Conway's Game of Life without objects I was wondering if anyone could provide some feedback on my solution to Conway's game of life. I tried to solve it without using objects. It returns a dictionary of living points on a grid and points that were alive at some point. I'm wondering if there is a better way to check the surrounding points or if storing the active points in a dictionary is a bad idea.

def list_to_check(x,y):
    return  [(x + 1, y),
             (x - 1, y),
             (x +1, y +1 ),
             (x -1, y -1 ),
             (x +1, y -1 ),
             (x -1, y +1 ),
             (x, y+1),
             (x, y-1)]

def neighbors_count(x,y,cells):
    count = 0
    for item in list_to_check(x,y):     
       if item in cells:
           count += cells[item]
    return count 


def next_state(cells):
    next_state = {}
    for point in cells.keys():
        alive = cells[point] == 1
        neighbors = neighbors_count(point[0],point[1],cells)
        if not alive and neighbors == 3:
            next_state[point] = 1
        elif neighbors &gt; 3 or neighbors &lt; 2:
            next_state[point] = 0
    return next_state


def add_to_grid_alive(cells):
    to_check = []
    to_add = {}
    for item in cells.keys():
        checking = [ x for x in list_to_check(item[0],item[1]) if x not in to_check and x not in cells ]
        to_check += checking

    for item in to_check:
        if neighbors_count(item[0],item[1],cells) == 3:
            to_add[(item[0],item[1])] = 1
    return to_add


def tick(cells):
    changes = next_state(cells)
    expansions = add_to_grid_alive(cells)
    cells.update(changes)
    cells.update(expansions)
    [ cells.pop(point,None) for point in cells.keys() if cells[point] == 0 ]


def print_grid( size, cells):
    grid = []
    for y in range(size):
        grid.append([])
        for x in range(size):
            grid[y].append(0)
    for point in cells.keys():
        if  0 &lt;= point[0] &lt; size and 0 &lt;= point[1] &lt; size:
            grid[point[1]][point[0]] = cells[point]  
    return grid

#testing

def load_list(inputs):
    point_dict = {}
    for point in inputs:
        point_dict[point] = 1
    return point_dict




inputs = [(1,2),(2,2),(3,2),(2,3),(3,3),(4,3),(13,2),(13,3),(12,3),(13,4)]



test_array_dict = load_list(inputs)


for x in range(20):

    print '&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;', x + 1


    for item in print_grid(20,test_array_dict)[::-1]:
        print item
    tick(test_array_dict)


Update

I added [ cells.pop(point,None) for point in cells.keys() if cells[point] == 0 ] to the tick function and it really sped up the the code on higher iterations
You're representing the cells mainly using a dictionary of coordinates, converting to a two-dimensional array only for the purposes of printing the output.  That is a good representation for sparse boards, but not so much for crowded boards.

The algorithm that you use, though, is cumbersome, particularly illustrated by these few lines:


for item in cells.keys():
    checking = [ x for x in list_to_check(item[0],item[1]) if x not in to_check and x not in cells ]
    to_check += checking



In other words, make a list of the neighbors of each cell, but make sure that each coordinate is listed just once.  Then, once you have that list to_check, what do you do with it?


for item in to_check:
    if neighbors_count(item[0],item[1],cells) == 3:
        to_add[(item[0],item[1])] = 1



You count each cell's neighbors!  But the number of neighbors is precisely the amount of overlapped processing that would have occurred had you not bothered to deduplicate the to_check list to begin with!  So, instead of drawing a list of interesting cells and counting their neighbors, why not just have each existing cell increment the neighbor count of each neighboring coordinate?

In addition to the change in algorithm, I also recommend:


Take advantage of list/set/dict comprehensions more.
Represent the board as a set of live cells, rather than a dict.
Rename list_to_check() to neighbors().
Rename print_grid() to grid(), since it actually doesn't print anything.
Construct the next state from scratch rather than by mutation: test_array_dict = tick(test_array_dict)


I'd rewrite three functions as follows (and eliminate neighbors_count(), next_state(), and add_to_grid_alive():

def tick(live_cells):
    """ Takes a set of coordinates of live cells, and returns a set of
    coordinates of the live cells in the next generation. """
    neighbor_count = {}
    for xy in live_cells:
        for neighbor in neighbors(*xy):
            neighbor_count[neighbor] = 1 + neighbor_count.get(neighbor, 0)
    return set(xy for xy in neighbor_count
                   if neighbor_count[xy] == 3
                   or xy in live_cells and neighbor_count[xy] == 2)

def grid(size, cells):
    return [[int((x, y) in cells) for x in range(size)] for y in range(size)]

def load_list(inputs):
    return set(inputs)


Actually, tick() would be even better written using a collections.Counter.  I don't know if you consider that "using an object".