Algorithm for solving Sudoku

Question

I want to write a code in python to solve a sudoku puzzle. Do you guys have any idea about a good algorithm for this purpose. I read somewhere in net about a algorithm which solves it by filling the whole box with all possible numbers, then inserts known values into the corresponding boxes.From the row and coloumn of known values the known value is removed.If you guys know any better algorithm than this please help me to write one. Also I am confused that how i should read the known values from the user. It is really hard to enter the values one by one through console. Any easy way for this other than using gui?

Answer 1

Here is my sudoku solver in python. It uses simple backtracking algorithm to solve the puzzle. For simplicity no input validations or fancy output is done. It's the bare minimum code which solves the problem.

Algorithm

1. Find all legal values of a given cell
2. For each legal value, Go recursively and try to solve the grid

Solution

It takes 9X9 grid partially filled with numbers. A cell with value 0 indicates that it is not filled.

Code

def findNextCellToFill(grid, i, j):         for x in range(i,9):                 for y in range(j,9):                         if grid[x][y] == 0:                                 return x,y         for x in range(0,9):                 for y in range(0,9):                         if grid[x][y] == 0:                                 return x,y         return -1,-1  def isValid(grid, i, j, e):         rowOk = all([e != grid[i][x] for x in range(9)])         if rowOk:                 columnOk = all([e != grid[x][j] for x in range(9)])                 if columnOk:                         # finding the top left x,y co-ordinates of the section containing the i,j cell                         secTopX, secTopY = 3 *(i//3), 3 *(j//3) #floored quotient should be used here.                          for x in range(secTopX, secTopX+3):                                 for y in range(secTopY, secTopY+3):                                         if grid[x][y] == e:                                                 return False                         return True         return False  def solveSudoku(grid, i=0, j=0):         i,j = findNextCellToFill(grid, i, j)         if i == -1:                 return True         for e in range(1,10):                 if isValid(grid,i,j,e):                         grid[i][j] = e                         if solveSudoku(grid, i, j):                                 return True                         # Undo the current cell for backtracking                         grid[i][j] = 0         return False 

Testing the code

  >>> input = [[5,1,7,6,0,0,0,3,4],[2,8,9,0,0,4,0,0,0],[3,4,6,2,0,5,0,9,0],[6,0,2,0,0,0,0,1,0],[0,3,8,0,0,6,0,4,7],[0,0,0,0,0,0,0,0,0],[0,9,0,0,0,0,0,7,8],[7,0,3,4,0,0,5,6,0],[0,0,0,0,0,0,0,0,0]] >>> solveSudoku(input) True >>> input [[5, 1, 7, 6, 9, 8, 2, 3, 4], [2, 8, 9, 1, 3, 4, 7, 5, 6], [3, 4, 6, 2, 7, 5, 8, 9, 1], [6, 7, 2, 8, 4, 9, 3, 1, 5], [1, 3, 8, 5, 2, 6, 9, 4, 7], [9, 5, 4, 7, 1, 3, 6, 8, 2], [4, 9, 5, 3, 6, 2, 1, 7, 8], [7, 2, 3, 4, 8, 1, 5, 6, 9], [8, 6, 1, 9, 5, 7, 4, 2, 3]]  

The above one is very basic backtracking algorithm which is explained at many places. But the most interesting and natural of the sudoku solving strategies I came across is this one from here

Answer 2

I also wrote a Sudoku solver in Python. It is a backtracking algorithm too, but I wanted to share my implementation as well.

Backtracking can be fast enough given that it is moving within the constraints and is choosing cells wisely. You might also want to check out my answer in this thread about optimizing the algorithm. But here I will focus on the algorithm and code itself.

