Rubrication

I was playing around with the Peano numerals in Haskell (as I often do), because I remember briefly reading about a generalisation of exponentiation in a paper once when I didn’t have time to read it. I decided I’d quickly look at it again now that I was in a mood to procrastinate.

> data Nat = Z | S Nat deriving Show

This data type is usually identified with the peano numerals. In fact as we will see, in haskell it isn’t quite the natural numbers. We take Z to mean 0 and take (Sⁿ Z) or (S (S (S … Z))) n times, to mean n. It’s basically a unary number system where the length of the string gives us the number.

It is pretty straightforward to write the basic arithmetic operations on the peano numerals.

> add x Z = x
> add x (S y) = S (add x y)

> mult x Z = Z
> mult x (S y) = add x (mult x y)

> expt x Z = (S Z)
> expt x (S y) = mult x (expt x y)

Here we can see an extension of this series fairly easily. The only real question we might have is what to do with the base case. It turns out that (S Z) or 1 is a good choice.

> exptexpt x Z = (S Z)
> exptexpt x (S y) = expt x (exptexpt x y)

Playing with exptexpt got me thinking about the fact that I wasn’t dealing with natural numbers and I was curious what I could prove about them. In order to see that we don’t actually have the natural numbers we can define the following:

> inf :: Nat
> inf = S (inf)

This element is clearly not a natural number since it is not a valid recursive equation in the sense that there is no decreasing quantity. It is however a valid corecursive equation. It is always productive in the sense that it always has it’s recursive reference inside of a constructor. We will always be able to “peel” something away from this function in order to reason about it. This will become obvious when we start looking at equational reasoning with infinity.

So this leads us to the question: what does inf+x equal? I’ll prove it in pseudo-Coq. Let us make the following conjecture:

∀ x, inf + x = inf

Theorem : ∀ x, inf + x = inf.
  By coinduction:

  case x = 0:
    1a) inf + 0 = inf
    2a) inf = inf
    * true by the reflexive property of =
  case x = S x'
    1b) inf + S x' = inf
    2b) S (inf + x') = inf
    3b) S (inf + x') = S inf
    4b) inf + x' = inf
    * true as this is the coinductive hypothesis!
Qed.

Now the equality that we are using here is actually not term based equality. It is a coinductive type of equality known as bisimulation. It also must follow the rule that we can never use recursion unless we are inside a constructor for the type of the function we are using. The recursion is the coinductive hypothesis. The constructor prior to the use of the coinductive hypothesis is from the definition of bisimulation. Namely:

∀ x y, x = y → S x = S y.

The full definition in Coq of the bisimulation which we denote above as “=” and below as “CoEq” is:

CoInductive CoEq : CoNat -> CoNat -> Prop :=
| coeq_base : forall x, CoEq Z Z
| coeq_next : forall x y, CoEq x y -> CoEq (S x) (S y).

The constructor coeq_next, which is used between step 3b and 4b above ensures that we satisfy our guardedness, or productivity condition and can safely “call” the coinductive hypothesis. What do I mean “call” and how is the “constructor” being used? We can see exactly by printing out the term in Coq that proves inhabitation of the type for our proposition.

add_x_inf =
cofix add_x_inf (x : CoNat) : x + inf ~= inf :=
  eq_ind_r (fun c : CoNat => x + c ~= c)
    (eq_ind_r (fun c : CoNat => c ~= S inf)
       (coeq_next (x + inf) inf (add_x_inf x)) (add_x_s x inf))
    (decomp_eql inf)
     : forall x : CoNat, x + inf ~= inf

Here we see in gory detail a corecursive function that allows us to show this type is inhabited along with the corecursive call shown embedded in the constructor coeq_next. Sorry if it isn’t very readable, it was constructed automagically from the proof script.

To see how this is all done in detail, check out the Coq proof script for the co-naturals. This might be interesting to those of you who want to see how to use coinductive reasoning on simple examples and how setoids work in Coq. You’ll notice that the reasoning is slightly more cumbersome then what I’ve used here. I think this mostly comes down to the “cofix” tactic being unwieldy in coq, and making automation a hassle. I’m going to think about ways to fix this since it really is unnecessarily inconvenient.

Here are some helper functions which convert from positive integers into our representation so you can play with the code more easily.

