How to improve performance of this code?
Question:
Thanks to some help from people here, I was able to get my code for Tasmanian camels puzzle working. However, it is horribly slow (I think. I’m not sure because this is my first program in Python). The example run in the bottom of the code takes a long time to be solved in my machine:
dumrat@dumrat:~/programming/python$ time python camels.py
[['F', 'F', 'F', 'G', 'B', 'B', 'B'], ['F', 'F', 'G', 'F', 'B', 'B', 'B'],
['F', 'F', 'B', 'F', 'G', 'B', 'B'], ['F', 'F', 'B', 'F', 'B', 'G', 'B'],
['F', 'F', 'B', 'G', 'B', 'F', 'B'], ['F', 'G', 'B', 'F', 'B', 'F', 'B'],
['G', 'F', 'B', 'F', 'B', 'F', 'B'], ['B', 'F', 'G', 'F', 'B', 'F', 'B'],
['B', 'F', 'B', 'F', 'G', 'F', 'B'], ['B', 'F', 'B', 'F', 'B', 'F', 'G'],
['B', 'F', 'B', 'F', 'B', 'G', 'F'], ['B', 'F', 'B', 'G', 'B', 'F', 'F'],
['B', 'G', 'B', 'F', 'B', 'F', 'F'], ['B', 'B', 'G', 'F', 'B', 'F', 'F'],
['B', 'B', 'B', 'F', 'G', 'F', 'F']]
real 0m20.883s
user 0m20.549s
sys 0m0.020s
Here’s the code:
import Queue
fCamel = 'F'
bCamel = 'B'
gap = 'G'
def solution(formation):
return len([i for i in formation[formation.index(fCamel) + 1:]
if i == bCamel]) == 0
def heuristic(formation):
fCamels, score = 0, 0
for i in formation:
if i == fCamel:
fCamels += 1;
elif i == bCamel:
score += fCamels;
else:
pass
return score
def getneighbors (formation):
igap = formation.index(gap)
res = []
# AB_CD --> A_BCD | ABC_D | B_ACD | ABD_C
def genn(i,j):
temp = list(formation)
temp[i], temp[j] = temp[j], temp[i]
res.append(temp)
if(igap > 0):
genn(igap, igap-1)
if(igap > 1):
genn(igap, igap-2)
if igap < len(formation) - 1:
genn(igap, igap+1)
if igap < len(formation) - 2:
genn(igap, igap+2)
return res
class node:
def __init__(self, a, g, p):
self.arrangement = a
self.g = g
self.parent = p
def astar (formation, heuristicf, solutionf, genneighbors):
openlist = Queue.PriorityQueue()
openlist.put((heuristicf(formation), node(formation, 0, None)))
closedlist = []
while 1:
try:
f, current = openlist.get()
except IndexError:
current = None
if current is None:
print "No solution found"
return None;
if solutionf(current.arrangement):
path = []
cp = current
while cp != None:
path.append(cp.arrangement)
cp = cp.parent
path.reverse()
return path
#arr = current.arrangement
closedlist.append(current)
neighbors = genneighbors(current.arrangement)
for neighbor in neighbors:
if neighbor in closedlist:
pass
else:
openlist.put((current.g + heuristicf(neighbor),
node(neighbor, current.g + 1, current)))
#sorted(openlist, cmp = lambda x, y : x.f > y.f)
def solve(formation):
return astar(formation, heuristic, solution, getneighbors)
print solve([fCamel, fCamel, fCamel, gap, bCamel, bCamel, bCamel])
#print solve([fCamel, fCamel, fCamel, fCamel, gap, bCamel, bCamel, bCamel, bCamel])
That is just for 3 camels each. I wanted to do this for 4 at least. That test case is still running (It’s been about 5 minutes now :(). I’ll update this if and when it finishes.
What should I do to improve this code? (Mostly performance-wise, but any other suggestions are welcome also).
Answers:
Replacing
class node:
def __init__(self, a, g, p):
self.arrangement = a
self.g = g
self.parent = p
with
node = collections.namedtuple('node', 'arrangement, g, parent')
dropped the times from around 340-600 msecs to 11.4 1.891 msecs on the input [fCamel, fCamel, gap, bCamel, bCamel]. It yielded the same output.
