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[[99_Haskell_exercises|Back to 99 Haskell exercises]]
 
 
 
__NOTOC__
 
__NOTOC__
   
These are Haskell translations of [http://www.ic.unicamp.br/~meidanis/courses/mc336/2006s2/funcional/L-99_Ninety-Nine_Lisp_Problems.html Ninety Nine Lisp Problems].
+
This is part of [[H-99:_Ninety-Nine_Haskell_Problems|Ninety-Nine Haskell Problems]], based on [https://prof.ti.bfh.ch/hew1/informatik3/prolog/p-99/ Ninety-Nine Prolog Problems] and [http://www.ic.unicamp.br/~meidanis/courses/mc336/2006s2/funcional/L-99_Ninety-Nine_Lisp_Problems.html 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.
 
 
 
 
== Problem 21 ==
 
== Problem 21 ==
   
 
Insert an element at a given position into a list.
 
Insert an element at a given position into a list.
  +
  +
Example:
   
 
<pre>
 
<pre>
Example:
 
 
* (insert-at 'alfa '(a b c d) 2)
 
* (insert-at 'alfa '(a b c d) 2)
 
(A ALFA B C D)
 
(A ALFA B C D)
  +
</pre>
  +
 
Example in Haskell:
 
Example in Haskell:
  +
<haskell>
 
P21> insertAt 'X' "abcd" 2
 
P21> insertAt 'X' "abcd" 2
 
"aXbcd"
 
"aXbcd"
</pre>
 
 
Solution:
 
<haskell>
 
insertAt :: a -> [a] -> Int -> [a]
 
insertAt x xs (n+1) = let (ys,zs) = split xs n in ys++x:zs
 
</haskell>
 
or
 
<haskell>
 
insertAt :: a -> [a] -> Int -> [a]
 
insertAt x ys 1 = x:ys
 
insertAt x (y:ys) n = y:insertAt x ys (n-1)
 
 
</haskell>
 
</haskell>
   
There are two possible simple solutions. First we can use <hask>split</hask> from problem 17 (or even <hask>splitAt</hask> from the Prelude) to split the list and insert the element. Second we can define a recursive solution on our own.
+
[[99 questions/Solutions/21 | Solutions]]
+
  +
 
== Problem 22 ==
 
== Problem 22 ==
   
 
Create a list containing all integers within a given range.
 
Create a list containing all integers within a given range.
  +
  +
Example:
   
 
<pre>
 
<pre>
Example:
 
 
* (range 4 9)
 
* (range 4 9)
 
(4 5 6 7 8 9)
 
(4 5 6 7 8 9)
  +
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
Prelude> [4..9]
 
[4,5,6,7,8,9]
 
</pre>
 
   
Solution:
 
 
<haskell>
 
<haskell>
range x y = [x..y]
+
Prelude> range 4 9
</haskell>
+
[4,5,6,7,8,9]
or
 
<haskell>
 
range = enumFromTo
 
</haskell>
 
or
 
<haskell>
 
range x y = take (y-x+1) $ iterate (+1) x
 
 
</haskell>
 
</haskell>
   
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.
+
[[99 questions/Solutions/22 | Solutions]]
+
 
== Problem 23 ==
 
== Problem 23 ==
   
 
Extract a given number of randomly selected elements from a list.
 
Extract a given number of randomly selected elements from a list.
  +
  +
Example:
   
 
<pre>
 
<pre>
Example:
 
 
* (rnd-select '(a b c d e f g h) 3)
 
* (rnd-select '(a b c d e f g h) 3)
 
(E D A)
 
(E D A)
  +
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
Prelude System.Random>rnd_select "abcdefgh" 3
 
Prelude System.Random>"eda"
 
   
</pre>
 
 
Solution:
 
 
<haskell>
 
<haskell>
import System.Random
+
Prelude System.Random>rnd_select "abcdefgh" 3 >>= putStrLn
+
eda
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]
 
 
</haskell>
 
</haskell>
   
In order to use getStdRandom and randomR here, we need import module System.Random.
+
[[99 questions/Solutions/23 | Solutions]]
  +
   
or using sequence all the way:
 
<haskell>
 
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)
 
</haskell>
 
 
 
== Problem 24 ==
 
== Problem 24 ==
   
 
Lotto: Draw N different random numbers from the set 1..M.
 
