How can two continuations cancel each other out?

Question

I'm reading through Some Tricks for List Manipulation, and it contains the following:

zipRev xs ys = foldr f id xs snd (ys,[])
  where
    f x k c = k (\((y:ys),r) -> c (ys,(x,y):r)) 

What we can see here is that we have two continuations stacked on top of each other. When this happens, they can often “cancel out”, like so:

zipRev xs ys = snd (foldr f (ys,[]) xs)
  where
    f x (y:ys,r) = (ys,(x,y):r)

I don't understand how you "cancel out" stacked continuations to get from the top code block to the bottom one. What pattern do you look for to make this transformation, and why does it work?

Cryptocrystalline

By the way: those versions of zipRev fail if the second list is shorter than the first. I believe the laziest possible version that works regardless of relative lengths is this one I just put together. – Salade

Answer 1

A function f :: a -> b can be "disguised" inside double continuations as a function f' :: ((a -> r1) -> r2) -> ((b -> r1) -> r2).

obfuscate :: (a -> b) -> ((a -> r1) -> r2) -> (b -> r1) -> r2
obfuscate f k2 k1 = k2 (k1 . f)

obfuscate has the nice property that it preserves function composition and identity: you can prove that obfuscate f . obfuscate g === obfuscate (f . g) and that obfuscate id === id in a few steps. That means that you can frequently use this transformation to untangle double-continuation computations that compose obfuscated functions together by factoring the obfuscate out of the composition. This question is an example of such an untangling.

The f in the top code block is the obfuscated version of the f in the bottom block (more precisely, top f x is the obfuscated version of bottom f x). You can see this by noticing how top f applies the outer continuation to a function that transforms its input and then applies the whole thing to the inner continuation, just like in the body of obfuscate.

So we can start to untangle zipRev:

zipRev xs ys = foldr f id xs snd (ys,[])
  where
    f x = obfuscate (\(y:ys,r) -> (ys,(x,y):r))

Since the action of foldr here is to compose a bunch of obfuscated functions with each other (and apply it all to id, which we can leave on the right), we can factor the obfuscate to the outside of the whole fold:

zipRev xs ys = obfuscate (\accum -> foldr f accum xs) id snd (ys,[])
  where
    f x (y:ys,r) = (ys,(x,y):r)

Now apply the definition of obfuscate and simplify:

zipRev xs ys = obfuscate (\accum -> foldr f accum xs) id snd (ys,[]) 
zipRev xs ys = id (snd . (\accum -> foldr f accum xs)) (ys,[])
zipRev xs ys = snd (foldr f (ys,[]) xs)

QED!

Sherrellsherrer

A footnote: for an immediate proof that obfuscate respects function composition, we just have to notice it is fmap for the double continuation type. – Boroughenglish

Answer 2

Given a function

g :: a₁ -> a₂

we can lift it to a function on continuations, switching the order:

lift g = (\c a₁ -> c (g a₁))
    :: (a₂ -> t) -> a₁ -> t

This transformation is a contravariant functor, which is to say that it interacts with function composition by switching its order:

g₁ :: a₁ -> a₂
g₂ :: a₂ -> a₃

lift g₁ . lift g₂
== (\c₁ a₁ -> c₁ (g₁ a₁)) . (\c₂ a₂ -> c₂ (g₂ a₂))
== \c₂ a₁ -> (\a₂ -> c₂ (g₂ a₂)) (g₁ a₁)
== \c₂ a₁ -> c₂ (g₂ (g₁ a₁)) 
== lift (g₂ . g₁)
    :: (a₃ -> t) -> a₁ -> t

lift id
== (\c a₁ -> c a₁)
== id
    :: (a₁ -> t) -> a₁ -> t

We can lift the lifted function again in the same way to a function on stacked continuations, with the order switched back:

lift (lift g)
== (\k c -> k ((\c a₁ -> c (g a₁)) c))
== (\k c -> k (\a₁ -> c (g a₁)))
    :: ((a₁ -> t) -> u) -> (a₂ -> t) -> u