The gist of the algorithm is to start iterating the grid and making decisions what to do - populate a cell, or try another digit for the same cell, or blank out a cell and move back to the previous cell, etc. It's important to note that there is no deterministic way to know how many steps or iterations you will need to solve the puzzle. Therefore, you really have two options - to use a while loop or to use recursion. Both of them can continue iterating until a solution is found or until a lack of solution is proven. The advantage of the recursion is that it is capable of branching out and generally supports more complex logics and algorithms, but the disadvantage is that it is more difficult to implement and often tricky to debug. For my implementation of the backtracking I have used a while loop because no branching is needed, the algorithm searches in a single-threaded linear fashion.

The logic goes like this:

While True: (main iterations)

1. If all blank cells have been iterated and the last blank cell iterated doesn't have any remaining digits to be tried - stop here because there is no solution.
2. If there are no blank cells validate the grid. If the grid is valid stop here and return the solution.
3. If there are blank cells choose the next cell. If that cell has at least on possible digit, assign it and continue to the next main iteration.
4. If there is at least one remaining choice for the current cell and there are no blank cells or all blank cells have been iterated, assign the remaining choice and continue to the next main iteration.
5. If none of the above is true, then it is time to backtrack. Blank out the current cell and enter the below loop.

While True: (backtrack iterations)

1. If there are no more cells to backtrack to - stop here because there is no solution.
2. Select the previous cell according to the backtracking history.
3. If the cell doesn't have any choices left, blank out the cell and continue to the next backtrack iteration.
4. Assign the next available digit to the current cell, break out from backtracking and return to the main iterations.

Some features of the algorithm:

• it keeps a record of the visited cells in the same order so that it can backtrack at any time

• it keeps a record of choices for each cell so that it doesn't try the same digit for the same cell twice

• the available choices for a cell are always within the Sudoku constraints (row, column and 3x3 quadrant)

• this particular implementation has a few different methods of choosing the next cell and the next digit depending on input parameters (more info in the optimization thread)

• if given a blank grid, then it will generate a valid Sudoku puzzle (use with optimization parameter "C" in order to generate random grid every time)

• if given a solved grid it will recognize it and print a message

The full code is:

