Python Program to find whether a no is power of two

Given a positive integer, write a function to find if it is a power of two or not.

Examples :

Input : n = 4
Output : Yes
22 = 4

Input : n = 7
Output : No

Input : n = 32
Output : Yes
25 = 32

1. A simple method for this is to simply take the log of the number on base 2 and if you get an integer then number is power of 2.



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# Python3 Program to find 
# whether a no is 
# power of two
import math
  
# Function to check
# Log base 2
def Log2(x):
    return (math.log10(x) / 
            math.log10(2));
  
# Function to check
# if x is power of 2
def isPowerOfTwo(n):
    return (math.ceil(Log2(n)) == math.floor(Log2(n)));
  
# Driver Code
if(isPowerOfTwo(31)):
    print("Yes");
else:
    print("No");
  
if(isPowerOfTwo(64)):
    print("Yes");
else:
    print("No");
      
# This code is contributed 
# by mits

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Output:

No
Yes

2. Another solution is to keep dividing the number by two, i.e, do n = n/2 iteratively. In any iteration, if n%2 becomes non-zero and n is not 1 then n is not a power of 2. If n becomes 1 then it is a power of 2.

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# Python program to check if given
# number is power of 2 or not 
  
# Function to check if x is power of 2
def isPowerOfTwo(n):
    if (n == 0):
        return False
    while (n != 1):
            if (n % 2 != 0):
                return False
            n = n // 2
              
    return True
  
# Driver code
if(isPowerOfTwo(31)):
    print('Yes')
else:
    print('No')
if(isPowerOfTwo(64)):
    print('Yes')
else:
    print('No')
  
# This code is contributed by Danish Raza

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Output:

No
Yes

3. All power of two numbers have only one bit set. So count the no. of set bits and if you get 1 then number is a power of 2. Please see Count set bits in an integer for counting set bits.

4. If we subtract a power of 2 numbers by 1 then all unset bits after the only set bit become set; and the set bit become unset.

For example for 4 ( 100) and 16(10000), we get following after subtracting 1
3 –> 011
15 –> 01111

So, if a number n is a power of 2 then bitwise & of n and n-1 will be zero. We can say n is a power of 2 or not based on value of n&(n-1). The expression n&(n-1) will not work when n is 0. To handle this case also, our expression will become n& (!n&(n-1)) (thanks to https://www.geeksforgeeks.org/program-to-find-whether-a-no-is-power-of-two/Mohammad for adding this case).
Below is the implementation of this method.

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# Python program to check if given
# number is power of 2 or not 
  
# Function to check if x is power of 2
def isPowerOfTwo (x):
  
    # First x in the below expression 
    # is for the case when x is 0 
    return (x and (not(x & (x - 1))) )
  
# Driver code
if(isPowerOfTwo(31)):
    print('Yes')
else:
    print('No')
      
if(isPowerOfTwo(64)):
    print('Yes')
else:
    print('No')
      
# This code is contributed by Danish Raza    

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Output:

No
Yes

Please refer complete article on Program to find whether a no is power of two for more details!



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