Stacking two contravariant functors gives us a (covariant) functor:

lift (lift g₁) . lift (lift g₂)
== lift (lift g₂ . lift g₁)
== lift (lift (g₁ . g₂))
    :: ((a₁ -> t) -> u) -> (a₃ -> t) -> u

lift (lift id)
== lift id
== id
    :: ((a₁ -> t) -> u) -> (a₁ -> t) -> u

This is exactly the transformation being reversed in your example, with g = \(y:ys, r) -> (ys, (x, y):r). This g is an endomorphism (a₁ = a₂), and the foldr is composing together a bunch of copies of it with various x. What we’re doing is replacing the composition of double-lifted functions with the double-lift of the composition of the functions, which is just an inductive application of the functor laws:

f :: x -> a₁ -> a₁
c :: (a₁ -> t) -> u
xs :: [x]

foldr (\x -> lift (lift (f x))) c xs
== lift (lift (\a₁ -> foldr f a₁ xs)) c
    :: (a₁ -> t) -> u

Toothbrush

Answer 3

Let's try to understand this code from an elementary point of view. What does it even do, one wonders?

zipRev xs ys = foldr f id xs snd (ys,[])
  where
     -- f x k c = k (\(y:ys, r) -> c (ys, (x,y):r))
        f x k c = k (g x c) 
     --         = (k . g x) c   -- so,
     -- f x k   =  k . g x

        g x   c       (y:ys, r) =  c (ys, (x,y):r)

Here we used lambda lifting to recover the g combinator.

So then because f x k = k . g x were k goes to the left of x, the input list is translated into a reversed chain of compositions,

foldr f id [x1, x2, x3, ..., xn]   where  f x k = k . g x
  ===>> 
   (((...(id . g xn) .  ...  . g x3) . g x2) . g x1)

and thus, it just does what a left fold would do,

zipRev [] ys = []
zipRev [x1, x2, x3, ..., xn] ys 
      = (id . g xn  .  ...  . g x3 . g x2 . g x1)  snd         (ys, [])
      = g xn (g xn1 (  ...  ( g x3 ( g x2 ( g x1   snd)))...)) (ys, [])
   where     ----c--------------------------------------------
        g x  c     (y:ys, r) = c (ys, (x,y):r)

So we went to the deep end of the xs list, and then we come back consuming the ys list left-to-right (i.e. top-down) on our way back right-to-left on the xs list (i.e. bottom-up). This is straightforwardly coded as a right fold with strict reducer, so the flow is indeed right-to-left on the xs. The bottom-most action (snd) in the chain is done last, so in the new code it becomes the topmost (still done last):

zipRev xs ys = snd (foldr h (ys,[]) xs)
  where
        h x        (y:ys, r) =   (ys, (x,y):r)

g x c was used as a continuation in the original code, with c as a second-tier continuation; but it's actually all just been a regular fold from the right, all along.

---

So indeed it zips the reversed first list with the second. It's also unsafe; it misses a clause:

        g x  c     ([],   r) = c ([], r)        -- or just `r`
        g x  c     (y:ys, r) = c (ys, (x,y):r)

---

(update:) The answers by duplode (and Joseph Sible) do the lambda lifting a bit differently, in a way which is better suited to the task. It goes like this:

zipRev xs ys = foldr f id xs  snd (ys,[])
  where
     f x k c = k      (\((y:ys), r) -> c (ys, (x,y):r)) 
             = k (c . (\((y:ys), r) ->   (ys, (x,y):r)) )
             = k (c . g x)
     g x     =        (\((y:ys), r) ->   (ys, (x,y):r))
  {- f x k c = k ((. g x) c) = (k . (. g x)) c = (. (. g x)) k c
     f x     =                                   (. (. g x))     -}

so then

foldr f id  [ x1,            x2,    ... ,         xn      ]  snd  (ys,[]) =
  = ( (. (. g x1)) $ (. (. g x2)) $ ... $ (. (. g xn)) id )  snd  (ys,[])  -- 1,2...n
  = ( id . (. g xn) .  ...  . (. g x2)  .    (. g x1)     )  snd  (ys,[])  -- n...2,1
  =      ( snd . g x1 .    g x2   . ... .       g xn            ) (ys,[])  -- 1,2...n!
  =        snd $ g x1 $    g x2   $ ... $       g xn              (ys,[])
  =        snd $  foldr g (ys,[])  [x1, x2, ...,  xn      ]

Simple. :) Flipping twice is no flipping at all.

