Re^2: Permutation of groups


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Re^2: Permutation of groups

by Limbic~Region (Chancellor)

on Nov 20, 2006 at 16:58 UTC ( [id://585086]=note: print w/replies, xml )

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in reply to Re: Permutation of groups
in thread Permutation of groups

blokhead,
I can see your point of view too. I admit that I focused on:

In fact, I want to get all possible teams:

While my solution addresses the above statement in isolation, taken in context with the rest of the post it is likely lacking. For instance:

For any given game, do all teams need to have the same number of players?
For any given game, can a player participate in more than one team?
For any given game, do all players need to play or can they "sit out"?

I am not sure how much of your code addresses those points either but I am pretty sure it misses how to convert a team to a "number" and its reciprocal. Any ideas on that front?

Cheers - L~R

Comment on Re^2: Permutation of groups

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Re^3: Permutation of groups by tomazos (Deacon) on Nov 20, 2006 at 20:16 UTC
Specifically for a given partitioning ever person must belong to exactly one team. Teams do not have to be of the same size. `eg For three people: 1. A B C 2. AB C 3. A BC 4. AC B 5. ABC` [download]	[reply] [d/l]
Re^4: Permutation of groups by Limbic~Region (Chancellor) on Nov 20, 2006 at 20:50 UTC
tomazos, Assuming it does the right thing, you are going to want to use the integer partitioning code blokhead refers to above and the binary counting I explained. You will not have a contigous range of team reference numbers but they will be guaranteed to be unique and will never be higher than 2 ** $num_of_people. If I find the time tonight or tomorrow, I will code this for you. If you end up coding it on your own, please post so others can benefit and I won't bother. Cheers - L~R	[reply]

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