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__NOTOC__
 
__NOTOC__
   
These are Haskell translations of [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://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].
 
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 11 ==
 
== Problem 11 ==
   
(*) Modified run-length encoding.
+
(*) Modified run-length encoding.
Modify the result of problem P10 in such a way that if an element has no duplicates it is simply copied into the result list. Only elements with duplicates are transferred as (N E) lists.
+
  +
Modify the result of problem 10 in such a way that if an element has no duplicates it is simply copied into the result list. Only elements with duplicates are transferred as (N E) lists.
  +
  +
Example:
   
 
<pre>
 
<pre>
Example:
 
 
* (encode-modified '(a a a a b c c a a d e e e e))
 
* (encode-modified '(a a a a b c c a a d e e e e))
 
((4 A) B (2 C) (2 A) D (4 E))
 
((4 A) B (2 C) (2 A) D (4 E))
  +
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
P11> encodeModified "aaaabccaadeeee"
 
[Multiple 4 'a',Single 'b',Multiple 2 'c',Multiple 2 'a',Single 'd',Multiple 4 'e']
 
</pre>
 
   
Solution:
 
 
<haskell>
 
<haskell>
data ListItem a = Single a | Multiple Int a
+
P11> encodeModified "aaaabccaadeeee"
+
[Multiple 4 'a',Single 'b',Multiple 2 'c',
encodeModified :: Eq a => [a] -> [ListItem a]
+
Multiple 2 'a',Single 'd',Multiple 4 'e']
encodeModified = map encodeHelper . encode
 
where
 
encodeHelper (1,x) = Single x
 
encodeHelper (n,x) = Multiple n x
 
 
</haskell>
 
</haskell>
   
Again, like in problem 7, we need a utility type because lists in haskell are homogeneous. Afterwards we use the <hask>encode</hask> function from problem 10 and map single instances of a list item to <hask>Single</hask> and multiple ones to <hask>Multiple</hask>.
+
[[99 questions/Solutions/11 | Solutions]]
+
 
== Problem 12 ==
 
== Problem 12 ==
   
(**) Decode a run-length encoded list.
+
(**) Decode a run-length encoded list.
Given a run-length code list generated as specified in problem P11. Construct its uncompressed version.
+
  +
Given a run-length code list generated as specified in problem 11. Construct its uncompressed version.
   
<pre>
 
 
Example in Haskell:
 
Example in Haskell:
P12> decodeModified [Multiple 4 'a',Single 'b',Multiple 2 'c',Multiple 2 'a',Single 'd',Multiple 4 'e']
 
"aaaabccaadeeee"
 
</pre>
 
   
Solution:
 
 
<haskell>
 
<haskell>
decodeModified :: [ListItem a] -> [a]
+
P12> decodeModified
decodeModified = concatMap decodeHelper
+
[Multiple 4 'a',Single 'b',Multiple 2 'c',
where
+
Multiple 2 'a',Single 'd',Multiple 4 'e']
decodeHelper (Single x) = [x]
+
"aaaabccaadeeee"
decodeHelper (Multiple n x) = replicate n x
 
 
</haskell>
 
</haskell>
   
We only need to map single instances of an element to a list containing only one element and multiple ones to a list containing the specified number of elements and concatenate these lists.
+
[[99 questions/Solutions/12 | Solutions]]
+
 
== Problem 13 ==
 
== Problem 13 ==
   
 
(**) Run-length encoding of a list (direct solution).
 
(**) Run-length encoding of a list (direct solution).
Implement the so-called run-length encoding data compression method directly. I.e. don't explicitly create the sublists containing the duplicates, as in problem P09, but only count them. As in problem P11, simplify the result list by replacing the singleton lists (1 X) by X.
 
   
<pre>
+
Implement the so-called run-length encoding data compression method directly. I.e. don't explicitly create the sublists containing the duplicates, as in problem 9, but only count them. As in problem P11, simplify the result list by replacing the singleton lists (1 X) by X.
  +
 
Example:
 
Example:
  +
  +
<pre>
 
* (encode-direct '(a a a a b c c a a d e e e e))
 
* (encode-direct '(a a a a b c c a a d e e e e))
 
((4 A) B (2 C) (2 A) D (4 E))
 
((4 A) B (2 C) (2 A) D (4 E))
  +
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
P13> encodeDirect "aaaabccaadeeee"
 
[Multiple 4 'a',Single 'b',Multiple 2 'c',Multiple 2 'a',Single 'd',Multiple 4 'e']
 
</pre>
 
   
Solution:
 