> to 0 = Z
> to (n+1) = (S (to n))

> from Z = 0
> from (S x) = 1+(from x)

A while ago I wrote a post about how one can use coq to make a proper monad module. I was just thinking today that it would be nice to have a tool for haskell that would allow one to write down conjectures and discharge them automatically with a theorem prover. Supercompilation makes a nice clean theorem prover for haskell since one could express the equations of interest in haskell itself. Below is an example of the list monad, and then the 3 monad laws written as conj1,conj2 and conj3. I prove the first law by manual supercompilation. The next two are left as an exercise for the interested reader.

equal xs ys = case xs of
                [] -> case ys of
                        [] -> True
                        _ -> False
                (x:xs') -> case ys of
                             [] -> False
                             (y:ys') -> case (eq x y) of
                                          False -> False
                                          True ->  equal xs' ys'

bind [] f = []
bind (h:t) f = (f h)++(bind t f)

ret a = [a]

conj1 a f = equal (bind (ret a) f) (f a)
conj2 m = equal (bind m (\a -> ret a)) m
conj3 m f g = equal (bind (bind m f) g) (bind m (\x -> (bind (f x) g)))

In order to define equality on lists we had to make reference to a different equality predicate on the elements of the list. We will make the assumption that this equality is decidable and supercompilation can prove the reflexive property for this equality predicate, that is “eq x x = True” will be taken as an assumption. It is somewhat hard to imagine a case where supercompilation would have a hard time with this because of the case substitution rule.

Now we take conj1 and attempt to prove it by semantics preserving transformations of the source code [1]. I’ve been brutally explicit in the steps used so that people who are interested can see exactly how to do these sorts of things themselves. I’ve found that program transformation techniques can be extremely useful in reasoning about code. Of course, it helps to assume that all functions are total, and I’ll be doing just that. I use the notation M[x:=y] to mean the substitution of y for x in M. Aside from that everything is just Haskell.

conj1 a f = equal (bind (ret a) f) (f a)

{- unfold equal -}
conj1 a f = case (bind (ret a) f) of
              [] -> case (f a) of
                      [] -> True
                      _ -> False
              (x:xs') -> case (f a) of
                           [] -> False
                           (y:ys') -> case eq x y of
                                        False -> False
                                        True -> equal xs' ys'

{- unfold bind -}
conj1 a f = case (case (ret a) of
                    [] -> []
                    (z:zs) -> (f h)++(bind zs f))
              [] -> case (f a) of
                      [] -> True
                      _ -> False
              (x:xs') -> case (f a) of
                           [] -> False
                           (y:ys') -> case eq x y of
                                        False -> False
                                        True -> equal xs' ys'

{- case distribution -}
conj1 a f = case (ret a) of
              [] -> case (f a) of
                      [] -> True
                      _ -> False
              (z:zs) -> case ((f h)++(bind zs f)) of
                          [] -> case (f a) of
                                  [] -> True
                                  _ -> False
                          (x:xs') -> case (f a) of
                                       [] -> False
                                       (y:ys') -> case eq x y of
                                                    False -> False
                                                    True -> equal xs' ys'

{- unfold ret -}
conj1 a f = case [a] of
              [] -> case (f a) of
                      [] -> True
                      _ -> False
              (z:zs) -> case ((f h)++(bind zs f)) of
                          [] -> case (f a) of
                                  [] -> True
                                  _ -> False
                          (x:xs') -> case (f a) of
                                       [] -> False
                                       (y:ys') -> case eq x y of
                                                    False -> False
                                                    True -> equal xs' ys'

{- case selection -}
conj1 a f = (case ((f z)++(bind zs f)) of
                      [] -> case (f a) of
                              [] -> True
                              _ -> False
                      (x:xs') -> case (f a) of
                                   [] -> False
                                   (y:ys') -> case eq x y of
                                                False -> False
                                                True -> equal xs' ys')
                                     [z := a,  zs := []]

{- substitution -}
conj1 a f = case ((f a)++(bind [] f)) of
              [] -> case (f a) of
                      [] -> True
                      _ -> False
              (x:xs') -> case (f a) of
                           [] -> False
                           (y:ys') -> case eq x y of
                                        False -> False
                                        True -> equal xs' ys'

{- unfold bind  -}
conj1 a f = case (case (f a) of
                    [] -> []
                    (z:zs') -> z:(zs'++[])) of
              [] -> case (f a) of
                      [] -> True
                      _ -> False
              (x:xs') -> case (f a) of
                           [] -> False
                           (y:ys') -> case eq x y of
                                        False -> False
                                        True -> equal xs' ys' 

{- case selection, substitution -}
conj1 a f = case (f a) of
              [] -> True
              (z:zs') -> case eq z z of
                           False -> False
                           True -> equal (zs'++[]) zs'

{- Assumption: eq z z = True -}
 conj1 a f = case (f a) of
              [] -> True
              (z:zs') -> equal (zs'++[]) zs'

