Difference between revisions of "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 [ |
+ | 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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+ | <div style="border-bottom:1px solid #eee">(*) Find the last element of a list. <span style="float:right"><small>[[99 questions/Solutions/1|Solutions]]</small></span> |
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+ | </div> |
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+ | <br> |
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− | ( |
+ | (Note that the Lisp transcription of this problem is incorrect.) |
− | |||
− | Example: |
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− | |||
− | <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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<haskell> |
<haskell> |
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− | + | λ> myLast [1,2,3,4] |
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− | + | 4 |
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− | + | λ> myLast ['x','y','z'] |
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− | + | 'z' |
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</haskell> |
</haskell> |
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− | Solution: |
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− | |||
− | <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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+ | <div style="border-bottom:1px solid #eee">(*) Find the last-but-one (or second-last) element of a list. <span style="float:right"><small>[[99 questions/Solutions/2|Solutions]]</small></span> |
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+ | </div> |
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+ | <br> |
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− | ( |
+ | (Note that the Lisp transcription of this problem is incorrect.) |
− | |||
− | Example: |
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− | |||
− | <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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<haskell> |
<haskell> |
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− | + | λ> myButLast [1,2,3,4] |
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+ | 3 |
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− | [3,4] |
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− | + | λ> myButLast ['a'..'z'] |
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+ | 'y' |
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− | "yz" |
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</haskell> |
</haskell> |
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− | Solution: |
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− | |||
− | <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 = 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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+ | <div style="border-bottom:1px solid #eee">(*) Find the K'th element of a list. <span style="float:right"><small>[[99 questions/Solutions/3|Solutions]]</small></span> |
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+ | </div> |
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+ | <br> |
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− | + | The first element in the list is number 1. |
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− | |||
Example: |
Example: |
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<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 |
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− | C |
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</pre> |
</pre> |
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Line 102: | Line 54: | ||
<haskell> |
<haskell> |
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− | + | λ> elementAt [1,2,3] 2 |
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2 |
2 |
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− | + | λ> elementAt "haskell" 5 |
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'e' |
'e' |
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</haskell> |
</haskell> |
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− | Solution: |
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− | |||
− | 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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+ | <div style="border-bottom:1px solid #eee">(*) Find the number of elements in a list. <span style="float:right"><small>[[99 questions/Solutions/4|Solutions]]</small></span> |
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− | |||
+ | </div> |
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− | (*) Find the number of elements of a list. |
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+ | <br> |
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Example in Haskell: |
Example in Haskell: |
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<haskell> |
<haskell> |
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− | + | λ> myLength [123, 456, 789] |
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3 |
3 |
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− | + | λ> myLength "Hello, world!" |
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13 |
13 |
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</haskell> |
</haskell> |
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− | Solution: |
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− | |||
− | <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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+ | <div style="border-bottom:1px solid #eee">(*) Reverse a list. <span style="float:right"><small>[[99 questions/Solutions/5|Solutions]]</small></span> |
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− | |||
+ | </div> |
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− | (*) Reverse a list. |
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+ | <br> |
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Example in Haskell: |
Example in Haskell: |
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<haskell> |
<haskell> |
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− | + | λ> myReverse "A man, a plan, a canal, panama!" |
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"!amanap ,lanac a ,nalp a ,nam A" |
"!amanap ,lanac a ,nalp a ,nam A" |
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− | + | λ> myReverse [1,2,3,4] |
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[4,3,2,1] |
[4,3,2,1] |
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</haskell> |
</haskell> |
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− | Solution: (defined in Prelude) |
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− | |||
− | <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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+ | <div style="border-bottom:1px solid #eee">(*) Find out whether a list is a palindrome. <span style="float:right"><small>[[99 questions/Solutions/6|Solutions]]</small></span> |
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+ | </div> |
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+ | <br> |
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− | + | Hint: A palindrome can be read forward or backward; e.g. (x a m a x). |
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Example in Haskell: |
Example in Haskell: |
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<haskell> |
<haskell> |