Lotto: Draw N different random numbers from the set 1..M.
  +
  +
Example:
   
 
<pre>
 
<pre>
Example:
 
 
* (rnd-select 6 49)
 
* (rnd-select 6 49)
 
(23 1 17 33 21 37)
 
(23 1 17 33 21 37)
  +
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
  +
  +
<haskell>
 
Prelude System.Random>diff_select 6 49
 
Prelude System.Random>diff_select 6 49
 
Prelude System.Random>[23,1,17,33,21,37]
 
Prelude System.Random>[23,1,17,33,21,37]
 
</pre>
 
 
Solution:
 
<haskell>
 
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)
 
 
</haskell>
 
</haskell>
   
The random numbers have to be distinct!
+
[[99 questions/Solutions/24 | Solutions]]
   
In order to use getStdRandom and randomR here, we need import module System.Random.
 
   
 
== Problem 25 ==
 
== Problem 25 ==
Line 123: Line 87:
 
Generate a random permutation of the elements of a list.
 
Generate a random permutation of the elements of a list.
   
<pre>
 
 
Example:
 
Example:
  +
  +
<pre>
 
* (rnd-permu '(a b c d e f))
 
* (rnd-permu '(a b c d e f))
 
(B A D C E F)
 
(B A D C E F)
+
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
Prelude>rnd_permu "abcdef"
 
Prelude>"badcef"
 
   
</pre>
 
 
Solution:
 
Above IMO is definitely NOT a permutation!
 
 
<haskell>
 
<haskell>
rnd_permu xs = diff_select' (length xs) xs
+
Prelude>rnd_permu "abcdef"
  +
Prelude>"badcef"
 
</haskell>
 
</haskell>
  +
  +
[[99 questions/Solutions/25 | Solutions]]
   
   
Line 145: Line 106:
   
 
(**) Generate the combinations of K distinct objects chosen from the N elements of a list
 
(**) 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
 
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.
 
well-known binomial coefficients). For pure mathematicians, this result may be great. But we want to really generate all the possibilities in a list.
   
<pre>
 
 
Example:
 
Example:
  +
  +
<pre>
 
* (combinations 3 '(a b c d e f))
 
* (combinations 3 '(a b c d e f))
 
((A B C) (A B D) (A B E) ... )
 
((A B C) (A B D) (A B E) ... )
  +
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
  +
  +
<haskell>
 
> combinations 3 "abcdef"
 
> combinations 3 "abcdef"
 
["abc","abd","abe",...]
 
["abc","abd","abe",...]
</pre>
+
</haskell>
   
Solution:
+
[[99 questions/Solutions/26 | Solutions]]
<haskell>
 
-- 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)
 
</haskell>
 
   
 
== Problem 27 ==
 
== Problem 27 ==
Line 189: Line 133:
 
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.
 
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.
   
<pre>
 
 
Example:
 
Example:
  +
  +
<pre>
 
* (group3 '(aldo beat carla david evi flip gary hugo ida))
 
* (group3 '(aldo beat carla david evi flip gary hugo ida))
 
( ( (ALDO BEAT) (CARLA DAVID EVI) (FLIP GARY HUGO IDA) )
 
( ( (ALDO BEAT) (CARLA DAVID EVI) (FLIP GARY HUGO IDA) )
Line 198: Line 143:
 
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.
 
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.
   
<pre>
 
 
Example:
 
Example:
  +
  +
<pre>
 
* (group '(aldo beat carla david evi flip gary hugo ida) '(2 2 5))
 
* (group '(aldo beat carla david evi flip gary hugo ida) '(2 2 5))
 
( ( (ALDO BEAT) (CARLA DAVID) (EVI FLIP GARY HUGO IDA) )
 
( ( (ALDO BEAT) (CARLA DAVID) (EVI FLIP GARY HUGO IDA) )
Line 209: Line 155:
 
You may find more about this combinatorial problem in a good book on discrete mathematics under the term "multinomial coefficients".
 
You may find more about this combinatorial problem in a good book on discrete mathematics under the term "multinomial coefficients".
   
<pre>
 
 
Example in Haskell:
 
Example in Haskell:
<example in Haskell>
 
   
  +
<haskell>
 
P27> group [2,3,4] ["aldo","beat","carla","david","evi","flip","gary","hugo","ida"]
 
P27> group [2,3,4] ["aldo","beat","carla","david","evi","flip","gary","hugo","ida"]
 
[[["aldo","beat"],["carla","david","evi"],["flip","gary","hugo","ida"]],...]
 