{- unfold of equal and ++ -}
conj1 a f = case (f a) of
              [] -> True
              (z:zs') -> case (case zs' of
                                 [] -> []
                                 (w:ws) -> w:(ws++[]) of
                           [] -> case zs' of
                                   [] -> True
                                   (y:ys) -> False
                           (x:xs) -> case zs' of
                                       [] -> False
                                       (y:ys) -> case (eq x y) of
                                                   False -> False
                                                   True -> equal xs ys

{- case selection -}
conj1 a f = case (f a) of
              [] -> True
              (z:zs') -> case zs' of
                           [] -> True
                           (w:ws) -> case (eq w w) of
                                       False -> False
                                       True -> equal (ws++[]) ws

{- Assumption: eq w w = True -}
conj1 a f = case (f a) of
              [] -> True
              (z:zs') -> case zs' of
                           [] -> True
                           (w:ws) -> equal (ws++[]) ws

{- fold, (we encountered a repeated instance of 'equal (ws++[]) ws') -}
conj1 a f = case (f a) of
              [] -> True
              (z:zs') -> let g = \ xs ->
                                 case xs of
                                   [] -> True
                                   (y:ys') -> g ys'
                         in g zs'

Now we have a function that can only return true, regardless of the value of (f a) and assuming that f is total, we can replace this term with True. The proof of this is simply by induction on (f a), but the principle can be built into a checker and is the principle used in the Poitín [3] theorem prover.

Something along these lines would be very light-weight in comparison to using a full theorem prover and could just issue a warning if some laws couldn’t be proved. The proof of this in Coq is actually more work and requires lemmas about the list type to be proven. I did in fact fudge the ((f a)++[]) situation there by replacing with (f a), and I’m not entirely sure what supercompilation does with it if you leave it in. I’ll have to try it later. UPDATE: This is fixed in the code above.

If we want to prove this result by hand we can also derive it in a much simpler way by not following the supercompilation method verbatim:

conj1 a f = equal (bind [a] f) (f a)
conj1 a f = equal (append (f a) (bind [] f)) (f a)
conj1 a f = equal (append (f a) []) (f a)
conj1 a f = equal (f a) (f a)

The advantage of the former is of course that it is entirely mechanical.

UPDATE: I went ahead and incorporated the definition of append and supercompiled it (by hand) with this definition and it reduces to exactly the same term, so in fact supercompilation, without any special knowledge, is capable of proving the conjecture. This is a real win over most proof assistants which would require this append lemma to be proved separately, or incorporated into the knowledge base. For a more visual and more compact depiction of the derivation we can draw a partial process tree with the append function definition included in the derivation.

And indeed the second monad law:

Is the third monad law for the list monad provable using supercompilation? I expect it is, but since I don’t have a supercompiler, and it is a bit of a bear to do by hand, I’m not sure.

Incidently, there is already a supercompiler being developed for haskell called Supero [2]. It probably wouldn’t be too much work to extend this with theorem proving capabilities using ideas from [3].

[1] A Transformation System for Developing Recursive Programs
[2] A Supercompiler for Core Haskell
[3] Poitın: Distilling Theorems From Conjectures

I’ve mentioned a number of times that given a sufficiently rich type theory we can use the type to provide a specification for the function, such that the function is correct by construction. I’d like to give a simple example of how this works in the Coq proof assistant.

The following code is an illustration of how the type theory of Coq allows you to fully specify the behaviour of functions. Kragen Sitaker made the comment that there weren’t too many reasonable interpretations of the function of the type given to the assoc function:

A -> list (A * B) -> option B

that is, a function taking an element of type A, a list of pairs of A and B, to the possibility of a B (and possible failure). That got me wondering how hard it would be to tighten the noose so that there was only one way to interpret the type but still possibly many implementations that satisfy the type, all of which will yield the same result.

The following code illustrates that this is indeed possible to do in Coq.

Require Import List.
Parameter A : Set.
Parameter B : Set.
Definition assoc_list := list (A * B)%type.
Parameter eq_dec : forall (x:A) (y:A),{x=y}+{x<>y}.

