Add draft of the first portion of day 8 Coq writeup.
This commit is contained in:
parent
250746e686
commit
60eb50737d
276
content/blog/01_aoc_coq.md
Normal file
276
content/blog/01_aoc_coq.md
Normal file
|
|
@ -0,0 +1,276 @@
|
||||||
|
---
|
||||||
|
title: "Advent of Code in Coq - Day 8"
|
||||||
|
date: 2020-12-31T17:55:25-08:00
|
||||||
|
tags: ["Advent of Code", "Coq"]
|
||||||
|
draft: true
|
||||||
|
---
|
||||||
|
|
||||||
|
Huh? We're on day 8? What happened to days 2 through 7?
|
||||||
|
|
||||||
|
Well, for the most part, I didn't think they were that interesting from the Coq point of view.
|
||||||
|
Day 7 got close, but not close enough to inspire me to create a formalization. Day 8, on the other
|
||||||
|
hand, is
|
||||||
|
{{< sidenote "right" "pl-note" "quite interesting," >}}
|
||||||
|
Especially to someone like me who's interested in programming languages!
|
||||||
|
{{< /sidenote >}} and took quite some time to formalize.
|
||||||
|
|
||||||
|
As before, here's an (abridged) description of the problem:
|
||||||
|
|
||||||
|
> Given a tiny assembly-like language, determine the state of its accumulator
|
||||||
|
> when the same instruction is executed twice.
|
||||||
|
|
||||||
|
Before we start on the Coq formalization, let's talk about an idea from
|
||||||
|
Programming Language Theory (PLT), _big step operational semantics_.
|
||||||
|
|
||||||
|
### Big Step Operational Semantics
|
||||||
|
What we have in Advent of Code's Day 8 is, undeniably, a small programming language.
|
||||||
|
We are tasked with executing this language, or, in PLT lingo, defining its _semantics_.
|
||||||
|
There are many ways of doing this - at university, I've been taught of [denotational](https://en.wikipedia.org/wiki/Denotational_semantics), [axiomatic](https://en.wikipedia.org/wiki/Axiomatic_semantics),
|
||||||
|
and [operational](https://en.wikipedia.org/wiki/Operational_semantics) semantics.
|
||||||
|
I believe that Coq's mechanism of inductive definitions lends itself very well
|
||||||
|
to operational semantics, so we'll take that route. But even "operational semantics"
|
||||||
|
doesn't refer to a concrete technique - we have a choice between small-step (structural) and
|
||||||
|
big-step (natural) operational semantics. The former describe the minimal "steps" a program
|
||||||
|
takes as it's being evaluated, while the latter define the final results of evaluating a program.
|
||||||
|
I decided to go with big-step operational semantics, since they're more intutive (natural!).
|
||||||
|
|
||||||
|
So, how does one go about "[defining] the final results of evaluating a program?" Most commonly,
|
||||||
|
we go about using _inference rules_. Let's talk about those next.
|
||||||
|
|
||||||
|
#### Inference Rules
|
||||||
|
Inference rules are a very general notion. The describe how we can determine (infer) a conclusion
|
||||||
|
from a set of assumption. It helps to look at an example. Here's a silly little inference rule:
|
||||||
|
|
||||||
|
{{< latex >}}
|
||||||
|
\frac
|
||||||
|
{\text{I'm allergic to cats} \quad \text{My friend has a cat}}
|
||||||
|
{\text{I will not visit my friend very much}}
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
It reads, "if I'm allergic to cats, and if my friend has a cat, then I will not visit my friend very much".
|
||||||
|
Here, "I'm allergic to cats" and "my friend has a cat" are _premises_, and "I will not visit my friend very much" is
|
||||||
|
a _conclusion_. An inference rule states that if all its premises are true, then its conclusion must be true.
|
||||||
|
Here's another inference rule, this time with some mathematical notation instead of words:
|
||||||
|
|
||||||
|
{{< latex >}}
|
||||||
|
\frac
|
||||||
|
{n < m}
|
||||||
|
{n + 1 < m + 1}
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
This one reads, "if \\(n\\) is less than \\(m\\), then \\(n+1\\) is less than \\(m+1\\)". We can use inference
|
||||||
|
rules to define various constructs. As an example, let's define what it means for a natural number to be even.
|
||||||
|
It takes two rules:
|
||||||
|
|
||||||
|
{{< latex >}}
|
||||||
|
\frac
|
||||||
|
{}
|
||||||
|
{0 \; \text{is even}}
|
||||||
|
\quad
|
||||||
|
\frac
|
||||||
|
{n \; \text{is even}}
|
||||||
|
{n+2 \; \text{is even}}
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
First of all, zero is even. We take this as fact - there are no premises for the first rule, so they
|
||||||
|
are all trivially true. Next, if a number is even, then adding 2 to that number results in another
|
||||||
|
even number. Using the two of these rules together, we can correctly determine whether any number
|
||||||
|
is or isn't even. We start knowing that 0 is even. Adding 2 we learn that 2 is even, and adding 2
|
||||||
|
again we see that 4 is even, as well. We can continue this to determine that 6, 8, 10, and so on
|
||||||
|
are even too. Never in this process will we visit the numbers 1 or 3 or 5, and that's good - they're not even!