import random, math, time  class Sudoku:     def __init__( self, _g=[] ):         self._input_grid = [] # store a copy of the original input grid for later use         self.grid = [] # this is the main grid that will be iterated         for i in _g: # copy the nested lists by value, otherwise Python keeps the reference for the nested lists             self._input_grid.append( i[:] )             self.grid.append( i[:] )      self.empty_cells = set() # set of all currently empty cells (by index number from left to right, top to bottom)     self.empty_cells_initial = set() # this will be used to compare against the current set of empty cells in order to determine if all cells have been iterated     self.current_cell = None # used for iterating     self.current_choice = 0 # used for iterating     self.history = [] # list of visited cells for backtracking     self.choices = {} # dictionary of sets of currently available digits for each cell     self.nextCellWeights = {} # a dictionary that contains weights for all cells, used when making a choice of next cell     self.nextCellWeights_1 = lambda x: None # the first function that will be called to assign weights     self.nextCellWeights_2 = lambda x: None # the second function that will be called to assign weights     self.nextChoiceWeights = {} # a dictionary that contains weights for all choices, used when selecting the next choice     self.nextChoiceWeights_1 = lambda x: None # the first function that will be called to assign weights     self.nextChoiceWeights_2 = lambda x: None # the second function that will be called to assign weights      self.search_space = 1 # the number of possible combinations among the empty cells only, for information purpose only     self.iterations = 0 # number of main iterations, for information purpose only     self.iterations_backtrack = 0 # number of backtrack iterations, for information purpose only      self.digit_heuristic = { 1:0, 2:0, 3:0, 4:0, 5:0, 6:0, 7:0, 8:0, 9:0 } # store the number of times each digit is used in order to choose the ones that are least/most used, parameter "3" and "4"     self.centerWeights = {} # a dictionary of the distances for each cell from the center of the grid, calculated only once at the beginning      # populate centerWeights by using Pythagorean theorem     for id in range( 81 ):         row = id // 9         col = id % 9         self.centerWeights[ id ] = int( round( 100 * math.sqrt( (row-4)**2 + (col-4)**2 ) ) )        # for debugging purposes     def dump( self, _custom_text, _file_object ):         _custom_text += ", cell: {}, choice: {}, choices: {}, empty: {}, history: {}, grid: {}\n".format(             self.current_cell, self.current_choice, self.choices, self.empty_cells, self.history, self.grid )         _file_object.write( _custom_text )      # to be called before each solve of the grid     def reset( self ):         self.grid = []         for i in self._input_grid:             self.grid.append( i[:] )          self.empty_cells = set()         self.empty_cells_initial = set()         self.current_cell = None         self.current_choice = 0         self.history = []         self.choices = {}          self.nextCellWeights = {}         self.nextCellWeights_1 = lambda x: None         self.nextCellWeights_2 = lambda x: None         self.nextChoiceWeights = {}         self.nextChoiceWeights_1 = lambda x: None         self.nextChoiceWeights_2 = lambda x: None          self.search_space = 1         self.iterations = 0         self.iterations_backtrack = 0          self.digit_heuristic = { 1:0, 2:0, 3:0, 4:0, 5:0, 6:0, 7:0, 8:0, 9:0 }      def validate( self ):         # validate all rows         for x in range(9):             digit_count = { 0:1, 1:0, 2:0, 3:0, 4:0, 5:0, 6:0, 7:0, 8:0, 9:0 }             for y in range(9):                 digit_count[ self.grid[ x ][ y ] ] += 1             for i in digit_count:                 if digit_count[ i ] != 1:                     return False          # validate all columns         for x in range(9):             digit_count = { 0:1, 1:0, 2:0, 3:0, 4:0, 5:0, 6:0, 7:0, 8:0, 9:0 }             for y in range(9):                 digit_count[ self.grid[ y ][ x ] ] += 1             for i in digit_count:                 if digit_count[ i ] != 1:                     return False          # validate all 3x3 quadrants         def validate_quadrant( _grid, from_row, to_row, from_col, to_col ):             digit_count = { 0:1, 1:0, 2:0, 3:0, 4:0, 5:0, 6:0, 7:0, 8:0, 9:0 }             for x in range( from_row, to_row + 1 ):                 for y in range( from_col, to_col + 1 ):                     digit_count[ _grid[ x ][ y ] ] += 1             for i in digit_count:                 if digit_count[ i ] != 1:                     return False             return True          for x in range( 0, 7, 3 ):             for y in range( 0, 7, 3 ):                 if not validate_quadrant( self.grid, x, x+2, y, y+2 ):                     return False         return True      def setCell( self, _id, _value ):         row = _id // 9         col = _id % 9         self.grid[ row ][ col ] = _value      def getCell( self, _id ):         row = _id // 9         col = _id % 9         return self.grid[ row ][ col ]      # returns a set of IDs of all blank cells that are related to the given one, related means from the same row, column or quadrant     def getRelatedBlankCells( self, _id ):         result = set()         row = _id // 9         col = _id % 9          for i in range( 9 ):             if self.grid[ row ][ i ] == 0: result.add( row * 9 + i )         for i in range( 9 ):             