couldn't you guess?). here's more information about it:

unfortunately, it mixes integers and strings ((i-j).. or a[j]) and arbitrary-size arithmetic (5**18), both of which are currently not supported by shed skin. the arbitrary-size arithmetic could be supported in this case, but I rewrote it for now. here's a version that works with shed skin CVS:

def r(a):
if not ~i: print a; exit()
[m in [a[j] for j in range(81) if not (i-j)%9*(i/9^j/9)*(i/27^j/27|i%9/3^j%9/3)] or r(a[:i]+m+a[i+1:]) for m in '3814697265625']
from sys import *;r(argv[1])

it's 211 characters, versus the 178 of the one above, so that's not too bad :) and it becomes 18 times faster here.


matthanlon said...

I believe the expression
is just a clever way to produce a string containing all the digits 1 through 9. You could replace it in your version with
and save a few characters with the same results.

srepmub said...

lol. it looks like they save one character that way?? :D thanks for letting me know.

srepmub said...

this looks nicer, imo:


and it would be more straightforward for shed skin to 'constant fold' the large number away..

Daivd said...

Why the silly '%d'% ? Backquotes are two characters less and look nicer. `5**18` Or am I missing something?

Daivd said...

Aha.. my suggestion will only work in a 64-bit environment. For those unfortunate 32-bitters, `5**18` prints an L at the end.

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