|
||||||
|
|
||||||
|
Let's now extend this notion to programming languages, starting with a simple arithmetic language.
|
||||||
|
This language is made up of natural numbers and the \\(\square\\) operation, which represents the addition
|
||||||
|
of two numbers. Again, we need two rules:
|
||||||
|
|
||||||
|
{{< latex >}}
|
||||||
|
\frac
|
||||||
|
{n \in \mathbb{N}}
|
||||||
|
{n \; \text{evaluates to} \; n}
|
||||||
|
\quad
|
||||||
|
\frac
|
||||||
|
{e_1 \; \text{evaluates to} \; n_1 \quad e_2 \; \text{evaluates to} \; n_2}
|
||||||
|
{e_1 \square e_2 \; \text{evaluates to} \; n_1 + n_2}
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
First, let me explain myself. I used \\(\square\\) to demonstrate two important points. First, languages can be made of
|
||||||
|
any kind of characters we want; it's the rules that we define that give these languages meaning.
|
||||||
|
Second, while \\(\square\\) is the addition operation _in our language_, \\(+\\) is the _mathematical addition operator_.
|
||||||
|
They are not the same - we use the latter to define how the former works.
|
||||||
|
|
||||||
|
Finally, writing "evaluates to" gets quite tedious, especially for complex languages. Instead,
|
||||||
|
PLT people use notation to make their semantics more concise. The symbol \\(\Downarrow\\) is commonly
|
||||||
|
used to mean "evaluates to"; thus, \\(e \Downarrow v\\) reads "the expression \\(e\\) evaluates to the value \\(v\\).
|
||||||
|
Using this notation, our rules start to look like the following:
|
||||||
|
|
||||||
|
{{< latex >}}
|
||||||
|
\frac
|
||||||
|
{n \in \mathbb{N}}
|
||||||
|
{n \Downarrow n}
|
||||||
|
\quad
|
||||||
|
\frac
|
||||||
|
{e_1 \Downarrow n_1 \quad e_2 \Downarrow n_2}
|
||||||
|
{e_1 \square e_2 \Downarrow n_1 + n_2}
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
If nothing else, these are way more compact! Though these may look intimidating at first, it helps to
|
||||||
|
simply read each symbol as its English meaning.
|
||||||
|
|
||||||
|
#### Encoding Inference Rules in Coq
|
||||||
|
Now that we've seen what inference rules are, we can take a look at how they can be represented in Coq.
|
||||||
|
We can use Coq's `Inductive` mechanism to define the rules. Let's start with our "is even" property.
|
||||||
|
|
||||||
|
```Coq
|
||||||
|
Inductive is_even : nat -> Prop :=
|
||||||
|
| zero_even : is_even 0
|
||||||
|
| plustwo_even : is_even n -> is_even (n+2).
|
||||||
|
```
|
||||||
|
|
||||||
|
The first line declares the property `is_even`, which, given a natural number, returns proposition.
|
||||||
|
This means that `is_even` is not a proposition itself, but `is_even 0`, `is_even 1`, and `is_even 2`
|
||||||
|
are all propositions.
|
||||||
|
|
||||||
|
The following two lines each encode one of our aforementioned inference rules. The first rule, `zero_even`,
|
||||||
|
is of type `is_even 0`. The `zero_even` rule doesn't require any arguments, and we can use it to create
|
||||||
|
a proof that 0 is even. On the other hand, the `plustwo_even` rule _does_ require an argument, `is_even n`.
|
||||||
|
To construct a proof that a number `n+2` is even using `plustwo_even`, we need to provide a proof
|
||||||
|
that `n` itself is even. From this definition we can see a general principle: we encode each inference
|
||||||
|
rule as constructor of an inductive Coq type. Each rule encoded in this manner takes as arguments
|
||||||
|
the proofs of its premises, and returns a proof of its conclusion.
|
||||||
|
|
||||||
|
For another example, let's encode our simple addition language. First, we have to define the language
|
||||||
|
itself:
|
||||||
|
|
||||||
|
```Coq
|
||||||
|
Inductive tinylang : Type :=
|
||||||
|
| number (n : nat) : tinylang
|
||||||
|
| box (e1 e2 : tinylang) : tinylang.
