# 99 questions/21 to 28

### From HaskellWiki

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 21

Insert an element at a given position into a list.

Example: * (insert-at 'alfa '(a b c d) 2) (A ALFA B C D) Example in Haskell: P21> insertAt 'X' "abcd" 2 "aXbcd"

Solution:

insertAt :: a -> [a] -> Int -> [a] insertAt x xs (n+1) = let (ys,zs) = split xs n in ys++x:zs

or

insertAt :: a -> [a] -> Int -> [a] insertAt x ys 1 = x:ys insertAt x (y:ys) n = y:insertAt x ys (n-1)

## 2 Problem 22

Create a list containing all integers within a given range.

Example: * (range 4 9) (4 5 6 7 8 9) Example in Haskell: Prelude> [4..9] [4,5,6,7,8,9]

Solution:

range x y = [x..y]

or

range = enumFromTo

or

range x y = take (y-x+1) $ iterate (+1) x

Since there's already syntactic sugar for ranges, there's usually no reason to define a function like 'range' in Haskell. In fact, the syntactic sugar is implemented using the enumFromTo function, which is exactly what 'range' should be.

## 3 Problem 23

Extract a given number of randomly selected elements from a list.

Example: * (rnd-select '(a b c d e f g h) 3) (E D A) Example in Haskell: Prelude System.Random>rnd_select "abcdefgh" 3 Prelude System.Random>"eda"

Solution:

import System.Random rnd_select :: [a]->Int->IO [a] rnd_select [] _ = return [] rnd_select l n | n<0 = error "N must be greater than zero." | otherwise = do pos<-sequence$replicate n$getStdRandom$randomR (0,(length l)-1) return [l!!p | p<-pos]

In order to use getStdRandom and randomR here, we need import module System.Random.

or using sequence all the way:

rnd_select xs n | n < 0 = error "N must be greater than zero." | otherwise = sequence $ replicate n rand where rand = do r <- randomRIO (0,(length xs) - 1) return (xs!!r)

## 4 Problem 24

Lotto: Draw N different random numbers from the set 1..M.

Example: * (rnd-select 6 49) (23 1 17 33 21 37) Example in Haskell: Prelude System.Random>rnd_select 6 49 Prelude System.Random>[23,1,17,33,21,37]

Solution:

import System.Random rnd_select :: Int->Int->IO [Int] rnd_select n m | n<0 = error "N must be a positive number." | m<1 = error "M must larger than 1." | otherwise = sequence$replicate n$getStdRandom$randomR$(1,m)

In order to use getStdRandom and randomR here, we need import module System.Random.

Above solution doesn't return DIFFERENT numbers from the set:

import System.Random diff_select :: Int -> Int -> IO [Int] diff_select n to = diff_select' n [1..to] diff_select' 0 _ = return [] diff_select' _ [] = error "too few elements to choose from" diff_select' n xs = do r <- randomRIO (0,(length xs)-1) let remaining = take r xs ++ drop (r+1) xs rest <- diff_select' (n-1) remaining return ((xs!!r) : rest)

I guess you should have a try on the first solution and have alook at how sequence works;-)

## 5 Problem 25

Generate a random permutation of the elements of a list.

Example: * (rnd-permu '(a b c d e f)) (B A D C E F) Example in Haskell: Prelude>rnd_permu "abcdef" Prelude>"badcef"

Solution:

rnd_permu :: [a]->IO [a] rnd_permu [] = [] rnd_permu l = do pos<-sequence$replicate (length l)$getStdRandom$randomR (0,(length l)-1) return [l!!p | p<-pos]

problem 23,24,25 have almost identical solution.And I believe there will be more elegant ways, feel free to modify them. Thank you:-)

Above IMO is definitely NOT a permutation!

rnd_permu xs = diff_select' (length xs) xs

## 6 Problem 26

(**) Generate the combinations of K distinct objects chosen from the N elements of a list In how many ways can a committee of 3 be chosen from a group of 12 people? We all know that there are C(12,3) = 220 possibilities (C(N,K) denotes the well-known binomial coefficients). For pure mathematicians, this result may be great. But we want to really generate all the possibilities in a list.

Example: * (combinations 3 '(a b c d e f)) ((A B C) (A B D) (A B E) ... ) Example in Haskell: > combinations 3 "abcdef" ["abc","abd","abe",...]

