Note: this is the completed version of lecture Foldr.

In class exercise: foldr

In HigherOrder we saw a few functions that you could write using the general purpose foldr function. This function captures the general pattern of list recursion and is also good practice for working with higher-order functions.

This set of exercises is intended to give you more practice with using foldr on lists. You should start with one at the right level for you and your partner (they are ordered in terms of difficulty).

If you finish this module, you should also look at variations in the definition of foldr shown in the module Sum.

> module Foldr where
> -- See: https://www.seas.upenn.edu/~cis5520/current/lectures/stub/02-higherorder/Foldr.html

> import Prelude hiding (all, filter, map, length, last, reverse, foldl, foldl1)

Length Example

This function counts the number of elements stored in a list. The recursive definition of the length function is

> -- >>> length1 "abc"
> -- 3
> length1 :: [Char] -> Int
> length1 []     = 0
> length1 (_:xs) = 1 + length1 xs

Now we can rewrite it in terms of foldr.

> length :: [Char] -> Int
> length = foldr (\_ n -> 1 + n) 0

and test it on some inputs

> -- >>> length "abc"
> -- 3
> -- >>> length ""
> -- 0

Once we have completed the foldr version, we can trace through its evaluation using the argument ['a','b','c'].

The evaluation starts out as:

  foldr (\_ n -> 1 + n) 0 ['a', 'b', 'c']
     = (_ n -> 1 + n) 'a' (foldr (\_ n -> 1 + n) 0 ['b', 'c'])
     = 1 + (foldr (\_ n -> 1 + n) 0 ['b', 'c'])

By unfolding the definition of foldr one step as above we can kind of see what it's doing. It's just adding 1 each time we encounter an element. The first argument in the anonymous function is just ignored, because you don't care what the elements of the list are when you're just counting them. The n is the accumulator, representing the rest of the fold (which in this case is the length of the tail). 1 + length of tail gives you the length of your list.

Unfolding this further we can see how the whole thing would evaluate:

     foldr (\_ n -> 1 + n) 0 ['a', 'b', 'c']
      = 1 + (foldr (\_ n -> 1 + n) 0 ['b', 'c'])
      = 1 + (1 + (foldr (\_ n -> 1 + n) 0 ['c']))
             -- Skipping to when the anonymous function is applied :)
      = 1 + (1 + (1 + (foldr (\_ n -> 1 + n) 0 [])))
      = 1 + (1 + (1 + 0))
      = 1 + (1 + 1)
      = 1 + 2
      = 3

Note, this evaluation can also be generated by the online tool.

Feel free to use this tool below, but understand that it is fairly simplistic. It won't be able to handle all of the examples that you throw at it.

All

Calculate whether all elements of a list satisfy a given predicate. The recursive definition is

> all1 :: (a -> Bool) -> [a] -> Bool
> all1 _ [] = True
> all1 p (x:xs) = p x && all1 p xs

Now implement using foldr

> -- >>> all (>10) ([1 .. 20] :: [Int])
> -- False
> -- >>> all (>0) ([1 .. 20] :: [Int])
> -- True
> all :: (a -> Bool) -> [a] -> Bool
> all p =  foldr (\x y -> p x && y) True

And trace through an evaluation all not [True,False]:

  foldr (\ x y -> not x && y) True [True, False]

  foldr (\ x y -> not x && y) True (True : [False])

  (\ x y -> not x && y) True
    (foldr (\ x y -> not x && y) True [False])

  not True && foldr (\ x y -> not x && y) True [False]

  False && foldr (\ x y -> not x && y) True [False]

  False

(Also available from online tool.)

Last

Find and return the last element of the lists (if the list is nonempty).

The recursive definition is

> last1 :: [Char] -> Maybe Char
> last1 [] = Nothing
> last1 (x:xs) = case last1 xs of
>                   Nothing -> Just x  -- xs does not have a last element, list must be `[x]`
>                   Just y  -> Just y  -- xs has a last element, keep it for `(x:xs)`

Now implement using foldr

> -- >>> last "abcd"
> -- Just 'd'
> -- >>> last ""
> -- Nothing
> last :: [Char] -> Maybe Char
> last =  foldr g Nothing where
>             g :: Char -> Maybe Char -> Maybe Char
>             g x Nothing = Just x
>             g x (Just y) = Just y

and trace through the evaluation last [1,2]

This can be calculated by the online tool.

Filter

The filter function selects items from a list that satisfy a given predicate. The output list should contain only the elements of the first list for which the input function returns True.

