骑士巡回问题的Warnsdorff算法

简介

骑士巡回问题指的是一个国际象棋中的棋子骑士(Knight)在棋盘上移动，使得最终走遍所有的格子，但每格仅能经过一次。Warnsdorff算法是一种启发式算法，用于解决骑士巡回问题，它可以在较短的时间内找到一个近似最优解。

算法步骤

1. 选择一个起始位置，将其标记为已访问。
2. 对于当前位置，计算出所有可以到达但未被访问过的下一个位置，按照每个位置所能到达未访问位置数量从小到大排序。
3. 选取上一步中可行位置中访问次数最少的位置作为下一步的位置，并将其标记为已访问。
4. 重复2-3步，直到没有未访问位置可到达。

代码实现

def warnsdorff(n: int, start: tuple) -> list:
    # 棋盘大小为n*n
    chessboard = [[0] * n for _ in range(n)]
    # 记录每个位置的可到达位置数
    move_count = [[0] * n for _ in range(n)]
    # 骑士移动的8种方向
    directions = [(2, 1), (1, 2), (-1, 2), (-2, 1), (-2, -1), (-1, -2), (1, -2), (2, -1)]
    
    # 计算从当前位置出发可以到达的未访问位置数量
    def count_moves(x: int, y: int):
        count = 0
        for dx, dy in directions:
            nx, ny = x + dx, y + dy
            if 0 <= nx < n and 0 <= ny < n and chessboard[nx][ny] == 0:
                count += 1
        return count
    
    # 标记当前位置已经访问过，并将访问顺序记录在solution中
    def visit(curr_x: int, curr_y: int, step: int):
        chessboard[curr_x][curr_y] = 1
        solution.append((curr_x, curr_y))
        return step + 1
    
    solution = []
    curr_x, curr_y = start
    step = 0
    
    while step < n * n:
        visit(curr_x, curr_y, step)
        move_count = [[count_moves(x, y) for y in range(n)] for x in range(n)] # 计算所有位置的可行度
        next_moves = [(curr_x + dx, curr_y + dy) for dx, dy in directions if 0 <= curr_x + dx < n and 0 <= curr_y + dy < n and chessboard[curr_x + dx][curr_y + dy] == 0] # 所有可以到达的位置
        next_moves.sort(key=lambda x: move_count[x[0]][x[1]])
        if next_moves:
            curr_x, curr_y = next_moves[0]
            step += 1
        else:
            break
            
    return solution

使用方法

将代码实现部分复制到自己的代码中，即可使用warnsdorff函数。函数接受一个棋盘大小n和一个起始位置(x, y)，返回一个按照访问顺序的位置列表。

例如，要访问一个大小为8*8的棋盘，从位置(0, 0)开始，可以这样调用函数：

solution = warnsdorff(8, (0, 0))
print(solution)

输出：

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

注意事项

1. Warnsdorff算法并不能保证得到最优解，但其解通常是非常接近最优解的。
2. 如果起始位置不同，得到的解可能会不同。