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python - 如何优化国际象棋引擎的对角线运动代码？

问题描述

我目前正在构建自己的国际象棋引擎，并且可以真正使用一些关于如何使这段代码更有效地计算对角线移动的建议。（这显然只适用于右上的对角线）

截至目前，我正在使用“Try-Except”迭代 1，然后我的 return 语句过滤掉任何非板载值。然而，这似乎是一种非常庞大的做事方式。

任何有关如何重构此代码的评论或建议将不胜感激。

import argparse, json

chessBoard = [[1, 1, 1, 1, 1, 1, 1, 1] for i in range(8)]

chess_map_from_alpha_to_index = {
    "a" : 0,
    "b" : 1,
    "c" : 2,
    "d" : 3,
    "e" : 4,
    "f" : 5,
    "g" : 6,
    "h" : 7
}

chess_map_from_index_to_alpha = {
    0: "a",
    1: "b",
    2: "c",
    3: "d",
    4: "e",
    5: "f",
    6: "g",
    7: "h"
}


def getBishopMoves(pos, chessBoard):
    column, row = list(pos.strip().lower())
    row = int(row) - 1
    column = chess_map_from_alpha_to_index[column]
    i,j = row, column
    solutionMoves = []

#Up-Right Diagonal
    try:
        temp = chessBoard[i + 1][j + 1]
        solutionMoves.append([i + 1, j + 1])
    except:
        pass
    try:
        temp = chessBoard[i + 2][j + 2]
        solutionMoves.append([i + 2, j + 2])
    except:
        pass
    try:
        temp = chessBoard[i + 3][j + 3]
        solutionMoves.append([i + 3, j + 3])
    except:
        pass
    try:
        temp = chessBoard[i + 4][j + 4]
        solutionMoves.append([i + 4, j + 4])
    except:
        pass
    try:
        temp = chessBoard[i + 5][j + 5]
        solutionMoves.append([i + 5, j + 5])
    except:
        pass
    try:
        temp = chessBoard[i + 6][j + 6]
        solutionMoves.append([i + 6, j + 6])
    except:
        pass
    try:
        temp = chessBoard[i + 7][j + 7]
        solutionMoves.append([i + 7, j + 7])
    except:
        pass    
    try:
        temp = chessBoard[i + 7][j + 7]
        solutionMoves.append([i + 7, j + 7])
    except:
        pass   

    temp = [i for i in solutionMoves if i[0] >=0 and i[1] >=0]
    solutionMoves = ["".join([chess_map_from_index_to_alpha[i[1]], str(i[0] + 1)]) for i in temp]
    solutionMoves.sort()
    return solutionMoves

标签： pythonloopsoptimizationchesstry-except

解决方案

对于主教，必须考虑四个不同的方向：右上、右下、左下、左上。对于所有四个选项，您可以找到每个选项的最大投影坐标，然后对所有选项进行迭代。如果在迭代过程中遇到另一个片段，则循环停止。该算法的可能实现如下：

board = [['-', '-', '-', '-', '-', '-', '-', '-'], 
         ['-', '-', '-', '-', '-', '-', '-', '-'], 
         ['-', '-', '-', '-', '-', '-', '-', '-'], 
         ['-', '-', '-', '-', '-', '-', '-', '-'], 
         ['-', '-', '-', 'b', '-', '-', '-', '-'], 
         ['-', '-', '-', '-', '-', '-', '-', '-'], 
         ['-', '-', '-', '-', '-', '-', '-', '-'], 
         ['-', '-', '-', '-', '-', '-', '-', '-']]
options = [lambda x, y:(x+1, y+1), lambda x, y:(x+1, y-1), lambda x, y:(x-1, y-1), lambda x, y:(x-1, y+1)]
projected = [lambda x, y:(x+min([x,y]), y+min([x,y])), lambda x, y:(x+min([x,y]), y-min([x,y])), lambda x, y:(x-min([x,y]), y-min([x,y])), lambda x, y:(x-min([x,y]), y+min([x,y]))]
def iterate_towards(current, to, func):
   while current != to:
      current = list(func(*current))
      if board[current[0]][current[-1]] != '-':
        break
      yield current
   yield to

def get_bishop_moves(current:list):
   for a, b in zip(options, projected):
      yield list(iterate_towards([3, 4], list(b(*current)), a))

print(list(get_bishop_moves([3, 4])))

输出：

[[[4, 5], [5, 6], [6, 7]], [[4, 3], [5, 2], [6, 1]], [[2, 3], [1, 2], [0, 1]], [[2, 5], [1, 6], [0, 7]]]

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