My task is to calculate number of 1-bits of interval [1,10^16]. Loop is obviously unusable for this case, and I've heard there exists an algorithm for this. Can anyone help?
More generally, an algorithm for number of 1-bits in an interval [1,n] would be nice.
If that helps, I figured that number of 1-bits of interval [1,2^n-1], n positive integer, is n*2^(n-1).
The number of 1-bits in interval [1,n] is the number of 1-bits in interval [1,2^m] plus the number of 1-bits in interval [1,n-2^m] plus n - 2^m.
m is ⌊log(n)/log2⌋.
Let's x = 2^k + b, f(1, x) = f(0, x) = f(0, 2^k + b) = f(0, 2^k - 1) + f(2^k, 2^k + b)
The key problem is f(2^k, 2^k + b)
2^k = 1 0 0 0 ... 0 0
2^k + 1 = 1 0 0 0 ... 0 1
2^k + 2 = 1 0 0 0 ... 1 0
2^k + 3 = 1 0 0 0 ... 0 1
... ...
2^k + b = 1 0 0 0 ... ? ?
Clearly, there is 1-bits in the first bit of each number from 2^k to 2^k + b. And there is (b+1) integers from 2^k to 2^k + b.
We can remove the first 1-bit. And it becomes below.
0 = 0 0 0 0 ... 0 0
1 = 0 0 0 0 ... 0 1
2 = 0 0 0 0 ... 1 0
3 = 0 0 0 0 ... 0 1
... ...
b = 0 0 0 0 ... ? ?
So, f(2^k, 2^k + b) = (b+1) + f(0, b).
f(0, x) = f(0, 2^k - 1) + f(2^k, 2^k + b) = f(0, 2^k - 1) + (b+1) + f(0, b)
Clearly, we have to recursively calculate f(0,b).
Give an example of step 4.
For f(1, 31) = 80, and 31 has 5 1-bits.
so f(1, 30) = 80 - 5 = 75;
Let's calculate f(1, 30) use the method of step 4.
f(1, 30) = f(0, 30)
= f(0, 15) + f(16, 30)
= 32 + 15 + f(0, 14)
= 47 + f(0, 14)
= 47 + f(0, 7) + f(8, 14)
= 47 + 12 + 7 + f(0, 6)
= 66 + f(0, 6)
= 66 + f(0, 3) + f(4, 6)
= 66 + 4 + 3 + f(0, 2)
= 73 + f(0, 2)
= 73 + f(0, 1) + f(2, 2)
= 74 + f(2, 2)
= 74 + 1 + f(0， 0)
= 75
