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

Answer 1

I wrote a simple program that solved the easy ones. It took its input from a file which was just a matrix with spaces and numbers. The datastructure to solve it was just a 9 by 9 matrix of a bit mask. The bit mask would specify which numbers were still possible on a certain position. Filling in the numbers from the file would reduce the numbers in all rows/columns next to each known location. When that is done you keep iterating over the matrix and reducing possible numbers. If each location has only one option left you're done. But there are some sudokus that need more work. For these ones you can just use brute force: try all remaining possible combinations until you find one that works.

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

Answer 3

a short attempt to achieve same algorithm using backtracking:

def solve(sudoku):
    #using recursion and backtracking, here we go.
    empties = [(i,j) for i in range(9) for j in range(9) if sudoku[i][j] == 0]
    predict = lambda i, j: set(range(1,10))-set([sudoku[i][j]])-set([sudoku[y+range(1,10,3)[i//3]][x+range(1,10,3)[j//3]] for y in (-1,0,1) for x in (-1,0,1)])-set(sudoku[i])-set(list(zip(*sudoku))[j])
    if len(empties)==0:return True
    gap = next(iter(empties))
    predictions = predict(*gap)
    for i in predictions:
        sudoku[gap[0]][gap[1]] = i
        if solve(sudoku):return True
        sudoku[gap[0]][gap[1]] = 0
    return False

Answer 4

I know I'm late, but this is my version:

board = [
    [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]
]


def solve(bo):
    find = find_empty(bo)
    if not find:  # if find is None or False
        return True
    else:
        row, col = find

    for num in range(1, 10):
        if valid(bo, num, (row, col)):
            bo[row][col] = num

            if solve(bo):
                return True

            bo[row][col] = 0

    return False


def valid(bo, num, pos):

    # Check row
    for i in range(len(bo[0])):
        if bo[pos[0]][i] == num and pos[1] != i:
            return False

    # Check column
    for i in range(len(bo)):
        if bo[i][pos[1]] == num and pos[0] != i:
            return False

    # Check box
    box_x = pos[1] // 3
    box_y = pos[0] // 3

    for i in range(box_y*3, box_y*3 + 3):
        for j in range(box_x*3, box_x*3 + 3):
            if bo[i][j] == num and (i, j) != pos:
                return False

    return True


def print_board(bo):
    for i in range(len(bo)):
        if i % 3 == 0:
            if i == 0:
                print(" ┎─────────┰─────────┰─────────┒")
            else:
                print(" ┠─────────╂─────────╂─────────┨")

        for j in range(len(bo[0])):
            if j % 3 == 0:
                print(" ┃ ", end=" ")

            if j == 8:
                print(bo[i][j], " ┃")
            else:
                print(bo[i][j], end=" ")

    print(" ┖─────────┸─────────┸─────────┚")


def find_empty(bo):
    for i in range(len(bo)):
        for j in range(len(bo[0])):
            if bo[i][j] == 0:
                return i, j  # row, column

    return None


print_board(board)
print('\n--------------------------------------\n')
solve(board)
print_board(board)

It uses backtracking. But is not coded by me, it's Tech With Tim's. That list contains the world hardest sudoku, and by implementing the timing function, the time is:

===========================
[Finished in 2.838 seconds]
===========================

But with a simple sudoku puzzle like:

board = [
    [7, 8, 0, 4, 0, 0, 1, 2, 0],
    [6, 0, 0, 0, 7, 5, 0, 0, 9],
    [0, 0, 0, 6, 0, 1, 0, 7, 8],
    [0, 0, 7, 0, 4, 0, 2, 6, 0],
    [0, 0, 1, 0, 5, 0, 9, 3, 0],
    [9, 0, 4, 0, 6, 0, 0, 0, 5],
    [0, 7, 0, 3, 0, 0, 0, 1, 2],
    [1, 2, 0, 0, 0, 7, 4, 0, 0],
    [0, 4, 9, 2, 0, 6, 0, 0, 7]
]

The result is :

===========================
[Finished in 0.011 seconds]
===========================

Pretty fast I can say.