I am struggling to understand why this code taken from the haskell.org exercise page typechecks (and works as a list reversal function):
myReverse :: [a] -> [a]
myReverse xs = foldr (\x fId empty -> fId (x : empty)) id xs []
My first point of confusion is that foldr accepts 3 arguments, not 4 :
foldr :: Foldable t => (a -> b -> b) -> b -> t a -> b
so I am guessing that myReverse is equivalent to:
myReverse xs = foldr ((\x fId empty -> fId (x : empty)) id) xs []
but then this should not work either since in the lambda, x is a list element rather than a function ...
Think of it this way. Every function accepts exactly one argument. It may return another function (that accepts one argument). The thing that looks like a multi-argument call
f a b c
is actually parsed as
((f a) b) c
that is, a chain of single-argument function applications. A function type
f :: a -> b -> c -> d
can be decomposed to
f :: a -> (b -> (c -> d))
i.e. a function returning a function returning a function. We usually regard it as a function of three arguments. But can it accept more than three? Yes, if d happens to be another function type.
This is exactly what happens with your fold example. The function that you pass as the first argument to foldr accepts three arguments, which is exactly the same as accepting two arguments and returning another function. Now the (simplified) type of foldr is
(a -> b -> b) -> b -> [a] -> b
but if you look at the first argument of it, you see it's a function of three arguments. Which is, as we have seen, exactly the same as a function that acceora two arguments and returns a function. So the b happens to be a function type. Since b is also the the return tuoe of foldr when applied to three arguments
foldr (\x fId empty -> fId (x : empty)) id
and it's a function, it can now be applied to another argument
(foldr (\x fId empty -> fId (x : empty)) id xs) []
I let you figure out what b actually is.
First of all the variables naming is atrocious. I always use r for the second argument to a foldr's reducer function, as a mnemonic for the "recursive result". "empty" is too overloaded with meaning; it is better to use some neutral name so it is easier to see what it is without any preconceived notions:
myReverse :: [a] -> [a]
myReverse xs = foldr (\x r n -> r (x : n)) id xs []
By virtue of foldr's definition,
foldr f z (x:xs) === f x (foldr f z xs)
i.e.
myReverse [a,b,c,...,z]
= foldr (\x r n -> r (x : n)) id [a,b,c,...,z] []
= (\x r n -> r (x : n)) a (foldr (\x r n -> r (x : n)) id [b,c,...,z]) []
= (\x r n -> r (x : n))
a
(foldr (\x r n -> r (x : n)) id [b,c,...,z])
[]
= let { x = a
; r = foldr (\x r n -> r (x : n)) id [b,c,...,z]
; n = []
}
in r (x : n)
= foldr (\x r n -> r (x : n)) id [b,c,...,z] (a : [])
= foldr (\x r n -> r (x : n)) id [b,c,...,z] [a]
= ....
= foldr (\x r n -> r (x : n)) id [c,...,z] (b : [a])
= foldr (\x r n -> r (x : n)) id [c,...,z] [b,a]
= ....
= foldr (\x r n -> r (x : n)) id [] [z,...,c,b,a]
= id [z,...,c,b,a]
I hope this illustration makes it clearer what is going on there. The extra argument is expected by the reducer function, which is pushed into action by foldr ... resulting in the operational equivalent of
= foldl (\n x -> (x : n)) [] [a,b,c,...,z]
As it turns out, myReverse implementation is using the equivalence
foldl (flip f) n xs === foldr (\x r -> r . f x) id xs n
The defined code is
fun foldl f e l = let
fun g(x, f'') = fn y => f''(f(x, y))
in foldr g (fn x => x) l e end
I don't understand how this works;
what is the purpose of g(x, f'')?
I also find a similar example in Haskell,
the definition is quite short
myFoldl f z xs = foldr step id xs z
where
step x g a = g (f a x)
Let's dissect the Haskell implementation of myFoldl and then take a look at the ocaml SML code. First, we'll look at some type signatures:
foldr :: (a -> b -> b) -- the step function
-> b -- the initial value of the accumulator
-> [a] -- the list to fold
-> b -- the result
It should be noted that although the foldr function accepts only three arguments we are applying it two four arguments:
foldr step id xs z
However, as you can see the second argument to foldr (i.e. the inital value of the accumulator) is id which is a function of the type x -> x. Therefore, the result is also of the type x -> x. Hence, it accepts four arguments.
Similarly, the step function is now of the type a -> (x -> x) -> x -> x. Hence, it accepts three arguments instead of two. The accumulator is an endofunction (i.e. a function whose domain and codomain is the same).
Endofunctions have a special property, they are composed from left to right instead of from right to left. For example, let's compose a bunch of Int -> Int functions:
inc :: Int -> Int
inc n = n + 1
dbl :: Int -> Int
dbl n = n * 2
The normal way to compose these functions is to use the function composition operator as follows:
incDbl :: Int -> Int
incDbl = inc . dbl
The incDbl function first doubles a number and then increments it. Note that this reads from right to left.
