nine-liner sudoku solver

s9.py

def complement(arr):
	return filter(lambda x:x not in arr,range(1,10))
def possibilities(a,arr):
	x,y=a%9,a/9
	return complement(
		map(lambda k:arr[k],
			range(x,81,9)+range(9*y,9*(y+1))+
			map(lambda z:z+(x-x%3)+(y-y%3)*9,[0,1,2,9,10,11,18,19,20])
			)
		)
def first(fn,arr):			#returns the first non empty value from applying every element of the list to fn
	return None if not arr else fn(arr[0]) or first(fn,arr[1:])
def recurr(arr):			#takes sudoku as a list of 81 integers with 0 indicationg an empty square
	return arr if 0 not in arr else \
	(lambda a:first(lambda x:recurr(arr[:a]+[x]+arr[a+1:]),possibilities(a,arr)))(arr.index(0))
print recurr(map(int,list("000600057600000000708100000005000104012405000000000000980001400026000009000390006")))