> filter :: (a -> Bool) -> [a] -> [a]
> filter p =  foldr (\x y -> if p x then x:y else y) []

> -- >>> filter (> 10) [1 .. 20]
> -- [11,12,13,14,15,16,17,18,19,20]

> -- >>> filter (\l -> sum l <= 42) [ [10,20], [50,50], [1..5] ] 
> -- [[10,20],[1,2,3,4,5]]

Trace the evaluation of filter (>2) [2,3]

Can be calculated via online tool.

   foldr (\ x y -> if (> 2) x then x : y else y) [] [2, 3]

   foldr (\ x y -> if (> 2) x then x : y else y) [] (2 : [3])

   (\ x y -> if (> 2) x then x : y else y) 2
     (foldr (\ x y -> if (> 2) x then x : y else y) [] [3])

   if (> 2) 2 then
     2 : foldr (\ x y -> if (> 2) x then x : y else y) [] [3] else
     foldr (\ x y -> if (> 2) x then x : y else y) [] [3]

   if 2 > 2 then
     2 : foldr (\ x y -> if (> 2) x then x : y else y) [] [3] else
     foldr (\ x y -> if (> 2) x then x : y else y) [] [3]

   if False then
     2 : foldr (\ x y -> if (> 2) x then x : y else y) [] [3] else
     foldr (\ x y -> if (> 2) x then x : y else y) [] [3]

   foldr (\ x y -> if (> 2) x then x : y else y) [] [3]

   foldr (\ x y -> if (> 2) x then x : y else y) [] (3 : [])

   (\ x y -> if (> 2) x then x : y else y) 3
     (foldr (\ x y -> if (> 2) x then x : y else y) [] [])

   if (> 2) 3 then
     3 : foldr (\ x y -> if (> 2) x then x : y else y) [] [] else
     foldr (\ x y -> if (> 2) x then x : y else y) [] []

   if 3 > 2 then
     3 : foldr (\ x y -> if (> 2) x then x : y else y) [] [] else
     foldr (\ x y -> if (> 2) x then x : y else y) [] []

   if True then
     3 : foldr (\ x y -> if (> 2) x then x : y else y) [] [] else
     foldr (\ x y -> if (> 2) x then x : y else y) [] []

   3 : foldr (\ x y -> if (> 2) x then x : y else y) [] []

   3 : []

Reverse

Reverse the elements appearing in the list.

Consider this linear time version that used direct recursion.

> reverse1 :: [Char] -> [Char]
> reverse1 l = aux l [] where
>   aux []     = id
>   aux (x:xs) = \ys -> aux xs (x : ys)

Now rewrite the aux function using foldr

> reverse :: [Char] -> [Char]
> reverse l = aux l [] where
>   aux =  foldr (\x f y -> f (x : y)) id

> -- >>> reverse "abcd"
> -- "dcba"

> -- >>> reverse ""
> -- ""

And trace through its evaluation on the list ['a','b','c']:

   foldr (\x f y -> f (x:y)) id ['a','b','c'] []

   =>   -- definition of foldr

   (\x f y -> f (x:y)) 'a' (foldr (\x f y -> f (x:y)) id ['b','c']) []

   =>   -- function application, substitute for x, f and y

   (foldr (\x f y -> f (x:y)) id ['b','c']) ('a':[])

   =>   -- definition of foldr

   (\x f y -> f (x:y)) 'b' (foldr (\x f y -> f (x:y)) id ['c']) ('a':[])

   =>   -- function application

   (foldr (\x f y -> f (x:y)) id ['c']) ('b':('a':[]))

   =>   -- definition of foldr

   (\x f y -> f (x:y)) 'c' (foldr (\x f y -> f (x:y)) id []) ('c':('b':('a':[])))

   =>   -- function application

   (foldr (\x f y -> f (x:y)) id []) ('c':('b':('a':[])))

   =>   -- definition of foldr

   id ('c':('b':('a':[])))

   =>   -- definition of id

   ('c':('b':('a':[])))

Intersperse

The intersperse function takes an element and a list and `intersperses' that element between the elements of the list. For example,

> -- >>> intersperse ',' "abcde"
> -- "a,b,c,d,e"

The recursive version looks like this:

> intersperse1 :: Char -> [Char] -> [Char]
> intersperse1 _ []     = [] 
> intersperse1 a (x:xs) = case xs of
>    [] -> [x]
>    _  -> x : a : intersperse1 a xs

Now rewrite using 'foldr'

> 
> intersperse :: Char -> [Char] -> [Char]
> intersperse a = foldr g [] where
>    g :: Char -> ([Char] -> [Char])
>    g x [] = [x]
>    g x ys = x : a : ys

> -- >>> intersperse ',' "abcde"
> -- "a,b,c,d,e"

> -- >>> intersperse ',' ""
> -- ""

and trace through an example of intersperse ',' "ab"

See online tool.

CHALLENGE: foldl

Here is the usual recursive definition for "fold left".

> foldl1 :: (b -> a -> b) -> b -> [a] -> b
> foldl1 _ z []     = z
> foldl1 f z (x:xs) = foldl1 f (z `f` x) xs

You can see that foldl1 is different than foldr by comparing the results on various examples:

> -- >>> foldl1 (flip (:)) [] [1,2,3]
> -- [3,2,1]
> -- >>> foldr  (:) [] [1,2,3]
> -- [1,2,3]

> -- >>> foldl1 (++) "x" ["1","2","3"]
> -- "x123"
> -- >>> foldr (++) "x" ["1","2","3"]
> -- "123x"

But, you can also define foldl in terms of foldr. Give it a try.

> foldl :: (b -> a -> b) -> b -> [a] -> b
> foldl f z xs =  foldr g id xs z where
>    g x f' = \z0 -> f' (z0 `f` x)

> -- >>> foldl (++) "x" ["1", "2", "3"] 
> -- "x123"

And trace through the test case above.