我目前正在解决this assignment中的第二个练习(这不是家庭作业,我实际上是在尝试解决this other problem)。我的解决方案使用BFS搜索“ Lights Out”问题的变体的最小解决方案,在该问题中,按下一个灯将翻转同一行和同一列上每盏灯的状态。
我认为我的实现是正确的,但是有点太慢了:目前在我的计算机上运行需要12秒钟以上的时间(这对我来说是不可接受的)。
from copy import deepcopy
from itertools import chain
from Queue import PriorityQueue
# See: http://www.seas.upenn.edu/~cis391/Homework/Homework2.pdf
class Puzzle(object):
def __init__(self, matrix):
self.matrix = matrix
self.dim = len(matrix)
def __repr__(self):
return str(self.matrix)
def solved(self):
return sum([sum(row) for row in self.matrix]) == 0
def move(self, i, j):
for k in range(self.dim):
self.matrix[i][k] = (self.matrix[i][k] + 1) % 2
self.matrix[k][j] = (self.matrix[k][j] + 1) % 2
self.matrix[i][j] = (self.matrix[i][j] + 1) % 2
return self
def copy(self):
return deepcopy(self)
def next(self):
result = []
for i in range(self.dim):
for j in range(self.dim):
result.append(self.copy().move(i, j))
return result
def solve(self):
q = PriorityQueue()
v = set()
q.put((0, self))
while True:
c = q.get()
if c[1].solved():
return c[0]
else:
for el in c[1].next():
t = el.tuple()
if t not in v:
v.add(t)
q.put((c[0] + 1, el))
def tuple(self):
return tuple(chain.from_iterable(self.matrix))
根据cProfile
的罪魁祸首似乎是deepcopy
呼叫。另一方面,我没有其他选择:我需要向队列中添加另一个Puzzle
对象,其中包含self.matrix
的新副本。
如何加快实施速度?
这是我正在使用的测试用例:
print Puzzle([
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
]).solve()
它应该返回1(我们只需要按右下角的灯)。
编辑:这是另一个过时的测试案例:
print Puzzle([
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0]
]).solve()
它的解决方案最多为14个:按对角线上所有已经点亮的灯。不幸的是,@ zch令人印象深刻的加速还不足以解决此问题,使我相信,由于分支系数高,BFS并不是解决此问题的正确方法。
python大神给出的解决方案
有许多优化要完成。
首先,避免deepcopy
,实现自己的复制(这本身对我来说快5倍):
class Puzzle(object):
def __init__(self, matrix):
self.matrix = [list(row) for row in matrix]
self.dim = len(matrix)
def copy(self):
return Puzzle(self.matrix)
其次,在BFS中,您不需要优先级队列,可以使用Queue
或实现自己的排队。这样可以加快速度。第三,在将其放入队列之前(而不是在取出东西之后)检查是否已解决。这样可以使您在可比的时间内更深一层:
def solve(self):
v = set()
q = [(0, self)]
i = 0
while True:
c = q[i]
i += 1
for el in c[1].next():
t = el.tuple()
if t not in v:
if el.solved():
return c[0] + 1
v.add(t)
q.append((c[0] + 1, el))
此外,使用位列表的列表在存储方面非常低效。您可以将所有位打包为一个整数,以获得更快的解决方案。此外,您可以为允许的移动预先计算蒙版:
def bits(iterable):
bit = 1
res = 0
for elem in iterable:
if elem:
res |= bit
bit <<= 1
return res
def mask(dim, i, j):
res = 0
for idx in range(dim * i, dim * (i + 1)):
res |= 1 << idx
for idx in range(j, dim * dim, dim):
res |= 1 << idx
return res
def masks(dim):
return [mask(dim, i, j) for i in range(dim) for j in range(dim)]
class Puzzle(object):
def __init__(self, matrix):
if isinstance(matrix, Puzzle):
self.matrix = matrix.matrix
self.dim = matrix.dim
self.masks = matrix.masks
else:
self.matrix = bits(sum(matrix, []))
self.dim = len(matrix)
self.masks = masks(len(matrix))
def __repr__(self):
return str(self.matrix)
def solved(self):
return self.matrix == 0
def next(self):
for mask in self.masks:
puzzle = Puzzle(self)
puzzle.matrix ^= mask
yield puzzle
def solve(self):
v = set()
q = [(0, self)]
i = 0
while True:
c = q[i]
i += 1
for el in c[1].next():
t = el.matrix
if t not in v:
if el.solved():
return c[0] + 1
v.add(t)
q.append((c[0] + 1, el))
最后,对于另外5倍,您可以传递一些位矩阵,而不是整个Puzzle
对象,并内联一些最常用的函数。
def solve(self):
v = set()
q = [(0, self.matrix)]
i = 0
while True:
dist, matrix = q[i]
i += 1
for mask in self.masks:
t = matrix ^ mask
if t not in v:
if t == 0:
return dist + 1
v.add(t)
q.append((dist + 1, t))
对我来说,这些优化相结合可以使速度提高约250倍。