 
<haskell>
 
<haskell>
encode' :: Eq a => [a] -> [(Int,a)]
+
P13> encodeDirect "aaaabccaadeeee"
encode' = foldr helper []
+
[Multiple 4 'a',Single 'b',Multiple 2 'c',
where
+
Multiple 2 'a',Single 'd',Multiple 4 'e']
helper x [] = [(1,x)]
 
helper x (y:ys)
 
| x == snd y = (1+fst y,x):ys
 
| otherwise = (1,x):y:ys
 
 
encodeDirect :: Eq a => [a] -> [ListItem a]
 
encodeDirect = map encodeHelper . encode'
 
where
 
encodeHelper (1,x) = Single x
 
encodeHelper (n,x) = Multiple n x
 
 
</haskell>
 
</haskell>
   
First of all we could rewrite the function <hask>encode</hask> from problem 10 in a way that is does not create the sublists. Thus, I decided to traverse the original list from right to left (using <hask>foldr</hask>) and to prepend each element to the resulting list in the proper way. Thereafter we only need to modify the function <hask>encodeModified</hask> from problem 11 to use <hask>encode'</hask>.
+
[[99 questions/Solutions/13 | Solutions]]
+
 
== Problem 14 ==
 
== Problem 14 ==
   
 
(*) Duplicate the elements of a list.
 
(*) Duplicate the elements of a list.
  +
  +
Example:
   
 
<pre>
 
<pre>
Example:
 
 
* (dupli '(a b c c d))
 
* (dupli '(a b c c d))
 
(A A B B C C C C D D)
 
(A A B B C C C C D D)
  +
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
  +
  +
<haskell>
 
> dupli [1, 2, 3]
 
> dupli [1, 2, 3]
 
[1,1,2,2,3,3]
 
[1,1,2,2,3,3]
</pre>
 
 
Solution:
 
<haskell>
 
dupli [] = []
 
dupli (x:xs) = x:x:dupli xs
 
 
</haskell>
 
</haskell>
   
or using the existance of the list monad:
+
[[99 questions/Solutions/14 | Solutions]]
<haskell>
+
dupli xs = xs >>= (\x -> [x,x])
+
</haskell>
 
 
 
== Problem 15 ==
 
== Problem 15 ==
   
 
(**) Replicate the elements of a list a given number of times.
 
(**) Replicate the elements of a list a given number of times.
   
<pre>
 
 
Example:
 
Example:
  +
  +
<pre>
 
* (repli '(a b c) 3)
 
* (repli '(a b c) 3)
 
(A A A B B B C C C)
 
(A A A B B B C C C)
  +
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
> repli "abc" 3
 
"aaabbbccc"
 
</pre>
 
   
Solution:
 
 
<haskell>
 
<haskell>
repli :: [a] -> Int -> [a]
+
> repli "abc" 3
repli as n = concatMap (replicate n) as
+
"aaabbbccc"
 
</haskell>
 
</haskell>
  +
  +
[[99 questions/Solutions/15 | Solutions]]
  +
   
 
== Problem 16 ==
 
== Problem 16 ==
  +
 
(**) Drop every N'th element from a list.
 
(**) Drop every N'th element from a list.
   
<pre>
 
 
Example:
 
Example:
  +
  +
<pre>
 
* (drop '(a b c d e f g h i k) 3)
 
* (drop '(a b c d e f g h i k) 3)
 
(A B D E G H K)
 
(A B D E G H K)
  +
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
*Main> drop = "abcdefghik" 3
 
"abdeghk"
 
</pre>
 
   
Solution:
 
 
<haskell>
 
<haskell>
drop xs n = drops xs (n-1) n
+
*Main> dropEvery "abcdefghik" 3
drops [] _ _ = []
+
"abdeghk"
drops (x:xs) 0 max = drops xs (max-1) max
 
drops (x:xs) (n+1) max = x:drops xs n max
 
 
</haskell>
 
</haskell>
   
Here, drops is a helper-function to drop. In drops, there is an index n that counts from max-1 down to 0, and removes the head element each time it hits 0.
+
[[99 questions/Solutions/16 | Solutions]]
   
Note that drop is one of the standard Haskell functions, so redefining it is generally not a good idea.
 
 
or using zip:
 
<haskell>
 
drop n = map snd . filter ((n/=) . fst) . zip (cycle [1..n])
 
</haskell>
 
 
 
 
== Problem 17 ==
 
== Problem 17 ==
Line 149: Line 130:
 
Do not use any predefined predicates.
 
Do not use any predefined predicates.
   
