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Given positive, integer N, what does this algorithm produce?
```let S and D = N
repeat
let D = ⌊D/2⌋
subtract D from S
until D=0
produce S as the result```
Note: ⌊x⌋ represents the integer part of x.

For non-programmers: start with a positive, integer number (say, 19). Divide it by 2, discarding remainders, until you get to 0. (In this case, you'd get 9, 4, 2, and 1.) Sum all the quotients. (9+4+2+1=16.) Subtract the sum from the original number. (19-16=3.) What's the result?

 See The Solution Submitted by e.g. Rating: 3.7143 (7 votes)

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 Recursive solution | Comment 3 of 9 |
Let's call the output of the algorithm, f(N). Obviously, f(0)=0.

If we run the algorithm with N=x and N=2X, the answer is the same: on the second run, the first pass gets D=x, which is subtracted from S, that started with 2x and thus ends with x, so the second pass starts the same way as with N=x. This allows us to write f(2x)=f(x).

A similar analysis shows that f(2x+1)=1+f(x), so we are now looking for a function that satisfies:

f(0)=0

f(2x)=f(x)

f(2x+1)=1+f(x)

Thinking in binary, 0 has no ones in its representation; 2x has the same number of ones as x; 2x+1 has one more one than x... so the answer is f(x)=number of ONEs in the binary representation of x.
 Posted by Federico Kereki on 2005-01-14 17:33:10

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