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− | + | λ> isPalindrome [1,2,3] |
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False |
False |
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− | + | λ> isPalindrome "madamimadam" |
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True |
True |
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− | + | λ> isPalindrome [1,2,4,8,16,8,4,2,1] |
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True |
True |
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</haskell> |
</haskell> |
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− | Solution: |
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− | |||
− | <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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+ | <div style="border-bottom:1px solid #eee">(**) Flatten a nested list structure. <span style="float:right"><small>[[99 questions/Solutions/7|Solutions]]</small></span> |
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− | |||
+ | </div> |
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− | (**) Flatten a nested list structure. |
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+ | <br> |
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Transform a list, possibly holding lists as elements into a `flat' list by replacing each list with its elements (recursively). |
Transform a list, possibly holding lists as elements into a `flat' list by replacing each list with its elements (recursively). |
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Line 213: | 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> |
<haskell> |
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+ | data NestedList a = Elem a | List [NestedList a] |
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− | *Main> flatten (Elem 5) |
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+ | </haskell> |
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+ | |||
+ | <haskell> |
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+ | λ> flatten (Elem 5) |
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[5] |
[5] |
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− | + | λ> flatten (List [Elem 1, List [Elem 2, List [Elem 3, Elem 4], Elem 5]]) |
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[1,2,3,4,5] |
[1,2,3,4,5] |
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− | + | λ> flatten (List []) |
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[] |
[] |
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</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] |
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− | 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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+ | <div style="border-bottom:1px solid #eee">(**) Eliminate consecutive duplicates of list elements. <span style="float:right"><small>[[99 questions/Solutions/8|Solutions]]</small></span> |
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− | |||
+ | </div> |
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− | (**) Eliminate consecutive duplicates of list elements. |
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+ | <br> |
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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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+ | λ> compress "aaaabccaadeeee" |
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− | compress :: Eq a => [a] -> [a] |
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+ | "abcade" |
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− | compress = map head . group |
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</haskell> |
</haskell> |
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− | We simply group equal values together (group), then take the head of each. |
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− | 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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== Problem 9 == |
== Problem 9 == |
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+ | <div style="border-bottom:1px solid #eee">(**) Pack consecutive duplicates of list elements into sublists. <span style="float:right"><small>[[99 questions/Solutions/9|Solutions]]</small></span> |
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+ | </div> |
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+ | <br> |
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− | (**) 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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+ | </pre> |
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− | <example in lisp> |
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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 |
+ | λ> pack ['a', 'a', 'a', 'a', 'b', 'c', 'c', 'a', |
− | + | 'a', 'd', 'e', 'e', 'e', 'e'] |
|
+ | ["aaaa","b","cc","aa","d","eeee"] |
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− | pack [] = [] |
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</haskell> |
</haskell> |
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+ | |||
− | 'group' is also in the Prelude, here's an implementation using 'span'. |
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− | |||
== Problem 10 == |
== Problem 10 == |
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+ | <div style="border-bottom:1px solid #eee">(*) Run-length encoding of a list. <span style="float:right"><small>[[99 questions/Solutions/10|Solutions]]</small></span> |
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+ | </div> |
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+ | <br> |
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+ | Use the result of Problem 9 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. |
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− | (*) Run-length encoding of a list. |
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− | 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. |
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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)) |
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− | + | ((4 A) (1 B) (2 C) (2 A) (1 D)(4 E)) |
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</pre> |
</pre> |
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Example in Haskell: |
Example in Haskell: |
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− | < |
+ | <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')] |
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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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− | 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]] |
Latest revision as of 05:27, 10 June 2023
This is part of Ninety-Nine Haskell Problems, based on Ninety-Nine Prolog Problems and Ninety-Nine Lisp Problems.
Problem 1
(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
(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
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'
Problem 4
Example in Haskell:
λ> myLength [123, 456, 789]
3
λ> myLength "Hello, world!"
13
Problem 5
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
Hint: 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
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 [])
[]
Problem 8
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
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
Use the result of Problem 9 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')]