A simple algorithm which uses a recursive function to solve the puzzle.
THE ALGORITHM
The credit for this algorithm must go to Richard Buckland: http://www.youtube.com/watch?v=bjObm0hxIYY&feature=autoplay&list=PL6B940F08B9773B9F&playnext=1
Takes a partially filled in grid, inserts the min value in a cell (could be a random cell, in this case the first free cell). If the min value is not legal it will increment until the max value is reached (number 9), checking each time if the incremented value is legal in that cell (ie does not clash with any already entered cells in square, col or row). If it is legal, it will call itself (the hasSolution function) thus using this slightly more filled in grid to find a new cell and check which value is legal in this next cell. If no values are legal in the next cell, it will clear the previous grid entry and try incrementing the value.
isLegal = does not conflict with any other numbers in the same row, column or square
Python, 119 lines
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import random
import os
# GLOBAL VARIABLES
grid_size = 81
def isFull (grid):
return grid.count('.') == 0
# can be used more purposefully
def getTrialCelli(grid):
for i in range(grid_size):
if grid[i] == '.':
print 'trial cell', i
return i
def isLegal(trialVal, trialCelli, grid):
cols = 0
for eachSq in range(9):
trialSq = [ x+cols for x in range(3) ] + [ x+9+cols for x in range(3) ] + [ x+18+cols for x in range(3) ]
cols +=3
if cols in [9, 36]:
cols +=18
if trialCelli in trialSq:
for i in trialSq:
if grid[i] != '.':
if trialVal == int(grid[i]):
print 'SQU',
return False
for eachRow in range(9):
trialRow = [ x+(9*eachRow) for x in range (9) ]
if trialCelli in trialRow:
for i in trialRow:
if grid[i] != '.':
if trialVal == int(grid[i]):
print 'ROW',
return False
for eachCol in range(9):
trialCol = [ (9*x)+eachCol for x in range (9) ]
if trialCelli in trialCol:
for i in trialCol:
if grid[i] != '.':
if trialVal == int(grid[i]):
print 'COL',
return False
print 'is legal', 'cell',trialCelli, 'set to ', trialVal
return True
def setCell(trialVal, trialCelli, grid):
grid[trialCelli] = trialVal
return grid
def clearCell( trialCelli, grid ):
grid[trialCelli] = '.'
print 'clear cell', trialCelli
return grid
def hasSolution (grid):
if isFull(grid):
print '\nSOLVED'
return True
else:
trialCelli = getTrialCelli(grid)
trialVal = 1
solution_found = False
while ( solution_found != True) and (trialVal < 10):
print 'trial valu',trialVal,
if isLegal(trialVal, trialCelli, grid):
grid = setCell(trialVal, trialCelli, grid)
if hasSolution (grid) == True:
solution_found = True
return True
else:
clearCell( trialCelli, grid )
print '++'
trialVal += 1
return solution_found
def main ():
#sampleGrid = ['2', '1', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '3', '1', '.', '.', '.', '.', '9', '4', '.', '.', '.', '.', '7', '8', '2', '5', '.', '.', '4', '.', '.', '.', '.', '.', '.', '6', '.', '.', '.', '.', '.', '1', '.', '.', '.', '.', '8', '2', '.', '.', '.', '7', '.', '.', '9', '.', '.', '.', '.', '.', '.', '.', '.', '3', '1', '.', '4', '.', '.', '.', '.', '.', '.', '.', '3', '8', '.']
#sampleGrid = ['.', '.', '3', '.', '2', '.', '6', '.', '.', '9', '.', '.', '3', '.', '5', '.', '.', '1', '.', '.', '1', '8', '.', '6', '4', '.', '.', '.', '.', '8', '1', '.', '2', '9', '.', '.', '7', '.', '.', '.', '.', '.', '.', '.', '8', '.', '.', '6', '7', '.', '8', '2', '.', '.', '.', '.', '2', '6', '.', '9', '5', '.', '.', '8', '.', '.', '2', '.', '3', '.', '.', '9', '.', '.', '5', '.', '1', '.', '3', '.', '.']
sampleGrid = ['.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '4', '6', '2', '9', '5', '1', '8', '1', '9', '6', '3', '5', '8', '2', '7', '4', '4', '7', '3', '8', '9', '2', '6', '5', '1', '6', '8', '.', '.', '3', '1', '.', '4', '.', '.', '.', '.', '.', '.', '.', '3', '8', '.']
printGrid(sampleGrid, 0)
if hasSolution (sampleGrid):
printGrid(sampleGrid, 0)
else: print 'NO SOLUTION'
if __name__ == "__main__":
main()
def printGrid (grid, add_zeros):
i = 0
for val in grid:
if add_zeros == 1:
if int(val) < 10:
print '0'+str(val),
else:
print val,
else:
print val,
i +=1
if i in [ (x*9)+3 for x in range(81)] +[ (x*9)+6 for x in range(81)] +[ (x*9)+9 for x in range(81)] :
print '|',
if add_zeros == 1:
if i in [ 27, 54, 81]:
print '\n---------+----------+----------+'
elif i in [ (x*9) for x in range(81)]:
print '\n'
else:
if i in [ 27, 54, 81]:
print '\n------+-------+-------+'
elif i in [ (x*9) for x in range(81)]:
print '\n'
|
Any suggestions to speed it up?
Good job this wasn't written in LISP as reading the title might be quite a challenge ;-)
I think you can use a shorter form of
isFullIn
hasSolutionI would return immediately theTrueresult (I'd moveprints somewhere else, they make code less readable)In this part:
I would use method
zfillfromstr. Don't reinvent the wheel ;-)You define the variable
grid_size, and then use it only once.point taken, updated.
what this program needs is a way of solving the obvious cells first. ie getTrialCell should return cells which can only be one possible value and then other random cells after that.
Any ideas?
You could check the first empty cell.
If it has >1 possible solution: Leave it empty and check the next empty cell. Elif it has 1 possible solution: Enter that solution and move to the next empty cell.
Repeat these checks through all empty cells over and over until you reach a point when no cells have only 1 possible solution.
At this point you have probably added definite values to quite a few cells (or maybe completed the whole grid).
Can you explain this code.
It seems you forgot parentheses. It won't let me run the code because of SyntaxError: Missing parentheses in call to 'print'. Could you please explain how to fix this?