Ok this is just counting now how to do the same to print all possibilities?
On Mon, Jun 7, 2010 at 1:14 PM, Raj N <rajn...@gmail.com> wrote: > @Dave: Hey i'm finding little difficulty in understanding the 3rd condition > > - p(*k*,*n*) = 0 if *k* > *n* > > - p(*k*,*n*) = 1 if *k* = *n* > > > - p(*k*+1,*n*)+p(*k*,*n*-*k*) otherwise > > Can you explain me p(k+1,n) partition. I understood p(k,n-k) > > > On Mon, Jun 7, 2010 at 6:16 AM, Dave <dave_and_da...@juno.com> wrote: > >> In number theory, a partition of a positive integer n is a way of >> writing n as a sum of positive integers. Two sums that differ only in >> the order of their summands are considered to be the same partition; >> if order matters then the sum becomes a composition. The number of >> partitions of n is given by the partition function p(n). You can >> compute p(n) recursively. See >> http://en.wikipedia.org/wiki/Partition_(number_theory)<http://en.wikipedia.org/wiki/Partition_%28number_theory%29> >> . >> >> Dave >> >> On Jun 6, 2:05 pm, Raj N <rajn...@gmail.com> wrote: >> > How do you count the number of ways a number can be expressed as a sum >> > of 2 or more numbers? >> > For eg. if the number is 5 , count=3 i.e 1+1+1+1+1, 4+1, 3+2 >> > note 2+3 is same as 3+2 >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to algoge...@googlegroups.com. >> To unsubscribe from this group, send email to >> algogeeks+unsubscr...@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 algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
