Question

我正在尝试编写一个程序，可以填充一个大小为2 ** nx 2 ** n的正方形网格，其中一个正方形被三角形移除（L形块占据三个正方形）。

函数的参数是n（电路板尺寸为2 ** n），removed_row和removed_column。该板是最初都设置为0的列表列表。然后我将删除的方块设置为全局计数变量（最初为1），然后递增计数。

所以我想要的是：如果n == 2那么在调用函数之后电路板应该是这样的：

[[4, 1, 3, 3]
 [4, 4, 2, 3]
 [5, 2, 2, 6]
 [5, 5, 6, 6]]

其中1是移除的正方形，2至6是三聚体。

我知道如何递归地解决这个问题，我只是将它放入代码中。

基本上我想将电路板分成4个象限1,2,3和4（1指的是右上象限，4指的是右下象限）。然后放一个三角形填充每个象限的正方形，这些象限没有填充或缺少正方形。然后在每个象限上重复此过程，直到所有方块都不等于零。

这是我认为应该有效的代码，但我显然在某处出错了，因为它适用的唯一案例是n == 1：

count = 1

def triominoes(n, removed_row=0, removed_column=0) :
    global count
    # initialization of the board:
    board = [ [0] * 2**n for i in range(2**n) ]

    # code for generating the board goes here
    # you can (and are encouraged to) use recursive helper function/s

    #Set the removed square to 1:
    board[removed_row][removed_column] = count
    count+=1    

    if (n==1):
        for row in range(len(board)) :
            for col in range(len(board[row])) :
                if (board[row][col] == 0) :
                    board[row][col] = count
        return board

    #Call the helper function on each sub-board
    triominoes_helper((n-1), removed_row, removed_column, 1, board) #Quadrant 1
    triominoes_helper((n-1), removed_row, removed_column, 2, board) #Quadrant 2
    triominoes_helper((n-1), removed_row, removed_column, 3, board) #Quadrant 3
    triominoes_helper((n-1), removed_row, removed_column, 4, board) #Quadrant 4

    return board




#def recursive_helper()
def triominoes_helper(n, removed_row, removed_column, location, board) :
    global count
    #Base Case
    if (n==1):
        for row in range(len(board)) :
            for col in range(len(board[row])) :
                if (board[row][col] == 0) :
                    board[row][col] = count
        return board

    #Recursive Case
    if (location == 1) :
        if (not is_removed(n, location, board)) :
            board[2**(n-1)-1][2**(n-1)] = count
    elif (location == 2) :
        if (not is_removed(n, location, board)) :
            board[2**(n-1)-1][2**(n-1)-1] = count
    elif (location == 3) :
        if (not is_removed(n, location, board)) :
            board[2**(n-1)][2**(n-1)-1] = count
    else :
        if (not is_removed(n, location, board)) :
            board[2**(n-1)][2**(n-1)] = count
    count+=1

    triominoes_helper(n-1, 2**(n-1)-1, 2**(n-1), 1, board)
    triominoes_helper(n-1, 2**(n-1)-1, 2**(n-1)-1, 2, board)
    triominoes_helper(n-1, 2**(n-1), 2**(n-1)-1, 3, board)
    triominoes_helper(n-1, 2**(n-1), 2**(n-1), 4, board)

#Function to see if a sub-board already has a square removed
def is_removed(n, location, board) :
    if (location == 1) :
        for r in range(0, 2**(n-1)-1) :
            for c in range(2**(n-1), 2**n-1) :              
                if (board[r][c] != 0) :
                    return True

    elif (location == 2) :
        for r in range(0, 2**(n-1)-1) :
            print(r)
            for c in range(0, 2**(n-1)-1) :
                print(c)

                if (board[r][c] != 0) :
                    return True

    elif (location == 3) :
        for r in range(2**(n-1), 2**n-1) :
            for c in range(0, 2**(n-1)-1) :
                if (board[r][c] != 0) :
                    return True

    else :
        for r in range(2**(n-1), 2**n-1) :
            for c in range(2**(n-1), 2**n-1) :
                if (board[r][c] != 0) :
                    return True

    return False




print(triominoes(2))

用递归三角形填充2 ** n方格？

0 个答案: