Haskell: Exercise #9

Haskell.org

@@ -3,7 +3,7 @@
 
 https://wiki.haskell.org/H-99:_Ninety-Nine_Haskell_Problems
 
-*** #1 Find the last element of a list. 
+*** #1 Find the last element of a list.
 
 #+begin_src haskell
 λ> myLast [1,2,3,4]
@@ -61,7 +61,7 @@ elementAt (x:xs) i = if i == 1 then x else elementAt xs (i - 1)
 #+end_src
 
 
-*** #4 Find the number of elements of a list. 
+*** #4 Find the number of elements of a list.
 
 #+begin_src haskell
 λ> myLength [123, 456, 789]
@@ -80,11 +80,162 @@ myLength xs = foldl (\acc _ -> acc + 1) 0 xs
 myLength' [] = 0
 myLength' [x] = 1
 myLength' (_:xs) = 1 + myLength xs
-
-myLength [1..5]
-myLength' [1..5]
 #+end_src
 
-#+RESULTS:
-: 5
+*** #5 Reverse a list.
 
+#+begin_src 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]
+#+end_src
+
+
+**** Solution
+
+#+begin_src haskell
+myReverse = go []
+  where go acc [] = acc
+        go acc (x:xs) = go (x:acc) xs
+#+end_src
+
+*** #6 Find out whether a list is a palindrome
+
+#+begin_src haskell
+λ> isPalindrome [1,2,3]
+False
+λ> isPalindrome "madamimadam"
+True
+λ> isPalindrome [1,2,4,8,16,8,4,2,1]
+True
+#+end_src
+
+**** Solution
+#+begin_src haskell
+isPalindrome xs = xs == reverse xs
+#+end_src
+*** #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:
+#+begin_src haskell
+,* (my-flatten '(a (b (c d) e)))
+(A B C D E)
+#+end_src
+
+Example in Haskell:
+
+We have to define a new data type, because lists in Haskell are homogeneous.
+
+#+begin_src haskell
+data NestedList a = Elem a | List [NestedList a]
+#+end_src
+
+#+begin_src haskell
+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 [])
+[]
+#+end_src
+
+**** Solution
+#+begin_src haskell
+flatten = reverse . go []
+  where go acc (List []) = acc
+        go acc (Elem x) = x:acc
+        go acc (List (x:xs)) = go (go acc x) (List xs)
+#+end_src
+*** #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:
+
+#+begin_src haskell
+,* (compress '(a a a a b c c a a d e e e e))
+(A B C A D E)
+#+end_src
+
+Example in Haskell:
+
+#+begin_src haskell
+λ> compress "aaaabccaadeeee"
+"abcade"
+#+end_src
+**** Solution
+
+This is not great...
+
+#+begin_src haskell
+compress xs = reverse $ go [] xs
+  where go acc [] = acc
+        go [] (x:xs) = go [x] xs
+        go [a] [x]
+            | a == x = [a]
+            | otherwise = [x,a]
+        go (a:as) [x]
+            | a == x = a:as
+            | otherwise = x:a:as
+        go [a] (x:xs)
+            | a == x = go [a] xs
+            | otherwise = go [x,a] xs
+        go (a:as) (x:xs)
+            | a == x = go (a:as) xs
+            | otherwise = go (x:a:as) xs
+#+end_src
+
+Especially when one of the solutions is the following;
+
+#+begin_src haskell
+import Data.List
+
+compress :: Eq a => [a] -> [a]
+compress = map head . group
+#+end_src
+*** #9 Pack consecutive duplicates of list elements into sublists
+
+If a list contains repeated elements they should be placed in separate sublists.
+
+Example:
+
+#+begin_src haskell
+,* (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))
+#+end_src
+
+Example in Haskell:
+
+#+begin_src haskell
+λ> pack ['a', 'a', 'a', 'a', 'b', 'c', 'c', 'a', 'a', 'd', 'e', 'e', 'e', 'e']
+["aaaa","b","cc","aa","d","eeee"]
+#+end_src
+**** Solution
+In python you can use ~Counter~
+#+begin_src python
+from collections import Counter
+def pack(arr):
+    return [[key] * val for key,val in Counter(arr).items()]
+#+end_src
+
+In Haskell, this is just ~group~
+
+#+begin_src haskell
+pack :: [a] -> [[a]]
+pack = group
+#+end_src
+
+Here's my naive implementation
+
+#+begin_src haskell
+pack :: Eq a => [a] -> [[a]]
+pack xs = reverse $ go [[]] xs
+  where
+    go acc [] = acc
+    go ([]:acc) (x:xs) = go ([x]:acc) xs
+    go ((a:as):acc) (x:xs)
+       | a == x = go ((x:a:as):acc) xs
+       | otherwise = go ([x]:(a:as):acc) xs
+#+end_src

README.org

@@ -7,12 +7,12 @@ challenge problems in the style of "99 Problems". See below for references. The
 intention is to practice these languages in the hopes of someday using them in
 anger. The solutions use a style of programming known as [[https://en.wikipedia.org/wiki/Literate_programming][Literate Programming]]
 with the help of Emacs, org-mode, and [[https://orgmode.org/worg/org-contrib/babel/intro.html][org-babel]]. All code is executable from
-within Emacs by just running ~C-c C-c~ while the cursor is in the source code
-block.
+within Emacs by invoking the ~C-c C-c~ keybinding while the cursor is in the
+source code block.
 
 * Progress
 - [16/99] OCaml
-- [4/99] Haskell
+- [7/99] Haskell
 - [0/99] Lisp (Soon)
 
 * References