<pre>
 
 
Example:
 
Example:
  +
  +
<pre>
 
* (split '(a b c d e f g h i k) 3)
 
* (split '(a b c d e f g h i k) 3)
 
( (A B C) (D E F G H I K))
 
( (A B C) (D E F G H I K))
  +
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
*Main> split "abcdefghik" 3
 
("abc", "defghik")
 
</pre>
 
   
Solution using take and drop:
 
 
<haskell>
 
<haskell>
split xs n = (take n xs, drop n xs)
+
*Main> split "abcdefghik" 3
  +
("abc", "defghik")
 
</haskell>
 
</haskell>
Note that this function, with the parameters in the other order, exists as <hask>splitAt</hask>.
 
   
+
[[99 questions/Solutions/17 | Solutions]]
  +
  +
 
== Problem 18 ==
 
== Problem 18 ==
   
 
(**) Extract a slice from a list.
 
(**) Extract a slice from a list.
   
Given two indices, i and k, the slice is the list containing the elements between the i'th and i'th element of the original list (both limits included). Start counting the elements with 1.
+
Given two indices, i and k, the slice is the list containing the elements between the i'th and k'th element of the original list (both limits included). Start counting the elements with 1.
   
<pre>
 
 
Example:
 
Example:
  +
  +
<pre>
 
* (slice '(a b c d e f g h i k) 3 7)
 
* (slice '(a b c d e f g h i k) 3 7)
 
(C D E F G)
 
(C D E F G)
  +
</pre>
   
 
Example in Haskell:
 
Example in Haskell:
*Main> slice ['a','b','c','d','e','f','g','h','i','k'] 3 7
 
</pre>
 
   
Solution:
 
 
<haskell>
 
<haskell>
slice xs (i+1) k = take (k-i) $ drop i xs
+
*Main> slice ['a','b','c','d','e','f','g','h','i','k'] 3 7
  +
"cdefg"
 
</haskell>
 
</haskell>
+
  +
[[99 questions/Solutions/18 | Solutions]]
  +
  +
 
== Problem 19 ==
 
== Problem 19 ==
   
Line 192: Line 169:
 
Hint: Use the predefined functions length and (++).
 
Hint: Use the predefined functions length and (++).
   
<pre>
 
 
Examples:
 
Examples:
  +
  +
<pre>
 
* (rotate '(a b c d e f g h) 3)
 
* (rotate '(a b c d e f g h) 3)
 
(D E F G H A B C)
 
(D E F G H A B C)
Line 199: Line 177:
 
* (rotate '(a b c d e f g h) -2)
 
* (rotate '(a b c d e f g h) -2)
 
(G H A B C D E F)
 
(G H A B C D E F)
  +
</pre>
   
 
Examples in Haskell:
 
Examples in Haskell:
  +
  +
<haskell>
 
*Main> rotate ['a','b','c','d','e','f','g','h'] 3
 
*Main> rotate ['a','b','c','d','e','f','g','h'] 3
 
"defghabc"
 
"defghabc"
Line 206: Line 187:
 
*Main> rotate ['a','b','c','d','e','f','g','h'] (-2)
 
*Main> rotate ['a','b','c','d','e','f','g','h'] (-2)
 
"ghabcdef"
 
"ghabcdef"
</pre>
 
 
Solution:
 
<haskell>
 
rotate [] _ = []
 
rotate l 0 = l
 
rotate (x:xs) (n+1) = rotate (xs ++ [x]) n
 
rotate l n = rotate l (length l + n)
 
 
</haskell>
 
</haskell>
   
There are two separate cases:
+
[[99 questions/Solutions/19 | Solutions]]
<br/>- If n > 0, move the first element to the end of the list n times.
 
<br/>- If n < 0, convert the problem to the equivalent problem for n > 0 by adding the list's length to n.
 
   
 
 
 
== Problem 20 ==
 
== Problem 20 ==
   
Remove the K'th element from a list.
+
(*) Remove the K'th element from a list.
  +
  +
Example in Prolog:
  +
  +
<pre>
  +
?- remove_at(X,[a,b,c,d],2,R).
  +
X = b
  +
R = [a,c,d]
  +
</pre>
  +
  +
Example in Lisp:
   
 
<pre>
 
<pre>
Example:
 
 
* (remove-at '(a b c d) 2)
 
* (remove-at '(a b c d) 2)
 
(A C D)
 
(A C D)
  +
</pre>
  +
  +
(Note that this only returns the residue list, while the Prolog version also returns the deleted element.)
   
