Personal tools

99 questions/11 to 20

From HaskellWiki

< 99 questions(Difference between revisions)
Jump to: navigation, search
(Added missing output in Problem 17 Haskell example.)
(Put arguments in right order in cycle-based solution to Problem 19.)
Line 245: Line 245:
or using cycle:
or using cycle:
rotate n xs = take len . drop (n `mod` len) . cycle $ xs
rotate xs n = take len . drop (n `mod` len) . cycle $ xs
     where len = length xs
     where len = length xs

Revision as of 11:47, 16 December 2006

This is part of Ninety-Nine Haskell Problems, based on Ninety-Nine Prolog Problems and Ninety-Nine Lisp Problems.

1 Problem 11

(*) Modified run-length encoding. Modify the result of problem 10 in such a way that if an element has no duplicates it is simply copied into the result list. Only elements with duplicates are transferred as (N E) lists.

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

Example in Haskell:
P11> encodeModified "aaaabccaadeeee"
[Multiple 4 'a',Single 'b',Multiple 2 'c',Multiple 2 'a',Single 'd',Multiple 4 'e']


data ListItem a = Single a | Multiple Int a
    deriving (Show)
encodeModified :: Eq a => [a] -> [ListItem a]
encodeModified = map encodeHelper . encode
      encodeHelper (1,x) = Single x
      encodeHelper (n,x) = Multiple n x
Again, like in problem 7, we need a utility type because lists in haskell are homogeneous. Afterwards we use the
function from problem 10 and map single instances of a list item to
and multiple ones to

The ListItem definition contains 'deriving (Show)' so that we can get interactive output.

2 Problem 12

(**) Decode a run-length encoded list. Given a run-length code list generated as specified in problem 11. Construct its uncompressed version.

Example in Haskell:
P12> decodeModified [Multiple 4 'a',Single 'b',Multiple 2 'c',Multiple 2 'a',Single 'd',Multiple 4 'e']


decodeModified :: [ListItem a] -> [a]
decodeModified = concatMap decodeHelper
      decodeHelper (Single x)     = [x]
      decodeHelper (Multiple n x) = replicate n x

We only need to map single instances of an element to a list containing only one element and multiple ones to a list containing the specified number of elements and concatenate these lists.

3 Problem 13

(**) Run-length encoding of a list (direct solution). Implement the so-called run-length encoding data compression method directly. I.e. don't explicitly create the sublists containing the duplicates, as in problem 9, but only count them. As in problem P11, simplify the result list by replacing the singleton lists (1 X) by X.

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

Example in Haskell:
P13> encodeDirect "aaaabccaadeeee"
[Multiple 4 'a',Single 'b',Multiple 2 'c',Multiple 2 'a',Single 'd',Multiple 4 'e']


encode' :: Eq a => [a] -> [(Int,a)]
encode' = foldr helper []
      helper x [] = [(1,x)]
      helper x (y:ys)
        | x == snd y   = (1+fst y,x):ys
        | otherwise    = (1,x):y:ys
encodeDirect :: Eq a => [a] -> [ListItem a]
encodeDirect = map encodeHelper . encode'
      encodeHelper (1,x) = Single x
      encodeHelper (n,x) = Multiple n x
First of all we could rewrite the function
from problem 10 in a way that is does not create the sublists. Thus, I decided to traverse the original list from right to left (using
) and to prepend each element to the resulting list in the proper way. Thereafter we only need to modify the function
from problem 11 to use

4 Problem 14

(*) Duplicate the elements of a list.

* (dupli '(a b c c d))
(A A B B C C C C D D)

Example in Haskell:
> dupli [1, 2, 3]


dupli [] = []
dupli (x:xs) = x:x:dupli xs

or using the list monad:

dupli xs = xs >>= (\x -> [x,x])

5 Problem 15

(**) Replicate the elements of a list a given number of times.

* (repli '(a b c) 3)
(A A A B B B C C C)

Example in Haskell:
> repli "abc" 3


repli :: [a] -> Int -> [a]
repli list n = concatMap (replicate n) list

6 Problem 16

(**) Drop every N'th element from a list.

* (drop '(a b c d e f g h i k) 3)
(A B D E G H K)

Example in Haskell:
*Main> drop = "abcdefghik" 3


drop xs n = drops xs (n-1) n
drops [] _ _ = []
drops (x:xs) 0 max = drops xs (max-1) max
drops (x:xs) (n+1) max = x:drops xs n max

Here, drops is a helper-function to drop. In drops, there is an index n that counts from max-1 down to 0, and removes the head element each time it hits 0.

Note that drop is one of the standard Haskell functions, so redefining it is generally not a good idea.

or using zip:

drop n = map snd . filter ((n/=) . fst) . zip (cycle [1..n])

7 Problem 17

(*) Split a list into two parts; the length of the first part is given.

Do not use any predefined predicates.

* (split '(a b c d e f g h i k) 3)
( (A B C) (D E F G H I K))

Example in Haskell:
*Main> split "abcdefghik" 3
("abc", "defghik")

Solution using take and drop:

split xs n = (take n xs, drop n xs)
Note that this function, with the parameters in the other order, exists as

8 Problem 18

(**) Extract a slice from a list.

Given two indices, i and k, the slice is the list containing the elements between the i'th and i'th element of the original list (both limits included). Start counting the elements with 1.

* (slice '(a b c d e f g h i k) 3 7)
(C D E F G)

Example in Haskell:
*Main> slice ['a','b','c','d','e','f','g','h','i','k'] 3 7


slice xs (i+1) k = take (k-i) $ drop i xs

9 Problem 19

(**) Rotate a list N places to the left.

Hint: Use the predefined functions length and (++).

* (rotate '(a b c d e f g h) 3)
(D E F G H A B C)

* (rotate '(a b c d e f g h) -2)
(G H A B C D E F)

Examples in Haskell:
*Main> rotate ['a','b','c','d','e','f','g','h'] 3

*Main> rotate ['a','b','c','d','e','f','g','h'] (-2)


rotate [] _ = []
rotate l 0 = l
rotate (x:xs) (n+1) = rotate (xs ++ [x]) n
rotate l n = rotate l (length l + n)

There are two separate cases:

  • If n > 0, move the first element to the end of the list n times.
  • If n < 0, convert the problem to the equivalent problem for n > 0 by adding the list's length to n.

or using cycle:

rotate xs n = take len . drop (n `mod` len) . cycle $ xs
    where len = length xs

10 Problem 20

(*) Remove the K'th element from a list.

* (remove-at '(a b c d) 2)
(A C D)

Example in Haskell:
*Main> removeAt 1 ['a','b','c','d']


removeAt k xs = take k xs ++ drop (k+1) xs

Simply use the take and drop functions from the Prelude to take k elements from the start of xs and prepend to the list of elements k+1 to the end. Note that the Lisp code treats 1 as the first element in the list, and it appends NIL elements to the end of the list if k is greater than the list length.