Solution:

-- Import the 'tails' function -- > tails [0,1,2,3] -- [[0,1,2,3],[1,2,3],[2,3],[3],[]] import Data.List (tails) -- The implementation first checks if there's no more elements to select, -- if so, there's only one possible combination, the empty one, -- otherwise we need to select 'n' elements. Since we don't want to -- select an element twice, and we want to select elements in order, to -- avoid combinations which only differ in ordering, we skip some -- unspecified initial elements with 'tails', and select the next element, -- also recursively selecting the next 'n-1' element from the rest of the -- tail, finally consing them together -- Using list comprehensions combinations :: Int -> [a] -> [[a]] combinations 0 _ = [ [] ] combinations n xs = [ y:ys | y:xs' <- tails xs , ys <- combinations (n-1) xs'] -- Alternate syntax, using 'do'-notation combinations :: Int -> [a] -> [[a]] combinations 0 _ = do return [] combinations n xs = do y:xs' <- tails xs ys <- combinations (n-1) xs' return (y:ys)

## 7 Problem 27

Group the elements of a set into disjoint subsets.

a) In how many ways can a group of 9 people work in 3 disjoint subgroups of 2, 3 and 4 persons? Write a function that generates all the possibilities and returns them in a list.

Example: * (group3 '(aldo beat carla david evi flip gary hugo ida)) ( ( (ALDO BEAT) (CARLA DAVID EVI) (FLIP GARY HUGO IDA) ) ... )

b) Generalize the above predicate in a way that we can specify a list of group sizes and the predicate will return a list of groups.

Example: * (group '(aldo beat carla david evi flip gary hugo ida) '(2 2 5)) ( ( (ALDO BEAT) (CARLA DAVID) (EVI FLIP GARY HUGO IDA) ) ... )

Note that we do not want permutations of the group members; i.e. ((ALDO BEAT) ...) is the same solution as ((BEAT ALDO) ...). However, we make a difference between ((ALDO BEAT) (CARLA DAVID) ...) and ((CARLA DAVID) (ALDO BEAT) ...).

You may find more about this combinatorial problem in a good book on discrete mathematics under the term "multinomial coefficients".

Example in Haskell: <example in Haskell> P27> group [2,3,4] ["aldo","beat","carla","david","evi","flip","gary","hugo","ida"] [[["aldo","beat"],["carla","david","evi"],["flip","gary","hugo","ida"]],...] (altogether 1260 solutions) 27> group [2,2,5] ["aldo","beat","carla","david","evi","flip","gary","hugo","ida"] [[["aldo","beat"],["carla","david"],["evi","flip","gary","hugo","ida"]],...] (altogether 756 solutions)

Solution:

combination :: Int -> [a] -> [([a],[a])] combination 0 xs = [([],xs)] combination n [] = [] combination n (x:xs) = ts ++ ds where ts = [ (x:ys,zs) | (ys,zs) <- combination (n-1) xs ] ds = [ (ys,x:zs) | (ys,zs) <- combination n xs ] group :: [Int] -> [a] -> [[[a]]] group [] _ = [[]] group (n:ns) xs = do (g,rs) <- combination n xs gs <- group ns rs return $ g:gs

And a way for those who like it shorter (but less comprehensive):

group :: [Int] -> [a] -> [[[a]]] group [] = const [[]] group (n:ns) = concatMap (uncurry $ (. group ns) . map . (:)) . combination n

## 8 Problem 28

Sorting a list of lists according to length of sublists

a) We suppose that a list contains elements that are lists themselves. The objective is to sort the elements of this list according to their length. E.g. short lists first, longer lists later, or vice versa.

Example: * (lsort '((a b c) (d e) (f g h) (d e) (i j k l) (m n) (o))) ((O) (D E) (D E) (M N) (A B C) (F G H) (I J K L)) Example in Haskell: Prelude>lsort ["abc","de","fgh","de","ijkl","mn","o"] Prelude>["o","de","de","mn","abc","fgh","ijkl"]

Solution:

import List lsort :: [[a]]->[[a]] lsort = sortBy (\x y->compare (length x) (length y))

This function also works for empty list. Import List to use sortBy.

b) Again, we suppose that a list contains elements that are lists themselves. But this time the objective is to sort the elements of this list according to their **length frequency**; i.e., in the default, where sorting is done ascendingly, lists with rare lengths are placed first, others with a more frequent length come later.

Example: * (lfsort '((a b c) (d e) (f g h) (d e) (i j k l) (m n) (o))) ((i j k l) (o) (a b c) (f g h) (d e) (d e) (m n)) Example in Haskell: lfsort ["abc", "de", "fgh", "de", "ijkl", "mn", "o"] ["ijkl","o","abc","fgh","de","de","mn"]

Solution:

import List comparing p x y = compare (p x) (p y) lfsort lists = sortBy (comparing frequency) lists where lengths = map length lists frequency list = length $ filter (== length list) lengths

What we need is a function that takes a sublist and counts the number of other sublists with the same length. To do this, we first construct a list containing the lengths of all the sublists (called lengths above). Then the function frequency can just count the number of times that the current sublist's length occurs in lengths.