 
Example in Haskell:
 
Example in Haskell:
*Main> removeAt 1 ['a','b','c','d']
 
"acd"
 
</pre>
 
   
Solution:
 
 
<haskell>
 
<haskell>
removeAt k xs = take k xs ++ drop (k+1) xs
+
*Main> removeAt 2 "abcd"
  +
('b',"acd")
 
</haskell>
 
</haskell>
   
Simply use the take and drop functions from the Prelude to take k elements from the start of xs and prepend to the list of elements k+1 to the end. Note that the Lisp code treats 1 as the first element in the list, and it appends NIL elements to the end of the list if k is greater than the list length.
+
[[99 questions/Solutions/20 | Solutions]]
  +
   
 
[[Category:Tutorials]]
 
[[Category:Tutorials]]

Latest revision as of 08:05, 26 May 2012


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

[edit] 1 Problem 11

(*) Modified run-length encoding.

Modify the result of problem 10 in such a way that if an element has no duplicates it is simply copied into the result list. Only elements with duplicates are transferred as (N E) lists.

Example:

* (encode-modified '(a a a a b c c a a d e e e e))
((4 A) B (2 C) (2 A) D (4 E))

Example in Haskell:

P11> encodeModified "aaaabccaadeeee"
[Multiple 4 'a',Single 'b',Multiple 2 'c',
 Multiple 2 'a',Single 'd',Multiple 4 'e']

Solutions

[edit] 2 Problem 12

(**) Decode a run-length encoded list.

Given a run-length code list generated as specified in problem 11. Construct its uncompressed version.

Example in Haskell:

P12> decodeModified 
       [Multiple 4 'a',Single 'b',Multiple 2 'c',
        Multiple 2 'a',Single 'd',Multiple 4 'e']
"aaaabccaadeeee"

Solutions

[edit] 3 Problem 13

(**) Run-length encoding of a list (direct solution).

Implement the so-called run-length encoding data compression method directly. I.e. don't explicitly create the sublists containing the duplicates, as in problem 9, but only count them. As in problem P11, simplify the result list by replacing the singleton lists (1 X) by X.

Example:

* (encode-direct '(a a a a b c c a a d e e e e))
((4 A) B (2 C) (2 A) D (4 E))

Example in Haskell:

P13> encodeDirect "aaaabccaadeeee"
[Multiple 4 'a',Single 'b',Multiple 2 'c',
 Multiple 2 'a',Single 'd',Multiple 4 'e']

Solutions

[edit] 4 Problem 14

(*) Duplicate the elements of a list.

Example:

* (dupli '(a b c c d))
(A A B B C C C C D D)

Example in Haskell:

> dupli [1, 2, 3]
[1,1,2,2,3,3]

Solutions


[edit] 5 Problem 15

(**) Replicate the elements of a list a given number of times.

Example:

* (repli '(a b c) 3)
(A A A B B B C C C)

Example in Haskell:

> repli "abc" 3
"aaabbbccc"

Solutions


[edit] 6 Problem 16

(**) Drop every N'th element from a list.

Example:

* (drop '(a b c d e f g h i k) 3)
(A B D E G H K)

Example in Haskell:

*Main> dropEvery "abcdefghik" 3
"abdeghk"

Solutions


[edit] 7 Problem 17

(*) Split a list into two parts; the length of the first part is given.

Do not use any predefined predicates.

Example:

* (split '(a b c d e f g h i k) 3)
( (A B C) (D E F G H I K))

Example in Haskell:

*Main> split "abcdefghik" 3
("abc", "defghik")

Solutions


[edit] 8 Problem 18

(**) Extract a slice from a list.

Given two indices, i and k, the slice is the list containing the elements between the i'th and k'th element of the original list (both limits included). Start counting the elements with 1.

Example:

* (slice '(a b c d e f g h i k) 3 7)
(C D E F G)

Example in Haskell:

*Main> slice ['a','b','c','d','e','f','g','h','i','k'] 3 7
"cdefg"

Solutions


[edit] 9 Problem 19

(**) Rotate a list N places to the left.

Hint: Use the predefined functions length and (++).

Examples:

* (rotate '(a b c d e f g h) 3)
(D E F G H A B C)

* (rotate '(a b c d e f g h) -2)
(G H A B C D E F)

Examples in Haskell:

*Main> rotate ['a','b','c','d','e','f','g','h'] 3
"defghabc"
 
*Main> rotate ['a','b','c','d','e','f','g','h'] (-2)
"ghabcdef"

Solutions


[edit] 10 Problem 20

(*) Remove the K'th element from a list.

Example in Prolog:

?- remove_at(X,[a,b,c,d],2,R).
X = b
R = [a,c,d]

Example in Lisp:

* (remove-at '(a b c d) 2)
(A C D)

(Note that this only returns the residue list, while the Prolog version also returns the deleted element.)

Example in Haskell:

*Main> removeAt 2 "abcd"
('b',"acd")

Solutions