Anand, do it in binary. Dave
On Jun 23, 1:29 pm, Anand <[email protected]> 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 <[email protected]> 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 <[email protected]> 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 <[email protected]> > > 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/%7Eseander/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 <[email protected]> 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 <[email protected]> wrote: > > > >> > find the no. of bits set in a no. in logn time > > > > >> -- > > > >> You received this message because you are subscribed to the Google > > Groups > > > >> "Algorithm Geeks" group. > > > >> To post to this group, send email to [email protected]. > > > >> To unsubscribe from this group, send email to > > > >> [email protected]<algogeeks%2bunsubscr...@googlegroups.com> > > <algogeeks%2bunsubscr...@googlegroups.com> > > > >> . > > > >> For more options, visit this group at > > > >>http://groups.google.com/group/algogeeks?hl=en. > > > > > -- > > > > You received this message because you are subscribed to the Google > > Groups > > > > "Algorithm Geeks" group. > > > > To post to this group, send email to [email protected]. > > > > To unsubscribe from this group, send email to > > > > [email protected]<algogeeks%2bunsubscr...@googlegroups.com> > > <algogeeks%2bunsubscr...@googlegroups.com> > > > > . > > > > For more options, visit this group at > > > >http://groups.google.com/group/algogeeks?hl=en.-Hide quoted text - > > > > - Show quoted text - > > > -- > > You received this message because you are subscribed to the Google Groups > > "Algorithm Geeks" group. > > To post to this group, send email to [email protected]. > > To unsubscribe from this group, send email to > > [email protected]<algogeeks%2bunsubscr...@googlegroups.com> > > . > > For more options, visit this group at > >http://groups.google.com/group/algogeeks?hl=en.- Hide quoted text - > > - Show quoted text - -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