This obviously won’t help with any algorithmic problems but as far as micro-optimizations go, it isn’t bad.
1 I had the wrong input. There was an extra fCamel that was making it run slower.
I’ve been tripped up by this before too. The bottleneck here is actually if neighbor in closedlist.
The in statement is so easy to use, you forget that it’s linear search, and when you’re doing linear searches on lists, it can add up fast. What you can do is convert closedlist into a set object. This keeps hashes of its items so the in operator is much more efficient than for lists. However, lists aren’t hashable items, so you will have to change your configurations into tuples instead of lists.
If the order of closedlist is crucial to the algorithm, you could use a set for the in operator and keep an parallel list around for your results.
I tried a simple implementation of this including aaronasterling’s namedtuple trick and it performed in 0.2 sec for your first example and 2.1 sec for your second, but I haven’t tried verifying the results for the second longer one.
The code below is using less than 1 second to solve this:
from itertools import permutations
GAP='G'
LEFT='F'
RIGHT='B'
BEGIN=('F','F','F','F','G','B','B','B','B')
END=('B','B','B','B','G','F','F','F','F')
LENGTH=len(BEGIN)
ALL=set(permutations(BEGIN))
def NextMove(seq):
g=seq.index(GAP)
ret = set()
def swap(n):
return seq[:n]+seq[n:n+2][::-1]+seq[n+2:]
if g>0 and seq[g-1]==LEFT:
ret.add(swap(g-1))
if g<LENGTH-1 and seq[g+1]==RIGHT:
ret.add(swap(g))
if g<LENGTH-2 and seq[g+1]==LEFT and seq[g+2]==RIGHT:
ret.add(seq[:g]+seq[g+2:g+3]+seq[g+1:g+2]+seq[g:g+1]+seq[g+3:])
if g>1 and seq[g-1]==RIGHT and seq[g-2]==LEFT:
ret.add(seq[:g-2]+seq[g:g+1]+seq[g-1:g]+seq[g-2:g-1]+seq[g+1:])
return ret
AllMoves={state:NextMove(state) for state in ALL}
def Solve(now,target):
if now==target:
return True
for state in AllMoves[now]:
if Solve(state,target):
print(now)
return True
return False
Solve(BEGIN,END)
Well, I can’t really say quite where your algorithm is running astray, but I just went ahead and made my own. In the interest of doing the simplest thing that could possibly work, I used a bastardized version of Dijkstra’s algorithm, where open nodes are visited in arbitrary order, without consideration of distance. This means I don’t have to come up with a heuristic.
""" notation: a game state is a string containing angle
brackets ('>' and '<') and blanks
'>>> <<<'
"""
def get_moves(game):
result = []
lg = len(game)
for i in range(lg):
if game[i] == '>':
if i < lg-1 and game[i+1] == ' ': # '> ' -> ' >'
result.append(game[:i]+' >'+game[i+2:])
if i < lg-2 and game[i+1] != ' ' and game[i+2] == ' ': # '>< ' -> ' <>'
result.append(game[:i]+' '+game[i+1]+'>'+game[i+3:])
if game[i] == '<':
if i >= 1 and game[i-1] == ' ': # ' <' -> '< '
result.append(game[:i-1]+'< '+game[i+1:])
if i >= 2 and game[i-1] != ' ' and game[i-2] == ' ': # ' ><' -> '<> '
result.append(game[:i-2]+'<'+game[i-1]+' '+game[i+1:])
return result
def wander(start, stop):
fringe = [start]
paths = {}
paths[start] = ()
def visit(state):
path = paths[state]
moves = [move for move in get_moves(state) if move not in paths]
for move in moves:
paths[move] = paths[state] + (state,)
fringe.extend(moves)
while stop not in paths:
visit(fringe.pop())
print "still open: ", len(fringe)
print "area covered: " , len(paths)
return paths[stop] + (stop,)
if __name__ == "__main__":
start = '>>>> <<<<'
stop = '<<<< >>>>'
print start, " --> ", stop
pathway = wander(start,stop)
print len(pathway), "moves: ", pathway
tkerwin is correct that you should be using a set for closedlist, which speeds things up a lot, but it is still kind of slow for 4 camels on each side. The next problem is that you are allowing a lot of solutions that aren’t possible because you are allowing fCamels to go backwards and bCamels to go forward. To fix this, replace the lines,
if(igap > 0):
genn(igap, igap-1)
if(igap > 1):
genn(igap, igap-2)
if igap < len(formation) - 1:
genn(igap, igap+1)
if igap < len(formation) - 2:
genn(igap, igap+2)
with
if(igap > 0 and formation[igap-1] == fCamel):
genn(igap, igap-1)
if(igap > 1 and formation[igap-2] == fCamel):
genn(igap, igap-2)
if (igap < len(formation) - 1) and formation[igap+1] == bCamel:
genn(igap, igap+1)
if (igap < len(formation) - 2) and formation[igap + 2] == bCamel:
genn(igap, igap+2)
then I get solution to the 4 camels on each side problem in like .05 seconds rather than 10 seconds. I also tried 5 camels on each side and it took 0.09 seconds. I also am using a set for closedlist and heapq rather than Queue.
