Find a positive number M such that gcd(N^M,N&M) is maximum in Python

Suppose we have a number N, we have to find a positive number M such that gcd(N^M, N&M) is as large as possible and m < n. We will also return the largest gcd thus obtained.

So, if the input is like 20, then the output will be 31

To solve this, we will follow these steps −

• if bit_count(n) is same as 0, then
  • for i in range 2 to int(square root of (n)) + 1, do
    • if n mod i is same as 0, then
      • return int(n / i)
• otherwise,
  • val := 0
  • p :=
  • dupn := n
  • while n is non-zero, do
    • if (n AND 1) is same as 0, then
      • val := val + p
    • p := p * 2
    • n := n >> 1
  • return gcd(val XOR dupn, val AND dupn)
• return 1

Example

Let us see the following implementation to get better understanding −

from math import gcd, sqrt
def bit_count(n):
   if (n == 0):
      return 0
   else:
      return (((n & 1) == 0) + bit_count(n >> 1))
def maximum_gcd(n):
   if (bit_count(n) == 0):
      for i in range(2, int(sqrt(n)) + 1):
         if (n % i == 0):
            return int(n / i)
   else:
      val = 0
      p = 1
      dupn = n
      while (n):
         if ((n & 1) == 0):
            val += p
         p = p * 2
         n = n >> 1
      return gcd(val ^ dupn, val & dupn)
   return 1
n = 20
print(maximum_gcd(n))

Input

20

Output

31

Arnab Chakraborty

Updated on: 28-Aug-2020

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