[[["aldo","beat"],["carla","david","evi"],["flip","gary","hugo","ida"]],...]
Line 220: Line 165:
 
[[["aldo","beat"],["carla","david"],["evi","flip","gary","hugo","ida"]],...]
 
[[["aldo","beat"],["carla","david"],["evi","flip","gary","hugo","ida"]],...]
 
(altogether 756 solutions)
 
(altogether 756 solutions)
</pre>
+
</haskell>
   
Solution:
+
[[99 questions/Solutions/27 | Solutions]]
<haskell>
 
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
 
</haskell>
 
   
First of all we acknowledge that we need something like <hask>combination</hask> from the above problem. Actually we need more than the elements we selected, we also need the elements we did not select. Therefore we cannot use the <hask>tails</hask> function because it throws too much information away. But in general this function works like the one above. In each step of the recursion we have to decide whether we want to take the first element of the list <hask>(x:xs)</hask> in the combination (we collect the possibilities for this choice in <hask>ts</hask>) or if we don't want it in the combination (<hask>ds</hask> collects the possibilities for this case).
 
   
Now we need a function <hask>group</hask> that does the needed work. First we denote that if we don't want any group there is only one solution: a list of no groups. But if we want at least one group with n members we have to select n elements of <hask>xs</hask> into a group <hask>g</hask> and the remaining elements into <hask>rs</hask>. Afterwards we group those remaining elements, get a list of groups <hask>gs</hask> and prepend <hask>g</hask> as the first group.
 
 
And a way for those who like it shorter (but less comprehensive):
 
<haskell>
 
group :: [Int] -> [a] -> [[[a]]]
 
group [] = const [[]]
 
group (n:ns) = concatMap (uncurry $ (. group ns) . map . (:)) . combination n
 
</haskell>
 
 
 
== Problem 28 ==
 
== Problem 28 ==
   
Line 248: Line 176:
   
 
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.
 
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:
   
 
<pre>
 
<pre>
Example:
 
 
* (lsort '((a b c) (d e) (f g h) (d e) (i j k l) (m n) (o)))
 
* (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))
 
((O) (D E) (D E) (M N) (A B C) (F G H) (I J K L))
  +
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
Prelude>lsort ["abc","de","fgh","de","ijkl","mn","o"]
 
Prelude>["o","de","de","mn","abc","fgh","ijkl"]
 
</pre>
 
   
Solution:
 
 
<haskell>
 
<haskell>
import List
+
Prelude>lsort ["abc","de","fgh","de","ijkl","mn","o"]
lsort :: [[a]]->[[a]]
+
Prelude>["o","de","de","mn","abc","fgh","ijkl"]
lsort = sortBy (\x y->compare (length x) (length y))
 
 
</haskell>
 
</haskell>
 
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 <b>length frequency</b>; 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.
 
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 <b>length frequency</b>; 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.
   
<pre>
 
 
Example:
 
Example:
  +
  +
<pre>
 
* (lfsort '((a b c) (d e) (f g h) (d e) (i j k l) (m n) (o)))
 
* (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))
 
((i j k l) (o) (a b c) (f g h) (d e) (d e) (m n))
  +
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
lfsort ["abc", "de", "fgh", "de", "ijkl", "mn", "o"]
 
["ijkl","o","abc","fgh","de","de","mn"]
 
</pre>
 
   
Solution:
 
 
<haskell>
 
<haskell>
import List
+
lfsort ["abc", "de", "fgh", "de", "ijkl", "mn", "o"]
comparing p x y = compare (p x) (p y)
+
["ijkl","o","abc","fgh","de","de","mn"]
lfsort lists = sortBy (comparing frequency) lists where
 
lengths = map length lists
 
frequency list = length $ filter (== length list) lengths
 
 
</haskell>
 
</haskell>
   
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.
+
[[99 questions/Solutions/28 | Solutions]]
  +
  +
 
[[Category:Tutorials]]
 
[[Category:Tutorials]]

Revision as of 13:25, 16 July 2010


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

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"

Solutions


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> range 4 9
[4,5,6,7,8,9]

Solutions

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 >>= putStrLn
eda

Solutions


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>diff_select 6 49
Prelude System.Random>[23,1,17,33,21,37]

Solutions


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"

Solutions


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",...]

Solutions


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:

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)

Solutions


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"]

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"]

Solutions