Magic number board puzzle solver in Python

The magic number board puzzle is yet another equation puzzle (similar to the first equation puzzle or Jodici) – but has a notable larger solution space.

Magic number board
A magic number board example

The magic number board is a quadratic board, consisting of 5×5 = 25 fields, which are organized as 5 horizontal rows and 5 vertical columns. Each field should contain a number in the range [5,29], which each number being used exactly once in the board. Each row, column and both diagonals should sum to 85 exactly. Some numbers are already given at start – the task is to fill in all remaining fields while sticking to the rules.

Solving the magic number board puzzle

We’re going to refer to rows/fields/diagonals as “groups” from now (e.g. if a fact or rule is true for all groups it’s true for all rows, columns and diagonals). The magic number puzzle contains 12 such groups, consequently we have 12 equations in total, and – given 8 of the 25 fields being filled at the start – 17 variables to fill in. Without additional rules, this would allow for multiple solutions. The additional rule of using each number once in the board reduces the amount of valid solutions to one, if the puzzle is designed correctly.

The solution space is notably larger than with the equation puzzle or Jodici. Given 8 initially filled fields we have 17 yet empty fields, for which we have 17! solutions (3.5*10^14 solutions or ~48bit entropy). Pure brute force would of course find the solution, but would require an arguable amount of time to do so with typical PCs or Laptops. Therefore, we are speeding up the process by using best first search and – more important – incorporating the following optimizations.

Successor generation (branching factor reduction)

When adding a number to the board, we don’t need to generate all successors (e.g. place every possible number in every still empty field). We know that we can still reach all possible solutions in our search tree, if we just fill in all possible numbers to a single field (generating multiple successors having this field filled, but all other previously empty field still being empty). Doing so causes us to skip different orders of filling numbers to fields (e.g. first filling 1 to field A, then 2 to field B, or first filling 2 to field B, then 1 to field A – which would result in the same state). Consequently, this reduces the branching factor and therefore the runtime. Given this optimization idea, the question is: how to determine the one field we’re going to fill numbers into? Which his brings us to:

Order of fields to be filled (heuristic search direction)

A human player would try to fill numbers to those groups that already contain many numbers. Why? Because usually, there are least possibilities of numbers to be filled into these groups (which again causes a lower branching factor, therefore less complexity to deal with for the human player). Hence, we always fill a field in the group with the least fields still empty (excluding completely filled groups of course). An example of using such an order: imagine we have a group which already contains 4 numbers. There is (at max) one possibility for the fifth number, as the five numbers have to sum to 85. A human player would immediately fill in that number – a behaviour we should also strive for in our approach.

Validity check – detecting dead ends early (branch cutting)

The third thought is about evaluating if a board is already invalid, which correlates to branch cutting in search. As we fill numbers into the board it’s very important to find out as soon as possible, if the board cannot yield a valid solution any more (therefore all possible successors can be excluded from search, which cuts parts of search tree, and which saves us lots of resources).  An example for such using such checks for groups containing 4 numbers: with the the fifth number all numbers of that group have to sum to 85. In case the number required to be filled into the field yet empty for the sum to be 85 is already used elsewhere, the current board cannot yield a valid solution any more – therefore the board can be excluded from search. We don’t only check groups already containing 4 numbers, but also groups containing 3 numbers. For these groups we try to fill in all permutations of numbers still available and check, if with at least one combination the group sums to 85. If this is the case, the group is counted as valid. If no combination allows for a sum of 85, the board is already invalid and we exclude it from search. Note: using permutations of numbers still available, conceptually it would be possible to check for groups with any amount of already filled in numbers (e.g. 2, 1 or even empty groups). The drawback is that the amount of permutations that need to be checked increases exponentially with the amount of fields of the group still being empty. Consequently, the gain of recognizing invalid boards earlier is easily eaten up by the additional complexity of generating and checking permutations.

Python implementation

# Magic number square solver implementation. The magic number square is a 5x5 number board in which each field must contain a nr [5,25]. Some numbers are given at the start, all others have to be found. In the end, each nr must be used once and all rows, columns and both diagonals have to sum to 85. 
# Rainhard Findling
# 2014/11
# Core concepts:
# Branch cutting: we look at each generated board: for each row (incl. row, col, diagonal) the remaining numbers must contain at least one permutation with which the row sums up exactly to 85, otherwise the board can only create invalid solutions and is skipped in future search.
# Heuristic (search direction): we prefer boards which are more filled over boards not so filled. We futher prefer rows with more filled cells over rows with less filled cells. In fact, we use a monotinic metric: 1 row with N filled cells is always worth more than [MAX] rows which have N-1 cell filled less. Therefore, we always try to fill the row with already has the most filled cells (except of completely filled rows). 
# Successor generation: of a board only those successors get created that fill a number to the row with most already filled cells (excluding completely filled rows). This is intuitive, but it seems it would not be necessary as we do heuristic sorting whcih would prefer "more filled" boards always. But it's also required for another reason: it is important in combination with branch cutting (it's one solution to the problem that would emerge otherwise). For a board with a row where only 1 cell is not filled yet, that cell could remain unfilled (even if we have a heuristic that would enforce this cell to be filled next) - simply because filling the cell makes the row valid, but there are no valid successors of the resulting board possible anymore. This would lead to the irrational behavior of trying this cell and discarding all successors, then trying the other candidates from the list which still have this one cell free, therefore leaving this cell free (for getting a still high bonus for having one nearly filled row) and trying to fill cells elsewhere. The approach of only filling the row with already highest filling status - except completely filled rows - ensures that there are no states that allow for "leaving out because resulting successors are invalid" scenarios.
import itertools
import copy
import numpy

