Difference between revisions of "99 questions/1 to 10"
(→Problem 1: try a different formatting, and add my simple head . reverse answer) 
m (→Problem 8) 

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__NOTOC__ 
__NOTOC__ 

−  This is part of [[H99:_NinetyNine_Haskell_ProblemsNinetyNine Haskell Problems]], based on [https:// 
+  This is part of [[H99:_NinetyNine_Haskell_ProblemsNinetyNine Haskell Problems]], based on [https://sites.google.com/site/prologsite/prologproblems NinetyNine Prolog Problems] and [http://www.ic.unicamp.br/~meidanis/courses/mc336/2006s2/funcional/L99_NinetyNine_Lisp_Problems.html NinetyNine 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 1 == 
== Problem 1 == 

Line 12:  Line 12:  
<haskell> 
<haskell> 

−  +  λ> myLast [1,2,3,4] 

4 
4 

−  +  λ> myLast ['x','y','z'] 

'z' 
'z' 

</haskell> 
</haskell> 

−  Solutions: 

+  [[99 questions/Solutions/1  Solutions]] 

−  <haskell> 

−  myLast :: [a] > a 

−  myLast [x] = x 

−  myLast (_:xs) = myLast xs 

−  
−  myLast' = foldr1 (const id) 

−  
−  myLast'' = head . reverse 

−  </haskell> 

−  The <hask>Prelude</hask> also provides the function <hask>last</hask>. 

== Problem 2 == 
== Problem 2 == 

Line 35:  Line 25:  
(*) Find the last but one element of a list. 
(*) Find the last but one element of a list. 

−  (Note 
+  (Note that the Lisp transcription of this problem is incorrect.) 
Example in Haskell: 
Example in Haskell: 

<haskell> 
<haskell> 

−  +  λ> myButLast [1,2,3,4] 

3 
3 

−  +  λ> myButLast ['a'..'z'] 

'y' 
'y' 

</haskell> 
</haskell> 

−  Solutions: 

+  [[99 questions/Solutions/2  Solutions]] 

−  <haskell> 

−  myButLast :: [a] > a 

−  myButLast = last . init 

−  </haskell> 

−  
−  <haskell> 

−  myButLast :: [a] > a 

−  myButLast [x,_] = x 

−  myButLast (_:xs) = myButLast xs 

−  </haskell> 

== Problem 3 == 
== Problem 3 == 

Line 67:  Line 47:  
<pre> 
<pre> 

* (elementat '(a b c d e) 3) 
* (elementat '(a b c d e) 3) 

−  C 

+  c 

</pre> 
</pre> 

Line 73:  Line 53:  
<haskell> 
<haskell> 

−  +  λ> elementAt [1,2,3] 2 

2 
2 

−  +  λ> elementAt "haskell" 5 

'e' 
'e' 

</haskell> 
</haskell> 

−  Solution: 

+  [[99 questions/Solutions/3  Solutions]] 

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

−  
−  <haskell> 

−  (!!) :: [a] > Int > a 

−  (x:_) !! 0 = x 

−  (_:xs) !! n = xs !! (n1) 

−  </haskell> 

−  
−  Except this doesn't quite work, because !! is zeroindexed, and elementat should be oneindexed. So: 

−  
−  <haskell> 

−  elementAt :: [a] > Int > a 

−  elementAt list i = list !! (i1) 

−  </haskell> 

== Problem 4 == 
== Problem 4 == 

Line 103:  Line 69:  
<haskell> 
<haskell> 

−  +  λ> myLength [123, 456, 789] 

3 
3 

−  +  λ> myLength "Hello, world!" 

13 
13 

</haskell> 
</haskell> 

−  Solutions: 

+  [[99 questions/Solutions/4  Solutions]] 

−  
−  <haskell> 

−  myLength :: [a] > Int 

−  myLength [] = 0 

−  myLength (_:xs) = 1 + myLength xs 

−  </haskell> 

−  
−  <haskell> 

−  myLength :: [a] > Int 

−  myLength = foldr (\x n > n + 1) 0 

−  </haskell> 

−  This is <hask>length</hask> in <hask>Prelude</hask>. 

== Problem 5 == 
== Problem 5 == 

Line 131:  Line 85:  
<haskell> 
<haskell> 

−  +  λ> myReverse "A man, a plan, a canal, panama!" 

