[Haskell-cafe] function unique

[Brent Yorgey]

> 2) The list might be infinite, and your function should work if you make
> only want to use the first part of it, so the following should return
> [1,2,3,4,5] in a finite amount of time:
>
> take 5 (unique [1..])


I don't think this is possible.  Perhaps you misread the original problem
description?  The unique function is supposed to return a list of those
elements which occur exactly once in the input list, which is impossible to
determine for an infinite input list (the only way to prove that a given
element occurs only once in a list, in the absence of any other information,
is to examine every element of the list).  Of course, a function that
behaves like the unix utility "uniq" (i.e. returning only one copy of every
list element) is possible to implement lazily in the manner you describe.

@Alexteslin: It probably would still be a useful exercise for you, though,
to get rid of the redundancy Dan describes in determining whether to keep
each element.  A function to count the occurrences of a given element
(although useful in its own right) is overkill here, since you only care
whether the element occurs once, or more than once.  You could implement a
function isUnique :: Int -> [Int] -> Bool which tells you whether the given
element occurs only once (True) or more than once (False), and doesn't
evaluate more of the list than necessary to determine this.  I doubt you'd
have much trouble implementing this function.

-Brent
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