global_sum = 85
global_nrs_range = range(5, 30)
global_branchcutting_min_amount_of_filled_cells_to_perform_permutation_check = 3
global_heuristics_row_fill_stat_hist_weights = [12**x for x in range(6)] # why? because with 5*5 we have 12 rows to validate in total. therefore, if we want to favor boards that have more filled or nearly filled rows, our base b for hist**b has to be 12 for the metric order to be monotone. why 6 weights if we have 5*5? because we weight the '0-fills' too.
_ = '__' # placeholder for empty fields

class Cell():
    """representation of a cells in the gameboard."""
    def __init__(self, nr=_): = nr
        #if( != _):
            #self.available_nrs = []
            #self.available_nrs = global_nrs_range

    def __str__(self):
        return '%2s' % str(

    def str_verbose(self):
        """str with more information about this cell than __str__."""
        return self.__str__() + ' (' + ','.join([ str(x) for x in self.available_nrs]) + ')'

    def __repr__(self):
        return self.__str__()
    def __deepcopy(self, memo):
        return Cell(

class Board():
    """Representation of a gameboard holding all numbers + helper infos for solving it."""
    def __init__(self, gameboard, parent, available_nrs=None):
        self.gameboard = gameboard
        self.parent = parent
        if(available_nrs != None):
            self.available_nrs = available_nrs
            self.available_nrs = self.generate_available_nrs_from_gameboard()

    def generate_available_nrs_from_gameboard(self):
        """derive still available numbers from gameboard."""
        return [x for x in global_nrs_range if not x in map(lambda, self.all_cells_as_vector())]
    def all_cells_as_vector(self):
        return reduce(lambda x,y:x+y, self.gameboard)
    def __str__(self):
        return self.str_gameboard(self.gameboard)
    def str_gameboard(self, gameboard):
        """string representation of a gameboard = board."""
        s = '\n================\n'
        for row in gameboard:
            s += '|' + ' '.join([x.__str__() for x in row]) + '|\n'
        s += '================\n'
        return s
    def __repr__(self):
        return self.__str__()
    def __deepcopy__(self, memo):
        # we do not/cannot deepcopy parent!
        return Board(copy.deepcopy(self.gameboard), self.parent, copy.deepcopy(self.available_nrs))
    def __eq__(self, other):
        if other == None:
            return False
        for self_row, other_row in zip(self.gameboard, other.gameboard):
            for self_cell, other_cell in zip(self_row, other_row):
                if not ==
                    return False                
        return True
    def __cmp__(self, other):
        # __cmp__ should return a negative integer if self < other, zero if self == other, and positive if self > other
        if self.heuristic() < other.heuristic():
            return -1
        if self.heuristic() == other.heuristic():
            return 0
            return 1
    def copy_transposed(self):
        """get a transposed copy of the gameboard: rows become columns, columns become rows."""
        return [[row[nr] for row in self.gameboard] for nr in range(len(self.gameboard))]
    def diagonal_topleft_bottomright(self):
        """get the topleft to bottomright diagonal 'row'"""
        return [self.gameboard[nr][nr] for nr in range(len(self.gameboard))]
    def diagonal_topright_bottomleft(self):
        """get the topright to bottomleft diagonal 'row'"""
        return [self.gameboard[len(self.gameboard)-nr-1][nr] for nr in range(len(self.gameboard))]
    def get_all_groups(self):
        """returns all groups (rows, cols, diagonals) that have to fulfill certain criteria."""
        return self.gameboard + self.copy_transposed() + [self.diagonal_topleft_bottomright()] + [self.diagonal_topright_bottomleft()]
    def is_valid(self):
        """checks if this gameboard is valid / can still be solved in a valid way."""
        for row in self.get_all_groups():
            free = len(filter(lambda == _, row))
            #print 'free', free
            if(len(self.gameboard) - free < global_branchcutting_min_amount_of_filled_cells_to_perform_permutation_check):
                # too many unfilled cells, we don't check if permutations can still make this row valid
                #print 'skipping row with', str(free), 'free cells during checks'
            if(free == 0):
                # is already filled. sanity check: is it's filled in valid way?
                if(reduce(lambda x,, row, 0) != global_sum):
                    return False
                    #raise Exception('Completely filled row does not sum to ' + str(global_sum) + ': ' + str(row))
            cur_sum = reduce(lambda x,y:x + ( if != _ else 0), row, 0) # sum of nrs currently in this row
            # check that from nrs still available it's still possible to get this row to the target sum
            sum_remainder = global_sum - cur_sum
            # does any combination of available nrs still fulfill this criteria?
            permutations = itertools.permutations(self.available_nrs, free)
            if not any([sum(x) == sum_remainder for x in permutations]):
                return False
        return True
    def is_filled(self):
        """checks if all fields are filled. valid + filled therefore means solved."""
        for row in self.gameboard:
            for cell in row:
                if == _:
                    return False
        return True
    def generate_successors(self):
        """generate all possible successor states of this board."""
        successors = []
        counts = [len(filter(lambda != _, row)) for row in self.get_all_groups()]
        row_nr = counts.index(max(filter(lambda x:x<len(self.gameboard),set(counts))))
        #for row_nr in range(len(self.gameboard)):
        row = self.get_all_groups()[row_nr]
        for cell_nr in range(len(row)):
            cell = row[cell_nr]
            if == _:
                # this cell can be set as next step
                for nr in self.available_nrs:
                    # generate clone
                    suc = copy.deepcopy(self)
                    # modify to show new state
                    suc.get_all_groups()[row_nr][cell_nr].nr = nr
                    suc.parent = self
                    if suc.is_valid():
        return successors
    def fill_stat_hist(self):
        """histogram of how row fill states: how much rows exist with have 0, 1, 2 etc. filled cells."""
        counts = [len(filter(lambda != _, row)) for row in self.get_all_groups()]
        #fill_stat_hist = [len(list(group)) for key,group in itertools.groupby(counts)]
        fill_stat_hist = [len(filter(lambda x:x==i, counts)) for i in range(len(self.gameboard)+1)]
        return fill_stat_hist

