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* (encode '(a a a a b c c a a d e e e e))
 
* (encode '(a a a a b c c a a d e e e e))
((4 A) (1 B) (2 C) (2 A) (1 D)(4 E))<Problem description>
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((4 A) (1 B) (2 C) (2 A) (1 D)(4 E))
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Example:
 
<example in lisp>
 
   
 
Example in Haskell:
 
Example in Haskell:
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encode "aaaabccaadeeee"
 
encode "aaaabccaadeeee"
 
[(4,'a'),(1,'b'),(2,'c'),(2,'a'),(1,'d'),(4,'e')]
 
[(4,'a'),(1,'b'),(2,'c'),(2,'a'),(1,'d'),(4,'e')]
 
 
</pre>
 
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Revision as of 14:07, 12 December 2006


These are Haskell translations of Ninety Nine Lisp Problems.

If you want to work on one of these, put your name in the block so we know someone's working on it. Then, change n in your block to the appropriate problem number, and fill in the <Problem description>,<example in lisp>,<example in Haskell>,<solution in haskell> and <description of implementation> fields.

1 Problem 1

(*) Find the last box of a list.

Example:

* (my-last '(a b c d))
(D)

Example in Haskell:

Prelude> last [1,2,3,4]
4
Prelude> last ['x','y','z']
'z'

Solution:

last :: [a] -> a
last [x] = x
last (_:xs) = last xs

This function is defined in Prelude.

2 Problem 2

(*) Find the last but one box of a list.

Example:
* (my-but-last '(a b c d))
(C D)

This can be done by dropping all but the last two elements of a list:

myButLast :: [a] -> [a]
myButLast list = drop ((length list) - 2) list

3 Problem 3

(*) Find the K'th element of a list. The first element in the list is number 1.

Example:
* (element-at '(a b c d e) 3)
C

This is (almost) the infix operator !! in Prelude, which is defined as:

(!!)                :: [a] -> Int -> a
(x:_)  !! 0         =  x
(_:xs) !! n         =  xs !! (n-1)

Except this doesn't quite work, because !! is zero-indexed, and element-at should be one-indexed. So:

elementAt :: [a] -> Int -> a
elementAt list i = list !! (i-1)

4 Problem 4

(*) Find the number of elements of a list.

This is "length" in Prelude, which is defined as:

length           :: [a] -> Int
length []        =  0
length (_:l)     =  1 + length l

5 Problem 5

(*) Reverse a list.

This is "reverse" in Prelude, which is defined as:

reverse          :: [a] -> [a]
reverse          =  foldl (flip (:)) []

The standard definition is concise, but not very readable. Another way to define reverse is:

reverse :: [a] -> [a]
reverse [] = []
reverse (x:xs) = reverse xs ++ [x]

6 Problem 6

(*) Find out whether a list is a palindrome. A palindrome can be read forward or backward; e.g. (x a m a x).

This is trivial, because we can use reverse:

isPalindrome :: (Eq a) => [a] -> Bool
isPalindrome xs = xs == (reverse xs)

7 Problem 7

(**) Flatten a nested list structure.

Transform a list, possibly holding lists as elements into a `flat' list by replacing each list with its elements (recursively).

Example:
* (my-flatten '(a (b (c d) e)))
(A B C D E)

This is tricky, because lists in Haskell are homogeneous. [1, [2, [3, 4], 5]] is a type error. We have to devise some way of represent a list that may (or may not) be nested:

data NestedList a = Elem a | List [NestedList a]
 
flatten :: NestedList a -> [a]
flatten (Elem x) = [x]
flatten (List []) = []
flatten (List (x:xs)) = flatten x ++ flatten (List xs)

Our NestedList datatype is either a single element of some type (Elem a), or a list of NestedLists of the same type. (List [NestedList a]). Let's try it out in ghci:

*Main> flatten (Elem 5)
[5]
*Main> flatten (List [Elem 1, List [Elem 2, List [Elem 3, Elem 4], Elem 5]])
[1,2,3,4,5]
*Main> flatten (List [])
[]

An equivalent definition for flatten is

flatten (Elem x) = x
flatten (List x) = concatMap flatten x

8 Problem 8

(**) Eliminate consecutive duplicates of list elements.

If a list contains repeated elements they should be replaced with a single copy of the element. The order of the elements should not be changed.

Example:
* (compress '(a a a a b c c a a d e e e e))
(A B C A D E)

Example in Haskell:
*Main> compress ['a','a','a','a','b','c','c','a','a','d','e','e','e','e']
['a','b','c','a','d','e']

Solution:

compress :: Eq a => [a] -> [a]
compress = map head . group

We simply group equal values together (group), then take the head of each. Note that (with GHC) we must give an explicit type to compress otherwise we get:

Ambiguous type variable `a' in the constraint:
      `Eq a'
	arising from use of `group'	
    Possible cause: the monomorphism restriction applied to the following:
      compress :: [a] -> [a]
    Probable fix: give these definition(s) an explicit type signature
		  or use -fno-monomorphism-restriction

We can circumvent the monomorphism restriction by writing compress this way (See: section 4.5.4 of the report):

compress xs = map head $ group xs

9 Problem 9

(**) Pack consecutive duplicates of list elements into sublists. If a list contains repeated elements they should be placed in separate sublists.


Example:
* (pack '(a a a a b c c a a d e e e e))
((A A A A) (B) (C C) (A A) (D) (E E E E))
<example in lisp>

Example in Haskell:

Solution:

group (x:xs) = let (first,rest) = span (==x) xs
               in (x:first) : group rest
group [] = []

'group' is also in the Prelude, here's an implementation using 'span'.

10 Problem 10

(*) Run-length encoding of a list. Use the result of problem P09 to implement the so-called run-length encoding data compression method. Consecutive duplicates of elements are encoded as lists (N E) where N is the number of duplicates of the element E.

Example:

    * (encode '(a a a a b c c a a d e e e e))
    ((4 A) (1 B) (2 C) (2 A) (1 D)(4 E))

Example in Haskell:

encode "aaaabccaadeeee"
[(4,'a'),(1,'b'),(2,'c'),(2,'a'),(1,'d'),(4,'e')]

Solution:

encode xs = map (\x -> (length x,head x)) (group xs)

Or writing it Pointfree:

encode :: Eq a => [a] -> [(Int, a)]
encode = map (\x -> (length x, head x)) . group