Talk:Combinations with repetitions

Original implementations needed; the linked site does not indicate any license or copyright notice, which leads it to default (at least where Rosetta Code is based out of) to a default of "all rights reserved." The task can likely go to non-draft once it's slightly cleaned up, and original implementations are provided. --Michael Mol 19:39, 16 November 2010 (UTC)

Suggestion: It is OK. We can remove the Java solution. The other solutions are original implementation. Do I must remove it, or do you do it? Pelci 19:18, 18 November 2010 (UTC)

I pulled it myself. However, I don't mind if others want to pull them first, particularly in cases of clear copyright issues. --Michael Mol 19:20, 18 November 2010 (UTC)

So we can put back the task into the tasks... Pelci 19:40, 18 November 2010 (UTC)

Task definition

At this stage, it's very hard for me to work out exactly what is to be implemented; the task gives very little guidance over what a “k-combination with repetition” is exactly (it's not the clearest of wikipedia pages that is linked to). At the very least, I'd expect it to use a simple small example (with as few elements as possible) to show exactly what is meant and how things differ from a standard combination-enumerator; it could then ask for the generation of the combinations for a larger input set. –Donal Fellows 23:31, 16 November 2010 (UTC)

Ditto on the lack of clarity.

What would be the result of:

   n=(1,2,3), k=2; 
   n=(1,1,2,3), k=2
   n=(1,1,1,2,3), k=2

And how do you compute the result in the general case? --Paddy3118 00:01, 17 November 2010 (UTC)

n=(1,2,3), k=1 would have the same result as n=(1,1,2,3), k=1. Similarly, (1,1,2,3), k=2 would be the same result as (1,1,1,2,3), k=2. One way of expressing the result would be: calculate all the possibilities and then eliminate the duplicate results. --Rdm 15:18, 17 November 2010 (UTC)