Additional speed-up
You can get an additional speed-up by using your heuristic correctly. Currently, you are using the line
openlist.put((current.g + heuristicf(neighbor), node(neighbor, current.g + 1, current)))
(or the heapq version of that) but you should change it to
openlist.put((heuristicf(neighbor), node(neighbor, current.g + 1, current)))
This doesn’t factor in the number of moves that has been needed, but that is okay. With this puzzle (and the screening out of moves that move camels in the wrong direction), you don’t need to worry about the number of moves it takes – either a move advances you towards the solution or it will come to a dead end. In other words, all possible solutions require the same number of moves. This one change takes the time to find the solution of the 12 camels on each side case from over 13 seconds (even using the heapq, set for closedlist, and the changes to find the neighbors above) to 0.389 seconds. That’s not bad.
By the way, a better way to find if you’ve found the solution is to check if the index of the first fCamel is equal to the length of the formation/2 + 1(using int division) and that the index before that is equal to the gap.
First let me tell you how to find the problem. Then I’ll tell you where it is:
I haven’t even bothered to try to figure out your code. I just ran it and took 3 random-time stack samples. I did that by typing control-C and looking at the resulting stacktrace.
One way to look at it is: if a statement appears on X% of random stack traces, then it is on the stack for about X% of the time, so that is what it’s responsible for. If you could avoid executing it, that is how much you would save.
OK, I took 3 stack samples. Here they are:
File "camels.py", line 87, in <module>
print solve([fCamel, fCamel, fCamel, gap, bCamel, bCamel, bCamel])
File "camels.py", line 85, in solve
return astar(formation, heuristic, solution, getneighbors)
File "camels.py", line 80, in astar
openlist.put((current.g + heuristicf(neighbor), node(neighbor, current.g + 1, current)))
File "camels.py", line 87, in <module>
print solve([fCamel, fCamel, fCamel, gap, bCamel, bCamel, bCamel])
File "camels.py", line 85, in solve
return astar(formation, heuristic, solution, getneighbors)
File "camels.py", line 80, in astar
openlist.put((current.g + heuristicf(neighbor), node(neighbor, current.g + 1, current)))
File "camels.py", line 87, in <module>
print solve([fCamel, fCamel, fCamel, gap, bCamel, bCamel, bCamel])
File "camels.py", line 85, in solve
return astar(formation, heuristic, solution, getneighbors)
File "camels.py", line 80, in astar
openlist.put((current.g + heuristicf(neighbor), node(neighbor, current.g + 1, current)))
Notice, in this case the stack samples are all identical. In other words, each one of these three lines is individually responsible for nearly all of the time. So look at them:
line 87: print solve([fCamel, fCamel, fCamel, gap, bCamel, bCamel, bCamel])
line solve: 85: return astar(formation, heuristic, solution, getneighbors)
line astar: 80: openlist.put((current.g + heuristicf(neighbor), node(neighbor, current.g + 1, current)))
Clearly line 87 is not one you can avoid executing, and probably not 85 either. That leaves 80, the openlist.put call. Now, you can’t tell if the problem is in the + operator, the heuristicf call, the node call, or in the put call. You could find out if you could split those out onto separate lines.
So I hope you pick up from this a quick and easy way to find out where your performance problems are.
