[algogeeks] Re: 1's counting

[Avik Mitra]

Thanks Dave

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[algogeeks] Re: 1's counting

[bittu]

.

 Is there a quick way to count the number of set bits in this number
 manually???

converting that no. in to binary as

int bin(int n)
{

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[algogeeks] Re: 1's counting

[Dave]

@Avik: Correcting/augmenting your post:

N = (3*4096+15*256+3*16+3)
   = 3*(2^12) + 15*(2^8) + 3*(2^4) + 3*(2^0)
   = (2+1)*(2^12) + (8+4+2+1)*(2^8) + (2+1)*(2^4) + (2+1)
   = (2^1+2^0)*(2^12) + (2^3+2^2+2^1+2^0)*(2^8) + (2^1+2^0)*(2^4) +
(2^1+2^0)
   = 2^13 + 2^12 + 2^11 + 2^10 + 2^9 + 2^8 + 2^5 + 2^4+ 2^1 + 2^0

So there are 10 1's in the binary representation of the number N.

Dave

On Aug 6, 12:42 am, Avik Mitra tutai...@gmail.com wrote:
 N = (3*4096+15*256+3*16+3)
    = 3* (2^10) + 15*( 2^8) + 3*(2^4) + 3* (2^0)
    = (1+2)*(2^10) + (1+2+2^2+ 2^3)*(2^8) + (1+2)*(2^4) + (1+2)
    = (2^10 + 2^11) + (2^8+2^9+2^10+2^11) + (2^4 + 2^6)+ (1+2)
    = 2^11+2^12+2^8+2^9+2^4+2^6+2+1
    = 1 + 2 + 2^4 + 2^6 + 2^8 + 2^9 + 2^11 + 2^12

 So there are 8 1's in the binary representation of the number N.

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[algogeeks] Re: 1's counting

[Avik Mitra]

N = (3*4096+15*256+3*16+3)
   = 3* (2^10) + 15*( 2^8) + 3*(2^4) + 3* (2^0)
   = (1+2)*(2^10) + (1+2+2^2+ 2^3)*(2^8) + (1+2)*(2^4) + (1+2)
   = (2^10 + 2^11) + (2^8+2^9+2^10+2^11) + (2^4 + 2^6)+ (1+2)
   = 2^11+2^12+2^8+2^9+2^4+2^6+2+1
   = 1 + 2 + 2^4 + 2^6 + 2^8 + 2^9 + 2^11 + 2^12

So there are 8 1's in the binary representation of the number N.

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[algogeeks] Re: 1's counting

[Dave]

The straightforward way is simply to write the number out in binary
and count the one-bits. It works for any number.

In this specific case, since the multiplications are by powers of 2,
resulting in shifts, and the resulting binary numbers don't overlap,
the number of bits is
bitcount(3*4096) + bitcount(15*256) + bitcount(3*16) + bitcount(3)
= bitcount(3) + bitcount(15) + bitcount(3) + bitcount(3)
= 2 + 4 + 2 + 2
= 10.

Dave

On Aug 3, 2:04 pm, amit amitjaspal...@gmail.com wrote:
 (3*4096+15*256+3*16+3). How many 1's are there in the binary
 representation of the result.

 Is there a quick way to count the number of set bits in this number
 manually???

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