[erlang-questions] omaha poker hand permutations

[Marc Worrell]

Can't you just do:

L4 = [ {A, B} || A <- L3, B <- L3, A < B].

This adds a strict ordering on the returned pairs.

- Marc



On 16 aug 2011, at 16:35, Dmitry Demeshchuk wrote:

> 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 <joelr1@REDACTED> 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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>> 
>> 
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> 
> 
> 
> -- 
> Best regards,
> Dmitry Demeshchuk
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