[Haskell-beginners] Beginners Digest, Vol 45, Issue 35

[franco00 at gmx.com]

Indeed the second snipper contains quite an obvious mistake. Thanks for noticing!

 It doesn't seem to me it utilises a lambda expression though? You mean the '.' operator for chaining function? If that's it, it could be rewritten

 unique :: [Integer] -> [Integer]
 unique [] = []
 unique (x:xs) | elem x xs = unique (filter (/= x) xs)
 | otherwise = x : unique xs





----- Original Message -----
From: Ramesh Kumar
Sent: 03/28/12 10:14 AM
To: franco00 at gmx.com, beginners at haskell.org
Subject: Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35

Thanks Franco, Your (first) solution is the only one which has worked so far although it utilizes a lambda expression.
The problem is indeed tricky. 

-----------------------------------------------------------------
From: "franco00 at gmx.com" <franco00 at gmx.com>
To: beginners at haskell.org
Sent: Wednesday, March 28, 2012 3:39 PM
Subject: Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35

gah sorry I obviously meant to reply to the "Unique integers in a list" message



----- Original Message -----
From: franco00 at gmx.com
Sent: 03/28/12 09:36 AM
To: beginners at haskell.org
Subject: Re: Beginners Digest, Vol 45, Issue 35

unique :: [Integer] -> [Integer]
 unique [] = []
 unique (x:xs) | elem x xs = (unique . filter (/= x)) xs
 | otherwise = x : unique xs

 -- This is a simpler to read version (albeit inefficient?)
 unique :: [Integer] -> [Integer]
 unique [] = []
 unique (x:xs) | elem x xs = unique xs
 | otherwise = x : unique xs
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