Reversion

btw I used similar technique in one of my recent answers: rltake here builds a chain of compositions first (just in a different direction), then applies it further; very much like zipRev here does. – Reversion

See my comment on the question. – Salade

what does fr stand for? – Reversion

It stands for "final result". – Salade

to me, z is more informative, invokes "zip", so adds something. "fr" adds nothing, just restates the given, and leaves me searching for the answer, futilely, because I search for the added meaning, and there's none. So that's why I was stumped. :) Names are important. :) – Reversion

Answer 4

Let's begin with a few cosmetic adjustments:

-- Note that `g x` is an endomorphism.
g :: a -> ([b], [(a,b)]) -> ([b], [(a,b)])
g x ((y:ys),r) = (ys,(x,y):r)

zipRev xs ys = foldr f id xs snd (ys,[])
  where
    f x k = \c -> k (c . g x)

f feeds a continuation (c . g x) to another function (k, a "double continuation", as user11228628 puts it).

While we might reasonably expect that repeated usage of f as the fold proceeds will somehow compose the g x endomorphisms made out of the elements of the list, the order in which the endomorphisms are composed might not be immediately obvious, so we'd better walk through a few fold steps to be sure:

-- x0 is the first element, x1 the second, etc.
f x0 k0 
\c -> k0 (c . g x0)
\c -> (f x1 k1) (c . g x0) -- k0 is the result of a fold step.
\c -> (\d -> k1 (d . g x1)) (c . g x0) -- Renaming a variable for clarity.
\c -> k1 (c . g x0 . g x1)
-- etc .
-- xa is the *last* element, xb the next-to-last, etc.
-- ka is the initial value passed to foldr.
\c -> (f xa ka) (c . g x0 . g x1 . . . g xb)
\c -> (\d -> ka (d . g xa)) (c . g x0 . g x1 . . . g xb)
\c -> ka (c . g x0 . g x1 . . . g xb . g xa)

ka, the initial value passed to foldr, is id, which makes things quite a bit simpler:

foldr f id xs = \c -> c . g x0 . g x1 . . . g xa

Since all we do with the c argument passed to foldr f id xs is post-composing it with the endomorphisms, we might as well factor it out of the fold:

zipRev xs ys = (snd . foldr h id xs) (ys,[])
  where
    h x e = g x . e

Note how we have gone from c . g x to g x . e. That can arguably be described as a collateral effect of the CPS trickery in the original implementation.

The final step is noticing how h x e = g x . e corresponds exactly to what we would do to implement foldr in terms of foldMap for the Endo monoid. Or, to put it more explicitly:

foldEndo g i xs = foldr g i xs  -- The goal is obtaining an Endo-like definition.

foldEndo _ i [] = i
foldEndo g i (x : xs) = g x (foldEndo g i xs)

foldEndo g i xs = go xs i
    where
    go [] = \j -> j
    go (x : xs) = \j -> g x (foldEndo g j xs)

foldEndo g i xs = go xs i
    where
    go [] = \j -> j
    go (x : xs) = \j -> g x (go xs j)

foldEndo g i xs = go xs i
    where
    go [] = id
    go (x : xs) = g x . go xs

foldEndo g i xs = go xs i
    where
    h x e = g x . e
    go [] = id
    go (x : xs) = h x (go xs)

foldEndo g i xs = go xs i
    where
    h x e = g x . e
    go xs = foldr h id xs

foldEndo g i xs = foldr h id xs i
    where
    h x e = g x . e

That finally leads us to what we were looking for:

zipRev xs ys = snd (foldr g (ys,[]) xs)

Boroughenglish

Answer 5

user11228628's answer led me to understanding. Here's a few insights I had while reading it, and some step-by-step transformations.