Another way to compose them is to use continuations (denoted by k):
inc' :: (Int -> Int) -> Int -> Int
inc' k n = k (n + 1)
dbl' :: (Int -> Int) -> Int -> Int
dbl' k n = k (n * 2)
Notice that the first argument is a continuation. If we want to recover the original functions then we can do:
inc :: Int -> Int
inc = inc' id
dbl :: Int -> Int
dbl = dbl' id
However, if we want to compose them then we do it as follows:
incDbl' :: (Int -> Int) -> Int -> Int
incDbl' = dbl' . inc'
incDbl :: Int -> Int
incDbl = incDbl' id
Notice that although we are still using the dot operator to compose the functions, it now reads from left to right.
This is the key behind making foldr behave as foldl. We fold the list from right to left but instead of folding it into a value, we fold it into an endofunction which when applied to an initial accumulator value actually folds the list from left to right.
Consider our incDbl function:
incDbl = incDbl' id
= (dbl' . inc') id
= dbl' (inc' id)
Now consider the definition of foldr:
foldr :: (a -> b -> b) -> b -> [a] -> b
foldr _ acc [] = acc
foldr fun acc (y:ys) = fun y (foldr fun acc ys)
In the basis case we simply return the accumulated value. However, in the inductive case we return fun y (foldr fun acc ys). Our step function is defined as follows:
step :: a -> (x -> x) -> x -> x
step x g a = g (f a x)
Here f is the reducer function of foldl and is of the type x -> a -> x. Notice that step x is an endofunction of the type (x -> x) -> x -> x which we know can be composed left to right.
Hence the folding operation (i.e. foldr step id) on a list [y1,y2..yn] looks like:
step y1 (step y2 (... (step yn id)))
-- or
(step y1 . step y2 . {dots} . step yn) id
Each step yx is an endofunction. Hence, this is equivalent to composing the endofunctions from left to right.
When this result is applied to an initial accumulator value then the list folds from left to right. Hence, myFoldl f z xs = foldr step id xs z.
Now consider the foldl function (which is written in Standard ML and not OCaml). It is defined as:
fun foldl f e l = let fun g (x, f'') = fn y => f'' (f (x, y))
in foldr g (fn x => x) l e end
The biggest difference between the foldr functions of Haskell and SML are:
In Haskell the reducer function has the type a -> b -> b.
In SML the reducer function has the type (a, b) -> b.
Both are correct. It's only a matter of preference. In SML instead of passing two separate arguments, you pass one single tuple which contains both arguments.
Now, the similarities:
The id function in Haskell is the anonymous fn x => x function in SML.
The step function in Haskell is the function g in SML which takes a tuple containing the first two arguments.
The step function is Haskell step x g a has been split into two functions in SML g (x, f'') = fn y => f'' (f (x, y)) for more clarity.
If we rewrite the SML function to use the same names as in Haskell then we have:
fun myFoldl f z xs = let step (x, g) = fn a => g (f (a, x))
in foldr step (fn x => x) xs z end
Hence, they are exactly the same function. The expression g (x, f'') simply applies the function g to the tuple (x, f''). Here f'' is a valid identifier.
Intuition
The foldl function traverses the list head to tail while operating elements with an accumulator:
(...(a⊗x1)⊗...⊗xn-1)⊗xn
And you want to define it via a foldr:
x1⊕(x2⊕...⊕(xn⊕e)...)
Rather unintuitive. The trick is that your foldr will not produce a value, but rather a function. The list traversal will operate the elements as to produce a function that, when applied to the accumulator, performs the computation you desire.
Lets see a simple example to illustrate how this works. Consider sum foldl (+) 0 [1,2,3] = ((0+1)+2)+3. We may calculate it via foldr as follows.
foldr ⊕ [1,2,3] id
-> 1⊕(2⊕(3⊕id))
-> 1⊕(2⊕(id.(+3))
-> 1⊕(id.(+3).(+2))
-> (id.(+3).(+2).(+1))
So when we apply this function to 0 we get
(id.(+3).(+2).(+1)) 0
= ((0+1)+2)+3
We began with the identity function and successively changed it as we traversed the list, using ⊕ where,
n ⊕ g = g . (+n)
Using this intuition, it isn't hard to define a sum with an accumulator via foldr. We built the computation for a given list via foldr ⊕ id xs. Then to calculate the sum we applied it to 0, foldr ⊕ id xs 0. So we have,
foldl (+) 0 xs = foldr ⊕ id xs 0
where n ⊕ g = g . (+n)
or equivalently, denoting n ⊕ g in prefix form by (⊕) n g and noting that (⊕) n g a = (g . (+n)) a = g (a+n),
foldl (+) 0 xs = foldr ⊕ id xs 0
where (⊕) n g a = g (a+n)
Note that the ⊕ is your step function, and that you can obtain the generic result you're looking for by substituting a function f for +, and accumulator a for 0.
