# Quest for most stupid Sudoku solver

Updated 0 Tallied Votes 838 Views

I had posted here already here to forum a beautiful code for solving Sudoku puzzles, so I was in kind of dilemma what to do, as I read the task to do a sudoku solve as one task in Project Euler.

To use the code from before and just take the correct corner numbers or really do something from cratch by myself?

I decided to do both. So I took the right answer from the super fine solver and entered the forum to see what kind of solutions the early posters had done for the task. I took as bench mark one Python code, which ran for around 25 seconds for the 50 Sudokus.

So I desided to take a filosophical stance and prove how stupid the sudoku solver can come and still solve all puzzles and reasonably give up for impossible ones.

My code is using only one method, no precomputed neighbour coordinates or localized propagation of values to keep as different as possible from the other solution.

``````Algorithm "Find the nail for the hammer"
Take alternative from stack (initialy the start position)
Stop if found solution.
Find the minimum length set of undesided values (typically two alternatives)
Assign the values possible to this place one by one and keep the alternatives which did not lead to immediate conflict
Push the correct alternatives to list used as stack.
``````

So the idea was to be as different from original advanced Sudoku solver as possible, but still I wanted to solve the Sudokus reasonable speed vs the 25 s bench mark solution (the advanced solution with double recursive constraint propagation takes all of 0.02 s per puzzle ie 1 s for 50).

Here is my code. One interesting point of this solution is that it solves the very hard random puzzles of Norvig also and that this program is really much faster with the Python 3.2rc2 thant with Python 2 including 2.6.6 with psyco module.

Finally the solution is reasonable fast, little under 8 seconds in my old computer with Python 3.2rc2, 12+ s with Python 2.6, one second slower than Python 3.2rc2 with Python 2.7.1.

If you want to speed it up little you can change the definition of possibilities to do inline the 2d array index calculation:

``````def possibilities(s):
s = [value if len(value)==1 else '.' for value in s]
return itertools.takewhile(lambda x: x, (''.join(sorted(remaining_neighbours(s, i, j)))
if s[i * 9 + j] in '.0' or len(s[i * 9 + j]) > 1 else s[i * 9 + j]
for i in range(9) for j in range(9)))
``````
``````from __future__ import print_function
import time
import itertools
import sys
import pretty

numbers, empty = set('123456789'), set('0.')

# without izip_longest to be python3 compatible. builtin zip is faster
def grouper(n, iterable): return zip(*((iter(iterable),) * n))

def sudoku_minisquare(s,x,y):
start = x * 3 * 9 + y * 3 ## starting corner
return (s[ss:ss+3] for ss in range(start, start+3*9, 9))

def sudoku_index(x,y): return x*9+y

def correct(s):
if not s: return False
s = ''.join(c for part in s for c in part if part and c in numbers)
ss = set(s)
return len(s) == len(ss) and ss.issubset(numbers)

def remaining(s): return numbers - set(c for item in s for c in item) - empty

def remaining_neighbours(s, i, j):
i9 = 9*i
return (remaining(sudoku_minisquare(s, i//3, j//3)) &
remaining(s[i9:i9+9])  &
remaining(s[j::9]))
def check(s):
if len(s) != 81 : return
s = tuple(c if len(c) == 1 else '.' for c in s)
value = (all(correct(sudoku_minisquare(s,i,j)) for i in range(3) for j in range(3))
and all(correct(gr) for gr in itertools.chain(grouper(9, s), (s[start::9] for start in range(9)) )))
return value

def possibilities(s):
s = [value if len(value)==1 else '.' for value in s]
return itertools.takewhile(lambda x: x, (''.join(sorted(remaining_neighbours(s, i, j)))
if s[sudoku_index(i,j)] in '.0' or len(s[sudoku_index(i,j)]) > 1 else s[sudoku_index(i,j)]
for i in range(9) for j in range(9)))

def solve(sudoku_list):
sol = [list(possibilities(sudoku_list))]
while sol:
base = sol.pop()
if all(len(item) == 1 for item in base):
return map(int, base)
shortest = tuple((ind,val) for ind, val in enumerate(base) if len(val)>1)
if shortest:
ind, prod = min(shortest, key = lambda x: len(x))
for v in prod:
under_test, under_test[ind] = base[:], v
under_test = list(possibilities(under_test))
if check(under_test): sol.append(under_test)

def solve_from_file(fn):
n, puzzle = 1, ''
print ('Puzzle set', fn.split('.') )
for sudoku_line in open(fn):
if not sudoku_line.startswith(tuple('G=#')): puzzle += sudoku_line.strip()
if len(puzzle) == 81:
print(n, end=',')
sys.stdout.flush()
yield tuple(grouper(9, solve(possibilities(puzzle))))
puzzle, n = '', n+1

if __name__=='__main__':
t0, count, fn = time.clock(), 0, 'sudoku.txt'
result, t0 = tuple(solve_from_file(fn)), time.clock()- t0
pretty.printer(result, indent_level = 10)
print('\nSolving time {t} s for {solved} puzzles.'.format(t=t0, solved=len(result)))``````
TrustyTony 888

I started to get little regret of my simpleton child doing such a hard work in my normal PPS (Post Posting Syndrome:icon_cheesygrin:)

So here is some doubled checked removed, and sudoku minisquare function also inlined in the two places where it used (pity for the repetition of code though, but that is life with interpreted language where function calls cost too much) and first extra check of possibilities removed.

Also I had removed the ability to give up when get too many failures. It is controlled by 'constant' FAIL_LIMIT. I have not found solvable sudoku harder than the limit 3000.

