ATBS Conway 的生命游戏代码似乎不对

如何解决ATBS Conway 的生命游戏代码似乎不对

我刚刚在 automatetheboringstuff.com 的帮助下开始学习 Python 我现在在第 4 章，试图理解康威的游戏代码。我了解游戏规则，但代码结果在我看来是错误的，我不确定代码是否真的错误或我不理解规则。

example (imgur picture) 为什么我认为这是错误的。 红点显示了一个死细胞，它有 3 个活着的邻居，这意味着它在下一步应该是活的，但它仍然是空的（死的）。黄点表示同样的问题。 Green 有 1 个活着的邻居，它保持活着，但不应该。

# Conway's Game of Life
import random,time,copy
WIDTH = 60
HEIGHT = 20

# Create a list of list for the cells:
nextCells = []
for x in range(WIDTH):
    column = [] # Create a new column.
    for y in range(HEIGHT):
        if random.randint(0,1) == 0:
            column.append('#') # Add a living cell.
        else:
            column.append(' ') # Add a dead cell.
    nextCells.append(column) # nextCells is a list of column lists.

while True: # Main program loop.
    print('\n\n\n\n\n') # Separate each step with newlines.
    currentCells = copy.deepcopy(nextCells)

    # Print currentCells on the screen:
    for y in range(HEIGHT):
        for x in range(WIDTH):
            print(currentCells[x][y],end='') # Print the # or space.
        print() # Print a newline at the end of the row.

    # Calculate the next step's cells based on current step's cells:
    for x in range(WIDTH):
        for y in range(HEIGHT):
            # Get neighboring coordinates:
            # `% WIDTH` ensures leftCoord is always between 0 and WIDTH - 1
            leftCoord  = (x - 1) % WIDTH
            rightCoord = (x + 1) % WIDTH
            aboveCoord = (y - 1) % HEIGHT
            belowCoord = (y + 1) % HEIGHT

            # Count number of living neighbors:
            numNeighbors = 0
            if currentCells[leftCoord][aboveCoord] == '#':
                numNeighbors += 1 # Top-left neighbor is alive.
            if currentCells[x][aboveCoord] == '#':
                numNeighbors += 1 # Top neighbor is alive.
            if currentCells[rightCoord][aboveCoord] == '#':
                numNeighbors += 1 # Top-right neighbor is alive.
            if currentCells[leftCoord][y] == '#':
                numNeighbors += 1 # Left neighbor is alive.
            if currentCells[rightCoord][y] == '#':
                numNeighbors += 1 # Right neighbor is alive.
            if currentCells[leftCoord][belowCoord] == '#':
                numNeighbors += 1 # Bottom-left neighbor is alive.
            if currentCells[x][belowCoord] == '#':
                numNeighbors += 1 # Bottom neighbor is alive.
            if currentCells[rightCoord][belowCoord] == '#':
                numNeighbors += 1 # Bottom-right neighbor is alive.

            # Set cell based on Conway's Game of Life rules:
            if currentCells[x][y] == '#' and (numNeighbors == 2 or
numNeighbors == 3):
                # Living cells with 2 or 3 neighbors stay alive:
                nextCells[x][y] = '#'
            elif currentCells[x][y] == ' ' and numNeighbors == 3:
                # Dead cells with 3 neighbors become alive:
                nextCells[x][y] = '#'
            else:
                # Everything else dies or stays dead:
                nextCells[x][y] = ' '
    time.sleep(1) # Add a 1-second pause to reduce flickering.

解决方法

在我看来，这个实现好像从上到下和从一边到另一边包裹着网格。因此，顶行中的单元格在底行中具有邻居（反之亦然），而最左侧列中的单元格在最右侧列中具有邻居（反之亦然）。所以：

• 红细胞没有活过来，因为它有五个活着的邻居：其中三个是你已经确定的细胞，另外两个在它上面并且在-左边。这两个其他两个都绕到了底行。

• 同样，黄色单元格有第四个邻居，在它的对角下方和左侧。这已绕到最右边的列。

• 存活的绿色单元格有两个邻居，一个在它的右边，一个在它的上面（再次环绕）。

最终，如果您要在有限网格上编写康威生命游戏的实现，您需要决定网格边缘会发生什么。看来这段代码的作者选择让网格环绕。