if self.grid[ i ][ col ] == 0: result.add( i * 9 + col )         for x in range( (row//3)*3, (row//3)*3 + 3 ):             for y in range( (col//3)*3, (col//3)*3 + 3 ):                 if self.grid[ x ][ y ] == 0: result.add( x * 9 + y )          return set( result ) # return by value      # get the next cell to iterate     def getNextCell( self ):         self.nextCellWeights = {}         for id in self.empty_cells:             self.nextCellWeights[ id ] = 0          self.nextCellWeights_1( 1000 ) # these two functions will always be called, but behind them will be a different weight function depending on the optimization parameters provided         self.nextCellWeights_2( 1 )          return min( self.nextCellWeights, key = self.nextCellWeights.get )      def nextCellWeights_A( self, _factor ): # the first cell from left to right, from top to bottom         for id in self.nextCellWeights:             self.nextCellWeights[ id ] += id * _factor      def nextCellWeights_B( self, _factor ): # the first cell from right to left, from bottom to top         self.nextCellWeights_A( _factor * -1 )      def nextCellWeights_C( self, _factor ): # a randomly chosen cell         for id in self.nextCellWeights:             self.nextCellWeights[ id ] += random.randint( 0, 999 ) * _factor      def nextCellWeights_D( self, _factor ): # the closest cell to the center of the grid         for id in self.nextCellWeights:             self.nextCellWeights[ id ] += self.centerWeights[ id ] * _factor      def nextCellWeights_E( self, _factor ): # the cell that currently has the fewest choices available         for id in self.nextCellWeights:             self.nextCellWeights[ id ] += len( self.getChoices( id ) ) * _factor      def nextCellWeights_F( self, _factor ): # the cell that currently has the most choices available         self.nextCellWeights_E( _factor * -1 )      def nextCellWeights_G( self, _factor ): # the cell that has the fewest blank related cells         for id in self.nextCellWeights:             self.nextCellWeights[ id ] += len( self.getRelatedBlankCells( id ) ) * _factor      def nextCellWeights_H( self, _factor ): # the cell that has the most blank related cells         self.nextCellWeights_G( _factor * -1 )      def nextCellWeights_I( self, _factor ): # the cell that is closest to all filled cells         for id in self.nextCellWeights:             weight = 0             for check in range( 81 ):                 if self.getCell( check ) != 0:                     weight += math.sqrt( ( id//9 - check//9 )**2 + ( id%9 - check%9 )**2 )      def nextCellWeights_J( self, _factor ): # the cell that is furthest from all filled cells         self.nextCellWeights_I( _factor * -1 )      def nextCellWeights_K( self, _factor ): # the cell whose related blank cells have the fewest available choices         for id in self.nextCellWeights:             weight = 0             for id_blank in self.getRelatedBlankCells( id ):                 weight += len( self.getChoices( id_blank ) )             self.nextCellWeights[ id ] += weight * _factor      def nextCellWeights_L( self, _factor ): # the cell whose related blank cells have the most available choices         self.nextCellWeights_K( _factor * -1 )        # for a given cell return a set of possible digits within the Sudoku restrictions     def getChoices( self, _id ):         available_choices = {1,2,3,4,5,6,7,8,9}         row = _id // 9         col = _id % 9          # exclude digits from the same row         for y in range( 0, 9 ):             if self.grid[ row ][ y ] in available_choices:                 available_choices.remove( self.grid[ row ][ y ] )          # exclude digits from the same column         for x in range( 0, 9 ):             if self.grid[ x ][ col ] in available_choices:                 available_choices.remove( self.grid[ x ][ col ] )          # exclude digits from the same quadrant         for x in range( (row//3)*3, (row//3)*3 + 3 ):             for y in range( (col//3)*3, (col//3)*3 + 3 ):                 if self.grid[ x ][ y ] in available_choices:                     available_choices.remove( self.grid[ x ][ y ] )          if len( available_choices ) == 0: return set()         else: return set( available_choices ) # return by value      def nextChoice( self ):         self.nextChoiceWeights = {}         for i in self.choices[ self.current_cell ]:             self.nextChoiceWeights[ i ] = 0          self.nextChoiceWeights_1( 1000 )         self.nextChoiceWeights_2( 1 )          self.current_choice = min( self.nextChoiceWeights, key = self.nextChoiceWeights.get )         self.setCell( self.current_cell, self.current_choice )         self.choices[ self.current_cell ].remove( self.current_choice )      def nextChoiceWeights_0( self, _factor ): # the lowest digit         for i in self.nextChoiceWeights:             self.nextChoiceWeights[ i ] += i * _factor      def nextChoiceWeights_1( self, _factor ): # the highest digit         self.nextChoiceWeights_0( _factor * -1 )      def nextChoiceWeights_2( self, _factor ): # a randomly chosen digit         for i in self.nextChoiceWeights:             self.nextChoiceWeights[ i ] += random.randint( 0, 999 ) * _factor      def nextChoiceWeights_3( self, _factor ): # heuristically, the least used digit across the board         self.digit_heuristic = { 1:0, 2:0, 3:0, 4:0, 5:0, 6:0, 7:0, 8:0, 9:0 }         for id in range( 81 ):             if self.getCell( id ) != 0: self.digit_heuristic[ self.getCell( id ) ] += 1         for i in self.nextChoiceWeights:             self.nextChoiceWeights[ i ] += self.digit_heuristic[ i ] * _factor      def nextChoiceWeights_4( self, _factor ): # heuristically, the most used digit across the board         self.nextChoiceWeights_3( _factor * -1 )      def nextChoiceWeights_5( self, _factor ): # the digit that will cause related blank cells to have the least number of choices available         cell_choices = {}         for id in self.getRelatedBlankCells( self.current_cell ):             cell_choices[ id ] = self.getChoices( id )          for c in self.nextChoiceWeights:             weight = 0             for id in cell_choices:                 weight += len( cell_choices[ id ] )                 if c in cell_choices[ id ]: weight -= 1             self.nextChoiceWeights[ c ] += weight * _factor      def nextChoiceWeights_6( self, _factor ): # the digit that will cause related blank cells to have the most number of choices available         self.nextChoiceWeights_5( _factor * -1 )      def nextChoiceWeights_7( self, _factor ): # the digit that is the least common available choice among related blank cells         cell_choices = {}         for id in self.getRelatedBlankCells( self.current_cell ):             cell_choices[ id ] = self.getChoices( id )          for c in self.nextChoiceWeights:             weight = 0             for id in cell_choices:                 if c in cell_choices[ id ]: weight += 1             self.nextChoiceWeights[ c ] += weight * _factor      def nextChoiceWeights_8( self, _factor ): # the digit that is the most common available choice among related blank cells         self.nextChoiceWeights_7( _factor * -1 )      def nextChoiceWeights_9( self, _factor ): # the digit that is the least common available choice across the board         cell_choices = {}         for id in range( 81 ):             if self.getCell( id ) == 0:                 cell_choices[ id ] = self.getChoices( id )          for c in self.nextChoiceWeights:             weight = 0             for id in cell_choices:                 if c in cell_choices[ id ]: weight += 1             self.nextChoiceWeights[ c ] += weight * _factor      def nextChoiceWeights_a( self, _factor ): # the digit that is the most common available choice across the board         self.nextChoiceWeights_9( _factor * -1 )        # the main function to be called     def solve( self, _nextCellMethod, _nextChoiceMethod, _start_time, _prefillSingleChoiceCells = False ):         s = self         s.reset()          # initialize optimization functions based on the optimization parameters provided         """         A - the first cell from left to right, from top to bottom         B - the first cell from right to left, from bottom to top         C - a randomly chosen cell         D - the closest cell to the center of the grid         E - the cell that currently has the fewest choices available         F - the cell that currently has the most choices available         G - the cell that has the fewest blank related cells         H - the cell that has the most blank related cells         I - the cell that is closest to all filled cells         J - the cell that is furthest from all filled cells         K - the cell whose related blank cells have the fewest available choices         L - the cell whose related blank cells have the most available choices         """         if _nextCellMethod[ 0 ] in "ABCDEFGHIJKLMN":             s.nextCellWeights_1 = getattr( s, "nextCellWeights_" + _nextCellMethod[0] )         elif _nextCellMethod[ 0 ] == " ":             s.nextCellWeights_1 = lambda x: None         else:             print( "(A) Incorrect optimization parameters provided" )             return False          if len( _nextCellMethod ) > 1:             if _nextCellMethod[ 1 ] in "ABCDEFGHIJKLMN":                 s.nextCellWeights_2 = getattr( s, "nextCellWeights_" + _nextCellMethod[1] )             elif _nextCellMethod[ 1 ] == " ":                 s.nextCellWeights_2 = lambda x: None             else:                 print( "(B) Incorrect optimization parameters provided" )                 return False         else:             s.nextCellWeights_2 = lambda x: None          # initialize optimization functions based on the optimization parameters provided         """         0 - the lowest digit         1 - the highest digit         2 - a randomly chosen digit         3 - heuristically, the least used digit across the board         4 - heuristically, the most used digit across the board         5 - the digit that will cause related blank cells to have the least number of choices available         6 - the digit that will cause related blank cells to have the most number of choices available         7 - the digit that is the least common available choice among related blank cells         8 - the digit that is the most common available choice among related blank cells         9 - the digit that is the least common available choice across the board         a - the digit that is the most common available choice across the board         """         if _nextChoiceMethod[ 0 ] in "0123456789a":             s.nextChoiceWeights_1 = getattr( s, "nextChoiceWeights_" + _nextChoiceMethod[0] )         elif _nextChoiceMethod[ 0 ] == " ":             s.nextChoiceWeights_1 = lambda x: None         else:             print( "(C) Incorrect optimization parameters provided" )             return False          if len( _nextChoiceMethod ) > 1:             if _nextChoiceMethod[ 1 ] in "0123456789a":                 s.nextChoiceWeights_2 = getattr( s, "nextChoiceWeights_" + _nextChoiceMethod[1] )             elif _nextChoiceMethod[ 1 ] == " ":                 s.nextChoiceWeights_2 = lambda