Next let us show that the above really is correct.
Formal derivation
Moving on to a more formal approach. It is useful, for simplicity, to be aware of the following universal property of foldr.
h [] = e
h (x:xs) = f x (h xs)
iff
h = foldr f e
This means that rather than defining foldr directly, we may instead and more simply define a function h in the form above.
We want to define such an h so that,
h xs a = foldl f a xs
or equivalently,
h xs = \a -> foldl f a xs
So lets determine h. The empty case is simple:
h [] = \a -> foldl f a []
= \a -> a
= id
The non-empty case results in:
h (x:xs) = \a -> foldl f a (x:xs)
= \a -> foldl f (f a x) xs
= \a -> h xs (f a x)
= step x (h xs) where step x g = \a -> g (f a x)
= step x (h xs) where step x g a = g (f a x)
So we conclude that,
h [] = id
h (x:xs) = step x (h xs) where step x g a = g (f a x)
satisfies h xs a = foldl f a xs
And by the universal property above (noting that the f in the universal property formula corresponds to step here, and e to id) we know that h = foldr step id. Therefore,
h = foldr step id
h xs a = foldl f a xs
-----------------------
foldl f a xs = foldr step id xs a
where step x g a = g (f a x)
Haskell newb here
I'm working on this problem in haskell:
(**) 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:
* (compress '(a a a a b c c a a d e e e e))
(A B C A D E)
The solution (which I had to look up) uses foldr:
compress' :: (Eq a) => [a] -> [a]
compress' xs = foldr (\x acc -> if x == (head acc) then acc else x:acc) [last xs] xs
This foldr, according to the solution, takes two parameters, x and acc. It would seem like all foldr's take these parameters; is there any exception to this? Like a foldr that takes 3 or more? If not, is this convention redundant and can the formula be written with less code?
foldr takes a function of 2 arguments, but this doesn't prevent it from taking a function of 3 arguments provided that function has the right type signature.
If we had a function
g :: x -> y -> z -> w
With
foldr :: (a -> b -> b) -> b -> [a] -> b
Where we want to pass g to foldr, then (a -> b -> b) ~ (x -> y -> z -> w) (where ~ is type equality). Since -> is right associative, this means we can write g's signature as
x -> y -> (z -> w)
and its meaning is the same. Now we've produced a function of two parameters that returns a function of one parameter. In order to unify this with the type a -> b -> b, we just need to line up the arguments:
a -> | x ->
b -> | y ->
b | (z -> w)
This means that b ~ z -> w, so y ~ b ~ z -> w and a ~ x so g's type really has to be
g :: x -> (z -> w) -> (z -> w)
implying
foldr g :: (z -> w) -> [x] -> (z -> w)
This is certainly not impossible, although more unlikely. Our accumulator is a function instead, and to me this begs to be demonstrated with DiffLists:
type DiffList a = [a] -> [a]
append :: a -> DiffList a -> DiffList a
append x dl = \xs -> dl xs ++ [x]
reverse' :: [a] -> [a]
reverse' xs = foldr append (const []) xs $ []
Note that foldr append (const []) xs returns a function which we apply to [] to reverse a list. In this case we've given an alias to functions of the type [a] -> [a] called DiffList, but it's really no different than having written
append :: a -> ([a] -> [a]) -> [a] -> [a]
which is a function of 3 arguments.
As with all things in haskell have a look at the types of things to guide your way you can do this for any function in ghci.
Looking at this for foldr we see:
Prelude> :t foldr
foldr :: (a -> b -> b) -> b -> [a] -> b
This slightly abstract string can be written in english as:
foldr is a function that takes
1 ) a function with two parameters one of type a and one of type b and returns something of type b
2 ) A value of type b
3 ) A list of values of type a
And returns a value of type b
Where a and b are type variables (see here for a good tutorial on them) which can be filled in with any type you like.
It turns out that you can solve your compress problem using a foldr with a three-argument function.
compress :: Eq a => [a] -> [a]
compress [] = []
compress (z:zs) = z : foldr f (const []) zs z
where f x k w | x==w = k x
| otherwise = x : k x
Let's dissect that. First, we can improve readability by changing the last two lines to
where f x k = \w -> if x==w then k x else x : k x
This makes it evident that a ternary function is nothing but a binary function returning a unary function. The advantage of looking at it in this way is that foldr is best understood when passed a binary function. Indeed, we are passing a binary function, which just happens to return a function.
Let's focus on types now:
f :: a -> (a -> [a]) -> (a -> [a])
f x k
So, x::a is the element of the list we are folding on. Function k is the result of the fold on the list tail. The result of f x k is something having the same type as k.
