[erlang-questions] omaha poker hand permutations

Dmitry Demeshchuk <>
Tue Aug 16 16:35:03 CEST 2011


Lis comprehensions are a combination of map and filter. But I think
you need lists:foldl/3 here, or just a custom reducing function. So
the answer is no.

On Tue, Aug 16, 2011 at 6:27 PM, Joel Reymont <> wrote:
> The problem can be reduced to the following:
>
> 5> L3 = ["AD", "AH", "AS", "QS"].
> ["AD","AH","AS","QS"]
>
> 6> L4 = [ {A, B} || A <- L3, B <- L3, A /= B].
> [{"AD","AH"},
>  {"AD","AS"},
>  {"AD","QS"},
>  {"AH","AD"},
>  {"AH","AS"},
>  {"AH","QS"},
>  {"AS","AD"},
>  {"AS","AH"},
>  {"AS","QS"},
>  {"QS","AD"},
>  {"QS","AH"},
>  {"QS","AS"}]
>
> Note the duplicate pairs of AD, AH and AH, AD.
>
> Can these be eliminated inside the list comprehension itself?
>
>        Thanks, Joel
>
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-- 
Best regards,
Dmitry Demeshchuk



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