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Equivalent unzippers - Lindsey Kuper [entries|archive|friends|userinfo]
Lindsey Kuper

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Equivalent unzippers [Oct. 21st, 2010|11:50 pm]
Lindsey Kuper

In an effort to become more familiar with Coq, I've been working my way through the Software Foundations book. So far, it's been more about proving things with Coq than it's been about programming with it (although the distinction is blurry). Yesterday, though, I got to this problem. As the text explains, combine, better known as zip, is a function that takes two lists and combines them into a list of pairs. (For instance, given [0, 1, 3, 18] and [true, false, false, true], it takes those two lists and "zips" them together, producing [(0, true), (1, false), (3, false), (18, true)].)

The problem asks you to do the inverse: given the list of pairs, unzip it to produce the pair of lists. At first, the only way I could think of to do it was with map:

Fixpoint map {X Y: Type} (fn : X -> Y) (lx : list X) : list Y :=
  match lx with
  | [] => []
  | x::tx => fn x :: map fn tx

Fixpoint split {X Y : Type} (lxy : list (X*Y)) 
          : (list X * list Y) :=
  (map fst lxy, map snd lxy).

Later, I realized that I could do it in one pass provided I have a couple of accumulator arguments:

Fixpoint split' {X Y : Type} (lxy : list (X*Y))
          : (list X * list Y) :=
  let fix f (res1 : list X) (res2 : list Y) (lxy : list (X*Y)) :=
    match lxy with
    | [] => (res1, res2)
    | xy::txy => f (res1 ++ [(fst xy)]) (res2 ++ [(snd xy)]) txy
  in f [] [] lxy.

So, that was fun. But the larger point is that since I'm doing it in Coq, pretty soon I'll be able to do a machine-assisted proof that split and split' are equivalent, not to mention a proof that split (or split') actually is the inverse of combine. That will be even better.


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From: simrob
2010-10-22 01:58 pm (UTC)

Because Chris showed-and-telled, I wrote it in Agda as well. Most of it is boilerplate because I 1) perversely insisted on using my own library and 2) I lazily haven't added lists to that library yet.

module Kuper where

  {- Bolierplate -}

  open import Prelude

  data List (A : Set) : Set where
    [] : List A
    _::_ : A → List A → List A

  [_] : ∀{A} → A → List A
  [ x ] = x :: []

  _++_ : ∀{A} → List A -> List A -> List A
  [] ++ ys = ys
  (x :: xs) ++ ys = x :: (xs ++ ys)

  append-cong2 : ∀{A} {a : A} {as bs : List A} 
     → as ≡ bs 
     → (a :: as) ≡ (a :: bs)
  append-cong2 Refl = Refl

  append-nil : ∀{A} {as : List A} → as ++ [] ≡ as
  append-nil {as = []} = Refl
  append-nil {as = a :: as} = append-cong2 (append-nil {as = as})

  assoc-append : ∀{A} {as bs cs : List A}
     → (as ++ (bs ++ cs)) ≡ ((as ++ bs) ++ cs)
  assoc-append {as = []} = Refl
  assoc-append {as = a :: as} = append-cong2 (assoc-append {as = as})
  map : ∀{A B} → (A → B) → List A → List B
  map f [] = []
  map f (x :: xs) = f x :: map f xs

  {- Functions -}

  split : ∀{A B} → List (A × B) → List A × List B
  split lxy = (map fst lxy , map snd lxy)

  split' : ∀{A B} → List (A × B) → List A → List B → List A × List B
  split' [] xs ys = (xs , ys)
  split' ((x , y) :: lxy) xs ys = split' lxy (xs ++ [ x ]) (ys ++ [ y ])

  {- Equivalence -}

  split-eq-lem : ∀{A B} (lx : List A) (ly : List B) (lxy : List (A × B))
     → (lx ++ map fst lxy , ly ++ map snd lxy) ≡ split' lxy lx ly
  split-eq-lem lx ly [] = Product.pair-cong append-nil append-nil
  split-eq-lem lx ly ((x , y) :: lxy) =
     Product.pair-cong (assoc-append {as = lx}) (assoc-append {as = ly}) 
     ≡≡ split-eq-lem (lx ++ [ x ]) (ly ++ [ y ]) lxy

  split-eq : ∀{A B} (lxy : List (A × B)) → split lxy ≡ split' lxy [] [] 
  split-eq lxy = split-eq-lem [] [] lxy

Question: the most natural way to express this function, in Agda, was neither of your options. Is the following definition naturally expressible in Coq? This comparison example has taught me at least as much about Coq as I think I knew before, which is to say I know very little.

  splitA : ∀{A B} → List (A × B) → List A × List B
  splitA [] = ([] , [])
  splitA ((x , y) :: lxy) = (x :: fst lxly) , (y :: snd lxly)
     where lxly = splitA lxy
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From: neelk
2010-10-22 02:24 pm (UTC)
Yikes. That program is a pretty strong argument for tactical theorem proving! (As an aside: your natural definition works, but makes you rewrite with the extensionality of pairs to deal with let-bindings showing up in mildly annoying fashion.)
Set Implicit Arguments.
Unset Strict Implicit.
Import Prenex Implicits.
Require Import List.

