Finding the total number of set-bits from 1 to n

Question

Write an algorithm to find F(n) the number of bits set to 1, in all numbers from 1 to n for any given value of n.

Complexity should be O(log n)

Answer 1

consider the below:

0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

If you want to find total number of set bits from 1 to say 14 (1110) Few Observations:

1. 0th bit (LSB) 1 bit appears once every two bit (see vertically) so number of set bits = n/2 +(1 if n's 0th bit is 1 else 0)
2. 1st bit : 2 consecutive 1s appear every four bits (see 1st bit vertically along all numbers) number of set bits in 1st bit position = (n/4 *2) + 1 (since 1st bit is a set, else 0)
3. 2nd bit: 4 consecutive 1s appear every 8 bits ( this one is a bit tricky) number of set bits in 2nd position = (n/8*4 )+ 1( since 2nd bit is set, else 0) + ((n%8)%(8/2)) The last term is to include the number of 1s that were outside first (n/8) group of bits (14/8 =1 considers only 1 group ie. 4 set bits in 8 bits. we need to include 1s found in last 14-8 = 6 bits)
4. 3rd bit: 8 consecutive 1s appear every 16 bits (similar to above) number of set bits in 3rd position = (n/16*8)+1(since 3rd bit is set, else 0)+ ((n%16)%(16/2))

so we do O(1) calculation for each bit of a number n. a number contains log2(n) bits. so when we iterate the above for all positions of n and add all the set bits at each step, we get the answer in O(logn) steps