Count valid unset bits in numbers from 1 to n in python

Python program for Count valid unset bits in numbers from 1 to n. Here more solutions.

import math
#   Python 3 program for
#   Count valid unset bits in all numbers from 1 to n
class BitCount :
    #  Returns number of set bits of given number
    def  countSetBit(self, num) :
        x = num
        x = x - ((x >> 1) & 1431655765)
        x = (x & 858993459) + ((x >> 2) & 858993459)
        x = (x + (x >> 4)) & 252645135
        x = x + (x >> 8)
        x = x + (x >> 16)
        x = x & 63
        #  Number of set bit
        return  x
    #  Handles the request of count the all 
    #  unset bits from (1 to n)
    def countUnsetBit(self, n) :
        result = 0
        #  Execute loop from 1 to n
        i = 1
        while (i <= n) :
            #  Total bits - Active bits
            result += (int((math.log(i) / math.log(2.0) + 1.0))) - \
              self.countSetBit(i)
            i += 1
        #  Display result
        print(" Unset bits from (1 to",n,") :",result)
def main() :
    task = BitCount()
    #  Test A
    #  number(1-7)  Un-setbit
    #  ----------------
    #  1  = 1         0
    #  2  = 10        1
    #  3  = 11        0   
    #  4  = 100       2
    #  5  = 101       1
    #  6  = 110       1 
    #  7  = 111       0
    #  --------------------
    #                 5
    task.countUnsetBit(7)
    #  Test B
    #  number(1-10)  Setbit
    #  ----------------
    #  1  = 1         0
    #  2  = 10        1
    #  3  = 11        0   
    #  4  = 100       2
    #  5  = 101       1
    #  6  = 110       1 
    #  7  = 111       0
    #  8  = 1000      3
    #  9  = 1001      2 
    #  10  = 1010     2 
    #  --------------------
    #                 12
    task.countUnsetBit(10)


if __name__=="__main__":
    main()

Output

 Unset bits from (1 to 7 ) : 5
 Unset bits from (1 to 10 ) : 12