Python 优化Knight';让我们在棋盘上游览
下面是我的代码 我想解决一个骑士之旅的小问题:在N*N棋盘上找到从a点到B点的最小移动次数 我创建了一个板,并使用了一个简单的算法:Python 优化Knight';让我们在棋盘上游览,python,algorithm,Python,Algorithm,下面是我的代码 我想解决一个骑士之旅的小问题:在N*N棋盘上找到从a点到B点的最小移动次数 我创建了一个板,并使用了一个简单的算法: 1. add point A to candidate list and start loop: 2. pop first element in candidate list and check it: 3. if end - return counter 4. else - add the candidate 's "sons" to end of ca
1. add point A to candidate list and start loop:
2. pop first element in candidate list and check it:
3. if end - return counter
4. else - add the candidate 's "sons" to end of candidate list
5. go to step 2 (counter is incremented after all previous level sons are popped)
该算法的工作原理与我预期的一样(在一些测试用例中使用),但速度非常慢:
调用f=Find_route(20,Tile(4,4),Tile(14,11))
(20是电路板尺寸,Tile(4,4)和Tile(14,11)分别是起始和结束位置)在到达答案之前检查了201590(!!)tiles
我尝试通过使用sorted(tiles,key=lambda e:abs(e.x-end.x)+abs(e.y-end.y))
对候选列表进行排序来优化它,其中tiles
是候选列表。这在某些情况下是有效的,但在某些情况下是无用的
有用的案例:
f=从459到309(~33%!!)查找路径(20,磁贴(1,4),磁贴(1,10)
f=找到从87738到79524(~10%:())的路线(20,磁贴(7,0),磁贴(1,11))
f=Find_路线(20,Tile(4,4),Tile(14,11)):从201891到201590
f=从2134到2111查找路径(20,磁贴(1,4),磁贴(1,11)
优化列表方法。有什么建议吗
代码
class Tile(object):
def __init__(self, x, y):
self.x = x
self.y = y
def __str__(self):
tmp = '({0},{1})'.format(self.x, self.y)
return tmp
def __eq__(self, new):
return self.x == new.x and self.y == new.y
def get_horse_jumps(self, max_x , max_y):
l = [(1,2), (1,-2), (-1,2), (-1,-2), (2,1), (2,-1), (-2,1), (-2,-1)]
return [Tile(self.x + i[0], self.y + i[1]) for i in l if (self.x + i[0]) >= 0 and (self.y + i[1]) >= 0 and (self.x + i[0]) < max_x and (self.y + i[1]) < max_y]
class Board(object):
def __init__(self, n):
self.dimension = n
self.mat = [Tile(x,y) for y in range(n) for x in range(n)]
def show_board(self):
print('-'*20, 'board', '-'*20)
n = self.dimension
s = ''
for i in range(n):
for j in range(n):
s += self.mat[i*n + j].__str__()
s += '\n'
print(s,end = '')
print('-'*20, 'board', '-'*20)
class Find_route(Board):
def __init__(self, n, start, end):
super(Find_route, self).__init__(n)
#self.show_board()
self.start = start
self.end = end
def optimize_list(self, tiles, end):
return sorted(tiles, key = lambda e : abs(e.x - end.x)+abs(e.y - end.y))
def find_shortest_path(self, optimize = False):
counter = 0
sons = [self.start]
next_lvl = []
num_of_checked = 0
while True:
curr = sons.pop(0)
num_of_checked += 1
if curr == self.end:
print('checked: ', num_of_checked)
return counter
else: # check sons
next_lvl += curr.get_horse_jumps(self.dimension, self.dimension)
# sons <- next_lvl (optimize?)
# next_lvl <- []
if sons == []:
counter += 1
if optimize:
sons = self.optimize_list(next_lvl, self.end)
else:
sons = next_lvl
next_lvl = []
optimize = True
f = Find_route(20, Tile(7,0), Tile(1,11))
print(f.find_shortest_path(optimize))
print(f.find_shortest_path())
我在计算跳转次数时遇到问题,因为我不知道何时增加计数器(可能在每次检查时?),但它似乎至少收敛得更快。此外,对于其他情况(例如f=Find_route(20,Tile(1,4),Tile(8,17))
)它根本没有改善(不确定是否停止…不要重新发明轮子
- 构建一个以分片为顶点的图。如果骑士可以一步从一个分片到另一个分片,则将分片与边连接
- 使用标准的路径查找算法。宽度优先搜索看起来是在未加权图中寻找最短路径的最佳选择
这就是我打算做的,我认为这应该是一种bfs。我不想构建一个图,因为我认为构建过程会很长,但看到我的运行时,我想我会重新考虑
if optimize == 2:
if sons == []:
#counter += 1
sons = self.optimize_list(next_lvl, self.end)
else:
sons = self.optimize_list(sons + next_lvl, self.end)
else:
if sons == []:
counter += 1
if optimize == 1:
sons = self.optimize_list(next_lvl, self.end)
else:
sons = next_lvl
next_lvl = []
optimize = 2
f = Find_route(20, Tile(1,4), Tile(8,18)) # from 103761 to 8 ( optimal!!! )
print(f.find_shortest_path(optimize))
print(f.find_shortest_path())