My other answer is rather long, so I decided to list this as a separate answer. This problem is really better suited to doing a depth-first search. I made a depth-first search solution and it is much, much faster than the optimized A-star method made with the changes outlined in my other answer (which is much, much faster than the OP code). For instance, here are the results for running both my A-star and my depth-first search methods on the 17 camels per side case.
A-star: 14.76 seconds
Depth-first search: 1.30 seconds
Here’s my depth-first method code if you are interested:
from sys import argv
fCamel = 'F'
bCamel = 'B'
gap = 'G'
def issolution(formlen):
def solution(formation):
if formation[formlen2] == gap:
return formation.index(fCamel) == x
return 0
x = formlen/2 + 1
formlen2 = formlen/2
return solution
def solve(formation):
def depthfirst(form, g):
if checksolution(form):
return [tuple(form)], g + 1
else:
igap = form.index(gap)
if(igap > 1 and form[igap-2] == fCamel):
form[igap-2],form[igap] = form[igap],form[igap-2]
res = depthfirst(form,g+1)
form[igap-2],form[igap] = form[igap],form[igap-2]
if res != 0:
return [tuple(form)]+res[0],res[1]
if (igap < flen - 2) and form[igap + 2] == bCamel:
form[igap+2],form[igap] = form[igap],form[igap+2]
res = depthfirst(form,g+1)
form[igap+2],form[igap] = form[igap],form[igap+2]
if res != 0:
return [tuple(form)]+res[0],res[1]
if(igap > 0 and form[igap-1] == fCamel):
form[igap-1],form[igap] = form[igap],form[igap-1]
res = depthfirst(form,g+1)
form[igap-1],form[igap] = form[igap],form[igap-1]
if res != 0:
return [tuple(form)]+res[0],res[1]
if (igap < flen - 1) and form[igap+1] == bCamel:
form[igap+1],form[igap] = form[igap],form[igap+1]
res = depthfirst(form,g+1)
form[igap+1],form[igap] = form[igap],form[igap+1]
if res != 0:
return [tuple(form)]+res[0],res[1]
return 0
flen = len(formation)
checksolution = issolution(flen)
res = depthfirst(list(formation), 0)
return res
L = lambda x: tuple([fCamel]*x + [gap] + [bCamel]*x)
print solve(L(int(argv[1])))
Thanks to some help from people here, I was able to get my code for Tasmanian camels puzzle working. However, it is horribly slow (I think. I’m not sure because this is my first program in Python). The example run in the bottom of the code takes a long time to be solved in my machine:
dumrat@dumrat:~/programming/python$ time python camels.py
[['F', 'F', 'F', 'G', 'B', 'B', 'B'], ['F', 'F', 'G', 'F', 'B', 'B', 'B'],
['F', 'F', 'B', 'F', 'G', 'B', 'B'], ['F', 'F', 'B', 'F', 'B', 'G', 'B'],
['F', 'F', 'B', 'G', 'B', 'F', 'B'], ['F', 'G', 'B', 'F', 'B', 'F', 'B'],
['G', 'F', 'B', 'F', 'B', 'F', 'B'], ['B', 'F', 'G', 'F', 'B', 'F', 'B'],
['B', 'F', 'B', 'F', 'G', 'F', 'B'], ['B', 'F', 'B', 'F', 'B', 'F', 'G'],
['B', 'F', 'B', 'F', 'B', 'G', 'F'], ['B', 'F', 'B', 'G', 'B', 'F', 'F'],
['B', 'G', 'B', 'F', 'B', 'F', 'F'], ['B', 'B', 'G', 'F', 'B', 'F', 'F'],
['B', 'B', 'B', 'F', 'G', 'F', 'F']]
real 0m20.883s
user 0m20.549s
sys 0m0.020s
Here’s the code:
import Queue
fCamel = 'F'
bCamel = 'B'
gap = 'G'
def solution(formation):
return len([i for i in formation[formation.index(fCamel) + 1:]
if i == bCamel]) == 0
def heuristic(formation):
fCamels, score = 0, 0
for i in formation:
if i == fCamel:
fCamels += 1;
elif i == bCamel:
score += fCamels;
else:
pass
return score
def getneighbors (formation):
igap = formation.index(gap)
res = []
# AB_CD --> A_BCD | ABC_D | B_ACD | ABD_C
def genn(i,j):
temp = list(formation)
temp[i], temp[j] = temp[j], temp[i]
res.append(temp)
if(igap > 0):
genn(igap, igap-1)
if(igap > 1):
genn(igap, igap-2)
if igap < len(formation) - 1:
genn(igap, igap+1)
if igap < len(formation) - 2:
genn(igap, igap+2)
return res
class node:
def __init__(self, a, g, p):
self.arrangement = a
self.g = g
self.parent = p
def astar (formation, heuristicf, solutionf, genneighbors):
openlist = Queue.PriorityQueue()
