我对以下问题感兴趣,主要是作为一种获取有关回溯算法的直觉的方式,因此,我没有在寻找不使用回溯的替代解决方案。
问题:找到所有n个元素向量,使得它们的元素之和小于或等于某个数K。向量中的每个元素都是整数。
示例:如果n = 3,并且K = 10,则[9,0,0]和[5,0,5]是解,而[3,1,8]不是。
[从this site开始,我已经改编了python代码以尝试实现解决方案。
这里是一般的“回溯引擎”功能:
def solve(values, safe_up_to, size):
solution = [None] * size
def extend_solution(position):
for value in values:
solution[position] = value
if safe_up_to(solution, position):
if position >= size-1 or extend_solution(position+1):
return solution
return None
return extend_solution(0)
这是检查解决方案是否“到目前为止安全”的功能:
def safe_up_to(partial_solution, target = 100):
partial_solution = np.array(partial_solution) # convert to np array
# replace None with NaN
partial_solution = np.where(partial_solution == None, np.nan, partial_solution)
if np.nansum(partial_solution) <= target:
return True
else:
return False
但是,当我同时运行这两个函数时,我只会得到一个全零的向量。
solve(values=range(10), safe_up_to=safe_up_to, size=5)
我应该如何修改此代码以获得所有可行的解决方案?
这里是您的代码的修改版本。我试图使它的工作尽可能少地改变:
import numpy as np
from functools import partial
def solve(values, safe_up_to, size):
solution = [None] * size
def extend_solution(position):
for value in values:
solution[position] = value
if safe_up_to(solution):
if position >= size-1:
yield np.array(solution)
else:
yield from extend_solution(position+1)
solution[position] = None
return extend_solution(0)
def safe_up_to(target, partial_solution):
partial_solution = np.array(partial_solution) # convert to np array
# replace None with NaN
partial_solution = np.where(partial_solution == None, np.nan, partial_solution)
if np.nansum(partial_solution) <= target:
return True
else:
return False
for sol in solve(values=range(10), safe_up_to=partial(safe_up_to,4), size=2):
print(sol,sol.sum())
打印:
[0 0] 0
[0 1] 1
[0 2] 2
[0 3] 3
[0 4] 4
[1 0] 1
[1 1] 2
[1 2] 3
[1 3] 4
[2 0] 2
[2 1] 3
[2 2] 4
[3 0] 3
[3 1] 4
[4 0] 4