8个难题计算邻居python

时间:2014-09-21 02:35:30

标签: python algorithm

我正在编写一个算法来解决这个难题。 https://en.wikipedia.org/wiki/15_puzzle

状态由元组(1,2,3,4,5,6,7,8,0)表示,其中0是空闲框(这相当于3 * 3矩阵)。

考虑到拼图的状态p =(1,3,2,5,4,6,0,7,8),我编写了一个计算邻居的函数。

def neighbours(p):
    p = np.array(p).reshape(3,3)
    ind0 = np.argwhere(p == 0).squeeze()
    x0, y0 = ind0

    ind_neig = [[ind0[0] - 1, ind0[1]], 
                [ind0[0] + 1, ind0[1]],
                [ind0[0], ind0[1] - 1],
                [ind0[0], ind0[1] + 1]]

    ind_neig = [ind for ind in ind_neig if min(ind) >= 0 and max(ind) < 3]

    neig = [ p.copy() for i in range(len(ind_neig))]

    for i in range(len(ind_neig)):
        x, y = ind_neig[i]
        neig[i][x0, y0] = p[x, y]
        neig[i][x, y] = 0

    neig = [tuple(np.ravel(i)) for i in neig]
    return neig

我想要一个更快版本的邻居函数,并且可能不需要numpy库。特别是我想要一个可以在拼图具有更大尺寸时使用的功能,例如15拼图

1 个答案:

答案 0 :(得分:3)

我认为您会发现以下实现非常简单,而且,它不使用numpy:

def neighbours(p):
    res = []
    ind = p.index(0)
    if not top(ind):
        res.append(up_neig(p, ind))
    if not bottom(ind):
        res.append(down_neig(p, ind))
    if not left(ind):
        res.append(left_neig(p, ind))
    if not right(ind):
        res.append(right_neig(p, ind))
    return res

def top(ind):
    return ind < 3

def bottom(ind):
    return ind > 5

def left(ind):
    return ind in [0, 3, 6]

def right(ind):
    return ind in [2, 5, 8]

def up_neig(p, ind):
    _p = list(p)
    _p[ind], _p[ind-3] = _p[ind-3], _p[ind]
    return tuple(_p)

def down_neig(p, ind):
    _p = list(p)
    _p[ind], _p[ind+3] = _p[ind+3], _p[ind]    
    return tuple(_p)

def left_neig(p, ind):
    _p = list(p)
    _p[ind], _p[ind-1] = _p[ind-1], _p[ind]
    return tuple(_p)

def right_neig(p, ind):
    _p = list(p)
    _p[ind], _p[ind+1] = _p[ind+1], _p[ind]
    return tuple(_p)

p = (1,3,2,5,4,6,0,7,8)
print neighbours(p)  # [(1, 3, 2, 0, 4, 6, 5, 7, 8), (1, 3, 2, 5, 4, 6, 7, 0, 8)]