Theorem assoc :
  forall (x:A) (l: assoc_list),
    {y | In (x,y) l} + {forall (y:B), ~In (x,y) l}.
  refine
    (fix assoc (x:A) (l:assoc_list) {struct l}
      : {y | In (x,y) l} + {forall (y:B), ~In (x,y) l}
      := match l
           return {y | In (x,y) l} + {forall (y:B), ~In (x,y) l}
           with
           | nil => inright _ (fun y => (fun p : In (x,y) nil => _))
           | ((x',y)::t) =>
             match eq_dec x x' with
               | left lft => inleft _ (exist _ y _)
               | right rgt => match assoc x t with
                                | inleft (exist y p) => inleft _ (exist _ y _)
                                | inright inrgt => inright _ _
                              end
             end
         end) ; clear assoc ; eauto ; subst. constructor. reflexivity.
  firstorder. intros. intro. inversion H ; clear H. firstorder. congruence.
  apply (inrgt y0). assumption.
Defined.

This isn’t the most beautiful code but what it lacks in beauty of implementation it makes up for in ease of reading. There are only a couple of lines that you need to understand to be assured that this is in fact the correct implementation.

Theorem assoc :
  forall (x:A) (l: assoc_list),
    {y | In (x,y) l} + {forall (y:B), ~In (x,y) l}.

This function has a type that takes an “x” of type “A”, a list of pairs and supplies a value “y” of type “B” with a proof that it came from a pair in which the x provided occurs *AND* that that pair exists in the list l. If the function can’t find a pair from which to supply a y, it must supply a proof that no such pair exists in the list.

First a bit of the notation so you can read this theorem properly

{ y | P y }

Can be read: “there exists a y such that the property P holds of y”.

The notation:

A + {B}

Means that we must supply an A or a proof of B.

The “In” predicate is an inductive predicate which specifies membership in a list and is provided by the Coq library.

With a little practice reading types, we can see that the two lines are correct. We can then run Coq on this code to check that the proof is correct and we can be assured of the correctness of the implementation WITHOUT EVER READING THE CODE! This means we have reduced the problem of correctness to two simple lines of completely declarative statements.

This function is more complicated than a normal assoc function because it carries all of these proof terms along with it. However, because of the clever people involved in writing Coq we have actually mixed two different type universes together. One (called Set) for values and one (called Prop) for proofs. Coq uses this fact to do something very clever. The {y|P y} existential type and the A+{B} type are designed such that the “P y” proof and the {B} proof disappear when we extract (compile) our code. As an illustration of this amazing fact look at the following ocaml code which has been automatically extracted from our definition:

let rec assoc x = function
  | Nil -> Inright
  | Cons (p, t) ->
      let Pair (x', y) = p in
      (match eq_dec x x' with
         | Left -> Inleft y
         | Right -> assoc x t)

Voila! Not only are the proof terms gone, but this is pretty much the code you probably would have written yourself. Notice that “Inright” is a constructor with no information (other than failure), basically implementing the “Nothing” of a Maybe type, and the “Inleft” as the “Just” of a Maybe type.

We have extracted a provably terminating, well specified, totally correct function that runs in ocaml! And to top it off we can also compile to Haskell!!

assoc :: A -> Assoc_list -> Sumor B
assoc x l =
  case l of
    Nil -> Inright
    Cons p t ->
      (case p of
         Pair x' y ->
           (case eq_dec x x' of
              Left -> Inleft y
              Right -> assoc x t))

So it isn’t as if this isn’t without some cost. There is a huge learning curve on Coq and I’m by no means an expert. The proofs can be arduous, and although this one only took me a couple of minutes, it took far longer than it would have to implement the function directly in ocaml. However as I gain experience with Coq, I increasingly believe the approach will scale better than traditional approaches to software design.

In addition to these caveats there is more than one function that satisfies this type. Namely there is an assoc which starts returning from the end of the list, instead of the beginning. We could get rid of this problem by specifying which of the elements should be returned, returning all of them, or restricting formation of assoc’s such that they are mappings. The latter is very elegant, but somewhat more involved than what we have done.

Functional programming reduces the difficulty of writing correct software by reducing the ways in which bugs can occur. However, most functional programming languages are unable to put complex constraints like “this is a sorted list” or “implements an assoc function” into the language. When other functions rely on the behaviour all kinds of funny things can happen.

I recently had a nasty non-termination bug in SML because a function *required* that the input list be sorted or the function wouldn’t terminate. My sort routine made use of a bogus less-than-equal predicate over a data-type which I had neglected to write properly. This caused hours of very difficult debugging.

I’m acting on the thesis that I can avoid these problems and I am rewriting a large program in Coq as an experiment. I’ll let you know how it goes.

I’ve been playing with the Coq proof assistant over the past few days, following closely on some frustrations that I’ve been having with using SML’s module system and a bit of toying with type-classes in Haskell.