\w -> if .... :: (a -> [a])
The overall idea behind this anonymous function is as follows. The parameter k plays the same role as acc in the OP code, except it is a function expecting the previous element w in the list before producing the accumulated compressed list.
Concretely, we use now k x when we used acc, passing on the current element to the next step, since by that time x will become the previous element w. At the top-level, we pass z to the function which is returned by foldr f (const []).
This compress variant is lazy, unlike the posted solution. In fact, the posted solution needs to scan the whole list before starting producing something: this is due to (\x acc -> ...) being strict in acc, and to the use of last xs. Instead, the above compress outputs list elements in a "streaming" fashion. Indeed, it works with infinite lists as well:
> take 10 $ compress [1..]
[1,2,3,4,5,6,7,8,9,10]
That being said, I think using a foldr here feels a bit weird: the code above is arguably less readable than the explicit recursion.
In Real World Haskell, Chapter 4. on Functional Programming:
Write foldl with foldr:
-- file: ch04/Fold.hs
myFoldl :: (a -> b -> a) -> a -> [b] -> a
myFoldl f z xs = foldr step id xs z
where step x g a = g (f a x)
The above code confused me a lot, and somebody called dps rewrote it with a meaningful name to make it a bit clearer:
myFoldl stepL zeroL xs = (foldr stepR id xs) zeroL
where stepR lastL accR accInitL = accR (stepL accInitL lastL)
Somebody else, Jef G, then did an excellent job by providing an example and showing the underlying mechanism step by step:
myFoldl (+) 0 [1, 2, 3]
= (foldR step id [1, 2, 3]) 0
= (step 1 (step 2 (step 3 id))) 0
= (step 1 (step 2 (\a3 -> id ((+) a3 3)))) 0
= (step 1 (\a2 -> (\a3 -> id ((+) a3 3)) ((+) a2 2))) 0
= (\a1 -> (\a2 -> (\a3 -> id ((+) a3 3)) ((+) a2 2)) ((+) a1 1)) 0
= (\a1 -> (\a2 -> (\a3 -> (+) a3 3) ((+) a2 2)) ((+) a1 1)) 0
= (\a1 -> (\a2 -> (+) ((+) a2 2) 3) ((+) a1 1)) 0
= (\a1 -> (+) ((+) ((+) a1 1) 2) 3) 0
= (+) ((+) ((+) 0 1) 2) 3
= ((0 + 1) + 2) + 3
But I still cannot fully understand that, here are my questions:
What is the id function for? What is the role of? Why should we need it here?
In the above example, id function is the accumulator in the lambda function?
foldr's prototype is foldr :: (a -> b -> b) -> b -> [a] -> b, and the first parameter is a function which need two parameters, but the step function in the myFoldl's implementation uses 3 parameters, I'm complelely confused!
Some explanations are in order!
What is the id function for? What is the role of? Why should we need it here?
id is the identity function, id x = x, and is used as the equivalent of zero when building up a chain of functions with function composition, (.). You can find it defined in the Prelude.
In the above example, id function is the accumulator in the lambda function?
The accumulator is a function that is being built up via repeated function application. There's no explicit lambda, since we name the accumulator, step. You can write it with a lambda if you want:
foldl f a bs = foldr (\b g x -> g (f x b)) id bs a
Or as Graham Hutton would write:
5.1 The foldl operator
Now let us generalise from the suml example and consider the standard operator foldl that processes the elements of a list in left-to-right order by using a function f to combine values, and a value v as the starting value:
foldl :: (β → α → β) → β → ([α] → β)
foldl f v [ ] = v
foldl f v (x : xs) = foldl f (f v x) xs
Using this operator, suml can be redefined simply by suml = foldl (+) 0. Many other functions can be defined in a simple way using foldl. For example, the standard function reverse can redefined using foldl as follows:
reverse :: [α] → [α]
reverse = foldl (λxs x → x : xs) [ ]
This definition is more efficient than our original definition using fold, because it avoids the use of the inefficient append operator (++) for lists.
A simple generalisation of the calculation in the previous section for the function suml shows how to redefine the function foldl in terms of fold:
foldl f v xs = fold (λx g → (λa → g (f a x))) id xs v
In contrast, it is not possible to redefine fold in terms of foldl, due to the fact that
foldl is strict in the tail of its list argument but fold is not. There are a number of useful ‘duality theorems’ concerning fold and foldl, and also some guidelines for deciding which operator is best suited to particular applications (Bird, 1998).
foldr's prototype is foldr :: (a -> b -> b) -> b -> [a] -> b
A Haskell programmer would say that the type of foldr is (a -> b -> b) -> b -> [a] -> b.
and the first parameter is a function which need two parameters, but the step function in the myFoldl's implementation uses 3 parameters, I'm complelely confused
This is confusing and magical! We play a trick and replace the accumulator with a function, which is in turn applied to the initial value to yield a result.
