There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

Shruthi Babu
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Joined: May 04, 2007
Posts: 54

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I am trying to use the combination rule of "n!/r!(n-r)!" but here is the pbm according to this if I wanted all the combinations for a particular number of pairs eg all two pair values ... eg 1, 2, 3, 4 the outcome would be (1,2) , (2,3) , (3,4) , (1,3) (1,4) , (2,4)

but I need all possible pairs say all 2 pairs, 3 pairs , 4 pairs etc... is it a good way to wrap it with a loop and pass the "r" value for each loop or is there a way to figure this out. Any help is highly appreciated.

if the order doesn't matter (i.e. (1,2) is the same as (2,1)), it's pretty simple. count from 0 to 2^(number of elements) - 1. Then, use a bit mask to figure out which to include.

in other words, assuming 3 elements:

[ May 22, 2007: Message edited by: Fred Rosenberger ]

subject: Iterating the hash map to form combinations