Difference between revisions of "99 questions/1 to 10"

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(Added an alternative solution to Problem 2 that is a lot more like the solution to Problem 1.)
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__NOTOC__
 
__NOTOC__
   
This is part of [[H-99:_Ninety-Nine_Haskell_Problems|Ninety-Nine Haskell Problems]], based on [http://www.hta-bi.bfh.ch/~hew/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].
+
This is part of [[H-99:_Ninety-Nine_Haskell_Problems|Ninety-Nine Haskell Problems]], based on [https://sites.google.com/site/prologsite/prolog-problems 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 1 ==
 
== Problem 1 ==
   
(*) Find the last box of a list.
+
(*) Find the last element of a list.
   
  +
(Note that the Lisp transcription of this problem is incorrect.)
Example:
 
 
<pre>
 
* (my-last '(a b c d))
 
(D)
 
</pre>
 
   
 
Example in Haskell:
 
Example in Haskell:
   
 
<haskell>
 
<haskell>
Prelude> myLast [1,2,3,4]
+
λ> myLast [1,2,3,4]
[4]
+
4
Prelude> myLast ['x','y','z']
+
λ> myLast ['x','y','z']
"z"
+
'z'
 
</haskell>
 
</haskell>
   
  +
[[99 questions/Solutions/1 | Solutions]]
Solution:
 
   
<haskell>
 
myLast :: [a] -> [a]
 
myLast [x] = [x]
 
myLast (_:xs) = myLast xs
 
</haskell>
 
 
Haskell also provides the function <hask>last</hask>.
 
   
 
== Problem 2 ==
 
== Problem 2 ==
   
(*) Find the last but one box of a list.
+
(*) Find the last but one element of a list.
   
  +
(Note that the Lisp transcription of this problem is incorrect.)
Example:
 
 
<pre>
 
* (my-but-last '(a b c d))
 
(C D)
 
</pre>
 
   
 
Example in Haskell:
 
Example in Haskell:
   
 
<haskell>
 
<haskell>
Prelude> myButLast [1,2,3,4]
+
λ> myButLast [1,2,3,4]
  +
3
[3,4]
 
Prelude> myButLast ['a'..'z']
+
λ> myButLast ['a'..'z']
  +
'y'
"yz"
 
 
</haskell>
 
</haskell>
   
  +
[[99 questions/Solutions/2 | Solutions]]
Solution:
 
   
<haskell>
 
myButLast :: [a] -> [a]
 
myButLast list = drop ((length list) - 2) list
 
</haskell>
 
 
This simply drops all the but last two elements of a list.
 
 
Some other options:
 
<haskell>
 
myButLast = reverse . take 2 . reverse
 
</haskell>
 
or
 
<haskell>
 
myButLast = last . liftM2 (zipWith const) tails (drop 1)
 
</haskell>
 
or
 
<haskell>
 
myButLast [a, b] = [a, b]
 
myButLast (_ : xs) = myButLast xs
 
</haskell>
 
(I'm very new to Haskell but this last one definitely seems to work -- bakert.)
 
 
Remark:
 
The Lisp solution is actually wrong, it should not be the last two elements; a correct Haskell solution is:
 
<haskell>
 
myButLast = last . init
 
Prelude> myButLast ['a'..'z']
 
'y'
 
</haskell>
 
See also the solution to problem 2 in the Prolog list.
 
   
 
== Problem 3 ==
 
== Problem 3 ==
Line 96: Line 47:
 
<pre>
 
<pre>
 
* (element-at '(a b c d e) 3)
 
* (element-at '(a b c d e) 3)
  +
c
C
 
 
</pre>
 
</pre>
   
Line 102: Line 53:
   
 
<haskell>
 
<haskell>
Prelude> elementAt [1,2,3] 2
+
λ> elementAt [1,2,3] 2
 
2
 
2
Prelude> elementAt "haskell" 5
+
λ> elementAt "haskell" 5
 
'e'
 
'e'
 
</haskell>
 
</haskell>
   
  +
[[99 questions/Solutions/3 | Solutions]]
Solution:
 
   
This is (almost) the infix operator !! in Prelude, which is defined as:
 
 
<haskell>
 
(!!) :: [a] -> Int -> a
 
(x:_) !! 0 = x
 
(_:xs) !! n = xs !! (n-1)
 
</haskell>
 
 
Except this doesn't quite work, because !! is zero-indexed, and element-at should be one-indexed. So:
 
 
<haskell>
 
elementAt :: [a] -> Int -> a
 
elementAt list i = list !! (i-1)
 
</haskell>
 
   
 
== Problem 4 ==
 
== Problem 4 ==
Line 132: Line 69:
   
 
<haskell>
 
<haskell>
Prelude> length [123, 456, 789]
+
λ> myLength [123, 456, 789]
 
3
 
3
Prelude> length "Hello, world!"
+
λ> myLength "Hello, world!"
 
13
 
13
 
</haskell>
 
</haskell>
   
  +
[[99 questions/Solutions/4 | Solutions]]
Solution:
 
 
<haskell>
 
length :: [a] -> Int
 
length [] = 0
 
length (_:l) = 1 + length l
 
</haskell>
 
   
This function is defined in Prelude.
 
