# 99 questions/1 to 10

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__NOTOC__ |
__NOTOC__ |
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− | 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. |
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== Problem 1 == |
== Problem 1 == |
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− | (*) Find the last box of a list. |
+ | (*) Find the last element of a list. |

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

− | |||

− | <pre> |
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− | * (my-last '(a b c d)) |
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− | (D) |
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− | </pre> |
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Example in Haskell: |
Example in Haskell: |
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Line 13: | Line 13: | ||

<haskell> |
<haskell> |
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Prelude> myLast [1,2,3,4] |
Prelude> myLast [1,2,3,4] |
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− | [4] |
+ | 4 |

Prelude> myLast ['x','y','z'] |
Prelude> myLast ['x','y','z'] |
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− | "z" |
+ | 'z' |

</haskell> |
</haskell> |
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− | Solution: |
+ | [[99 questions/Solutions/1 | Solutions]] |

− | <haskell> |
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− | myLast :: [a] -> [a] |
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− | myLast [x] = [x] |
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− | myLast (_:xs) = myLast xs |
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− | </haskell> |
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− | |||

− | Haskell also provides the function <hask>last</hask>. |
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== Problem 2 == |
== Problem 2 == |
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− | (*) Find the last but one box of a list. |
+ | (*) Find the last but one element of a list. |

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

− | |||

− | <pre> |
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− | * (my-but-last '(a b c d)) |
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− | (C D) |
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− | </pre> |
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Example in Haskell: |
Example in Haskell: |
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Line 38: | Line 31: | ||

<haskell> |
<haskell> |
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Prelude> myButLast [1,2,3,4] |
Prelude> myButLast [1,2,3,4] |
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− | [3,4] |
+ | 3 |

Prelude> myButLast ['a'..'z'] |
Prelude> myButLast ['a'..'z'] |
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− | "yz" |
+ | 'y' |

</haskell> |
</haskell> |
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− | Solution: |
+ | [[99 questions/Solutions/2 | Solutions]] |

− | <haskell> |
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− | myButLast :: [a] -> [a] |
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− | myButLast list = drop ((length list) - 2) list |
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− | </haskell> |
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− | |||

− | This simply drops all the but last two elements of a list. |
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− | |||

− | Some other options: |
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− | <haskell> |
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− | myButLast = reverse . take 2 . reverse |
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− | </haskell> |
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− | or |
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− | <haskell> |
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− | myButLast list = snd $ splitAt (length list - 2) list |
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− | </haskell> |
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− | or |
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− | <haskell> |
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− | myButLast = last . liftM2 (zipWith const) tails (drop 1) |
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− | </haskell> |
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− | or |
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− | <haskell> |
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− | myButLast [a, b] = [a, b] |
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− | myButLast (_ : xs) = myButLast xs |
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− | </haskell> |
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− | (I'm very new to Haskell but this last one definitely seems to work -- bakert.) |
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− | |||

− | Remark: |
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− | The Lisp solution is actually wrong, it should not be the last two elements; a correct Haskell solution is: |
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− | <haskell> |
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− | myButLast = last . init |
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− | Prelude> myButLast ['a'..'z'] |
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− | 'y' |
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− | </haskell> |
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− | See also the solution to problem 2 in the Prolog list. |
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== Problem 3 == |
== Problem 3 == |
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Line 88: | Line 47: | ||

<pre> |
<pre> |
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* (element-at '(a b c d e) 3) |
* (element-at '(a b c d e) 3) |
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− | C |
+ | c |

</pre> |
</pre> |
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Line 100: | Line 59: | ||

</haskell> |
</haskell> |
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− | Solution: |
+ | [[99 questions/Solutions/3 | Solutions]] |

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

− | <haskell> |
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− | (!!) :: [a] -> Int -> a |
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− | (x:_) !! 0 = x |
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− | (_:xs) !! n = xs !! (n-1) |
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− | </haskell> |
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− | |||

− | Except this doesn't quite work, because !! is zero-indexed, and element-at should be one-indexed. So: |
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− | |||