Graham Hutton explains the trick to turn foldl into foldr in the above article. We start by writing down a recursive definition of foldl:
foldl :: (a -> b -> a) -> a -> [b] -> a
foldl f v [] = v
foldl f v (x : xs) = foldl f (f v x) xs
And then refactor it via the static argument transformation on f:
foldl :: (a -> b -> a) -> a -> [b] -> a
foldl f v xs = g xs v
where
g [] v = v
g (x:xs) v = g xs (f v x)
Let's now rewrite g so as to float the v inwards:
foldl f v xs = g xs v
where
g [] = \v -> v
g (x:xs) = \v -> g xs (f v x)
Which is the same as thinking of g as a function of one argument, that returns a function:
foldl f v xs = g xs v
where
g [] = id
g (x:xs) = \v -> g xs (f v x)
Now we have g, a function that recursively walks a list, apply some function f. The final value is the identity function, and each step results in a function as well.
But, we have handy already a very similar recursive function on lists, foldr!
2 The fold operator
The fold operator has its origins in recursion theory (Kleene, 1952), while the use
of fold as a central concept in a programming language dates back to the reduction operator of APL (Iverson, 1962), and later to the insertion operator of FP (Backus,
1978). In Haskell, the fold operator for lists can be defined as follows:
fold :: (α → β → β) → β → ([α] → β)
fold f v [ ] = v
fold f v (x : xs) = f x (fold f v xs)
That is, given a function f of type α → β → β and a value v of type β, the function
fold f v processes a list of type [α] to give a value of type β by replacing the nil
constructor [] at the end of the list by the value v, and each cons constructor (:) within the list by the function f. In this manner, the fold operator encapsulates a simple pattern of recursion for processing lists, in which the two constructors for lists are simply replaced by other values and functions. A number of familiar functions on lists have a simple definition using fold.
This looks like a very similar recursive scheme to our g function. Now the trick: using all the available magic at hand (aka Bird, Meertens and Malcolm) we apply a special rule, the universal property of fold, which is an equivalence between two definitions for a function g that processes lists, stated as:
g [] = v
g (x:xs) = f x (g xs)
if and only if
g = fold f v
So, the universal property of folds states that:
g = foldr k v
where g must be equivalent to the two equations, for some k and v:
g [] = v
g (x:xs) = k x (g xs)
From our earlier foldl designs, we know v == id. For the second equation though, we need
to calculate the definition of k:
g (x:xs) = k x (g xs)
<=> g (x:xs) v = k x (g xs) v -- accumulator of functions
<=> g xs (f v x) = k x (g xs) v -- definition of foldl
<= g' (f v x) = k x g' v -- generalize (g xs) to g'
<=> k = \x g' -> (\a -> g' (f v x)) -- expand k. recursion captured in g'
Which, substituting our calculated definitions of k and v yields a
definition of foldl as:
foldl :: (a -> b -> a) -> a -> [b] -> a
foldl f v xs =
foldr
(\x g -> (\a -> g (f v x)))
id
xs
v
The recursive g is replaced with the foldr combinator, and the accumulator becomes a function built via a chain of compositions of f at each element of the list, in reverse order (so we fold left instead of right).
This is definitely somewhat advanced, so to deeply understand this transformation, the universal property of folds, that makes the transformation possible, I recommend Hutton's tutorial, linked below.
References
Haskell Wiki: Foldl as foldr
A tutorial on the universality and expressiveness of fold, Graham Hutton, J. Functional Programming 9 (4): 355–372, July 1999.
Malcolm, G. Algebraic data types and program transformation., PhD thesis, Groningen University.
Consider the type of foldr:
foldr :: (b -> a -> a) -> a -> [b] -> a
Whereas the type of step is something like b -> (a -> a) -> a -> a. Since step is getting passed to foldr, we can conclude that in this case the fold has a type like (b -> (a -> a) -> (a -> a)) -> (a -> a) -> [b] -> (a -> a).
Don't be confused by the different meanings of a in different signatures; it's just a type variable. Also, keep in mind that the function arrow is right associative, so a -> b -> c is the same thing as a -> (b -> c).
So, yes, the accumulator value for the foldr is a function of type a -> a, and the initial value is id. This makes some sense, because id is a function that doesn't do anything--it's the same reason you'd start with zero as the initial value when adding all the values in a list.
As for step taking three arguments, try rewriting it like this:
step :: b -> (a -> a) -> (a -> a)
step x g = \a -> g (f a x)
Does that make it easier to see what's going on? It takes an extra parameter because it's returning a function, and the two ways of writing it are equivalent. Note also the extra parameter after the foldr: (foldr step id xs) z. The part in parentheses is the fold itself, which returns a function, which is then applied to z.
