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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
 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119``` ```import re 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?

David Adler (author) 11 years, 11 months ago

Good job this wasn't written in LISP as reading the title might be quite a challenge ;-)

Patrycja Szabłowska 11 years, 10 months ago

I think you can use a shorter form of `isFull`

``````def isFull (grid):
return grid.count('.') == 0
``````

In `hasSolution` I would return immediately the `True` result (I'd move `print`s somewhere else, they make code less readable)

``````def hasSolution (grid):
if isFull(grid):
return True
``````

In this part:

``````if int(val) < 10:
print '0'+str(val),
``````

I would use method `zfill` from `str`. Don't reinvent the wheel ;-)

You define the variable `grid_size`, and then use it only once.

David Adler (author) 11 years, 10 months ago

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?

Kelly 11 years, 10 months ago

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).

prat 8 years, 4 months ago

Can you explain this code.

Braden Townsend 7 years, 8 months ago

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?

 Created by David Adler on Sat, 19 May 2012 (MIT)