---

Insights

• The continuations don't directly cancel out. They can only eventually be canceled (by beta-reducing) because it's possible to factor them out.
• The pattern you're looking for to do this transformation is \k c -> k (c . f) (or if you love unreadable pointfree, (. (. f))) for any f (note that the f isn't a parameter to the lambda).
• As duplode points out in a comment, continuation-passing style functions can be considered a functor, and obfuscate is their definition of fmap.
• The trick of pulling a function like this out of foldr works for any function that could be a valid fmap.

---

Full transformation from the first code block to the second

zipRev xs ys = foldr f id xs snd (ys,[])
  where
    f x k c = k (\((y:ys),r) -> c (ys,(x,y):r))

Pull c out of the lambda

zipRev xs ys = foldr f id xs snd (ys,[])
  where
    f x k c = k (c . \((y:ys),r) -> (ys,(x,y):r))

Substitute obfuscate for its definition

zipRev xs ys = foldr f id xs snd (ys,[])
  where
    f x = obfuscate (\((y:ys),r) -> (ys,(x,y):r))

Pull obfuscate out of the lambda

zipRev xs ys = foldr f id xs snd (ys,[])
  where
    f = obfuscate . \x ((y:ys),r) -> (ys,(x,y):r)

Pull obfuscate out of f

zipRev xs ys = foldr (obfuscate . f) id xs snd (ys,[])
  where
    f x ((y:ys),r) = (ys,(x,y):r)

Since obfuscate follows the Functor laws, we can pull it out of foldr

zipRev xs ys = obfuscate (flip (foldr f) xs) id snd (ys,[])
  where
    f x ((y:ys),r) = (ys,(x,y):r)

Inline obfuscate

zipRev xs ys = (\k c -> k (c . flip (foldr f) xs)) id snd (ys,[])
  where
    f x ((y:ys),r) = (ys,(x,y):r)

Beta-reduce

zipRev xs ys = (id (snd . flip (foldr f) xs)) (ys,[])
  where
    f x ((y:ys),r) = (ys,(x,y):r)

Simplify

zipRev xs ys = snd (foldr f (ys,[]) xs)
  where
    f x (y:ys,r) = (ys,(x,y):r)

---

Justification for pulling functions that are valid fmaps out of foldr

foldr (fmap . f) z [x1,x2,...,xn]

Expand the foldr

(fmap . f) x1 . (fmap . f) x2 . ... . (fmap . f) xn $ z

Inline the inner .s

fmap (f x1) . fmap (f x2) . ... . fmap (f xn) $ z

Apply the Functor laws

fmap (f x1 . f x2 . ... . f xn) $ z

Eta-expand the section in parentheses

fmap (\z2 -> f x1 . f x2 . ... . f xn $ z2) z

Write the lambda body in terms of foldr

fmap (\z2 -> foldr f z2 [x1,x2,...,xn]) z

Write the lambda body in terms of flip

fmap (flip (foldr f) [x1,x2,...,xn]) z

---

Bonus: Justification for pulling functions that are valid contramaps out of foldr

foldr (contramap . f) z [x1,x2,...,xn]

Expand the foldr

(contramap . f) x1 . (contramap . f) x2 . ... . (contramap . f) xn $ z

Inline the inner .s

contramap (f x1) . contramap (f x2) . ... . contramap (f xn) $ z

Apply the Contravariant laws

contramap (f xn . ... . f x2 . f x1) $ z

Eta-expand the section in parentheses

contramap (\z2 -> f xn . ... . f x2 . f x1 $ z2) z

Write the lambda body in terms of foldr

contramap (\z2 -> foldr f z2 [xn,...,x2,x1]) z

Write the lambda body in terms of flip

contramap (flip (foldr f) [xn,...,x2,x1]) z

Apply foldr f z (reverse xs) = foldl (flip f) z xs

contramap (flip (foldl (flip f)) [x1,x2,...,xn]) z

Cryptocrystalline