The approach of adding one edge connecting a connected point to an unconnected point will produce every possible set of n-1 edges which connect all the points, but it will include duplicates, because it can generate all of those edges in any order. There are (n-1)! possible orders that the same set of edges can be generated, so to get the number of unique sets of edges, you have to divide by (n-1)!.
Don On Feb 8, 11:43 am, sravanreddy001 <sravanreddy...@gmail.com> wrote: > @Don: > > I had the similar approach, but I didn't think of "dividing by (n-1)!" > Why is this needed? -- I think this is to avoid the cases in which, just > the order of picking the nodes is different and lines drawn are same. > > How is this (n-1)! -- i might be missing the the very basic thing.. plz > correct. -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@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.
