康威的生命博弈——邻里关系问题

Question

我正试图让一个“生活游戏”在python中运行，但我似乎无法让程序的邻居计数方面正常工作。我正在尝试使用多个if语句来避免数组中出现任何“越界”错误，这已经奏效了。然而，该项目正在寻找不存在的“邻居”，导致该项目在不应该的情况下到处产生活细胞。任何帮助都将不胜感激，我真的被困在这件事上了

额外信息：列表为10x10，单元格检查所有八个方向

谢谢

def checkNeighbours(newWorld, row, column):
   neighboursCount = 0
   if(column > 0):
       if(newWorld[row][column-1] == ALIVE): #checks cell to left
           neighboursCount = neighboursCount + 1
   if(row < 9):
       if(newWorld[row+1][column] == ALIVE): #checks cell below
           neighboursCount = neighboursCount + 1
   if(row < 9 and column < 9):
       if(newWorld[row+1][column+1] == ALIVE): #checks cell to bottom right
           neighboursCount = neighboursCount + 1
   if(row < 9 and column > 0):
       if(newWorld[row+1][column-1] == ALIVE): #checks cell to bottom left
           neighboursCount = neighboursCount + 1
   if(column < 9):
       if(newWorld[row][column+1] == ALIVE): #checks cell to right
           neighboursCount = neighboursCount + 1
   if(row > 0 and column > 0):
       if(newWorld[row-1][column-1] == ALIVE): #checks cell to top left
           neighboursCount= neighboursCount + 1
   if(row > 0):
       if(newWorld[row-1][column] == ALIVE): #checks cell above
           neighboursCount = neighboursCount + 1
   if(row != 0 and column != 9):
       if(newWorld[row-1][column+1] == ALIVE): #checks cell to top right
           neighboursCount = neighboursCount + 1
   return(neighboursCount)


def birthsAndDeaths(newWorld, row, column, neighboursCount):
   if(neighboursCount == 3 and newWorld[row][column] == DEAD):
       return(ALIVE)
   elif(neighboursCount == 2 or neighboursCount == 3 and newWorld[row][column] == ALIVE):
       return(ALIVE)
   elif(neighboursCount <= 1):
       return(DEAD)
   elif(neighboursCount >= 4):
       return(DEAD) 

def turnChanger(turn, newWorld):
   turn = turn + 1
   row = 0
   while (row < SIZE): # Each iteration accesses a single row 
       column = 0
       while (column < SIZE):  # Each iteration accesses a single column in the given row
           neighboursCount = checkNeighbours(newWorld, row, column)
           print(neighboursCount, "for cell", row, column)
           newWorld[row][column] = birthsAndDeaths(newWorld, row, column, neighboursCount)
           column = column + 1
       row = row + 1  
   return(turn, newWorld)

def start():
   choice = 0
   choice = selection()
   world = worldChooser(choice)

   oldWorld = []
   newWorld = []
   turn = 0

   oldWorld = world
   newWorld = world
   
   turn, newWorld = turnChanger(turn, newWorld)
   display(turn, oldWorld, newWorld)   
   
   world = newWorld
start()

Answer 1

问题就在birthsAndDeaths函数中，在这一行：

elif(neighboursCount == 2 or neighboursCount == 3 and newWorld[row][column] == ALIVE):

由于操作的顺序，必须确保首先计算or，因此在该部分周围使用括号

elif((neighboursCount == 2 or neighboursCount == 3) and newWorld[row][column] == ALIVE):

任何and运算符都是在or之前计算的，因此在代码中，当任何单元格旁边有两个相邻单元格时，它都会将单元格设置为活动的

根据你当前的代码，假设我们有两个邻居在一个死囚的旁边，它就会消失

elif(neighboursCount == 2 or neighboursCount == 3 and newWorld[row][column] == ALIVE):

然后

elif(true or false and false):

然后

elif(true or false):

然后

elif(true):

并且elif的主体返回一个ALIVE结果

在or周围使用括号会导致首先计算这些括号的内容。在这种情况下，内容是一个or操作。使用括号意味着我们将得到

elif((neighboursCount == 2 or neighboursCount == 3) and newWorld[row][column] == ALIVE):

然后

elif((true or false) and false):

然后

elif(true and false):

然后

elif(false):

这是您在这里询问的特定问题的解决方案。然而，Carcigenicate指出了另一个主要问题。如果你试图解决这个问题，但仍然有问题，你可以发布一个新问题