(quickly skim through my answers [1], [2], [3], [4] to make sure you understand Haskell's syntax, higher-order functions, currying, function composition, $ operator, infix/prefix operators, sections and lambdas)
Universal property of fold
A fold is just a codification of certain kinds of recursion. And universality property simply states that, if your recursion conforms to a certain form, it can be transformed into fold according to some formal rules. And conversely, every fold can be transformed into a recursion of that kind. Once again, some recursions can be translated into folds that give exactly the same answer, and some recursions can't, and there is an exact procedure to do that.
Basically, if your recursive function works on lists an looks like on the left, you can transform it to fold one the right, substituting f and v for what actually is there.
g [] = v ⇒
g (x:xs) = f x (g xs) ⇒ g = foldr f v
For example:
sum [] = 0 {- recursion becomes fold -}
sum (x:xs) = x + sum xs ⇒ sum = foldr 0 (+)
Here v = 0 and sum (x:xs) = x + sum xs is equivalent to sum (x:xs) = (+) x (sum xs), therefore f = (+). 2 more examples
product [] = 1
product (x:xs) = x * product xs ⇒ product = foldr 1 (*)
length [] = 0
length (x:xs) = 1 + length xs ⇒ length = foldr (\_ a -> 1 + a) 0
Exercise:
Implement map, filter, reverse, concat and concatMap recursively, just like the above functions on the left side.
Convert these 5 functions to foldr according to a formula above, that is, substituting f and v in the fold formula on the right.
Foldl via foldr
How to write a recursive function that sums numbers up from left to right?
sum [] = 0 -- given `sum [1,2,3]` expands into `(1 + (2 + 3))`
sum (x:xs) = x + sum xs
The first recursive function that comes to find fully expands before even starts adding up, that's not what we need. One approach is to create a recursive function that has accumulator, that immediately adds up numbers on each step (read about tail recursion to learn more about recursion strategies):
suml :: [a] -> a
suml xs = suml' xs 0
where suml' [] n = n -- auxiliary function
suml' (x:xs) n = suml' xs (n+x)
Alright, stop! Run this code in GHCi and make you sure you understand how it works, then carefully and thoughtfully proceed. suml can't be redefined with a fold, but suml' can be.
suml' [] = v -- equivalent: v n = n
suml' (x:xs) n = f x (suml' xs) n
suml' [] n = n from function definition, right? And v = suml' [] from the universal property formula. Together this gives v n = n, a function that immediately returns whatever it receives: v = id. Let's calculate f:
suml' (x:xs) n = f x (suml' xs) n
-- expand suml' definition
suml' xs (n+x) = f x (suml' xs) n
-- replace `suml' xs` with `g`
g (n+x) = f x g n
Thus, suml' = foldr (\x g n -> g (n+x)) id and, thus, suml = foldr (\x g n -> g (n+x)) id xs 0.
foldr (\x g n -> g (n + x)) id [1..10] 0 -- return 55
Now we just need to generalize, replace + by a variable function:
foldl f a xs = foldr (\x g n -> g (n `f` x)) id xs a
foldl (-) 10 [1..5] -- returns -5
Conclusion
Now read Graham Hutton's A tutorial on the universality and expressiveness of fold. Get some pen and paper, try to figure everything that he writes until you get derive most of the folds by yourself. Don't sweat if you don't understand something, you can always return later, but don't procrastinate much either.
Here's my proof that foldl can be expressed in terms of foldr, which I find pretty simple apart from the name spaghetti the step function introduces.
The proposition is that foldl f z xs is equivalent to
myfoldl f z xs = foldr step_f id xs z
where step_f x g a = g (f a x)
The first important thing to notice here is that the right hand side of the first line is actually evaluated as
(foldr step_f id xs) z
since foldr only takes three parameters. This already hints that the foldr will calculate not a value but a curried function, which is then applied to z. There are two cases to investigate to find out whether myfoldl is foldl:
Base case: empty list
myfoldl f z []
= foldr step_f id [] z (by definition of myfoldl)
= id z (by definition of foldr)
= z
foldl f z []
= z (by definition of foldl)
Non-empty list
myfoldl f z (x:xs)
= foldr step_f id (x:xs) z (by definition of myfoldl)
= step_f x (foldr step_f id xs) z (-> apply step_f)
= (foldr step_f id xs) (f z x) (-> remove parentheses)
= foldr step_f id xs (f z x)
= myfoldl f (f z x) xs (definition of myfoldl)
foldl f z (x:xs)
= foldl f (f z x) xs
Since in 2. the first and the last line have the same form in both cases, it can be used to fold the list down until xs == [], in which case 1. guarantees the same result. So by induction, myfoldl == foldl.
There is no Royal Road to Mathematics, nor even through Haskell. Let
h z = (foldr step id xs) z where
step x g = \a -> g (f a x)
What the heck is h z? Assume that xs = [x0, x1, x2].