openlist.put((heuristicf(formation), node(formation, 0, None)))
closedlist = []
while 1:
try:
f, current = openlist.get()
except IndexError:
current = None
if current is None:
print "No solution found"
return None;
if solutionf(current.arrangement):
path = []
cp = current
while cp != None:
path.append(cp.arrangement)
cp = cp.parent
path.reverse()
return path
#arr = current.arrangement
closedlist.append(current)
neighbors = genneighbors(current.arrangement)
for neighbor in neighbors:
if neighbor in closedlist:
pass
else:
openlist.put((current.g + heuristicf(neighbor),
node(neighbor, current.g + 1, current)))
#sorted(openlist, cmp = lambda x, y : x.f > y.f)
def solve(formation):
return astar(formation, heuristic, solution, getneighbors)
print solve([fCamel, fCamel, fCamel, gap, bCamel, bCamel, bCamel])
#print solve([fCamel, fCamel, fCamel, fCamel, gap, bCamel, bCamel, bCamel, bCamel])
That is just for 3 camels each. I wanted to do this for 4 at least. That test case is still running (It’s been about 5 minutes now :(). I’ll update this if and when it finishes.
What should I do to improve this code? (Mostly performance-wise, but any other suggestions are welcome also).
Replacing
class node:
def __init__(self, a, g, p):
self.arrangement = a
self.g = g
self.parent = p
with
node = collections.namedtuple('node', 'arrangement, g, parent')
dropped the times from around 340-600 msecs to 11.4 1.891 msecs on the input [fCamel, fCamel, gap, bCamel, bCamel]. It yielded the same output.
This obviously won’t help with any algorithmic problems but as far as micro-optimizations go, it isn’t bad.
1 I had the wrong input. There was an extra fCamel that was making it run slower.
I’ve been tripped up by this before too. The bottleneck here is actually if neighbor in closedlist.
The in statement is so easy to use, you forget that it’s linear search, and when you’re doing linear searches on lists, it can add up fast. What you can do is convert closedlist into a set object. This keeps hashes of its items so the in operator is much more efficient than for lists. However, lists aren’t hashable items, so you will have to change your configurations into tuples instead of lists.
If the order of closedlist is crucial to the algorithm, you could use a set for the in operator and keep an parallel list around for your results.
I tried a simple implementation of this including aaronasterling’s namedtuple trick and it performed in 0.2 sec for your first example and 2.1 sec for your second, but I haven’t tried verifying the results for the second longer one.
The code below is using less than 1 second to solve this:
from itertools import permutations
GAP='G'
LEFT='F'
RIGHT='B'
BEGIN=('F','F','F','F','G','B','B','B','B')
END=('B','B','B','B','G','F','F','F','F')
LENGTH=len(BEGIN)
ALL=set(permutations(BEGIN))
def NextMove(seq):
g=seq.index(GAP)
ret = set()
def swap(n):
return seq[:n]+seq[n:n+2][::-1]+seq[n+2:]
if g>0 and seq[g-1]==LEFT:
ret.add(swap(g-1))
if g<LENGTH-1 and seq[g+1]==RIGHT:
ret.add(swap(g))
if g<LENGTH-2 and seq[g+1]==LEFT and seq[g+2]==RIGHT:
ret.add(seq[:g]+seq[g+2:g+3]+seq[g+1:g+2]+seq[g:g+1]+seq[g+3:])
if g>1 and seq[g-1]==RIGHT and seq[g-2]==LEFT:
ret.add(seq[:g-2]+seq[g:g+1]+seq[g-1:g]+seq[g-2:g-1]+seq[g+1:])
return ret
AllMoves={state:NextMove(state) for state in ALL}
def Solve(now,target):
if now==target:
return True
for state in AllMoves[now]:
if Solve(state,target):
print(now)
return True
return False
Solve(BEGIN,END)
Well, I can’t really say quite where your algorithm is running astray, but I just went ahead and made my own. In the interest of doing the simplest thing that could possibly work, I used a bastardized version of Dijkstra’s algorithm, where open nodes are visited in arbitrary order, without consideration of distance. This means I don’t have to come up with a heuristic.
