## 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. []

### 9 Responses to “Expressing foldl in terms of foldr”

1. Ryan Ingram says:

Here’s a derivation I like, using operators from Control.Arrow:

foldl f z [a,b,c,d]
= f (f (f (f z a) b) c) d
= flip f d (flip f c (flip f b (flip f a z)))
= flip f d . flip f c . fip f b . flip f a \$ z
= flip f d <<< flip f c <<< flip f b <<>> flip f b >>> flip f c >>> flip f d \$ z
= flip f a >>> flip f b >>> flip f c >>> flip f d >>> id \$ z
= foldr (\x k -> flip f x >>> k) id [a,b,c,d] z

Evaluating the anonymous function a bit:

\x k -> flip f x >>> k
= \x k v -> k . flip f x
= \x k v -> k (flip f x v)
= \x k v -> k (f v x)

Therefore,

foldl f z xs = foldr (\x k v -> k (f v x)) id xs z

2. Jan Stolarek says:

I’m yet to learn arrows, so I don’t understand this yet, but what happens with “a” when you move from function composition to arrows?

3. Ryan Ingram says:

Seems to be a bug in the HTML comment markup. Pretend ] is > and [ is flip f x ]]] k) id [a,b,c,d]) z

4. Ryan Ingram says:

ARGH. Trying again.

foldl f z [a,b,c,d]
f (f (f (f z a) b) c) d
— reverse order of arguments
flip f d (flip f c (flip f b (flip f a z)))
— change to function composition pipeline
— function composition is associative so we
— can ignore parentheses
(flip f d . flip f c . flip f b . flip f a) z
— ({{{) is just (.)
(flip f d {{{ flip f c {{{ flip f b {{{ flip f a) z
— (}}}) is ({{{) with arguments reversed
(flip f a }}} flip f b }}} flip f c }}} flip f d) z
— id is the monoid identity for function composition
(flip f a }}} flip f b }}} flip f c }}} flip f d }}} id) z
— this is now in the form of a right fold
(foldr (\x k -> flip f x }}} k) id [a,b,c,d]) z

5. Andre says:

Thank you for your article. I really appreciated the thorough derivation.
As I wanted to challenge myself, I tried to find a solution before reading your article. Here is what I came up with. Hopefully it helps to build some intuition (it did for me!).

To make foldr and foldl look more concrete, I rewrote what foldr and foldl do with an arbitrary list [x1, …, xn]:
foldr f v [x1, …, xn] ? (f x1 ? … ? f xn) v
foldl f v [x1, …, xn] ? (flip f xn ? … ? flip f x1) v

Their overall structure is quite similar. They differ mostly in two aspects:
1. The operator’s arguments are flipped.
2. Composition happens in reverse order.
Now the question is, which g and w satisfy foldr g w ? foldl f v for arbitrary but well-typed f and v? This function g has to flip the operator’s arguments and compose in reverse order, that is: g ? \y h -> h ? flip f y.
h is the function that was composed in the previous step. Since h is the first argument of ?, h will always be evaluated after the flip f y.
To give the complete solution:
foldl f v xs = foldr (\y h -> h ? flip f y) id xs v

Here an example for an arbitrary list with three elements.
foldr (\y h -> h ? flip f y) id [x1, x2, x3] v
? ((\y h -> h ? flip f y) x1 ? (\y h -> h ? flip f y) x2 ? (\y h -> h ? flip f y) x3) id v
? ((\h -> h ? flip f x1) ? (\h -> h ? flip f x2) ? (\h -> h ? flip f x3)) id v
? ((\h -> h ? flip f x1) ? (\h -> h ? flip f x2)) (id ? flip f x3) v
? ((\h -> h ? flip f x1)) (id ? flip f x3 ? flip f x2) v
? (id ? flip f x3 ? flip f x2 ? flip f x1) v
? (flip f x3 ? flip f x2 ? flip f x1) v
? foldl f v [x1, x2, x3]

By the way, my derivation yields the same function. By using flip, ?-reduction and function composition \y h w -> h (f w y) equates \y h -> h ? flip f y.

6. Jan Stolarek says:

There’s a third duality theorem (see Bird & Wadler mentioned in my post) that states:

foldr f a xs = foldl (flip f) a (reverse xs)

which is the same thing that you wrote. Andre, thanks for pointing out that \y h w -> h (f w y) is the same \y h -> h ? flip f y. This is of course obvious once you see it, but this transformation allows to bridge a gap between my derivation and the third duality theorem.

7. Matt Skalecki says:

A lot of intuition can be gained by factoring out the map to endomorphisms. ((>>>) = \f g x -> g (f x), which is just reverse function composition)

foldl2 f v xs = foldr (\y h w -> h (f w y)) id xs v
= foldr ((>>>) . flip f) id xs v
= foldr (>>>) id (map (flip f) xs) v

All the foldr does in this case is compose all the endomorphisms together in reverse order. Incidentally, since function composition is associative, it doesn’t particularly matter which flavor of fold we use. Predictably, using normal composition leads to an analogous definition of foldr.

foldr2 f v xs = foldr (.) id (map f xs) v

In this case, it does actually matter which fold we use, but only to preserve laziness.

8. csoroz says:

Andre solution in general, by induction:

foldl f v xs = foldr (\y h -> h ° flip f y) id xs v

BASE:
foldl f v []
=> v {by foldl definition}

foldr (\y h -> h ° flip f y) id [] v
=> id v {by foldr definition}
=> v {application of id}

INDUCTION:

foldl f v (x:xs)
=> foldl f (f v x) xs {by foldl definition}

foldr (\y h -> h ° flip f y) id (x:xs) v
{by foldr definition}
=> (\y h -> h ° flip f y) x (foldr (\y h -> h ° flip f y) id xs) v
{g => (\w -> g w)}
=> (\y h -> h ° flip f y) x (\w foldr (\y h -> h ° flip f y) id xs w) v
{by induction step}
=> (\y h -> h ° flip f y) x (\w foldl f w xs) v
=> ((\w foldl f w xs) ° flip f x) v
=> (\w foldl f w xs) (flip f x v)
=> (\w foldl f w xs) (f v x)
=> foldl f (f v x) xs

9. csoroz says:

Errata: “\w” should be “\w ->”

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