if we do it in binary. Here is how it goes

Number : 15
Step 1   1010 :  0111 & 0101 + 1111 & 0101.
Step2:   0100:   0010 & 0011 + 1010 & 0011
Step3:  0100:    0000 & 0F0F + 0100 & 0F0F;
Step4:  0100:    0000 & FFFF +0100 &FFFF;

Final Answer: 4.Stiil cud not get the logic



On Wed, Jun 23, 2010 at 12:18 PM, Dave <dave_and_da...@juno.com> wrote:

> Anand, do it in binary.
>
> Dave
>
> On Jun 23, 1:29 pm, Anand <anandut2...@gmail.com> wrote:
> > @Dave I tried your logic on 15 it got converted to 10, 4, 4,4. But still
> > could not understand the logic could you please explain?
> >
> >
> >
>  > On Tue, Jun 22, 2010 at 9:34 PM, Dave <dave_and_da...@juno.com> wrote:
> > > Did you actually try the code by hand on a number to see what it does?
> > > If you do, you will see that the first statement replaces the bits in
> > > each pair of bit positions with the number of bits in those positions.
> > > The second adds the bits in each pair of pairs, replacing each nibble
> > > with the number of bits originally set in that nibble. Etc.
> >
> > > Dave
> >
> > > On Jun 22, 10:54 am, divya jain <sweetdivya....@gmail.com> wrote:
> > > > @ dave
> > > > how did u come to this formula... m nt getting the logic..
> >
> > > > @ sathaiah
> > > > yes u r rite
> >
> > > > On 22 June 2010 19:32, chitta koushik <koushik.infin...@gmail.com>
> > > wrote:
> >
> > > > > For more such problems and solns
> >
> > > > >http://graphics.stanford.edu/~seander/bithacks.html<
> http://graphics.stanford.edu/%7Eseander/bithacks.html<http://graphics.stanford.edu/~seander/bithacks.html>
> >
> > > <http://graphics.stanford.edu/%7Eseander/bithacks.html<http://graphics.stanford.edu/~seander/bithacks.html>
> >
> >
> > > > > for (c = 0; v; c++)
> > > > > {
> > > > >   v &= v - 1; // clear the least significant bit set
> > > > > }
> >
> > > > > O(k)   -- no. of bits set in the number
> >
> > > > > --Koushik C
> >
> > > > > On Tue, Jun 22, 2010 at 7:16 PM, Dave <dave_and_da...@juno.com>
> wrote:
> >
> > > > >> Assuming 32 bit integers:
> > > > >> n = ((x >> 1) & 0x55555555) + (x & 0x55555555)
> > > > >> n = ((n >> 2) & 0x33333333) + (n % 0x33333333)
> > > > >> n = ((n >> 4) & 0x0F0F0F0F) + (n & 0x0F0F0F0F)
> > > > >> n = ((n >> 8) & 0x00FF00FF) + (n & 0x00FF00FF)
> > > > >> n = ((n >>16) & 0x0000FFFF) + (n & 0x0000FFFF)
> >
> > > > >> n now is the number of bits set in x. Time is O(1).
> >
> > > > >> Dave
> >
> > > > >> On Jun 22, 6:26 am, divya <sweetdivya....@gmail.com> wrote:
> > > > >> > find the no. of bits set in a no. in logn time
> >
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