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Danila Fedorin 4d24e7095b Make some more progress on the UCC evaluator article

2021-11-28 01:48:01 -08:00

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Changes Since Last Time

In trying to write and verify this evaluator, I decided to make changes to the way I formalized the UCC. Remember how we used a predicate, IsValue, to tag expressions that were values? It turns out that this is a very cumbersome approach. For one thing, this formalization allows for the case in which the exact same expression is a value for two different reasons. Although IsValue has only one constructor (Val_quote), it's actually {{< sidenote "right" "hott-note" "not provable" >}} Interestingly, it's also not provable that any two proofs of a = b are equal, even though equality only has one constructor, \text{eq_refl}\ :\ a \rightarrow (a = a) . Under the homotopic interpretation of type theory, this corresponds to the fact that two paths from a to b need not be homotopic (continuously deformable) to each other.

As an intuitive example, imagine wrapping a string around a pole. Holding the ends of the string in place, there's nothing you can do to "unwrap" the string. This string is thus not deformable into a string that starts and stops at the same points, but does not go around the pole. {{< /sidenote >}} that any two proofs of IsValue e are equal. I ended up getting into a lot of losing arguments with the Coq runtime about whether or not two stacks are actually the same. Also, some of the semantic rules expected a value on the stack with a particular proof for IsValue, and rejected the exact same stack with a generic value.

Thus, I switched from our old implementation:

To the one I originally had in mind:

I then had the following function to convert a value back into an equivalent expression:

I replaced instances of projT1 with instances of value_to_expr.

Where We Are

At the end of my previous article, we ended up with a Coq encoding of the big-step operational semantics of UCC, as well as some proofs of correctness about the derived forms from the article (like \(\text{quote}_3\) and \(\text{rotate}_3\)). The trouble is, despite having our operational semantics, we can't make our Coq code run anything. This is for several reasons:

Our definitions only let us to confirm that given some initial stack, a program ends up in some other final stack. We even have a little Ltac2 tactic to help us automate this kind of proof. However, in an evaluator, the final stack is not known until the program finishes running. We can't confirm the result of evaluation, we need to find it.
To address the first point, we could try write a function that takes a program and an initial stack, and produces a final stack, as well as a proof that the program would evaluate to this stack under our semantics. However, it's quite easy to write a non-terminating UCC program, whereas functions in Coq have to terminate! We can't write a terminating function to run non-terminating code.

So, is there anything we can do? No, there isn't. Writing an evaluator in Coq is just not possible. The end, thank you for reading.

Just kidding -- there's definitely a way to get our code evaluating, but it will look a little bit strange.

A Step-by-Step Evaluator

The trick is to recognize that program evaluation occurs in steps. There may well be an infinite number of steps, if the program is non-terminating, but there's always a step we can take. That is, unless an invalid instruction is run, like trying to clone from an empty stack, or unless the program finished running. You don't need a non-terminating function to just give you a next step, if one exists. We can write such a function in Coq.

Here's a new data type that encodes the three situations we mentioned in the previous paragraph. Its constructors (one per case) are as follows:

err - there are no possible evaluation steps due to an error.
middle e s - the evaluation took a step; the stack changed to s, and the rest of the program is e.
final s - there are no possible evaluation steps because the evaluation is complete, leaving a final stack s.

We can now write a function that tries to execute a single step given an expression.

Most intrinsics, by themselves, complete after just one step. For instance, a program consisting solely of \(\text{swap}\) will either fail (if the stack doesn't have enough values), or it will swap the top two values and be done. We list only "correct" cases, and resort to a "catch-all" case on line 26 that returns an error. The one multi-step intrinsic is \(\text{apply}\), which can evaluate an arbitrary expression from the stack. In this case, the "one step" consists of popping the quoted value from the stack; the "remaining program" is precisely the expression that was popped.

