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Always interested in offers/projects/new ideas. Eclectic experience in fields like: numerical computing; Python web; Java enterprise; functional languages; GPGPU; SQL databases; etc. Based in Santiago, Chile; telecommute worldwide. CV; email.

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Solving Grid Puzzle in Python

From: andrew cooke <andrew@...>

Date: Thu, 30 Aug 2012 06:22:42 -0400

A neat little puzzle I spent some rainy holiday time on is described at

[This post has been updated since first posting]

The code below is my final attempt, which solves the puzzle in under 3
secs.  I'm posting it here rather than updating the SO post as it's
more complex and not going to help people trying to understand the
basic idea.

One new feature here is that I restrict the search over all solutions
to a single symmetry (so there are four times as many solutions as
found, which can be generated by flipping the solution vertically,
horizontally, or both).  Finding the best way to enforce this was the
hardest part of the entire problem - I finally hit upon requiring the
largest value to be at a certain corner.



nx, ny = 4, 5
values = [1,2,3,4,5,6,7,8,9,10,12,18,20,21,24,27,30,35,36,40]
# grid[x][y] so it is a list of columns (prints misleadingly!)
grid = [[0 for _ in range(ny)] for _ in range(nx)]
# cache these to avoid re-calculating
xy_moves = {}
debug = False

def edges(grid, x, y):
    'coordinates of vertical/horizontal neighbours'
    return [(x-1,y),(x+1,y),(x,y-1),(x,y+1)]

def corners(grid, x, y):
    'coordinates of vertical/horizontal neighbours'
    return [(x-1,y-1),(x+1,y-1),(x-1,y+1),(x+1,y+1)]

def inside(coords):
    'filter coordinates inside the grid'
    return ((x, y) for (x, y) in coords 
            if x > -1 and x < nx and y > -1 and y < ny)

def filled(grid, coords):
    'filter coords to give only filled cells'
    return filter(lambda xy: grid[xy[0]][xy[1]], coords)

def count_neighbours(grid, x, y):
    '''use this to find most-constrained location
    including corners makes the global search with symmetry removal slightly
    slower (2m40s v 2m20s), but the (2,2)=10 search faster (2s v 6s),
    presumably because edges alone hits the symmetry test sooner.'''
#    return sum(1 for _ in filled(grid, inside(edges(grid, x, y))))
    return sum(1 for _ in filled(grid, inside(edges(grid, x, y)))) + \
        sum(0.5 for _ in filled(grid, inside(corners(grid, x, y))))

def cluster(grid, depth):
    '''given a certain depth in the search, where should we move next?  
       choose a place with lots of neighbours so that we have good 
       constraints (and so can reject bad moves)'''
    if depth not in xy_moves:
        best, x, y = 0, 0, 0 # default matches symmetry check
        for xx in range(nx):
            for yy in range(ny):
                if not grid[xx][yy]:
                    count = count_neighbours(grid, xx, yy)
                    if count > best:
                        best, x, y = count, xx, yy
        xy_moves[depth] = (x, y)
        if debug: print('next move for %d is %d,%d' % (depth, x, y))
    return xy_moves[depth]

def to_corners(grid, depth):
    '''alternative move sequence, targetting corners first.
    much slower - 110m for all values.'''
    if depth not in xy_moves:
        if depth >= 2*(nx+ny) - 4:
            cluster(grid, depth)
            d = depth
            if d < nx: xy_moves[depth] = (d, 0)
                d -= nx
                if d+1 < ny: xy_moves[depth] = (0, 1+d)
                    d -= ny-1
                    if d+1 < nx: xy_moves[depth] = (1+d, ny-1)
                        d -= nx-1
                        xy_moves[depth] = (nx-1, d+1)
            if debug: 
                print('next move for %d is %s' % (depth, xy_moves[depth])) 
    return xy_moves[depth]

def drop_value(value, values):
    'remove value from the values'
    return [v for v in values if v != value]

def copy_grid(grid, x, y, value):
    'copy grid, replacing the value at x,y'
    return [[value if j == y else grid[i][j] for j in range(ny)]
            if x == i else grid[i]
            for i in range(nx)]

def move_ok(grid, x, y, value):
    'are all neighbours multiples?'
    for (xx, yy) in filled(grid, inside(edges(grid, x, y))):
        g = grid[xx][yy]
        if (g > value and g % value) or (g < value and value % g):
            if debug: 
                print('fail: %d at %d,%d in %s' % (value, x, y, grid))
            return False
    return True

def always_ok(grid):
    'dummy test to allow all solutions'
    return True

def check_corners(grid):
    '''remove symmetrically-identical solutions by requiring the largest
    corner to be top right (took a long time to think of this constraint)'''
    return grid[0][0] >= max(grid[0][ny-1], grid[nx-1][0], grid[nx-1][ny-1])

def search(grid, values, next_xy=cluster, symmetry_ok=always_ok, depth=0):
    'search over all values, backtracking on failure'
    if symmetry_ok(grid):
        if values:
            (x, y) = next_xy(grid, depth)
            for value in values:
                if move_ok(grid, x, y, value):
                    if debug: print('add %d to %d,%d' % (value, x, y))
                    for result in search(copy_grid(grid, x, y, value),
                                         drop_value(value, values), 
                                         next_xy, symmetry_ok, depth+1):
                        yield result
            yield grid

# run the search, knowing that (2,2) (which is (1,1) for zero-indexing)
# has the value 10.
for result in search(copy_grid(grid, 1, 1, 10), drop_value(10, values)):

# how many solutions in total?
xy_moves = {} # reset cache
for (n, solution) in enumerate(search(grid, values, next_xy=to_corners,
for (n, solution) in enumerate(search(grid, values, next_xy=cluster,
    print('%d: %s' % (n, solution))

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