    def fill_stat_hist_weighted(self):
        """the same hist weighted by how much we think it's worth to have more of 'such' row fill states."""
        return map(lambda x:x[0]*x[1], zip(self.fill_stat_hist(), global_heuristics_row_fill_stat_hist_weights))
    def heuristic(self):
        """heuristic of how good this field is. bigger means better."""
        return sum(self.fill_stat_hist_weighted())
    def path(self):
        """evolution chain from beginning to this state - for getting detailled insight."""
        return [self] if self.parent == None else self.parent.path() + [self]

def generate_board(board):
    return Board([[Cell(x) for x in row] for row in board], None)

# generate board
b = generate_board([    [ _, _, _, _,15],
                        [ _, _,25,17, _], 
                        [ _,26, _, 5, _], 
                        [ _,18, _, _,11], 
                        [ _, _,21, _, _]])
print b

# best first search
nodes = [b]
closed = []
solutions = []
while(len(nodes) > 0):
    # sort: bigger heuristic = best = last in list
    node = nodes.pop()
#    print 'current board:', node
    #print 'nodes remaining in list:', len(nodes)
    #print 'fill_stat_hist:', str(node.fill_stat_hist())
    #print 'fill_stat_hist weighted:', str(node.fill_stat_hist_weighted())
#    print 'solutions:', len(solutions)
    # produce some kiddies
    successors = node.generate_successors()
    suc_finished = filter(lambda suc:suc.is_filled(), successors)
    if(len(suc_finished) > 0):
        #print 'solution(s):', map(lambda x:x.path(), suc_finished)
        solutions += suc_finished
        print 'solutions:', len(solutions)
    suc_unfinished = filter(lambda suc:not suc.is_filled(), successors)
    suc_to_add = filter(lambda suc:not suc in nodes and not suc in closed, suc_unfinished)
    nodes += suc_to_add
print map(lambda x:x.path(), solutions)
print 'found solutions:', len(solutions)
print 'finished.'

When running the solver on the magic number board example from above, we get presented the single possible, valid solution (along with the steps it took to get there, which are not shown here for layouting reasons ):

|__ __ __ __ 15|
|__ __ 25 17 __|
|__ 26 __  5 __|
|__ 18 __ __ 11|
|__ __ 21 __ __|

solutions: 1
|__ __ __ __ 15|
|__ __ 25 17 __|
|__ 26 __  5 __|
|__ 18 __ __ 11|
|__ __ 21 __ __|
|28 24  6 12 15|
|13  7 25 17 23|
| 8 26 19  5 27|
|20 18 14 22 11|
|16 10 21 29  9|
found solutions: 1

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