``````from __future__ import print_function
import time
import itertools
import sys
import pretty

numbers=set('123456789')
FAIL_LIMIT = 3000
# without izip_longest to be python3 compatible. builtin zip is faster
def grouper(n, iterable): return zip(*((iter(iterable),) * n)) if iterable else ()

def correct(s):
if not s: return False
s = ''.join(c for part in s for c in part if part and c in numbers)
ss = set(s)
return len(s) == len(ss)

def remaining(s): return numbers - set(c for item in s for c in item) # if c in numbers)# does not need as s is checked

def remaining_neighbours(s, i, j):
start =  i//3 * 27 + j//3 * 3 # start corner of minisquare where i,j belongs
return (remaining(s[ss:ss+3] for ss in range(start, start + 27, 9)) &
remaining(s[9*i:9*i+9])  &
remaining(s[j::9]))

def check(s):
if len(s) == 81 :
s = tuple(c if len(c) == 1 else '.' for c in s)
value = (all(correct(s[ss:ss+3] for ss in range(start, start + 27, 9)) for start in range(0,81,27))
and all(correct(gr) for gr in itertools.chain(grouper(9, s), (s[start::9] for start in range(9)) )))
return value

def possibilities(s):
s = [value if len(value)==1 else '.' for value in s]
return itertools.takewhile(lambda x: x, (''.join(sorted(remaining_neighbours(s, i, j)))
if s[i * 9 + j] in '.0' or len(s[i * 9 + j]) > 1 else s[i * 9 + j]
for i in range(9) for j in range(9)))

def solve(sudoku_list):
sol, fails = [list(possibilities(sudoku_list))], 0
while sol and fails < FAIL_LIMIT:
base = sol.pop()
if all(len(item) == 1 for item in base):
return map(int, base)
shortest = tuple((ind,val) for ind, val in enumerate(base) if len(val)>1)
if shortest:
ind, prod = min(shortest, key = lambda x: len(x))
for v in prod:
under_test, under_test[ind] = base[:], v
under_test = list(possibilities(under_test))
if check(under_test): sol.append(under_test)
else: fails += 1
return ()

def solve_from_file(fn):
n, puzzle = 1, ''
print ('Puzzle set', fn.split('.') )
for sudoku_line in open(fn):
if not sudoku_line.startswith(tuple('G=#')): puzzle += sudoku_line.strip()
if len(puzzle) == 81:
print(n, end=',')
sys.stdout.flush()
yield tuple(grouper(9, solve(puzzle)))
puzzle, n = '', n+1

if __name__=='__main__':
t0, count, fn = time.clock(), 0, 'hardest.txt'
result, t0 = tuple(solve_from_file(fn)), time.clock()- t0
pretty.printer(result, indent_level = 10)
print('\nSolving time {t:0.2f} s for {solved} out of {count} puzzles.'.format(t=t0, solved = sum(1 for r in result if r), count = len(result)))``````