""" notation: a game state is a string containing angle
brackets ('>' and '<') and blanks
'>>> <<<'
"""
def get_moves(game):
result = []
lg = len(game)
for i in range(lg):
if game[i] == '>':
if i < lg-1 and game[i+1] == ' ': # '> ' -> ' >'
result.append(game[:i]+' >'+game[i+2:])
if i < lg-2 and game[i+1] != ' ' and game[i+2] == ' ': # '>< ' -> ' <>'
result.append(game[:i]+' '+game[i+1]+'>'+game[i+3:])
if game[i] == '<':
if i >= 1 and game[i-1] == ' ': # ' <' -> '< '
result.append(game[:i-1]+'< '+game[i+1:])
if i >= 2 and game[i-1] != ' ' and game[i-2] == ' ': # ' ><' -> '<> '
result.append(game[:i-2]+'<'+game[i-1]+' '+game[i+1:])
return result
def wander(start, stop):
fringe = [start]
paths = {}
paths[start] = ()
def visit(state):
path = paths[state]
moves = [move for move in get_moves(state) if move not in paths]
for move in moves:
paths[move] = paths[state] + (state,)
fringe.extend(moves)
while stop not in paths:
visit(fringe.pop())
print "still open: ", len(fringe)
print "area covered: " , len(paths)
return paths[stop] + (stop,)
if __name__ == "__main__":
start = '>>>> <<<<'
stop = '<<<< >>>>'
print start, " --> ", stop
pathway = wander(start,stop)
print len(pathway), "moves: ", pathway
tkerwin is correct that you should be using a set for closedlist, which speeds things up a lot, but it is still kind of slow for 4 camels on each side. The next problem is that you are allowing a lot of solutions that aren’t possible because you are allowing fCamels to go backwards and bCamels to go forward. To fix this, replace the lines,
if(igap > 0):
genn(igap, igap-1)
if(igap > 1):
genn(igap, igap-2)
if igap < len(formation) - 1:
genn(igap, igap+1)
if igap < len(formation) - 2:
genn(igap, igap+2)
with
if(igap > 0 and formation[igap-1] == fCamel):
genn(igap, igap-1)
if(igap > 1 and formation[igap-2] == fCamel):
genn(igap, igap-2)
if (igap < len(formation) - 1) and formation[igap+1] == bCamel:
genn(igap, igap+1)
if (igap < len(formation) - 2) and formation[igap + 2] == bCamel:
genn(igap, igap+2)
then I get solution to the 4 camels on each side problem in like .05 seconds rather than 10 seconds. I also tried 5 camels on each side and it took 0.09 seconds. I also am using a set for closedlist and heapq rather than Queue.
Additional speed-up
You can get an additional speed-up by using your heuristic correctly. Currently, you are using the line
openlist.put((current.g + heuristicf(neighbor), node(neighbor, current.g + 1, current)))
(or the heapq version of that) but you should change it to
openlist.put((heuristicf(neighbor), node(neighbor, current.g + 1, current)))
This doesn’t factor in the number of moves that has been needed, but that is okay. With this puzzle (and the screening out of moves that move camels in the wrong direction), you don’t need to worry about the number of moves it takes – either a move advances you towards the solution or it will come to a dead end. In other words, all possible solutions require the same number of moves. This one change takes the time to find the solution of the 12 camels on each side case from over 13 seconds (even using the heapq, set for closedlist, and the changes to find the neighbors above) to 0.389 seconds. That’s not bad.
By the way, a better way to find if you’ve found the solution is to check if the index of the first fCamel is equal to the length of the formation/2 + 1(using int division) and that the index before that is equal to the gap.
First let me tell you how to find the problem. Then I’ll tell you where it is:
I haven’t even bothered to try to figure out your code. I just ran it and took 3 random-time stack samples. I did that by typing control-C and looking at the resulting stacktrace.
One way to look at it is: if a statement appears on X% of random stack traces, then it is on the stack for about X% of the time, so that is what it’s responsible for. If you could avoid executing it, that is how much you would save.