Quoting an expression also always completes in one step (it simply places the quoted expression on the stack). The really interesting case is composition. Expressions are evaluated left-to-right, so we first determine what kind of step the left expression (e1) can take. We may need more than one step to finish up with e1, so there's a good chance it returns a "rest program" e1' and a stack vs'. If this happens, to complete evaluation of \(e_1\ e_2\), we need to first finish evaluating \(e_1'\), and then evaluate \(e_2\). Thus, the "rest of the program" is \(e_1'\ e_2\), or e_comp e1' e2. On the other hand, if e1 finished evaluating, we still need to evaluate e2, so our "rest of the program" is \(e_2\), or e2. If evaluating e1 led to an error, then so did evaluating e_comp e1 e2, and we return err.

Extracting Code

Just knowing a single step is not enough to run the code. We need something that repeatedly tries to take a step, as long as it's possible. However, this part is once again not possible in Coq, as it brings back the possibility of non-termination. So if we can't use Coq, why don't we use another language? Coq's extraction mechanism allows us to do just that.

I added the following code to the end of my file:

Coq happily produces a new file, UccGen.hs with a lot of code. It's not exactly the most aesthetic; here's a quick peek:

data Intrinsic =
   Swap
 | Clone
 | Drop
 | Quote
 | Compose
 | Apply

data Expr =
   E_int Intrinsic
 | E_quote Expr
 | E_comp Expr Expr

data Value =
   V_quote Expr

-- ... more

All that's left is to make a new file, Ucc.hs. I use a different file so that changes I make aren't overridden by Coq each time I run extraction. In this file, we place the "glue code" that tries running one step after another:

Finally, loading up GHCi using ghci Ucc.hs, I can run the following commands:

ghci> fromList = foldl1 E_comp
ghci> test = eval [] $ fromList [true, false, UccGen.or]
ghci> :f test
test = Just [V_quote (E_comp (E_int Swap) (E_int Drop))]

That is, applying or to true and false results a stack with only true on top. As expected, and proven by our semantics!

Proving Equivalence

Okay, so true false or evaluates to true, much like our semantics claims. However, does our evaluator always match the semantics? So far, we have not claimed or verified that it does. Let's try giving it a shot.

The first thing we can do is show that if we have a proof that e takes initial stack vs to final stack vs', then each eval_step "makes progress" towards vs' (it doesn't simply return vs', since it only takes a single step and doesn't always complete the evaluation). We also want to show that if the semanics dictates e finishes in stack vs', then eval_step will never return an error. Thus, we have two possibilities:

eval_step returns final. In this case, for it to match our semantics, the final stack must be the same as vs'. Here's the relevant section from the Coq file:

{{< codelines "Coq" "dawn/DawnEval.v" 30 30 >}}
eval_step returns middle. In this case, the "rest of the program" needs to evaluate to vs' according to our semantics (otherwise, taking a step has changed the program's final outcome, which should not happen). We need to quantify the new variables (specifically, the "rest of the program", which we'll call ei, and the "stack after one step", vsi), for which we use Coq's exists clause. The whole relevant statement is as follows:

{{< codelines "Coq" "dawn/DawnEval.v" 31 33 >}}

The whole theorem claim is as follows:

I have the Coq proof script for this (in fact, you can click the link at the top of the code block to view the original source file). However, there's something unsatisfying about this statement. In particular, how do we prove that an entire sequence of steps evaluates to something? We'd have to examine the first step, checking if it's a "final" step or a "middle" step; if it's a "middle" step, we'd have to move on to the "rest of the program" and repeat the process. Each time we'd have to "retrieve" ei and vsi from eval_step_correct, and feed it back to eval_step.