BTW the problem solves the sudoku '.....6....59.....82....8....45........3........6..3.54...325..6..................' in 0.13 s which was very hard for the recurvive search and propagation version. http://norvig.com/sudoku.html

TrustyTony 888

Sorry about bombing you on more versions, this I finally deem as version 1.0. Got rid of some last 2d calculations and added fancy histogram. Timing improved to little less than 6.5 s for 50 easy puzzles (which are often more hard than hard ones for my simpistic approach. Imposible case give up time is around 6 s with limit of failures set as 2500.

impossible case from norwig:
# one unsolvable for testing
.....5.8....6.1.43..........1.5........1.6...3.......553.....61........4.........

``````""" Stupid Sudoku 1.0
As original as possible Sudoku solver for Project Euler solution
Keep this comment if you use this code
Tony Veijalainen 2011, posted in Daniweb as user tonyjv

"""

from __future__ import print_function
import time
import itertools
import pretty

numbers = set('123456789')
FAIL_LIMIT = 2500

# without izip_longest to be python3 compatible (see the manual for itertools version)
def grouper(n, iterable): return zip(*((iter(iterable),) * n)) if iterable else ()

def correct(s):
s = [c for part in s for c in part if part and c in numbers]
ls, lss = len(s), len(set(s))
return ls == lss

def remaining(s): return numbers - set(c for item in s for c in item)

def remaining_neighbours(s, index):
# start corner of minisquare where index belongs:
# minisquare row (3*9 lines per minisquare row) and offset by column (integer)divided by 3 * 3
start =  (index // 27 * 27) + (index % 9 // 3)*3
return (remaining(s[ss:ss+3] for ss in range(start, start + 27, 9)) &
remaining(s[index - index % 9:index + 9 - index % 9])  &
remaining(s[index % 9::9]))

def check(s):
if len(s) == 81 :
s = tuple(c if len(c) == 1 else '.' for c in s)
return (all(correct(s[ss:ss+3]
for ss in range(start, start + 27, 9))
for start in range(0, 81, 27)) and
all(correct(gr)
for gr in itertools.chain(grouper(9, s), (s[start::9] for start in range(9)))))

def possibilities(s):
s = [value if len(value)==1 else '.' for value in s]
return itertools.takewhile(lambda x: x, (''.join(sorted(remaining_neighbours(s, index)))
if s[index] in '.0' or len(s[index]) > 1 else s[index]
for index in range(81)))

def solve(sudoku_to_solve, histogram_scale = 50):
t0, sol, fails = time.clock(), [list(possibilities(c for c in sudoku_to_solve if c in numbers or c in '0.'))], 0

while sol and fails < FAIL_LIMIT:
base = sol.pop()

if all(len(item) == 1 for item in base):
t0 = 1000*(time.clock() - t0)
print('Success with {0:4} fails, time {1:-7.2f} ms:\t{histogram:50}'.format(fails, t0,
histogram = int(t0/histogram_scale)*'*' or '|'))
return map(int, base)

shortest = tuple((ind,val) for ind, val in enumerate(base) if len(val)>1)
if shortest:
ind, prod = min(shortest, key = lambda x: len(x))
for v in prod:
under_test, under_test[ind] = base[:], v
under_test = list(possibilities(under_test))
if check(under_test):
sol.append(under_test)
else:
fails += 1
return ()

def solve_from_file(fn):
n, puzzle = 1, ''
print ('Puzzle set', fn.split('.') ,'\n')
for sudoku_line in open(fn):
if sudoku_line.strip() and not sudoku_line.startswith(tuple('G=#')):
puzzle += sudoku_line.strip()
if len(puzzle) == 81:
print('Sudoku {0:2}'.format(n), end=': ')
yield tuple(grouper(9, solve(puzzle)))
puzzle, n = '', n+1

if __name__=='__main__':
t0, count, fn = time.clock(), 0, 'sudoku.txt'
result, t0 = tuple(solve_from_file(fn)), time.clock()- t0
pretty.printer(result, indent_level = 10)
print('\nSolving time {t:0.2f} s for {solved} out of {count} puzzles.'.format(t=t0,
solved = sum(1 for r in result if r), count = len(result)))``````
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