"!amanap ,lanac a ,nalp a ,nam A" 
"!amanap ,lanac a ,nalp a ,nam A" 

−  +  λ> myReverse [1,2,3,4] 

[4,3,2,1] 
[4,3,2,1] 

</haskell> 
</haskell> 

−  Solution: (defined in Prelude) 

+  [[99 questions/Solutions/5  Solutions]] 

−  <haskell> 

−  reverse :: [a] > [a] 

−  reverse = foldl (flip (:)) [] 

−  </haskell> 

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

−  
−  <haskell> 

−  reverse :: [a] > [a] 

−  reverse [] = [] 

−  reverse (x:xs) = reverse xs ++ [x] 

−  </haskell> 

−  
−  However this definition is more wasteful than the one in Prelude as it repeatedly reconses the result as it is accumulated. The following variation avoids that, and thus computationally closer to the Prelude version. 

−  
−  <haskell> 

−  reverse :: [a] > [a] 

−  reverse list = reverse' list [] 

−  where 

−  reverse' [] reversed = reversed 

−  reverse' (x:xs) reversed = reverse' xs (x:reversed) 

−  </haskell> 

== Problem 6 == 
== Problem 6 == 

Line 169:  Line 101:  
<haskell> 
<haskell> 

−  +  λ> isPalindrome [1,2,3] 

False 
False 

−  +  λ> isPalindrome "madamimadam" 

True 
True 

−  +  λ> isPalindrome [1,2,4,8,16,8,4,2,1] 

True 
True 

</haskell> 
</haskell> 

−  Solution: 

+  [[99 questions/Solutions/6  Solutions]] 

−  <haskell> 

−  isPalindrome :: (Eq a) => [a] > Bool 

−  isPalindrome xs = xs == (reverse xs) 

−  </haskell> 

== Problem 7 == 
== Problem 7 == 

Line 199:  Line 127:  
Example in Haskell: 
Example in Haskell: 

+  We have to define a new data type, because lists in Haskell are homogeneous. 

<haskell> 
<haskell> 

−  *Main> flatten (Elem 5) 

+  data NestedList a = Elem a  List [NestedList a] 

+  </haskell> 

+  
+  <haskell> 

+  λ> flatten (Elem 5) 

[5] 
[5] 

−  +  λ> flatten (List [Elem 1, List [Elem 2, List [Elem 3, Elem 4], Elem 5]]) 

[1,2,3,4,5] 
[1,2,3,4,5] 

−  +  λ> flatten (List []) 

[] 
[] 

</haskell> 
</haskell> 

−  Solution: 

−  <haskell> 

−  data NestedList a = Elem a  List [NestedList a] 

−  flatten :: NestedList a > [a] 

+  [[99 questions/Solutions/7  Solutions]] 

−  flatten (Elem x) = [x] 

−  flatten (List x) = concatMap flatten x 

−  </haskell> 

−  
−  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]). 

== Problem 8 == 
== Problem 8 == 

Line 230:  Line 150:  
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. 
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: 

<pre> 
<pre> 

−  Example: 

* (compress '(a a a a b c c a a d e e e e)) 
* (compress '(a a a a b c c a a d e e e e)) 

(A B C A D E) 
(A B C A D E) 

+  </pre> 

Example in Haskell: 
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'] 

−  </pre> 

−  Solution: 

<haskell> 
<haskell> 

−  compress :: Eq a => [a] > [a] 

+  λ> compress "aaaabccaadeeee" 

−  compress = map head . group 

+  "abcade" 

</haskell> 
</haskell> 

−  We simply group equal values together (group), then take the head of each. 

+  [[99 questions/Solutions/8  Solutions]] 

−  Note that (with GHC) we must give an explicit type to ''compress'' otherwise we get: 

−  
−  <haskell> 

−  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 fnomonomorphismrestriction 

−  </haskell> 

−  
−  We can circumvent the monomorphism restriction by writing ''compress'' this way (See: section 4.5.4 of [http://haskell.org/onlinereport the report]): 

−  
−  <haskell>compress xs = map head $ group xs</haskell> 

−  
−  An alternative solution is 

−  
−  <haskell> 

−  compress [] = [] 

−  compress [a] = [a] 

−  compress (x : y : xs) = (if x == y then [] else [x]) ++ compress (y : xs) 

−  </haskell> 

== Problem 9 == 
== Problem 9 == 

Line 276:  Line 171:  
(**) Pack consecutive duplicates of list elements into sublists. 
(**) Pack consecutive duplicates of list elements into sublists. 

If a list contains repeated elements they should be placed in separate sublists. 
If a list contains repeated elements they should be placed in separate sublists. 