Apply the definition of foldr:
h z = (step x0 (step x1 (step x2 id))) z
Apply the definition of step:
= (\a0 -> (\a1 -> (\a2 -> id (f a2 x2)) (f a1 x1)) (f a0 x0)) z
Substitute into the lambda functions:
= (\a1 -> (\a2 -> id (f a2 x2)) (f a1 x1)) (f z x0)
= (\a2 -> id (f a2 x2)) (f (f z x0) x1)
= id (f (f (f z x0) x1) x2)
Apply definition of id :
= f (f (f z x0) x1) x2
Apply definition of foldl :
= foldl f z [x0, x1, x2]
Is it a Royal Road or what?
I'm posting the answer for those people who might find this approach better suited to their way of thinking. The answer possibly contains redundant information and thoughts, but it is what I needed in order to tackle the problem. Furthermore, since this is yet another answer to the same question, it's obvious that it has substantial overlaps with the other answers, however it tells the tale of how I could grasp this concept.
Indeed I started to write down this notes as a personal record of my thoughts while trying to understand this topic. It took all the day for me to touch the core of it, if I really have got it.
My long way to understanding this simple exercise
Easy part: what do we need to determine?
What happens with the following example call
foldl f z [1,2,3,4]
can be visualized with the following diagram (which is on Wikipedia, but I first saw it on another answer):
_____results in a number
/
f f (f (f (f z 1) 2) 3) 4
/ \
f 4 f (f (f z 1) 2) 3
/ \
f 3 f (f z 1) 2
/ \
f 2 f z 1
/ \
z 1
(As a side note, when using foldl each applications of f is not performed, and the expressions are thunked just the way I wrote them above; in principle, they could be computed as you go bottom-top, and that's exactly what foldl' does.)
The exercise essentially challenges us to use foldr instead of foldl by appropriately changing the step function (so we use s instead of f) and the initial accumulator (so we use ? instead of z); the list stays the same, otherwise what are we talking about?
The call to foldr has to look like this:
foldr s ? [1,2,3,4]
and the corresponding diagram is this:
_____what does the last call return?
/
s
/ \
1 s
/ \
2 s
/ \
3 s
/ \
4 ? <--- what is the initial accumulator?
The call results in
s 1 (s 2 (s 3 (s 4 ?)))
What are s and ?? And what are their types? It looks like s it's a two argument function, much like f, but let's not jump to conclusions. Also, let's leave ? aside for a moment, and let's observe that z has to come into play as soon as 1 comes into play; however, how can z come into play in the call to the maybe-two-argument s function, namely in the call s 1 (…)? We can solve this part of the enigma by choosing an s which takes 3 arguments, rather than the 2 we mentioned earlier, so that the outermost call s 1 (…) will result in a function taking one argument, which we can pass z to!
This means that we want the original call, which expands to
f (f (f (f z 1) 2) 3) 4
to be equivalent to
s 1 (s 2 (s 3 (s 4 ?))) z
or, in other words, we want the partially applied function
s 1 (s 2 (s 3 (s 4 ?)))
to be equivalent to the following lambda function
(\z -> f (f (f (f z 1) 2) 3) 4)
Again, the "only" pieces we need are s and ?.
Turning point: recognize function composition
Let's redraw the previous diagram and write on the right what we want each call to s be equivalent to:
s s 1 (…) == (\z -> f (f (f (f z 1) 2) 3) 4)
/ \
1 s s 2 (…) == (\z -> f (f (f z 2) 3) 4)
/ \
2 s s 3 (…) == (\z -> f (f z 3) 4)
/ \
3 s s 4 ? == (\z -> f z 4)
/ \
4 ? <--- what is the initial accumulator?
I hope it's clear from the structure of the diagram that the (…) on each line is the right hand side of the line below it; better, it is the function returned from the previous (below) call to s.
It should be also clear that a call to s with arguments x and y is the (full) application of y to the partial application of f to the only argument x (as its second argument). Since the partial application of f to x can be written as the lambda (\z -> f z x), fully applying y to it results in the lambda (\z -> y (f z x)), which in this case I would rewrite as y . (\z -> f z x); translating the words into an expression for s we get
s x y = y . (\z -> f z x)
(This is the same as s x y z = y (f z x), which is the same as the book, if you rename the variables.)
The last bit is: what is the initial "value" ? of the accumulator? The above diagram can be rewritten by expanding the nested calls to make them composition chains:
s s 1 (…) == (\z -> f z 4) . (\z -> f z 3) . (\z -> f z 2) . (\z -> f z 1)
/ \
1 s s 2 (…) == (\z -> f z 4) . (\z -> f z 3) . (\z -> f z 2)
/ \
2 s s 3 (…) == (\z -> f z 4) . (\z -> f z 3)
/ \
3 s s 4 ? == (\z -> f z 4)
/ \
4 ? <--- what is the initial accumulator?
We here see that s simply "piles up" successive partial applications of f, but the y in s x y = y . (\z -> f z x) suggests that the interpretation of s 4 ? (and, in turn, all the others) misses a leading function to be composed with the leftmost lambda.
That's just our ? function: it's time to give it a reason for its existence, beside occupying a place in the call to foldr. What can we choose it to be, in order not to change the resulting functions? Answer: id, the identity function, which is also the identity element with respect to the composition operator (.).
s s 1 (…) == id . (\z -> f z 4) . (\z -> f z 3) . (\z -> f z 2) . (\z -> f z 1)
/ \
1 s s 2 (…) == id . (\z -> f z 4) . (\z -> f z 3) . (\z -> f z 2)
/ \
2 s s 3 (…) == id . (\z -> f z 4) . (\z -> f z 3)
/ \
3 s s 4 id == id . (\z -> f z 4)
/ \
4 id
So the sought function is
myFoldl f z xs = foldr (\x g a -> g (f a x)) id xs z
foldr step zero (x:xs) = step x (foldr step zero xs)
foldr _ zero [] = zero
myFold f z xs = foldr step id xs z
where step x g a = g (f a x)
myFold (+) 0 [1, 2, 3] =
foldr step id [1, 2, 3] 0
-- Expanding foldr function
step 1 (foldr step id [2, 3]) 0
step 1 (step 2 (foldr step id [3])) 0
step 1 (step 2 (step 3 (foldr step id []))) 0
-- Expanding step function if it is possible
step 1 (step 2 (step 3 id)) 0
step 2 (step 3 id) (0 + 1)
step 3 id ((0 + 1) + 2)
id (((0 + 1) + 2) + 3)
Well, at least, this helped me. Even it is not quite right.
This might help, I tried expanding in a different way.
myFoldl (+) 0 [1,2,3] =
foldr step id [1,2,3] 0 =
foldr step (\a -> id (a+3)) [1,2] 0 =
foldr step (\b -> (\a -> id (a+3)) (b+2)) [1] 0 =
foldr step (\b -> id ((b+2)+3)) [1] 0 =
foldr step (\c -> (\b -> id ((b+2)+3)) (c+1)) [] 0 =
foldr step (\c -> id (((c+1)+2)+3)) [] 0 =
(\c -> id (((c+1)+2)+3)) 0 = ...
This answer makes the definition below easily understood in three step.
-- file: ch04/Fold.hs
myFoldl :: (a -> b -> a) -> a -> [b] -> a
myFoldl f z xs = foldr step id xs z
where step x g a = g (f a x)
Step 1. transform the fold of function evaluation to function combination
foldl f z [x1 .. xn] = z & f1 & .. & fn = fn . .. . f1 z. in which fi = \z -> f z xi.
(By using z & f1 & f2 & .. & fn it means fn ( .. (f2 (f1 z)) .. ).)
Step 2. express the function combination in a foldr manner
foldr (.) id [f1 .. fn] = (.) f1 (foldr (.) id [f2 .. fn]) = f1 . (foldr (.) id [f2 .. fn]). Unfold the rest to get foldr (.) id [f1 .. fn] = f1 . .. . fn.
Noticing that the sequence is reversed, we should use the reversed form of (.). Define rc f1 f2 = (.) f2 f1 = f2 . f1, then foldr rc id [f1 .. fn] = rc f1 (foldr (.) id [f2 .. fn]) = (foldr (.) id [f2 .. fn]) . f1. Unfold the rest to get foldr rc id [f1 .. fn] = fn . .. . f1.
Step 3. transform the fold on function list to the fold on operand list
Find step that makes foldr step id [x1 .. xn] = foldr rc id [f1 .. fn]. It is easy to find step = \x g z -> g (f z x).
In 3 steps, the definition of foldl using foldr is clear:
foldl f z xs
= fn . .. . f1 z
= foldr rc id fs z
= foldr step id xs z
Prove the correctness:
foldl f z xs = foldr (\x g z -> g (f z x)) id xs z
= step x1 (foldr step id [x2 .. xn]) z
= s1 (foldr step id [x2 .. xn]) z
= s1 (step x2 (foldr step id [x3 .. xn])) z
= s1 (s2 (foldr step id [x3 .. xn])) z
= ..
= s1 (s2 (.. (sn (foldr step id [])) .. )) z
= s1 (s2 (.. (sn id) .. )) z
= (s2 (.. (sn id) .. )) (f z x1)
= s2 (s3 (.. (sn id) .. )) (f z x1)
= (s3 (.. (sn id) .. )) (f (f z x1) x2)
= ..
= sn id (f (.. (f (f z x1) x2) .. ) xn-1)
= id (f (.. (f (f z x1) x2) .. ) xn)
= f (.. (f (f z x1) x2) .. ) xn
in which xs = [x1 .. xn], si = step xi = \g z -> g (f z xi)
If you find anything to be unclear, please add a comment. :)