The gist of the problem is this. Although you can define type-classes and modules such that external users of these modules/type-classes see a uniform interface, consistency is left as an exercise for the implementer. This is not really acceptable in my view. When you are writing software, often times *you* are the implementer. What you really want is for these modules not just to provide a consistent interface to outsiders, but to guarantee the correctness of the implementation! Isn’t that the whole point of types? If we can’t do that, why are we using types?

Ok, so in Coq I *can* get the properties I’ve been wanting out of SML’s module system. For instance take the following implementation of the Monad signature:

Module Type MONAD.
Set Implicit Arguments. 

Parameter M : forall (A : Type), Type.
Parameter bind : forall (A B : Type),
  M A -> (A -> M B) -> M B.
Parameter ret : forall (A : Type),
  A -> M A.

Infix ">>=" := bind (at level 20, left associativity) : monad_scope.
Open Scope monad_scope.

Axiom left_unit : forall (A B : Type) (f : A -> M B) (a : A),
  (ret a) >>= f = f a.
Axiom right_unit : forall (A B : Type) (m : M A),
  m >>= (fun a : A => ret a) = m.
Axiom bind_assoc : forall (A B C : Type) (m : M A) (f : A -> M B) (g : B -> M C) (x : B),
  (m >>= f) >>= g = m >>= (fun x => (f x) >>= g).

End MONAD.

This signature describes something much like the monad that is given by the type-class in haskell. I neglected some stuff like implementing join from bind etc, but we can safely ignore that for now. The point is that users of the MONAD signature can’t just fake a monad by supplying an implementation that is nominally the same. i.e. In order to implement this MONAD you actually have to have the right signature for “>>=” *AND* you have to satisfy the monad laws. So what does an implementation look like? Here is an example:

Module ListMonad < : MONAD. 

Require Import List.

Set Implicit Arguments.

Definition M := list.

Fixpoint bind (A : Type) (B : Type) (l : M A) (f : A -> M B) {struct l} : M B :=
  match l with
    | nil => nil
    | h::t => (f h)++(bind t f)
  end.

Infix “>>=” := bind (at level 20, left associativity) : monad_scope.
Open Scope monad_scope.

Definition ret (A : Type) := fun a : A => a::nil.

Lemma left_unit : forall (A B : Type) (f : A -> M B) (a : A),
 (ret a) >>= f = f a.
Proof.
  intros. simpl. rewrite app_nil_end. reflexivity.
Defined. 

Lemma right_unit : forall (A B : Type) (m : M A),
  m >>= (fun a : A => ret a) = m.
Proof.
  simple induction m.
    simpl. reflexivity.
    intros. simpl.
    cut (bind l (fun a0 : A => ret a0) = l).
      intros. rewrite H0. reflexivity.
      exact H.
Defined. 

Lemma bind_assoc : forall (A B C : Type) (m : M A) (f : A -> M B) (g : B -> M C) (x : B),
  (m >>= f) >>= g = m >>= (fun x => (f x) >>= g).
Proof.
  simple induction m.
    intros. simpl. reflexivity.
    intros. simpl.
    cut (l >>= f >>= g = l >>= (fun x0 : A => f x0 >>= g)).
      intros. rewrite < - H0.
      induction (f a).
        simpl. reflexivity.
        simpl. rewrite IHm0. rewrite app_ass. reflexivity.
      apply H. exact x.
Defined.

End ListMonad.

(* Example *)
Import ListMonad.
Require Import Peano.
Require Import List.

Fixpoint downfrom (n : nat) {struct n} : (list nat) :=
  match n with
    | 0 => n::nil
    | S m => n::(downfrom m)
  end.

Eval compute in (1::2::3::4::nil) >>= downfrom.
  = 1 :: 0 :: 2 :: 1 :: 0 :: 3 :: 2 :: 1 :: 0 :: 4 :: 3 :: 2 :: 1 :: 0 :: nil
     : M nat

Ok, That took me about an hour to write. I’m not really that good at using Coq, so presumably you could do this more elegantly and in less time. In any case it would be nice if the proofs could be automated a bit more. That aside this is a *much* better situation than we have in SML and Haskell. We have provided a monad that is guaranteed to actually be one!

I’m of the growing opinion that software that is forced to meet specifications will end up being less trouble in the end than the current state of free-wheeling wild-west style implementation.

Coq gives a civilized alternative to the current free-for-all. Coq can help us make good on the promise that “well typed programs can’t go wrong”.

I just got “ML for the working programmer” in the mail a few days ago,
and worked through it at a breakneck pace since receiving it. It
turns out that a lot of the stuff from the “Total Functional
Programming” book is also in this one. Paulson goes through the use
of structural recursion and extends the results by showing techniques
for proving a large class of programs to be terminating. Work with
co-algebras and bi-simulation didn’t quite make it in, except for a
brief mention about co-variant types leading to the possibility of a
type: D := D → D which is the type of programs in the untyped lambda
calculus, and hence liable to lead one into trouble.