+  
+  Example: 

<pre> 
<pre> 

−  Example: 

* (pack '(a a a a b c c a a d e e e e)) 
* (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)) 
((A A A A) (B) (C C) (A A) (D) (E E E E)) 

−  <example in lisp> 

+  </pre> 

Example in Haskell: 
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"] 

−  </pre> 

−  Solution: 

<haskell> 
<haskell> 

−  pack (x:xs) = let (first,rest) = span (==x) xs 

+  λ> pack ['a', 'a', 'a', 'a', 'b', 'c', 'c', 'a', 

−  +  'a', 'd', 'e', 'e', 'e', 'e'] 

−  +  ["aaaa","b","cc","aa","d","eeee"] 

</haskell> 
</haskell> 

−  A more verbose solution is 

+  [[99 questions/Solutions/9  Solutions]] 

−  <haskell> 

−  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 

−  </haskell> 

−  
−  This is implemented as <hask>group</hask> in <hask>Data.List</hask>. 

== Problem 10 == 
== Problem 10 == 

Line 317:  Line 196:  
Example: 
Example: 

<pre> 
<pre> 

−  +  * (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)) 

</pre> 
</pre> 

Example in Haskell: 
Example in Haskell: 

−  <pre> 

+  <haskell> 

−  encode "aaaabccaadeeee" 
+  λ> 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> 

−  
−  Solution: 

−  <haskell> 

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

</haskell> 
</haskell> 

−  which can also be expressed as a list comprehension: 

+  [[99 questions/Solutions/10  Solutions]] 

−  <haskell> 

−  [(length x, head x)  x < group xs] 

−  </haskell> 

−  Or writing it [[Pointfree]]: 

−  
−  <haskell> 

−  encode :: Eq a => [a] > [(Int, a)] 

−  encode = map (\x > (length x, head x)) . group 

−  </haskell> 

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

−  
−  <haskell> 

−  encode :: Eq a => [a] > [(Int, a)] 

−  encode xs = map (length &&& head) $ group xs 

−  </haskell> 

[[Category:Tutorials]] 
[[Category:Tutorials]] 
Latest revision as of 09:06, 9 February 2019
This is part of NinetyNine Haskell Problems, based on NinetyNine Prolog Problems and NinetyNine Lisp Problems.
Problem 1
(*) Find the last element of a list.
(Note that the Lisp transcription of this problem is incorrect.)
Example in Haskell:
λ> myLast [1,2,3,4]
4
λ> myLast ['x','y','z']
'z'
Problem 2
(*) Find the last but one element of a list.
(Note that the Lisp transcription of this problem is incorrect.)
Example in Haskell:
λ> myButLast [1,2,3,4]
3
λ> myButLast ['a'..'z']
'y'
Problem 3
(*) Find the K'th element of a list. The first element in the list is number 1.
Example:
* (elementat '(a b c d e) 3) c
Example in Haskell:
λ> elementAt [1,2,3] 2
2
λ> elementAt "haskell" 5
'e'
Problem 4
(*) Find the number of elements of a list.
Example in Haskell:
λ> myLength [123, 456, 789]
3
λ> myLength "Hello, world!"
13
Problem 5
(*) Reverse a list.
Example in Haskell:
λ> myReverse "A man, a plan, a canal, panama!"
"!amanap ,lanac a ,nalp a ,nam A"
λ> myReverse [1,2,3,4]
[4,3,2,1]
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:
λ> isPalindrome [1,2,3]
False
λ> isPalindrome "madamimadam"
True
λ> isPalindrome [1,2,4,8,16,8,4,2,1]
True
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:
* (myflatten '(a (b (c d) e))) (A B C D E)
Example in Haskell:
We have to define a new data type, because lists in Haskell are homogeneous.
data NestedList a = Elem a  List [NestedList a]
λ> flatten (Elem 5)
[5]
λ> flatten (List [Elem 1, List [Elem 2, List [Elem 3, Elem 4], Elem 5]])
[1,2,3,4,5]
λ> flatten (List [])
[]
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:
λ> compress "aaaabccaadeeee"
"abcade"
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 Haskell:
λ> pack ['a', 'a', 'a', 'a', 'b', 'c', 'c', 'a',
'a', 'd', 'e', 'e', 'e', 'e']
["aaaa","b","cc","aa","d","eeee"]
Problem 10
(*) Runlength encoding of a list. Use the result of problem P09 to implement the socalled runlength 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:
λ> encode "aaaabccaadeeee"
[(4,'a'),(1,'b'),(2,'c'),(2,'a'),(1,'d'),(4,'e')]