我知道其他一些类似的线程，但是我不确定如何将其应用于我的示例。
我正在尝试通过蛮力算法来攻击倒数游戏。这是我对此类事情的首次尝试，那么有关如何加快该过程的任何提示？
我正在测试在给定初始数字的情况下答案无法解决的情况。最终将与完整游戏的tkinter界面配对。
from datetime import datetime
import itertools
import math
starttime = datetime.now()
def permutationG(input, s):
if len(s) == len(input): yield s
for i in input:
if i in s: continue
s=s+i
for x in permutationG(input, s): yield x
s=s[:-1]
def op(operator, number1,number2):
string=str(number1)+str(operator)+str(number2)
return eval(string)
a=11
b=10
c=9
d=8
e=7
f=6
targetnumber = 101234
listofnumbers = ['a','b','c','d','e','f']
listprep = ['+','-','*','/']
stringofnumbers = ''.join(str(e) for e in listofnumbers)
numberlocations =[]
for item in permutationG(listofnumbers,''):
numberlocations.append(item)
numberlocations = set(numberlocations)
myarray = itertools.combinations_with_replacement(listprep, 5)
operatorlist = []
for item in myarray:
#for all different numbers find the different permutations
temp = list(itertools.permutations(item))
operatorlist.extend(temp)
#remove duplicates
finaloplist = list(set(operatorlist))
dist=[math.inf]
currentclosestnumber = 0
count=0
looptime=datetime.now()
print('Starting Loop')
for item in numberlocations:
for ops in finaloplist:
initial_value = op(ops[0],item[0],item[1])
for i in range(2,len(item)):
intcheck2 = int(initial_value) - initial_value
if initial_value != targetnumber and initial_value >= 0 and intcheck2 == 0:
newvalue = op(ops[i-1], initial_value, item[i])
else:
break
initial_value = newvalue
attempt = initial_value
intcheck = int(attempt) - attempt
distance = targetnumber - initial_value
if abs(distance) < abs(dist[0]) and intcheck == 0:
currentclosestnumber = attempt
dist[0]=distance
print(attempt)
if targetnumber == attempt:
break
if targetnumber == attempt:
break
endtime = datetime.now()
stringtime= endtime-starttime
#print('Loops: ', count)
if targetnumber == attempt:
print('FOUNDIT!! Target Number = %s Closest Number = %s Time Elapsed = %s' %(targetnumber, currentclosestnumber, stringtime))
elif targetnumber!=attempt:
print('Heres how close: Target Number = %s Closest Number = %s Time Elapsed = %s' %(targetnumber, currentclosestnumber, stringtime))
这将输出大约一分半钟的时间。
另一个问题是由于我使用的方法（使用eval字符串操作），我不知道在打印时要显示括号在最终公式中的位置，也不知道如何将eval放入拉链的末尾。显示数字而不是字母。
真的很感谢任何指导。
注意：我已使用最新版本的代码对帖子进行了编辑。这样可以将计算时间从1:30减少到0:45。主要的变化是，我为每个顺序操作创建了一个for循环，而不是一长串计算，并使用了if语句来确保如果当前值是负数或十进制数，它将中断。
这大大减少了所需的计算次数。
答案 0 :(得分：0)
这是一个较慢的版本，但没有错误。
我没有时间尝试进行优化，但是我的想法是，大多数问题在于在拆分子集时创建/销毁垃圾。
如果我真的希望这样快，那么我认为您可以使用动态编程方法。
def subset_splits (a_list):
if 0 == len(a_list):
yield ([], [])
else:
for sublist1, sublist2 in subset_splits(a_list[1:]):
yield ([a_list[0]] + sublist1, sublist2)
yield ([a_list[0]] + sublist2, sublist1)
def reachable (numbers):
if 1 == len(numbers):
yield (numbers[0], numbers[0])
else:
for list1, list2 in subset_splits(numbers):
if 0 == len(list2):
continue
for x1, expr1 in reachable(list1):
for x2, expr2 in reachable(list2):
yield x1+x2, (expr1, '+', expr2)
yield x1*x2, (expr1, '*', expr2)
yield x1-x2, (expr1, '-', expr2)
yield x2-x1, (expr2, '-', expr1)
if 0 != x2:
yield x1/x2, (expr1, '/', expr2)
if 0 != x1:
yield x2/x1, (expr2, '/', expr1)
numbers = [1, 2, 3, 4, 5, 6]
target = 10000
best = numbers[0]
if best == target:
print(("Answer: ", numbers[0], numbers[0]))
else:
print(("Best: ", numbers[0], numbers[0]))
done = False;
for s, t in subset_splits(numbers):
if done:
break
for x, expr in reachable(s):
if x == target:
print(("Answer: ", x, expr))
done = True
break
elif abs(target-x) < abs(target-best):
print(("Best: ", x, expr))
best = x
if done:
break
for x, expr in reachable(t):
if x == target:
print(("Answer: ", x, expr))
done = True
break
elif abs(target-x) < abs(best-x):
print(("Best: ", x, expr))
best = x