我的代码
我正试图解决一些骑士之旅问题:在N * N国际象棋棋盘上找到从A点到B点的最小移动次数
我创建了一个电路板,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 candidate list
5. go to step 2 (counter is incremented after all previous level sons are popped)
这个算法按照我的预期工作(在一些测试用例中使用它),但它非常慢:
电话f = Find_route(20, Tile(4,4), Tile(14,11))
(20是电路板尺寸,Tile(4,4)和Tile(14,11)分别是开始和结束位置)已检查201590( !!)瓦片在到达答案之前。
我尝试通过sorted(tiles, key = lambda e : abs(e.x - end.x)+abs(e.y - end.y))
对候选人列表进行排序来优化它,其中tiles
是候选人列表。这适用于某些情况,但对某些情况来说它有点无用。
有用的案例:
f = Find_route(20, Tile(1,4), Tile(1,10)) from 459 to 309 (~33% !!)
f = Find_route(20, Tile(7,0), Tile(1,11)) from 87738 to 79524 (~10% :( )
无益的案例:
f = Find_route(20, Tile(4,4), Tile(14,11)): from 201891 to 201590
f = Find_route(20, Tile(1,4), Tile(1,11)) from 2134 to 2111
我希望最终得到一个近端案例列表,算法会从中确切地知道要做什么,(类似于5个tile的半径),我认为这可能有所帮助,但我对如何改进optimize_list
方法更感兴趣。有什么提示吗?
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())
修改
我添加了另一个优化级别 - 在任何新的候选切片插入时优化列表,在某些情况下它似乎像魅力一样:
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())
我有计算跳数的问题,因为我不知道何时递增计数器(可能在每次检查?),但它似乎至少会收敛得更快。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(不确定它是否会停止......)
答案 0 :(得分:1)
不要重新发明轮子。
使用切片作为顶点构建图形。如果骑士可以一步从一个瓷砖到另一个瓷砖,则用边缘连接瓷砖。
使用标准路径查找算法。广度优先搜索看起来像是您在未加权图表中寻找最短路径的最佳选择。