− | <haskell> |
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− | elementAt :: [a] -> Int -> a |
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− | elementAt list i = list !! (i-1) |
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− | </haskell> |
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== Problem 4 == |
== Problem 4 == |
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Line 124: | Line 69: | ||

<haskell> |
<haskell> |
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− | Prelude> length [123, 456, 789] |
+ | Prelude> myLength [123, 456, 789] |

3 |
3 |
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− | Prelude> length "Hello, world!" |
+ | Prelude> myLength "Hello, world!" |

13 |
13 |
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</haskell> |
</haskell> |
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− | Solution: |
+ | [[99 questions/Solutions/4 | Solutions]] |

− | |||

− | <haskell> |
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− | length :: [a] -> Int |
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− | length [] = 0 |
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− | length (_:l) = 1 + length l |
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− | </haskell> |
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− | This function is defined in Prelude. |
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== Problem 5 == |
== Problem 5 == |
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Line 147: | Line 91: | ||

</haskell> |
</haskell> |
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− | Solution: (defined in Prelude) |
+ | [[99 questions/Solutions/5 | Solutions]] |

− | <haskell> |
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− | reverse :: [a] -> [a] |
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− | reverse = foldl (flip (:)) [] |
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− | </haskell> |
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− | |||

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

− | <haskell> |
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− | reverse :: [a] -> [a] |
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− | reverse [] = [] |
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− | reverse (x:xs) = reverse xs ++ [x] |
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− | </haskell> |
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== Problem 6 == |
== Problem 6 == |
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Line 177: | Line 109: | ||

</haskell> |
</haskell> |
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− | Solution: |
+ | [[99 questions/Solutions/6 | Solutions]] |

− | <haskell> |
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− | isPalindrome :: (Eq a) => [a] -> Bool |
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− | isPalindrome xs = xs == (reverse xs) |
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− | </haskell> |
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== Problem 7 == |
== Problem 7 == |
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Line 198: | Line 126: | ||

Example in Haskell: |
Example in Haskell: |
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+ | |||

+ | We have to define a new data type, because lists in Haskell are homogeneous. |
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+ | <haskell> |
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+ | data NestedList a = Elem a | List [NestedList a] |
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+ | </haskell> |
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<haskell> |
<haskell> |
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Line 208: | Line 141: | ||

</haskell> |
</haskell> |
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− | Solution: |
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− | <haskell> |
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− | data NestedList a = Elem a | List [NestedList a] |
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− | flatten :: NestedList a -> [a] |
+ | [[99 questions/Solutions/7 | Solutions]] |

− | flatten (Elem x) = [x] |
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− | flatten (List x) = concatMap flatten x |
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− | </haskell> |
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− | |||

− | We have to define a new data type, because lists in Haskell are homogeneous. |
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− | [1, [2, [3, 4], 5]] is a type error. Therefore, we must have a way of |
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− | representing a list that may (or may not) be nested. |
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− | |||

− | Our NestedList datatype is either a single element of some type (Elem a), or a |
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− | list of NestedLists of the same type. (List [NestedList a]). |
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== Problem 8 == |
== Problem 8 == |
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Line 220: | 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. |
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+ | |||

+ | Example: |
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<pre> |
<pre> |
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− | Example: |
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* (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)) |
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(A B C A D E) |
(A B C A D E) |
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+ | </pre> |
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Example in Haskell: |
Example in Haskell: |
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− | *Main> compress ['a','a','a','a','b','c','c','a','a','d','e','e','e','e'] |
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− | ['a','b','c','a','d','e'] |
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− | </pre> |
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− | Solution: |
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<haskell> |
<haskell> |
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− | compress :: Eq a => [a] -> [a] |
+ | > compress ["a","a","a","a","b","c","c","a","a","d","e","e","e","e"] |

− | compress = map head . group |
+ | ["a","b","c","a","d","e"] |

</haskell> |
</haskell> |
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− | 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: |
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− | |||

− | <haskell> |
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− | Ambiguous type variable `a' in the constraint: |
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− | `Eq a' |
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− | arising from use of `group' |
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− | Possible cause: the monomorphism restriction applied to the following: |
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− | compress :: [a] -> [a] |
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− | Probable fix: give these definition(s) an explicit type signature |
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− | or use -fno-monomorphism-restriction |
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− | </haskell> |
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− | |||