OK, I took 3 stack samples. Here they are:
File "camels.py", line 87, in <module>
print solve([fCamel, fCamel, fCamel, gap, bCamel, bCamel, bCamel])
File "camels.py", line 85, in solve
return astar(formation, heuristic, solution, getneighbors)
File "camels.py", line 80, in astar
openlist.put((current.g + heuristicf(neighbor), node(neighbor, current.g + 1, current)))
File "camels.py", line 87, in <module>
print solve([fCamel, fCamel, fCamel, gap, bCamel, bCamel, bCamel])
File "camels.py", line 85, in solve
return astar(formation, heuristic, solution, getneighbors)
File "camels.py", line 80, in astar
openlist.put((current.g + heuristicf(neighbor), node(neighbor, current.g + 1, current)))
File "camels.py", line 87, in <module>
print solve([fCamel, fCamel, fCamel, gap, bCamel, bCamel, bCamel])
File "camels.py", line 85, in solve
return astar(formation, heuristic, solution, getneighbors)
File "camels.py", line 80, in astar
openlist.put((current.g + heuristicf(neighbor), node(neighbor, current.g + 1, current)))
Notice, in this case the stack samples are all identical. In other words, each one of these three lines is individually responsible for nearly all of the time. So look at them:
line 87: print solve([fCamel, fCamel, fCamel, gap, bCamel, bCamel, bCamel])
line solve: 85: return astar(formation, heuristic, solution, getneighbors)
line astar: 80: openlist.put((current.g + heuristicf(neighbor), node(neighbor, current.g + 1, current)))
Clearly line 87 is not one you can avoid executing, and probably not 85 either. That leaves 80, the openlist.put call. Now, you can’t tell if the problem is in the + operator, the heuristicf call, the node call, or in the put call. You could find out if you could split those out onto separate lines.
So I hope you pick up from this a quick and easy way to find out where your performance problems are.
My other answer is rather long, so I decided to list this as a separate answer. This problem is really better suited to doing a depth-first search. I made a depth-first search solution and it is much, much faster than the optimized A-star method made with the changes outlined in my other answer (which is much, much faster than the OP code). For instance, here are the results for running both my A-star and my depth-first search methods on the 17 camels per side case.
A-star: 14.76 seconds
Depth-first search: 1.30 seconds
Here’s my depth-first method code if you are interested:
from sys import argv
fCamel = 'F'
bCamel = 'B'
gap = 'G'
def issolution(formlen):
def solution(formation):
if formation[formlen2] == gap:
return formation.index(fCamel) == x
return 0
x = formlen/2 + 1
formlen2 = formlen/2
return solution
def solve(formation):
def depthfirst(form, g):
if checksolution(form):
return [tuple(form)], g + 1
else:
igap = form.index(gap)
if(igap > 1 and form[igap-2] == fCamel):
form[igap-2],form[igap] = form[igap],form[igap-2]
res = depthfirst(form,g+1)
form[igap-2],form[igap] = form[igap],form[igap-2]
if res != 0:
return [tuple(form)]+res[0],res[1]
if (igap < flen - 2) and form[igap + 2] == bCamel:
form[igap+2],form[igap] = form[igap],form[igap+2]
res = depthfirst(form,g+1)
form[igap+2],form[igap] = form[igap],form[igap+2]
if res != 0:
return [tuple(form)]+res[0],res[1]
if(igap > 0 and form[igap-1] == fCamel):
form[igap-1],form[igap] = form[igap],form[igap-1]
res = depthfirst(form,g+1)
form[igap-1],form[igap] = form[igap],form[igap-1]
if res != 0:
return [tuple(form)]+res[0],res[1]
if (igap < flen - 1) and form[igap+1] == bCamel:
form[igap+1],form[igap] = form[igap],form[igap+1]
res = depthfirst(form,g+1)
form[igap+1],form[igap] = form[igap],form[igap+1]
if res != 0:
return [tuple(form)]+res[0],res[1]
return 0
flen = len(formation)
checksolution = issolution(flen)
res = depthfirst(list(formation), 0)
return res
L = lambda x: tuple([fCamel]*x + [gap] + [bCamel]*x)
print solve(L(int(argv[1])))