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Re: [Haskell-cafe] function unique

Subject: Re: [Haskell-cafe] function unique
From: "Brent Yorgey"
Date: Wed, 11 Jul 2007 14:41:36 -0400

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.

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