   
 
== Problem 5 ==
 
== Problem 5 ==
Line 155: Line 85:
   
 
<haskell>
 
<haskell>
Prelude> reverse "A man, a plan, a canal, panama!"
+
λ> myReverse "A man, a plan, a canal, panama!"
 
"!amanap ,lanac a ,nalp a ,nam A"
 
"!amanap ,lanac a ,nalp a ,nam A"
Prelude> reverse [1,2,3,4]
+
λ> myReverse [1,2,3,4]
 
[4,3,2,1]
 
[4,3,2,1]
 
</haskell>
 
</haskell>
   
  +
[[99 questions/Solutions/5 | Solutions]]
Solution: (defined in Prelude)
 
   
<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>
 
   
 
== Problem 6 ==
 
== Problem 6 ==
Line 183: Line 101:
   
 
<haskell>
 
<haskell>
*Main> isPalindrome [1,2,3]
+
λ> isPalindrome [1,2,3]
 
False
 
False
*Main> isPalindrome "madamimadam"
+
λ> isPalindrome "madamimadam"
 
True
 
True
*Main> isPalindrome [1,2,4,8,16,8,4,2,1]
+
λ> isPalindrome [1,2,4,8,16,8,4,2,1]
 
True
 
True
 
</haskell>
 
</haskell>
   
  +
[[99 questions/Solutions/6 | Solutions]]
Solution:
 
   
<haskell>
 
isPalindrome :: (Eq a) => [a] -> Bool
 
isPalindrome xs = xs == (reverse xs)
 
</haskell>
 
   
 
== Problem 7 ==
 
== Problem 7 ==
Line 212: Line 126:
   
 
Example in Haskell:
 
Example in Haskell:
  +
  +
We have to define a new data type, because lists in Haskell are homogeneous.
  +
<haskell>
  +
data NestedList a = Elem a | List [NestedList a]
  +
</haskell>
   
 
<haskell>
 
<haskell>
*Main> flatten (Elem 5)
+
λ> flatten (Elem 5)
 
[5]
 
[5]
*Main> flatten (List [Elem 1, List [Elem 2, List [Elem 3, Elem 4], Elem 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]
*Main> flatten (List [])
+
λ> flatten (List [])
 
[]
 
[]
 
</haskell>
 
</haskell>
   
Solution:
 
   
<haskell>
 
data NestedList a = Elem a | List [NestedList a]
 
   
  +
[[99 questions/Solutions/7 | Solutions]]
flatten :: NestedList a -> [a]
 
flatten (Elem x) = [x]
 
flatten (List x) = concatMap flatten x
 
</haskell>
 
 
We have to defined 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 244: 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>
  +
λ> compress "aaaabccaadeeee"
compress :: Eq a => [a] -> [a]
 
  +
"abcade"
compress = map head . group
 
 
</haskell>
 
</haskell>
   
  +
[[99 questions/Solutions/8 | Solutions]]
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:
 
 
<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 -fno-monomorphism-restriction
 
</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>
 
   
 
== Problem 9 ==
 
== Problem 9 ==
Line 283: Line 172:
 
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))
  +
</pre>
<example in lisp>
 
   
 
Example in Haskell:
 
Example in Haskell:
</pre>
 
   
Solution:
 
 
<haskell>
 
<haskell>
  +
λ> pack ['a', 'a', 'a', 'a', 'b', 'c', 'c', 'a',
group (x:xs) = let (first,rest) = span (==x) xs
 
in (x:first) : group rest
+
'a', 'd', 'e', 'e', 'e', 'e']
  +
["aaaa","b","cc","aa","d","eeee"]
group [] = []
 
 
</haskell>
 
</haskell>
   
  +
[[99 questions/Solutions/9 | Solutions]]
'group' is also in the Prelude, here's an implementation using 'span'.
 
  +
 
 
== Problem 10 ==
 
== Problem 10 ==
   
Line 310: Line 196:
 
Example:
 
Example:
 
<pre>
 
<pre>
* (encode '(a a a a b c c a a d e e e e))
+
* (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))
+
((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>
   
  +
[[99 questions/Solutions/10 | Solutions]]
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]]

Revision as of 09:06, 9 February 2019


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

Solutions


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'

Solutions


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:

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

Solutions


Problem 4

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

Example in Haskell:

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

Solutions


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]

Solutions


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

Solutions


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:

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 [])
[]


Solutions

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"

Solutions

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"]

Solutions

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:

λ> encode "aaaabccaadeeee"
[(4,'a'),(1,'b'),(2,'c'),(2,'a'),(1,'d'),(4,'e')]

Solutions