# 99 questions/1 to 10

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== Problem 6 == | == Problem 6 == |

## Revision as of 15:15, 13 July 2010

This is part of Ninety-Nine Haskell Problems, based on Ninety-Nine Prolog Problems and 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 element of a list.

(Note that the Lisp transcription of this problem is incorrect.)

Example in Haskell:

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

## 2 Problem 2

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

(Note that the Lisp transcription of this problem is incorrect.)

Example in Haskell:

Prelude> myButLast [1,2,3,4] 3 Prelude> myButLast ['a'..'z'] 'y'

## 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

Example in Haskell:

Prelude> elementAt [1,2,3] 2 2 Prelude> elementAt "haskell" 5 'e'

## 4 Problem 4

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

Example in Haskell:

Prelude> myLength [123, 456, 789] 3 Prelude> myLength "Hello, world!" 13

## 5 Problem 5

(*) Reverse a list.

Example in Haskell:

Prelude> reverse "A man, a plan, a canal, panama!" "!amanap ,lanac a ,nalp a ,nam A" Prelude> reverse [1,2,3,4] [4,3,2,1]

## 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).

Example in Haskell:

*Main> isPalindrome [1,2,3] False *Main> isPalindrome "madamimadam" True *Main> isPalindrome [1,2,4,8,16,8,4,2,1] True

Solution:

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)

Example in Haskell:

*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 []) []

Solution:

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

We have to define a new data type, because lists in Haskell are homogeneous. [1, [2, [3, 4], 5]] is a type error. Therefore, we must have a way of representing a list that may (or may not) be nested.

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]).

## 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

An alternative solution is

compress [] = [] compress [a] = [a] compress (x : y : xs) = (if x == y then [] else [x]) ++ compress (y : 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: *Main> pack ['a', 'a', 'a', 'a', 'b', 'c', 'c', 'a', 'a', 'd', 'e', 'e', 'e', 'e'] ["aaaa","b","cc","aa","d","eeee"]

Solution:

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

A more verbose solution is

pack :: Eq a => [a] -> [[a]] pack [] = [] pack (x:xs) = (x:first) : pack rest where getReps [] = ([], []) getReps (y:ys) | y == x = let (f,r) = getReps ys in (y:f, r) | otherwise = ([], (y:ys)) (first,rest) = getReps xs

## 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)

which can also be expressed as a list comprehension:

[(length x, head x) | x <- group xs]

Or writing it Pointfree:

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

Or (ab)using the "&&&" arrow operator for tuples:

encode :: Eq a => [a] -> [(Int, a)] encode xs = map (length &&& head) $ group xs