x: None             else:                 print( "(D) Incorrect optimization parameters provided" )                 return False         else:             s.nextChoiceWeights_2 = lambda x: None          # fill in all cells that have single choices only, and keep doing it until there are no left, because as soon as one cell is filled this might bring the choices down to 1 for another cell         if _prefillSingleChoiceCells == True:             while True:                 next = False                 for id in range( 81 ):                     if s.getCell( id ) == 0:                         cell_choices = s.getChoices( id )                         if len( cell_choices ) == 1:                             c = cell_choices.pop()                             s.setCell( id, c )                             next = True                 if not next: break          # initialize set of empty cells         for x in range( 0, 9, 1 ):             for y in range( 0, 9, 1 ):                 if s.grid[ x ][ y ] == 0:                     s.empty_cells.add( 9*x + y )         s.empty_cells_initial = set( s.empty_cells ) # copy by value          # calculate search space         for id in s.empty_cells:             s.search_space *= len( s.getChoices( id ) )          # initialize the iteration by choosing a first cell         if len( s.empty_cells ) < 1:             if s.validate():                 print( "Sudoku provided is valid!" )                 return True             else:                 print( "Sudoku provided is not valid!" )                 return False         else: s.current_cell = s.getNextCell()          s.choices[ s.current_cell ] = s.getChoices( s.current_cell )         if len( s.choices[ s.current_cell ] ) < 1:             print( "(C) Sudoku cannot be solved!" )             return False            # start iterating the grid         while True:             #if time.time() - _start_time > 2.5: return False # used when doing mass tests and don't want to wait hours for an inefficient optimization to complete              s.iterations += 1              # if all empty cells and all possible digits have been exhausted, then the Sudoku cannot be solved             if s.empty_cells == s.empty_cells_initial and len( s.choices[ s.current_cell ] ) < 1:                 print( "(A) Sudoku cannot be solved!" )                 return False              # if there are no empty cells, it's time to validate the Sudoku             if len( s.empty_cells ) < 1:                 if s.validate():                     print( "Sudoku has been solved! " )                     print( "search space is {}".format( self.search_space ) )                     print( "empty cells: {}, iterations: {}, backtrack iterations: {}".format( len( self.empty_cells_initial ), self.iterations, self.iterations_backtrack ) )                     for i in range(9):                         print( self.grid[i] )                     return True              # if there are empty cells, then move to the next one             if len( s.empty_cells ) > 0:                  s.current_cell = s.getNextCell() # get the next cell                 s.history.append( s.current_cell ) # add the cell to history                 s.empty_cells.remove( s.current_cell ) # remove the cell from the empty queue                 s.choices[ s.current_cell ] = s.getChoices( s.current_cell ) # get possible choices for the chosen cell                  if len( s.choices[ s.current_cell ] ) > 0: # if there is at least one available digit, then choose it and move to the next iteration, otherwise the iteration continues below with a backtrack                     s.nextChoice()                     continue              # if all empty cells have been iterated or there are no empty cells, and there are still some remaining choices, then try another choice             if len( s.choices[ s.current_cell ] ) > 0 and ( s.empty_cells == s.empty_cells_initial or len( s.empty_cells ) < 1 ):                  s.nextChoice()                 continue              # if none of the above, then we need to backtrack to a cell that was previously iterated             # first, restore the current cell...             s.history.remove( s.current_cell ) # ...by removing it from history             s.empty_cells.add( s.current_cell ) # ...adding back to the empty queue             del s.choices[ s.current_cell ] # ...scrapping all choices             s.current_choice = 0             s.setCell( s.current_cell, s.current_choice ) # ...and blanking out the cell              # ...and then, backtrack to a previous cell             while True:                 s.iterations_backtrack += 1                  if len( s.history ) < 1:                     print( "(B) Sudoku cannot be solved!" )                     return False                  s.current_cell = s.history[ -1 ] # after getting the previous cell, do not recalculate all possible choices because we will lose the information about has been tried so far                  if len( s.choices[ s.current_cell ] ) < 1: # backtrack until a cell is found that still has at least one unexplored choice...                     s.history.remove( s.current_cell )                     s.empty_cells.add( s.current_cell )                     s.current_choice = 0                     del s.choices[ s.current_cell ]                     s.setCell( s.current_cell, s.current_choice )                     continue                  # ...and when such cell is found, iterate it                 s.nextChoice()                 break # and break out from the backtrack iteration but will return to the main iteration 

