## Why foldr works for infinite lists and foldl doesn’t

About two months ago I wrote a post about expressing `foldl` in terms of `foldr`. Back then I left one question open – why does `foldr` work for infinite lists, while `foldl` doesn’t? I finally found some time sit down and find the answer.

First of all the fact that `foldl` doesn’t work for infinite lists, while `foldr` does, was counter-intuitive for me. I thought that since `foldl` consumes the list from the beginning it should be able to stop at some point and return the result. On the other hand `foldr` was explained to me as consuming a list from the right, that is from the end. Since infinite lists have no end it seemed to me that `foldr` shouldn’t be able to handle infinite lists.

Here are the definitions of folds from Haskell 2010 Language Report:

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

These two definitions supported my incorrect intuition. After all they show clearly that `foldl` processes the first argument of a list immediately, while `foldr` needs to process whole list with `foldr` before it can pass the result to `f`. Or at least I thought that they show this.

As I already said all the above intuitions are wrong. My mistake became clear to me when I explicitly wrote how recursion looks for each fold. For `foldl` the recursion is:

`f (... (f ( f (f z x1) x2) x3) ...) xn`

For `foldr` recursion looks like this:

`f x1 (f x2 (f x3 (...(f xn z) ...)))`

Now you can see that for `foldl` you need to get to the end of the list to make the most outer call. In case of `foldr` you need the first element of the list and the result of processing the rest of the list with `foldr`. Unless you can determine the value of `f` without the need for its second parameter! This is in fact the case for some operators, logical conjunction for example – if first parameter is False then we can conclude that the whole expression is False, without the need to evaluate the second argument. Therefore `foldr` will work for infinite lists if the accumulating function is lazy in its second argument. One might ask if `foldl` will work for infinite lists if the accumulating function is lazy in its first argument. The answer is no – you still need the last element of a list to calculate the value of first call and there is no last element for infinite lists.

Looking at the fold definitions given earlier I made one embarrassing omission. Recursion in `foldl` is unconditional! The recursive call is being made no matter what. The only way to stop the recursion is getting to the end of a list, but for infinite lists this ain’t gonna happen. In case of `foldr` recursion is conditional – it depends on the first argument of `f`, assuming of course that `f` is lazy for its second argument. Moreover, looking at the implementation of `foldr` given above you can see that it in fact works from the left! My intuition about `foldr` was so fixed on it consuming the list from the right that I even missed this obvious fact when writing this post. Big thanks to nh2, who pointed that out in his comment. So in the end consuming a list from the right is about grouping of terms with parentheses.

As a final remark let me note that definitions of `foldl` and `foldr` in Haskell libraries are slightly different from those given in the Haskell report. GHC.Base defines `foldr` as:

```foldr :: (a -> b -> b) -> b -> [a] -> b -- foldr _ z [] = z -- foldr f z (x:xs) = f x (foldr f z xs) {-# INLINE [0] foldr #-} -- Inline only in the final stage, after the foldr/cons rule has had a chance -- Also note that we inline it when it has *two* parameters, which are the -- ones we are keen about specialising! foldr k z = go where go [] = z go (y:ys) = y `k` go ys```

While Data.List contains following definition of `foldl`:

```foldl :: (a -> b -> a) -> a -> [b] -> a foldl f z0 xs0 = lgo z0 xs0 where lgo z [] = z lgo z (x:xs) = lgo (f z x) xs```

Semantics are of course identical with folds defined in the report.

## Expressing foldl in terms of foldr

In chapter 4 of Real World Haskell there is a challenge for the reader: express `foldl` using `foldr`. Authors warn that this is not trivial and I must admit that I have not attempted this exercise leaving it for later. Yesterday I read Graham Hutton’s “Tutorial on the universality and expressiveness of fold” and it happens that, among other things, it presents an approach that can be applied to solve this problem. Hutton first presents a specific case in which `sum` function is defined using `foldl` expressed with `foldr`. Then he gives a general formula for expressing `foldl` with `foldr`. Despite having a derivation for one specific case and a ready result (without its derivation) it wasn’t straightforward to provide my own derivation of the general solution. This was a mind bending exercise and I’d like to go through the details here.

