Part 1 Chapter 7 Exercise 6, part 4.

Author: PiotrJustyna

part1_chapter7_exercise6/Main.hs

@@ -19,11 +19,17 @@ main = do
     putStrLn . show $ chop8' [1, 2, 3, 4, 5, 6, 7, 8, 9]
 
     putStrLn "map (+1) [1, 2, 3]"
-    putStrLn . show $ map (+1) [1, 2, 3]
+    putStrLn . show $ map show [1, 2, 3]
 
     putStrLn "map' (+1) [1, 2, 3]"
     putStrLn . show $ map' (+1) [1, 2, 3]
 
+    putStrLn "take 5 $ iterate (+1) 2"
+    putStrLn . show . take 5 $ iterate (+1) 2
+
+    putStrLn "take 5 $ iterate' (+1) 2"
+    putStrLn . show . take 5 $ iterate' (+1) 2
+
 unfold :: (a -> Bool) -> (a -> a) -> (a -> a) -> a -> [a]
 unfold predicate headFunction tailFunction x
     | predicate x = []
@@ -50,17 +56,24 @@ chop8' = unfold null (take 8) (drop 8)
 -- Now with that in mind, does it really makes sense to apply "unfold" here? Is the function not supposed to build a list
 -- using the starting value and the functions provided? If not, I'm not sure what really is required. If so, though,
 -- the type signature defined by the regular "map" is going to make:
--- the implementation unnecessarily complicated
--- map and map' are not going to be interchangable.
+-- * the implementation unnecessarily complicated
+-- * map and map' are not going to be interchangable.
+-- Additionally, "unfold" requires the function f to be of type (a -> a) while "map" requires the function f to be of type (a -> a)
+-- so we know already that we won't be able to create an exact, interchangeable replica of the original "map".
+-- E.g. we would not be able to use the "map'" like this:
+--
+-- map' show [1, 2, 3]
+--
+-- which would be perfectly fine with the original "map".
 --
 -- My initial thought was to implement function:
 --
--- map' :: (a -> b) -> [a] -> [b]
+-- map' :: (a -> a) -> [a] -> [a]
 --
--- with the signature identical to the regular "map" (just a different implementation) so that both can be used interchangeably.
+-- with the signature similar to the regular "map" (just a different implementation) so that both can be used interchangeably.
 -- It took me a while before I realized that it is probably more tricky than it should be.
 -- This is when I started looking for other solutions, just to get an idea how people interpret the task.
--- What I found mademe even more confused. Examples:
+-- What I found made me even more confused. Examples:
 --
 -- 1. https://github.com/evturn's implementaion: https://github.com/evturn/programming-in-haskell/blob/master/07-higher-order-functions/07.9-exercises.hs
 -- 
@@ -70,20 +83,48 @@ chop8' = unfold null (take 8) (drop 8)
 --
 -- map' :: (a -> Bool) -> (a -> a) -> a -> [a]
 --
--- So as you can see, it doesn't match the original "map"'s type and the input argument (second last "a") gets "unfolded" into [a].
+-- It doesn't match the original "map"'s type very precisely and the input argument (second last "a") gets "unfolded" into [a].
 --
 -- 2. Another example I found is the one by https://github.com/RoccoMathijn: https://github.com/RoccoMathijn/programming-in-haskell/blob/master/chapter07.hs
 --
 -- map' :: (a -> b) -> [a] -> [b]
 -- map' f = unfold (null) (f.head) (tail)
 --
 -- But this again is not intercangeable with the original "map" as b is simply [a], so the result is really [[a]].
+-- Furthermore, I'm not sure how it even builds as "unfold" operates on (a -> a) and not (a -> b) as the function f. I'm using GHC 8.6.3 and getting the following:
+--
+-- $ ghci
+-- GHCi, version 8.6.3: http://www.haskell.org/ghc/  :? for help
+-- Prelude> :load Main
+-- [1 of 1] Compiling Main             ( Main.hs, interpreted )
+
+-- Main.hs:95:26: error:
+--     * Couldn't match type `b' with `[a]'
+--       `b' is a rigid type variable bound by
+--         the type signature for:
+--           map'' :: forall a b. (a -> b) -> [a] -> [b]
+--         at Main.hs:94:1-31
+--       Expected type: [a] -> [a]
+--         Actual type: [a] -> b
+--     * In the second argument of `unfold', namely `(f . head)'
+--       In the expression: unfold (null) (f . head) (tail)
+--       In an equation for map'': map'' f = unfold (null) (f . head) (tail)
+--     * Relevant bindings include
+--         f :: a -> b (bound at Main.hs:95:7)
+--         map'' :: (a -> b) -> [a] -> [b] (bound at Main.hs:95:1)
+--    |
+-- 95 | map'' f = unfold (null) (f.head) (tail)
+--    |                          ^^^^^^
+-- Failed, no modules loaded.
 --
--- After more time than I'd like to admit, I finally figured out the implementation which is interchangeable with the original "map".
+-- After more time than I'd like to admit, I finally figured out the implementation which is almost interchangeable with the original "map".
 -- Terribly inefficient, but I believe it answers the exercise's question precisely.
 
-map' :: (a -> b) -> [a] -> [b]
+map' :: (a -> a) -> [a] -> [a]
 map' f = concat . unfold null (listify . f . head) (tail)
 
 listify :: a -> [a]
-listify x = [x]
+listify x = [x]
+
+iterate' :: (a -> a) -> a -> [a]
+iterate' = unfold (\x -> False) id