I'll do you one better: how do we even say that an expression "evaluates step-by-step to final stack vs'"? For one step, we can say:

eval_step vs e = final vs'

For two steps, we'd have to assert the existence of an intermediate expression (the "rest of the program" after the first step):

exists ei vsi,
    eval_step vs e = middle ei vsi /\ (* First step to intermediate expression. *)
    eval_step vsi ei = final vs' (* Second step to final state. *)

For three steps:

exists ei1 ei2 vsi1 vsi2,
    eval_step vs e = middle ei1 vsi1 /\ (* First step to intermediate expression. *)
    eval_step vsi1 ei1 = middle ei2 vsi2 /\ (* Second intermediate step *)
    eval_step vsi2 ei2 = final vs' (* Second step to final state. *)

This is awful! Not only is this a lot of writing, but it also makes various sequences of steps have a different "shape". This way, we can't make proofs about evalutions of an arbitrary number of steps. Not all is lost, though: if we squint a little at the last example (three steps), a pattern starts to emerge. First, let's re-arrange the exists qualifiers:

exists ei1 vsi1, eval_step vs e = middle ei1 vsi1 /\  (* Cons *)
exists ei2 vsi2, eval_step vsi1 ei1 = middle ei2 vsi2 /\ (* Cons *)
eval_step vsi2 ei2 = final vs' (* Nil *)

If you squint at this, it kind of looks like a list! The "empty" part of a list is the final step, while the "cons" part is a middle step. The analogy holds up for another reason: an "empty" sequence has zero intermediate expressions, while each "cons" introduces a single new intermediate program. Perhaps we can define a new data type that matches this? Indeed we can!

The new data type is parameterized by the initial and final stacks, as well as the expression that starts in the former and ends in the latter. Then, consider the following type:

eval_chain nil (e_comp (e_comp true false) or) (true :: nil)

This is the type of sequences (or chains) of steps corresponding to the evaluation of the program \(\text{true}\ \text{false}\ \text{or}\), starting in an empty stack and evaluating to a stack with only \(\text{true}\) on top. Thus to say that an expression evaluates to some final stack vs', in some unknown number of steps, it's sufficient to write:

eval_chain vs e vs'

Evaluation chains have a couple of interesting properties. First and foremost, they can be "concatenated". This is analagous to the Sem_e_comp rule in our original semantics: if an expression e1 starts in stack vs and finishes in stack vs', and another expression starts in stack vs' and finishes in stack vs'', then we can compose these two expressions, and the result will start in vs and finish in vs''.

The proof is very short. We go {{< sidenote "right" "concat-note" "by induction on the left evaluation" >}} It's not a coincidence that defining something like (++) (list concatenation) in Haskell typically starts by pattern matching on the left list. In fact, proofs by induction actually correspond to recursive functions! It's tough putting code blocks in sidenotes, but if you're playing with the DawnEval.v file, try running Print eval_chain_ind, and note the presence of fix, the fixed point combinator used to implement recursion. {{< /sidenote >}} chain (the one for e1). Coq takes care of most of the rest with auto. The key to this proof is that whatever P is cotained within a "node" in the left chain determines how eval_step (e_comp e1 e2) behaves. Whether e1 evaluates to final or middle, the composition evaluates to middle (a composition of two expressions cannot be done in one step), so we always {{< sidenote "left" "cons-note" "create a new "cons" node." >}} This is unlike list concatenation, since we typically don't create new nodes when appending to an empty list. {{< /sidenote >}} via chain_middle. Then, the two cases are as follows:

If the step was final, then the rest of the steps use only e2, and good news, we already have a chain for that!
Otherwise, the step was middle, an we stll have a chain for some intermediate program ei that starts in some stack vsi and ends in vs'. By induction, we know that this chain can be concatenated with the one for e2, resulting in a chain for e_comp ei e2. All that remains is to attach to this sub-chain the one step from (vs, e1) to (vsi, ei), which is handled by chain_middle.

{{< todo >}}A drawing would be nice here!{{< /todo >}}

The merge operation is reversible; chains can be split into two pieces, one for each composed expression:

While interesting, this particular fact isn't used anywhere else in this post, and I will omit the proof here.

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