在Python中计算字符串的长度时,“递归深度超出错误”

问题描述 投票:-1回答:2

我正在尝试实现python 2.7中的课程中提到的Karatsuba算法。这是我目前的代码:

# Karatsuba multiplication implementation in python

import numpy as np
import sys

# x = 10^(n/2)*a + b and y = 10^(n/2)*c + d
# x.y = 10^n*(ac) + 10^(n/2)*(ad + bc) + bd
# now recursively compute ac, ad, bc and bd

sys.setrecursionlimit(15000)

def algo_recurs(val1, val2):
  # Assuming that the length of both the multiplier and multiplicand is 
    same
  # Currently employing numbers which are of length 2^n
  n = len(str(val1))            # n = 4
  print(n)
  divVal    = 10**(n/2)
  a = val1 / divVal         # a = 12
  b = val1 % divVal         # b = 34
  c = val2 / divVal         # c = 43
  d = val2 % divVal         # d = 21
  # let the example case be 1234 * 4321

  if(len(str(val1)) == 2):
    prob1 = a * c
    prob2 = b * d
    prob3 = (a+b)*(c+d) - prob1 - prob2
    finalResult = prob1*(divVal*divVal)+prob3*divVal+prob2
    return(finalResult)
  else:
    prob1 = algo_recurs(a,c)
    prob2 = algo_recurs(b,d)
    prob3 = algo_recurs((a+b),(c+d)) - prob1 -prob2
    finalResult = prob1*(divVal*divVal)+prob3*divVal+prob2
    #print(finalResult)
    return(finalResult)
#Enter the inputs

multiplicand    = input("Enter the multiplicand:")
multiplier      = input("Enter the multiplier:")
output = algo_recurs(multiplicand, multiplier)  
print(output)

上面的代码适用于长度为4或更短的数字。但是当我超越它时,它会引发以下错误:

File "Karatsuba.py", line 31, in algo_recurs
  prob1 = algo_recurs(a,c)
File "Karatsuba.py", line 31, in algo_recurs
  prob1 = algo_recurs(a,c)
File "Karatsuba.py", line 31, in algo_recurs
  prob1 = algo_recurs(a,c)
File "Karatsuba.py", line 15, in algo_recurs
  n = len(str(val1))            # n = 4
RuntimeError: maximum recursion depth exceeded while getting the str of an object

我也增加了递归限制,认为它可能是问题。但这也没有解决它。

如果你能指出我在实施中可能做错了什么,我将不胜感激。

python algorithm
2个回答
0
投票

无论您设置递归限制有多高,您的算法都不会终止。这是因为一旦val1达到单个数字,参数a和c将始终保持不变,因为那时n为1而10**(n/2)也为1。


-1
投票

更改递归限制是危险的,因为通常当您超过递归限制时,这是因为您的程序包含错误或设计决策不佳。递归总是可以用相等或更低内存成本的迭代替换。

除非你的算法坚持你这样做,因为知道在某些时候你会收到一个结果,你可以在每次调用函数时改变最大递归深度,但我再也不建议你因为你的程序超过1500次递归调用对此而言,这是相当过分的。

# Karatsuba multiplication implementation in python

import numpy as np
import sys

def algo_recurs(val1, val2):
    sys.setrecursionlimit(sys.getrecursionlimit() + 1)  # Changes the recursion limit every time
    n = len(str(val1))
    #print(n)
    divVal = 10**(n/2)
    a = val1 / divVal         # a = 12
    b = val1 % divVal         # b = 34
    c = val2 / divVal         # c = 43
    d = val2 % divVal         # d = 21

    if(len(str(val1)) == 2):
        prob1 = a * c
        prob2 = b * d
        prob3 = (a+b)*(c+d) - prob1 - prob2
        finalResult = prob1*(divVal*divVal)+prob3*divVal+prob2
        return(finalResult)
    else:
        prob1 = algo_recurs(a,c)
        prob2 = algo_recurs(b,d)
        prob3 = algo_recurs((a+b),(c+d)) - prob1 -prob2
        finalResult = prob1*(divVal*divVal)+prob3*divVal+prob2
        return(finalResult)

multiplicand    = int(input("Enter the multiplicand:"))
multiplier      = int(input("Enter the multiplier:"))
output = algo_recurs(multiplicand, multiplier)  
print(output)
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