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

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

− | An alternative solution is |
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− | |||

− | <haskell> |
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− | compress [] = [] |
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− | compress [a] = [a] |
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− | compress (x : y : xs) = (if x == y then [] else [x]) ++ compress (y : xs) |
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− | </haskell> |
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− | |||

− | Another alternative is to use the Data.Set module to construct a Set. This solution also works if the repeated elements are not consecutive. This solution is potentially incorrect in the spirit of this quiz as it creates a data structure of a new type (Set), but still addresses the problem in question. |
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− | |||

− | <pre> |
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− | Prelude> let r = ["a","a","b","b","c","c","c"] |
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− | Prelude> :m Data.Set |
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− | Prelude Data.Set> let s = fromList r |
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− | Prelude Data.Set> s |
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− | fromList ["a","b","c"] |
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− | </pre> |
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== Problem 9 == |
== Problem 9 == |
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Line 243: | Line 171: | ||

(**) Pack consecutive duplicates of list elements into sublists. |
(**) Pack consecutive duplicates of list elements into sublists. |
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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. |
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+ | |||

+ | Example: |
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<pre> |
<pre> |
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− | Example: |
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* (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)) |
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((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)) |
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− | <example in lisp> |
+ | </pre> |

Example in Haskell: |
Example in Haskell: |
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− | *Main> pack ['a', 'a', 'a', 'a', 'b', 'c', 'c', 'a', 'a', 'd', 'e', 'e', 'e', 'e'] |
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− | ["aaaa","b","cc","aa","d","eeee"] |
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− | </pre> |
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− | Solution: |
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<haskell> |
<haskell> |
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− | 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'] |

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

</haskell> |
</haskell> |
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− | 'group' is also in the Prelude, here's an implementation using 'span'. |
+ | [[99 questions/Solutions/9 | Solutions]] |

− | + | ||

== Problem 10 == |
== Problem 10 == |
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Line 271: | Line 196: | ||

Example: |
Example: |
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<pre> |
<pre> |
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− | * (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> |
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Example in Haskell: |
Example in Haskell: |
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− | <pre> |
+ | <haskell> |

encode "aaaabccaadeeee" |
encode "aaaabccaadeeee" |
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[(4,'a'),(1,'b'),(2,'c'),(2,'a'),(1,'d'),(4,'e')] |
[(4,'a'),(1,'b'),(2,'c'),(2,'a'),(1,'d'),(4,'e')] |
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− | </pre> |
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− | |||

− | Solution: |
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− | <haskell> |
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− | encode xs = map (\x -> (length x,head x)) (group xs) |
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</haskell> |
</haskell> |
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− | which can also be expressed as a list comprehension: |
+ | [[99 questions/Solutions/10 | Solutions]] |

− | <haskell> |
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− | [(length x, head x) | x <- group xs] |
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− | </haskell> |
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− | Or writing it [[Pointfree]]: |
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− | |||

− | <haskell> |
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− | encode :: Eq a => [a] -> [(Int, a)] |
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− | encode = map (\x -> (length x, head x)) . group |
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− | </haskell> |
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− | |||

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

− | <haskell> |
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− | encode :: Eq a => [a] -> [(Int, a)] |
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− | encode xs = map (length &&& head) $ group xs |
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− | </haskell> |
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[[Category:Tutorials]] |
[[Category:Tutorials]] |

## Revision as of 16:10, 5 October 2012

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

## 1 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'

## 2 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'

## 3 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'

## 4 Problem 4

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

Example in Haskell:

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

## 5 Problem 5

(*) Reverse a list.

Example in Haskell:

Prelude> reverse "A man, a plan, a canal, panama!" "!amanap ,lanac a ,nalp a ,nam A" Prelude> reverse [1,2,3,4] [4,3,2,1]

## 6 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

## 7 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 []) []

## 8 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 ["a","a","a","a","b","c","c","a","a","d","e","e","e","e"] ["a","b","c","a","d","e"]

## 9 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"]

## 10 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')]