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

Revision as of 08:17, 27 February 2013

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:

Prelude> myLast [1,2,3,4]
4
Prelude> 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:

Prelude> myButLast [1,2,3,4]
3
Prelude> 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:

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

Solutions

Problem 4

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

Example in Haskell:

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

Solutions

Problem 5

(*) Reverse a list.

Example in Haskell:

Prelude> myReverse "A man, a plan, a canal, panama!"
"!amanap ,lanac a ,nalp a ,nam A"
Prelude> 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:

*Main> isPalindrome [1,2,3]
False
*Main> isPalindrome "madamimadam"
True
*Main> 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]

*Main> flatten (Elem 5)
[5]
*Main> flatten (List [Elem 1, List [Elem 2, List [Elem 3, Elem 4], Elem 5]])
[1,2,3,4,5]
*Main> 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:

> 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:

*Main> 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:

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

Solutions

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

Revision as of 08:17, 27 February 2013

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

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@@ Line 1: / Line 1: @@
 __NOTOC__
-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].
+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 ==
@@ Line 20: / Line 18: @@
 </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 ==
@@ Line 48: / Line 36: @@
 </haskell>
-Solutions:
+[[99 questions/Solutions/2 | Solutions]]
-<haskell>
-myButLast :: [a] -> a
-myButLast = last . init
-myButLast' x = reverse x !! 1
-myButLast'' [x,_]  = x
-myButLast'' (_:xs) = myButLast xs
-</haskell>
 == Problem 3 ==
@@ Line 68: / Line 47: @@
 <pre>
 * (element-at '(a b c d e) 3)
+c
-C
 </pre>
@@ Line 80: / Line 59: @@
 </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 ==
@@ Line 110: / Line 75: @@
 </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 ==
@@ Line 132: / Line 85: @@
 <haskell>
-Prelude> reverse "A man, a plan, a canal, panama!"
+Prelude> myReverse "A man, a plan, a canal, panama!"
 "!amanap ,lanac a ,nalp a ,nam A"
-Prelude> reverse [1,2,3,4]
+Prelude> myReverse [1,2,3,4]
 [4,3,2,1]
 </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>
-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 ==
@@ Line 178: / Line 109: @@
 </haskell>
+[[99 questions/Solutions/6 | Solutions]]
-Solution:
-<haskell>
-isPalindrome :: (Eq a) => [a] -> Bool
-isPalindrome xs = xs == (reverse xs)
-</haskell>
 == Problem 7 ==
@@ Line 199: / Line 126: @@
 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>
@@ Line 209: / Line 141: @@
 </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 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 ==
@@ Line 231: / 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.
+Example:
 <pre>
-Example:
 * (compress '(a a a a b c c a a d e e e e))
 (A B C A D E)
+</pre>
 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>
+> compress "aaaabccaadeeee"
-compress :: Eq a => [a] -> [a]
+"abcade"
-compress = map head . group
 </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>
-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 ==
@@ Line 277: / Line 171: @@
 (**) Pack consecutive duplicates of list elements into sublists.
 If a list contains repeated elements they should be placed in separate sublists.
+Example:
 <pre>
-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))
+</pre>
-<example in lisp>
 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>
-pack (x:xs) = let (first,rest) = span (==x) xs
+*Main> pack ['a', 'a', 'a', 'a', 'b', 'c', 'c', 'a',
-               in (x:first) : pack rest
+             'a', 'd', 'e', 'e', 'e', 'e']
+["aaaa","b","cc","aa","d","eeee"]
-pack [] = []
 </haskell>
+[[99 questions/Solutions/9 | Solutions]]
-A more verbose solution is
-<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 ==
@@ Line 318: / Line 196: @@
 Example:
 <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>
 Example in Haskell:
-<pre>
+<haskell>
 encode "aaaabccaadeeee"
 [(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>
+[[99 questions/Solutions/10 | Solutions]]
-which can also be expressed as a list comprehension:
-<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]]