I have to say that having finished the book, this is the single most
interesting programming book that I’ve read since “Paradigms of
Artificial Intelligence Programming” and “Structure and Interpretation
of Computer Programs”. In fact, I would rate this one above the other
two, though I think it might be because of my current interests.

The conjunction of “Total Functional Programming” and “ML for the
Working Programmer” have completely changed my world view of automated
deduction. Interestingly, my advisers work looks even more appealing
from this perspective.

The basic crux of the issue is this. Given that we restrict ourselves
to total functional programming, as described in the aforementioned
article, we will have the ability to prove theorems directly by
manipulation of the program source.

My adviser told me about this the first time that I met him. It at
first seemed like a curiosity. It has however, blossomed into a
completely different way of thinking about how proofs should be
performed.

I’ve spent the last few days converting proofs into program source,
which I then manipulate equationally. I have a few examples of
inductive proofs that can be written quite naturally as structurally
recursive functions.

datatype nat = Z | S of nat

fun lt n m =
    case m of
	Z => false
      | (S m') =>
	case n of
	    Z => true
	  | (S n') => lt n' m'

fun divide n d =
    case n of
	Z => (Z,Z)
      | (S n') =>
	let val (q',r') = divide n' d in
	    case lt (S r') d of
		true => (q',S r')
	      | false => (S q',Z)
	end

Both of these functions are direct “extractions” of a program from a
proofs that I wrote. The second one is a proof with two existential
variables. The two existential variables present themselves as return
values. In the case of “lt” it simply computes a truth value.

The datatype nat = Z | S of nat defines natural numbers with the
correspondence Z=0, S(Z)=1, S(S(Z))=2 etc…

The proof is a proposition that goes like this:

∀n,d:nat. d>0 ∃q,r:nat. n=dq+r ∧ 0≤r<d

You might notice I dropped the restriction on d in the actual source
code. That can be encoded with an auxiliary function in any case.

The proof proceeds inductively, basically you can turn the divide
function inside out to recover the inductive proof.

The base case is n=0 which gives 0=dq+r which gives r=0 and q=0 (since
d is greater than 0). The first case is exemplified in the case
structure above. (Z,Z) = (q,r) the two existential variables are
returned with a valid result for n=0.

The inductive case is this.

n: ∃q,r. n=dq+r ∧ 0≤r<d
(n+1): ∃q’,r’. n+1=dq’+r’ ∧ 0≤r’<d
if(r+1)<d
then n+1=dq+r+1

We see that this gives us that r’=r+1, q’=q giving us a solution to r’ and q’ from r and q (and hence solving the n case, gives us some information about n+1)

else n+1 = d(q+1)+r’

which we can rewrite as: n+1 = d(q+1)+r’ = dq+d+r’=dq+r+1 so d+r’=r+1,
but r+1 must be at least d otherwise we wouldn’t have failed the test
r+1<d, and r’ can be no less than zero, so we must choose r’=0
this results in (finally!)

n+1 = d(q+1) with r’=0,q’=q+1

We have a proof! For the extracted program you can look at the source
and see that in fact it mirrors the proof directly, short of a few
simplifications done by hand. Making use of the r,q in the proof of
r’ and q’ is the same as my use of r and q in my proof of the n+1 case.

I’m looking forward to a deeper understanding of the correspondences
between proofs and programs with my new found capacity to produce actual functional program proofs by extraction manually. I’m also
curious to try my hand at generating termination proofs directly, and
also at making a syntax that more directly supports total functional
programming.

Recently on Lambda the Ultimate there was a post called “Total Functional Programming”. The title didn’t catch my eye particularly, but I tend to read the abstract of any paper that is posted there that doesn’t sound terribly boring. I’ve found that this is a fairly good strategy since I tend to get very few false positives this way and I’m too busy for false negatives.

The paper touches directly and indirectly on subjects I’ve posted about here before. The idea is basically to eschew the current dogma that programming languages should be Turing-complete, and run with the alternative to the end of supplying
“Total Functional Programming”.

At first glance this might seem to be a paper aimed at “hair-shirt” wearing functional programming weenies. My reading was somewhat different.

Most hobbiest mathematicians have some passing familiarity with “Turing’s Halting Problem” and the related “Goedel’s Incompleteness Theorem”. What this paper is trying to do is say “So what!”. The basic idea is that restriction to a syntacticly restricted language can restrict the semantics in such a way that these problems do not apply. The resulting (*huge*) class of programs can then solve almost everything that we think is important.