The solution is based on using the so called universal property of `foldr`1. This property states that if we have some function `g` defined as:

```g [] = v g (x:xs) = f x (g xs)```

then

`g = foldr f v`

Indeed, if we substitute `foldr f v` into definition of `g` we get a definition of `foldr`:

```foldr f v [] = v foldr f v (x:xs) = f x (foldr f v xs)```

Moreover, `foldr f v` is a unique solution to the defining equations of `g`.

Recall that `foldl` is defined as:

```foldl f v [] = v foldl f v (x:xs) = foldl f (f v x) xs```

The base case of `foldr` and `foldl` is identical, but the recursive one is not. Moreover, the recursive case of `foldl` cannot be rewritten in the form `f x (g xs)`. This means that we need to apply some transformation to definition of `foldl` so it can be rewritten in that form. Let’s create a function `foldl2` defined as:

```foldl2 f [] v = v foldl2 f (x:xs) v = foldl2 f xs (f v x)```

Nothing special so far. We just made a function that is the same as `foldl`, but has the last two parameters swapped. We can rewrite the base case as:

`foldl2 f [] v = id v`

`id` is the identity function that accepts one parameter and returns that parameter unchanged. Now we remove the `v` parameter:

`foldl2 f [] = id`

Such transformation is known as ?-reduction2. Let us now concentrate on the recursive case. It can be rewritten as

`foldl2 f (x:xs) v = (\w -> foldl2 f xs (f w x)) v`

We created an anonymous function with one of the parameters of `f` factored out. This expression can also be ?-reduced:

`foldl2 f (x:xs) = \w -> foldl2 f xs (f w x)`

Let’s factor out second parameter of function `f` in the same manner:

`foldl2 f (x:xs) = (\y w -> foldl2 f xs (f w y)) x`

And finally let’s factor out `foldl2 f xs` and just pass it as another parameter to the lambda expression:

`foldl2 f (x:xs) = (\y h w -> h (f w y)) x (foldl2 f xs)`

We’re almost there. Recall that the universal property requires function of the form3:

`g (x:xs) = k x (g xs)`

And it so happens that we just converted `foldl2` to that form, where `k = \y h w -> h (f w y)`. Comparing last two equation we see that `g = foldl2 f`, but from the universal property we also know that `g = foldr k v`, which means that `foldl2 f = foldr k v`. The notation here might be a bit confusing. From the base case we determined the value of `v` in the last equality to be equal to `id` function, which yields `foldl2 f = foldr k id`. Substituting the value of `k` we get:

`foldl2 f = foldr (\y h w -> h (f w y)) id`

Original definition of `foldl2` had two more parameters, but they were removed by ?-reductions. Let’s restore these two parameters by adding them to both lhs and rhs:

`foldl2 f xs v = foldr (\y h w -> h (f w y)) id xs v`

Recall that `foldl2` was only a convenience function we used for the derivation. Going back to the original `foldl` function yields the final result:

`foldl f v xs = foldr (\y h w -> h (f w y)) id xs v`

OK, formulas don’t lie, but this result is definitely not an intuitive one and deserves some good explanation. You may be surprised that four parameters are passed into `foldr`, but this should become clear in a moment. We will play with it to get some intuition on how this works.

Let us begin with verifying that this expression is type-correct. Type of `foldr` is:

```ghci> :t foldr foldr :: (a -> b -> b) -> b -> [a] -> b```

So the first parameter to `foldr` should be of type `(a -> b -> b)`. Lambda that we pass to `foldr` as the first parameter uses `f`. This is the function that is passed as first parameter to `foldl`. Since `foldl` has type:

