I took a liking to Conway's Game of Life and began to try and write it in python. At this moment I have yet to write any code for the borders of the program so I am just asking for help with what I have right now. I seem to have trouble when initialization a "blinker" formation. Instead of oscillating like it should, it seems to turn itself into a cube.
#File: gameoflife.py
#Description: This is a basic "life" simulation that follows three distinct rules:
#1.Any live cell with fewer than two live neighbours dies
#2.Any live cell with two or three live neighbours lives
#3.Any live cell with more than three live neighbours dies
#4.Any dead cell with exactly three live neighbours becomes a live cell
#A neighbor is deemed as any cell directly horizantal/vertical/diagonal
#meaning there are 9 neighbors to a cell at any time
from graphics import *
import random
from time import sleep
def initial(D,M,win):
#Creates the white board background
for i in range (11):
m = [] # rectangle list
for j in range (11):
rec = Rectangle(Point(6 + 4 * i, 6 + 4 * j), Point(10 + 4 * i, 10 + 4 * j))
D[i][j] = 0
rec.setFill("white")
rec.draw(win)
m.append(rec)
M.append(m)
def check(x,y,D):
#Checks all 9 adjacent neihbors to see if "alive" and checks this number
#means the cell should stay alive(1), die(0), or if already dead come
#back alive(2)
counter = 0
if D[x+1][y] == 1:
counter += 1
if D[x-1][y] == 1:
counter += 1
if D[x][y+1] == 1:
counter += 1
if D[x][y-1] == 1:
counter +=1
if D[x+1][y+1] == 1:
counter+=1
if D[x+1][y-1] == 1:
counter+= 1
if D[x-1][y-1] == 1:
counter += 1
if D[x-1][y+1] == 1:
counter +=1
if counter<2 or counter>3:
return 0
if counter == 2:
return 1
if counter == 3:
return 2
def main():
win = GraphWin("Game of Life", 700, 600)
win. setCoords(0, 0, 70, 60)
#Initialize two dimesion arrays.
#D records color of grids, M records rectangles
D = []
M = []
C = []
#initialize the grids to create all "white"
for i in range(11):
d = []
c = []
for j in range(11):
d.append(0)
c.append(0)
D.append(d)
C.append(c)
initial(D,M,win)
#Initialzes three "alive" units
D[5][5],D[4][5] ,D[6][5]= 1,1,1
C[5][5],C[4][5] ,C[6][5]= 1,1,1
M[5][5].setFill("Black")
M[4][5].setFill("Black")
M[6][5].setFill("Black")
#Contiually checking
while True:
#Purposfully not checking the "Borders" of the array
for i in range (len(D)-1):
for j in range(len(D[i])-1):
#If the cell is alive
if D[i][j] == 1:
#If the cell should die
if check(i,j,D) == 0:
sleep(1)
#Sets a temporary list to white
C[i][j] = 0
#Fills the cell white
M[i][j].setFill("White")
#if the cell is dead
if D[i][j] == 0:
#If the cell should be revived
if check(i,j,D) == 2:
sleep(1)
#Sets a temporary list to black
C[i][j] = 1
#Fills the cell black
M[i][j].setFill("Black")
#Sets the main list = to the temporary list
D = C
main()
You will need to swap D and C, and not just assign C to D. As it stands now, D and C will be referring to the same list after the first iteration.
Here is a simple algorithm to do Conway's Game of Life in python using a numpy
array of arbitrary 2D size:
import numpy
# this function does all the work
def play_life(a):
xmax, ymax = a.shape
b = a.copy() # copy grid & Rule 2
for x in range(xmax):
for y in range(ymax):
n = numpy.sum(a[max(x - 1, 0):min(x + 2, xmax), max(y - 1, 0):min(y + 2, ymax)]) - a[x, y]
if a[x, y]:
if n < 2 or n > 3:
b[x, y] = 0 # Rule 1 and 3
elif n == 3:
b[x, y] = 1 # Rule 4
return(b)
# replace (5, 5) with the desired dimensions
life = numpy.zeros((5, 5), dtype=numpy.byte)
# place starting conditions here
life[2, 1:4] = 1 # a simple "spinner"
# now let's play
print(life)
for i in range(3):
life = play_life(life)
print(life)
This is not very efficient, but will certainly get the job done. Replace print(life)
with whatever graphical calls you prefer.
来源:https://stackoverflow.com/questions/8484802/conways-game-of-life-with-python