Fixpoint split (X Y : Type) (xys : list (X * Y)) :=
  match xys with
    | nil => (nil, nil)
    | cons (x,y) xys' =>
      let (xs', ys') := split xys' in (cons x xs', cons y ys')

Fixpoint zip (X Y : Type) (xs : list X) (ys : list Y) {struct xs} :=
  match xs, ys with
    | nil, _ => nil
    | _, nil => nil
    | (cons x xs'), (cons y ys') => cons (x,y) (zip xs' ys')

Lemma pair_ext : forall (A B : Type) (a : A * B), a = (fst a, snd a). 
 intros A B [u v]; simpl; auto. (* Why isn't this in the stdlib??? *)

Lemma zipsplit : forall (X Y : Type) (xys : list (X * Y)), 
   let (xs, ys) := split xys in zip xs ys = xys.
  intros X Y; simpl; induction xys; [simpl; auto | unfold split; fold split].
  rewrite (pair_ext a); rewrite (pair_ext (split xys)) in IHxys |- *.
  simpl; rewrite IHxys; reflexivity.
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[User Picture]From: lindseykuper
2010-10-22 03:24 pm (UTC)
Question: the most natural way to express this function, in Agda, was neither of your options. Is the following definition naturally expressible in Coq?

I would hesitate to say that either of my options were the most natural way to express it in Coq, either. Well, the first one isn't bad. But the "most natural way" probably involves, first of all, letting the type inferencer do more work, and second, relying on libraries more. So, more like Neel did it, probably.
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From: simrob
2010-10-22 09:25 pm (UTC)
Dan pointed out that if I want to prove equivalence of split and splitA the proof is actually very, very nice. I needed an annoying lemma whose proof is Refl = refl in this setup, but the proof is dead simple if that is factored out.

  cong-lem : ∀{A B} {x : A} {y : B} {xs1 xs2 : List A} {ys1 ys2 : List B}
     → (xs1 , ys1) ≡ (xs2 , ys2) 
     → IdT {A = List A × List B} (x :: xs1 , y :: ys1) (x :: xs2 , y :: ys2)
  cong-lem Refl = refl

  split-eqA : ∀{A B} → (lxy : List (A × B)) → split lxy ≡ splitA lxy
  split-eqA [] = refl
  split-eqA ((x , y) :: lxy) = cong-lem (split-eqA lxy)
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[User Picture]From: lindseykuper
2010-10-22 02:56 pm (UTC)

Ooh, me next, me next!

Here it is in miniKanren:

(import (minikanren vanilla))
(load "matche.scm")

(define zipo
  (lambda (pairoflists listofpairs)
    (matche `(,pairoflists ,listofpairs)
      [( (() ()) () )]
      [( ((,X . ,Xs) (,Y . ,Ys)) ((,X ,Y) . ,XYs) )
       (zipo `(,Xs ,Ys) XYs)])))

> (run* (q) (zipo '((1 3 5) (2 4 6)) q))
(((1 2) (3 4) (5 6)))
> (run* (q) (zipo q '((1 2) (3 4) (5 6))))
(((1 3 5) (2 4 6)))

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[User Picture]From: sstrickl
2010-10-22 06:48 pm (UTC)

Meanwhile I've taken a try at it in Isabelle/HOL using Isar. I make no claim on how "native" this is, given my relative lack of experience in the system.

theory Kuper
imports Main


fun ksplit :: "('a * 'b) list ⇒ 'a list * 'b list"
where "ksplit ABs = (map fst ABs, map snd ABs)"

fun ksplit' :: "('a * 'b) list ⇒ 'a list * 'b list"
  "ksplit' [] = ([], [])" |
  "ksplit' ((a, b) # ABs) =
    (case ksplit' ABs of
      (As, Bs) ⇒ (a # As, b # Bs))"

theorem ksplit_ksplit' : "ksplit l = ksplit' l"
proof (induct l rule: ksplit'.induct)
  case 1 thus ?case by simp
  case (2 a b ABs)
  assume H : "ksplit ABs = ksplit' ABs"
  then have "ksplit' ABs = (map fst ABs, map snd ABs)" by simp
  then have "ksplit' ((a, b) # ABs) = (a # map fst ABs, b # map snd ABs)" by simp
  thus ?case by simp
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(Deleted comment)
From: aleffert
2010-10-22 06:02 am (UTC)
Do you actually get to prove that two coq functions are equivalent? How do you take apart the construction of split and split' to reason about them? I didn't think you could deal with meta level functions like that, but I've never really used Coq.
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[User Picture]From: lindseykuper
2010-10-22 03:12 pm (UTC)
Yeah, you do. How you "take apart" the definitions depends, but it's a matter of showing that given the same arguments they produce the same results. Look at how Neel did it up above. His split' (the one that uses map) is basically my split, while his split is sort of like my split' but prettier. Mine might be harder to prove stuff about -- I'm not really sure yet.
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[User Picture]From: pmb
2010-10-22 06:04 pm (UTC)
The real problem is that zip() is of arity 2, and not arity n. An n-arity zip() is its own inverse! :)
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[User Picture]From: sstrickl
2010-10-22 11:35 pm (UTC)
Eh, I posted the ACL2 version, and given the lack of higher-order functions, it's even less interesting (especially if you had to define your own mapcar and mapcdr).
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From: markluffel
2010-10-26 06:52 pm (UTC)

Proof language for non-PL-researchers?

Hey Lindsey et al!
If I wanted to learn one of these wild proof languages, say for doing geometric/computer-graphics proofs, what language would you recommend? I started working through a tutorial for that Omega language, which I understand is rather different from what the folks in this thread are using... but the documentation made it seem to be a better choice for a beginner.

If I'm concerned mostly with ease of use (and libraries for dealing with real numbers, I guess) what should I use?
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[User Picture]From: lindseykuper
2010-10-26 08:04 pm (UTC)

Re: Proof language for non-PL-researchers?

Hm. Have you looked at GeoProof? Apparently it talks to Coq.

In any case, I'd recommend whichever one has the best library for geometric proofs, since if it doesn't already exist you'll be spending a lot of time building up scaffolding before you can even get started.

To my knowledge, Omega's less a proof assistant than it is a general-purpose language with dependent types. Having said that, the line between the two is blurry, and I think a lot of people like Agda precisely because it straddles that line pretty well.
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