我如何知道国王在棋盘上移动而不走到对面的所有有效位置

Question

所以我有一个国际象棋棋盘，表示为一个大小为 64 的数组，左上角为 0，右下角为 63。我有这个函数，它给出了国王所有可能的移动。

current_pos = i
arr = np.array([i-9, i-8, i-7, i-1, i+1, i+7, i+8, 1+9])
return arr

.
.
.

if selected position is in arr:
    move king

其中 i 是国王当前所在方格的编号。

如果国王不在棋盘边缘，则此方法有效。

但如果国王在右下角，即 63 号，该函数将左下角 56 号作为国王移动的有效位置。

是否有任何有效的方法可以知道国王将走向另一边并且不是有效的举动？

我几乎所有的棋子都有同样的问题，该功能允许棋子在棋盘的另一侧移动，但我认为国王的移动是最简单的问题。

Answer 1

一维列表比二维 8x8 列表快得多，所以我喜欢您使用这种方法。

处理方法是使用 10x12 板，其中底部和顶部多出 2 行，左右多出一列：

然后在您的生成移动功能中，您只需检查您正在查看的方块是否在棋盘内。如果不是，则跳到循环中的下一个方块。

请在https://www.chessprogramming.org/10x12_Board 上阅读更多相关信息。它也是一个很好的网站，提供有关国际象棋编程的一般信息。

Answer 2

这是使用表查找的一种方法。

代码

piece_offsets = {
    'n': [-17, -15, -10, -6, 6, 10, 15, 17],
    'b': [ -9,  -7,  9, 7],
    'r': [ -8,  -1,  8, 1],
    'q': [ -9,  -8, -7, -1, 9,  8,  7,  1],
    'k': [ -9,  -8, -7, -1, 9,  8,  7,  1]
}

sqdist = [[0 for x in range(64)] for y in range(64)]

pseudo_legal = {    
    'n': [[] for y in range(64)],
    'b': [[] for y in range(64)],
    'r': [[] for y in range(64)],    
    'q': [[] for y in range(64)],
    'k': [[] for y in range(64)],
}


def distance(sq1, sq2):
   file1 = sq1 & 7
   file2 = sq2 & 7
   rank1 = sq1 >> 3
   rank2 = sq2 >> 3
   rank_distance = abs(rank2 - rank1)
   file_distance = abs(file2 - file1)
   return max(rank_distance, file_distance)


def print_board():
    for i in range(64):
        print(f'{i:02d} ', end='')
        if (i+1)%8 == 0:
            print()


def on_board(s):
    return s >= 0 and s < 64


def init_board():    
    for sq1 in range(64):
        for sq2 in range(64):
            sqdist[sq1][sq2] = distance(sq1, sq2)
            
    for pt in ['n', 'b', 'r', 'q', 'k']:
        for s in range(64):
            for offset in piece_offsets[pt]:
                to = s + offset
                if pt in ['k', 'n']:
                    if on_board(to) and sqdist[s][to] < 4:
                        pseudo_legal[pt][s].append(to)
                else:  # sliders
                    s1 = s
                    while True:
                        to1 = s1 + offset
                        if on_board(to1) and sqdist[s1][to1] < 4:
                            pseudo_legal[pt][s].append(to1)
                            s1 = to1
                        else:
                            break


def main():
    init_board()  # build sqdist and pseudo_legal_to tables
    
    print_board()
    print()  
    
    for pt in ['n', 'b', 'r', 'q', 'k']:
        for s in [0, 63, 36]:
            print(f'pt: {pt}, from: {s}: to: {pseudo_legal[pt][s]}')
        print()
        
    # pseudo_legal_sq = pseudo_legal['b'][61]
    # print(pseudo_legal_sq)
    
    
main()

输出

00 01 02 03 04 05 06 07 
08 09 10 11 12 13 14 15 
16 17 18 19 20 21 22 23 
24 25 26 27 28 29 30 31 
32 33 34 35 36 37 38 39 
40 41 42 43 44 45 46 47 
48 49 50 51 52 53 54 55 
56 57 58 59 60 61 62 63 

pt: n, from: 0: to: [10, 17]
pt: n, from: 63: to: [46, 53]
pt: n, from: 36: to: [19, 21, 26, 30, 42, 46, 51, 53]

pt: b, from: 0: to: [9, 18, 27, 36, 45, 54, 63]
pt: b, from: 63: to: [54, 45, 36, 27, 18, 9, 0]
pt: b, from: 36: to: [27, 18, 9, 0, 29, 22, 15, 45, 54, 63, 43, 50, 57]

pt: r, from: 0: to: [8, 16, 24, 32, 40, 48, 56, 1, 2, 3, 4, 5, 6, 7]
pt: r, from: 63: to: [55, 47, 39, 31, 23, 15, 7, 62, 61, 60, 59, 58, 57, 56]
pt: r, from: 36: to: [28, 20, 12, 4, 35, 34, 33, 32, 44, 52, 60, 37, 38, 39]

pt: q, from: 0: to: [9, 18, 27, 36, 45, 54, 63, 8, 16, 24, 32, 40, 48, 56, 1, 2, 3, 4, 5, 6, 7]
pt: q, from: 63: to: [54, 45, 36, 27, 18, 9, 0, 55, 47, 39, 31, 23, 15, 7, 62, 61, 60, 59, 58, 57, 56]
pt: q, from: 36: to: [27, 18, 9, 0, 28, 20, 12, 4, 29, 22, 15, 35, 34, 33, 32, 45, 54, 63, 44, 52, 60, 43, 50, 57, 37, 38, 39]

pt: k, from: 0: to: [9, 8, 1]
pt: k, from: 63: to: [54, 55, 62]
pt: k, from: 36: to: [27, 28, 29, 35, 45, 44, 43, 37]