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 ( 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 de?nitions 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