```ghci> :t foldl foldl :: (a -> b -> a) -> a -> [b] -> a```

We require that `f` has type `a -> b -> a`. Let’s define simplest possible function that has that type and then check the type of lambda passed to `foldr`:

```ghci> let f = \a b -> a ghci> :t (\y h w -> h (f w y)) (\y h w -> h (f w y)) :: t2 -> (t1 -> t) -> t1 -> t```

Recall that `->` is right-associative, which means that above type is equivalent to `t2 -> (t1 -> t) -> (t1 -> t)`. Parentheses at the end can be added and the meaning is the same. This corresponds to our expected type of `(a -> b -> b)`. Here, the value of `b` is assumed to be `t1 -> t`. If we substitute (`t1 -> t`) for `b` in the type signature of `foldr` we get

`(a -> (t1 -> t) -> (t1 -> t)) -> (t1 -> t) -> [a] -> (t1 -> t)`

Note that last parentheses can be dropped, which result in function that has four parameters:

`(a -> (t1 -> t) -> (t1 -> t)) -> (t1 -> t) -> [a] -> t1 -> t`

We already verified that lambda passed to `foldr` is of type `(a -> (t1 -> t) -> (t1 -> t))`. The second parameter, `id` function, is of type `(a -> a)`, which corresponds to `(t1 -> t)` in the type signature. Therefore usage of `id` imposes additional restriction that `t1` ~ `t`4, which means that type signature can be rewritten as:

`(a -> (t -> t) -> (t -> t)) -> (t -> t) -> [a] -> t -> t`

`[a]` corresponds to parameter `xs`, the list that we are folding. `t` corresponds to initial value of accumulator and the last `t` is the return type.

Now that we have verified type-correctness of the solution, let’s see how it works in practice. Let’s say we want to fold a list of three elements using `(+)` as folding function and `0` as initial value of the accumulator. In other words, we want to calculate the sum of elements in a list. If we use `foldr`, the evaluation process will look like this:

```foldr (+) 0 [1,2,3] = (+) 1 (foldr (+) 0 [2,3]) = (+) 1 ((+) 2 (foldr (+) 0 [3])) = (+) 1 ((+) 2 ((+) 3 (foldr (+) 0 []))) = (+) 1 ((+) 2 ((+) 3 0 )) = (+) 1 ((+) 2 3) = (+) 1 5 = 6```

If we instead use `foldl`, evaluation will look like this:

```foldl (+) 0 [1,2,3] = foldl ((+) 0 1) [2,3] = foldl ((+) ((+) 0 1) 2) [3] = foldl ((+) ((+) ((+) 0 1) 2) 3) [] = ((+) ((+) ((+) 0 1) 2) 3) = ((+) ((+) 1 2) 3) = ((+) 3 3) = 6```

Both folds produce the same result, which is a direct consequence of first duality theorem for folds5. Now let’s see how evaluation will proceed if we use `foldl` expressed using `foldr`:

```foldl (+) 0 [1,2,3] = foldr (\y h w -> h ((+) w y)) id [1,2,3] 0 = (\h w -> h ((+) w 1)) (foldr (\y h w -> h ((+) w y)) id [2,3]) 0 = (\h w -> h ((+) w 1)) (\h w -> h ((+) w 2)) (foldr (\y h w -> h ((+) w y)) id [3]) 0 = (\h w -> h ((+) w 1)) (\h w -> h ((+) w 2)) (\h w -> h ( (+) w 3) ) (foldr (\y h w -> h ( (+) w y) ) id []) 0 = (\h w -> h ((+) w 1)) (\h w -> h ((+) w 2)) (\h w -> h ((+) w 3)) id 0 = (\h w -> h ((+) w 1)) (\h w -> h ((+) w 2)) (w -> id ((+) w 3)) 0 = (\h w -> h ((+) w 1)) (w -> id ((+) ((+) w 2) 3)) 0 = (w -> id ((+) ((+) ((+) w 1) 2) 3)) 0 = id ((+) ((+) ((+) 0 1) 2) 3))```

We’ve reached expression that is the same as the one we reached when evaluating `foldl`. Well, in fact that is what we expected. After all this is also `foldl`! So the whole trick is based on using `foldr` to generate a function that accepts initial value of the accumulator and produces the same expression we would get when using `foldl` (plus the identity function).

I hope that this post made it clear how to express `foldl` using `foldr`. This is of course by no means an exhaustive treatment of the subject of folds. There’s a lot more. I think that Hutton’s paper is a good starting point. Bird’s and Wadler’s “Introduction to Functional Programming” also seems to be a very valuable resource, though I’ve read only chapter about folds6. There’s still some more stuff to figure out about folds, like the difference in behaviour of `foldr` and `foldl` for infinite lists or expressing `foldr` using `foldl` (for finite lists only).

1. Graham Hutton in his paper uses the name fold to denote `foldr`. I’m sticking with `foldr` so my derivation remains consistent with Haskell []
2. ? is pronounced eta. []
3. Notice that I renamed the name of the first function from `f` to `k` to prevent the name clash. []
4. ~ notation means that two types are the same. []
5. See: Bird, R., Wadler, P. “Introduction to functional programming”, 1st ed., p. 68 []
6. There’s also a second edition of this book authored only by Bird, but I wasn’t able to find it. []

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