The “Bazooka” of the paper is a line describing how every program for which we have a complexity upper bound, is in fact inside the restricted language. This realisation is profound. It means that all of the algorithms that anyone has bothered to find upper bounds for (a huge class of programs mind you!) is accessible to this technique.

The paper even deals with infinite data (co-data) in the framework and mechanisms to properly account for streams and other (possibly) infinite structures.

The idea of working in syntactically restricted languages that are sub-Turing (for any number of reasons) is not new. In fact I think I’ve mentioned DataLog (wikipedia entry) previously. In the case of DataLog however, there are serious deficiencies in the expressive power.

Definitely the best paper I’ve read this year. I’m not sure how accessible it is to lay-people, but it should be accessible to anyone with a CS B.S. or people who have some familiarity with functional programming.

I just spent a couple of very cold weeks in Anchorage Alaska. I went out with a bunch of friends to a bar on the night prior to leaving, where I met up with an old friend (and listened to two other friends play music). I started describing the lambda calculus to him since it is related to my work and he was curious what I was up to. In any case I found myself unable to produce addition of the church numerals off the top of my head, which I found pretty disturbing since it shows that I probably didn’t quite understand it in the first place. Therefor I sat down this morning to see if I couldn’t reconstruct it from first principles.

It turns out that it is relatively easy. First, let us start with a bit of notation, and a description of the lambda calculus. Since wikipedia describes the lambda calculus in detail, I’ll just show how one procedes with calculations. As examples let us start with the church numerals.

1 = (λf x.f x)
2 = (λf x.f f x)
3 = (λf x.f f f x)
…
n = (λf x.fⁿ x)

where fⁿ means “f f … f” n times.

The idea is that each numeral is represented by a lambda term. The lambda term describes the arguments (f and x in this case) and the “body” in which we substitute the arguments when the are passed in. An example of an “application” of a lambda term to an argument would be:

1 g = (λf x . f x) g => (λx . g x)

We pass in g, and replace every occurance of f in the body with g, and delete f from the list of variables. we can continue to apply this term to another term

(λx . g x) y => g y

Which leads us to conclude that

1 g y => g y

Ok, so now that you’ve seen a few calculations, let us try to construct addition. The first thing I tried to do is to construct the function +1. Clearly, we want +1 n => (n+1). n+1 is (λf x. f⁽ⁿ⁺¹⁾ x) which is also (λf x. f fⁿ x). Since n = (λf x. fⁿ x) we need to somehow add another f. The trick is to get the arguments to use the same symbols, and to remove the lambda abstraction. We can do this by applying the church numeral to the symbols we want to use, in this case f and x.

(λf x. fⁿ x) f x => fⁿ x

So now we know how to get rid of the lambda abstraction. Now we can add an f on the front, which will get us closer to n+1.

f (λf x. fⁿ x) f x => f fⁿ x

We are almost there. Now we need to have the λf x part at the beggining so that we look like a church numeral.

(λf x. f (λf x. fⁿ x) f x) => (λf x. f fⁿ x) = (λf x. f⁽ⁿ⁺¹⁾ x)

Looking great so far! Now we just need to be able to take the whole (λf x. fⁿ x) form as an argument. This turns out to be really easy, we just put a variable in it’s place and add it to the front of the list of lambda arguments.

(λa f x. f a f x) (λf x . fⁿ x) => (λf x. f (λf x. fⁿ x) f x) => (λf x. f fⁿ x)

This shows that (λa f x. f a f x) is the +1 function!

+1 = (λa f x. f a f x)

Ok, so now that we have +1 can we get a function that takes an n and returns +n? This would get us a long way towards addition. We can call this function n->+n.

Ok, so we can guess what +n will look like easily. It’s going to look just like +1 except with n leading f’s.

+n = (λa f x. fⁿ a f x)

So we should try to figure out how to take the fⁿ out of a church numeral, and place it there. Well, we should apply the same trick of applying the church numeral to f and x so that we can extract the body.

(λf x . fⁿ x) f x => fⁿ x

ok, but really we want a to follow, based on +n, so instead of using x, let us apply the form to a.

(λf x . fⁿ x) f a => fⁿ a

Great! Look how close we are. now we just need to abstract the a,f,x and place f x following it.
(λa f x. (λf x . fⁿ x) f a f x) => (λa f x. fⁿ a f x)

Ok, so now that we know how to take an n, and able to take the entire church numeral n as an a beta reduce it to n+1, let us abstract the n as an argument

(λb a f x. b f a f x)

Let us verify quickly that this function works.