Example call using the world's hardest Sudoku as per this article http://www.telegraph.co.uk/news/science/science-news/9359579/Worlds-hardest-sudoku-can-you-crack-it.html

hardest_sudoku = [     [8,0,0,0,0,0,0,0,0],     [0,0,3,6,0,0,0,0,0],     [0,7,0,0,9,0,2,0,0],     [0,5,0,0,0,7,0,0,0],     [0,0,0,0,4,5,7,0,0],     [0,0,0,1,0,0,0,3,0],     [0,0,1,0,0,0,0,6,8],     [0,0,8,5,0,0,0,1,0],     [0,9,0,0,0,0,4,0,0]]  mySudoku = Sudoku( hardest_sudoku ) start = time.time() mySudoku.solve( "A", "0", time.time(), False ) print( "solved in {} seconds".format( time.time() - start ) ) 

And example output is:

Sudoku has been solved! search space is 9586591201964851200000000000000000000 empty cells: 60, iterations: 49559, backtrack iterations: 49498 [8, 1, 2, 7, 5, 3, 6, 4, 9] [9, 4, 3, 6, 8, 2, 1, 7, 5] [6, 7, 5, 4, 9, 1, 2, 8, 3] [1, 5, 4, 2, 3, 7, 8, 9, 6] [3, 6, 9, 8, 4, 5, 7, 2, 1] [2, 8, 7, 1, 6, 9, 5, 3, 4] [5, 2, 1, 9, 7, 4, 3, 6, 8] [4, 3, 8, 5, 2, 6, 9, 1, 7] [7, 9, 6, 3, 1, 8, 4, 5, 2] solved in 1.1600663661956787 seconds