(λb a f x. b f a f x) n =>
(λa f x. n f a f x) =>
(λa f x. (λf x. fⁿ x) f a f x) =>
(λa f x. fⁿ a f x)

Now, we can verify that this works on m, to obtain n+m as the +n function should:

(λa f x. fⁿ a f x) m =>
(λf x. fⁿ (λf x. f^m x) f x) =>
(λf x. fⁿ f^m x) = (λf x. f^(n+m) x) = n+m

Hooray! Not only did we find n->+n, but we have obtained the function “addition” for free. Since, once we have the +n function we can apply it to m. So we now have the function:

add = (λb a f x. b f a f x)

I haven’t tried multiplication and exponentiation yet, but you are welcome to try!

Probably one of the best papers I’ve read on the relationship between CBN, CBV and the Curry-Howard correspondance is the paper Call-by-value is dual to call-by-name by Wadler. The calculus he develops for describing the relationship shows an obvious schematic duality that is very visually appealing.

After reading the paper that I mentioned earlier on Socially Responsive, Environmentally Friendly Logic (which shall henceforth be called SREF Logic), it struck me that it would be interesting to see what a CPS (Continuation-passing Style) like construction looks like in SREF logic, so I went back to the Wadler paper to see if I could figure out how to mimic the technique for multi-player logic. It looks like the formulation by Wadler comes out directly by thinking about logic as a two player game! I’m excited to see what happens with n-player logic.

This has been a big diversion from what I’m actually suppose to be working on but I didn’t want to forget about it :).

Once again the internet comes to the rescue with a Systematic Search for Lambda Expressions. This is the answer to yesterdays question of whether we can iterate over isomorphic proofs exhaustively in order to extract all programs of a specification which in this case is realised as a “type”. Hooray for computer science!

I have a bunch of ideas that I don’t have time to follow up on but I’d like to retain in some fashion so here goes.

1.

A long time ago I had a question about the shortest assembly program (parameterised by the instruction set of course!) that could encode the generators of the quaternions under multiplication. It is a very easy thing to program this, but as I was doing it, I found myself being highly doubtful that I was choosing the “right” encoding. In fact I almost certainly wasn’t. Would there not be some very tight encoding in terms of “normal” assembly language operators? This has been a persistent question in my mind that comes up frequently in different and often more general forms. If you want to make a fast compiler, how do you know what instructions to output? If you have a highlevel specification, how can you avoid the ridiculously high overhead usually associated with extremely high level languages?

Today I found this fabulous paper (from 1987!) that deals with some of the basic issues involved in finding the shortest program for a given problem. It’s called “Superoptimizer — A look at the smallest program”. I’d link to it, but I got it off ACM so if you are one of the few people that has a subscription you’ll know where to find it and otherwise I don’t want to give the money grubbers the link-love.

Some other thoughts that are in this region of my thought-space. Can you take a given specification with a proof and systematically transform it to enumerate the space of satisfying programs? If not, why not? Even if you can’t, might it not be interesting to just perform large searches of this space? Are there methods of using the transformation operators such that programs are only decreasing or minimally increasing? If so then perhaps we can call the search good at some size threshold since very large programs are unlikely to be good because of locality issues. Also Automatic Design of Algorithms Through Evolution is in a very similar vein.

2.

Concurrency is a nasty problem. It doesn’t have a nice formalism that everyone in the CS world can agree on. There must be like 500 different formalisms. All of them better (easier, not neccessarily faster) than the ones that we actually use (locking, threads, condition variables) but none of them stand out as “the right thing”.

I recently found a paper called:
Socially Responsive and Environmentally Friendly Logic. I love the title But aside from that, the formalism is very nice. It is something that I’ve contemplated a bit but never had the drive to actually try to work out formally. The basic idea comes from the knowledge that Classical Logic and Intuisionistic Logic can be viewed as 2 player games. This game is pretty simple. If I have a proof phi then to win I have to prove it. If I have a proof ¬φ, then to win my oponent has to fail to prove it. If I have φ ∧ ψ then my partern gets to pick a formula and I have to prove it. If I have ∀x then my oponent gets to pick any stand-in for x that he would like. You can probably guess the rest (or look it up). This alternate logic breaks the essential two person nature of the logic. One interesting practical feature of negation in the traditional logics, is that they give rise to Continuations in the Curry-Howard Correspondance. So what does this give rise to in the N-player games defined by Abramsky? I’m not sure, but I suspect it might give process migration! Something worth looking in to.