|
||||||
|
```
|
||||||
|
|
||||||
|
This defines the two elements of our example language: `number n` corresponds to \\(n\\), and `box e1 e2` corresponds
|
||||||
|
to \\(e_1 \square e_2\\). Finally, we define the inference rules:
|
||||||
|
|
||||||
|
```Coq {linenos=true}
|
||||||
|
Inductive tinylang_sem : tinylang -> nat -> Prop :=
|
||||||
|
| number_sem : forall (n : nat), tinylang_sem (number n) n
|
||||||
|
| box_sem : forall (e1 e2 : tinylang) (n1 n2 : nat),
|
||||||
|
tinylang_sem e1 n1 -> tinylang_sem e2 n2 ->
|
||||||
|
tinylang_sem (box e1 e2) (n1 + n2).
|
||||||
|
```
|
||||||
|
|
||||||
|
When we wrote our rules earlier, by using arbitrary variables like \\(e_1\\) and \\(n_1\\), we implicitly meant
|
||||||
|
that our rules work for _any_ number or expression. When writing Coq we have to make this assumption explicit
|
||||||
|
by using `forall`. For instance, the rule on line 2 reads, "for any number `n`, the expression `n` evaluates to `n`".
|
||||||
|
|
||||||
|
#### Semantics of Our Language
|
||||||
|
|
||||||
|
We've now written some example big-step operational semantics, both "on paper" and in Coq. Now, it's time to take a look at
|
||||||
|
the specific semantics of the language from Day 8! Our language consists of a few parts.
|
||||||
|
|
||||||
|
First, there are three opcodes: \\(\texttt{jmp}\\), \\(\\texttt{nop}\\), and \\(\\texttt{add}\\). Opcodes, combined
|
||||||
|
with an integer, make up an instruction. For example, the instruction \\(\\texttt{add} \\; 3\\) will increase the
|
||||||
|
content of the accumulator by three. Finally, a program consists of a sequence of instructions; They're separated
|
||||||
|
by newlines in the puzzle input, but we'll instead separate them by semicolons. For example, here's a complete program.
|
||||||
|
|
||||||
|
{{< latex >}}
|
||||||
|
\texttt{add} \; 0; \; \texttt{nop} \; 2; \; \texttt{jmp} \; -2
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
Now, let's try evaluating this program. Starting at the beginning and with 0 in the accumulator,
|
||||||
|
it will add 0 to the accumulator (keeping it the same),
|
||||||
|
do nothing, and finally jump back to the beginning. At this point, it will try to run the addition instruction again,
|
||||||
|
which is not allowed; thus, the program will terminate.
|
||||||
|
|
||||||
|
Did you catch that? The semantics of this language will require more information than just our program itself (which we'll denote \\(p\\)).
|
||||||
|
* First, to evaluate the program we will need a program counter, \\(\\textit{c}\\). This program counter
|
||||||
|
will tell us the position of the instruction to be executed next. It can also point past the last instruction,
|
||||||
|
which means our program terminated successfully.
|
||||||
|
* Next, we'll need the accumulator \\(a\\). Addition instructions can change the accumulator, and we will be interested
|
||||||
|
in the number that ends up in the accumulator when our program finishes executing.
|
||||||
|
* Finally, and more subtly, we'll need to keep track of the states we visited. For instance,
|
||||||
|
in the course of evaluating our program above, we encounter the \\((c, a)\\) pair of \\((0, 0)\\) twice: once
|
||||||
|
at the beginning, and once at the end. However, whereas at the beginning we have not yet encountered the addition
|
||||||
|
instruction, at the end we have, so the evaluation behaves differently. To make the proofs work better in Coq,
|
||||||
|
we'll use a set \\(v\\) of
|
||||||
|
{{< sidenote "right" "allowed-note" "allowed (valid) program counters (as opposed to visited program counters)." >}}
|
||||||
|
Whereas the set of "visited" program counters keeps growing as our evaluation continues,
|
||||||
|
the set of "allowed" program counters keeps shrinking. Because the "allowed" set never stops shrinking,
|
||||||
|
assuming we're starting with a finite set, our execution will eventually terminate.
|
||||||
|
{{< /sidenote >}}
|
||||||
|
|
||||||
|
Now we have all the elements of our evaluation. Let's define some notation. A program starts at some state,
|
||||||
|
and terminates in another, possibly different state. In the course of a regular evaluation, the program
|
||||||
|
never changes; only the state does. So I propose this (rather unorthodox) notation:
|
||||||
|
|
||||||
|
{{< latex >}}
|
||||||
|
(c, a, v) \Rightarrow_p (c', a', v')
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
This reads, "after starting at program counter \\(c\\), accumulator \\(a\\), and set of valid addresses \\(v\\),
|
||||||
|
the program \\(p\\) terminates with program counter \\(c'\\), accumulator \\(a'\\), and set of valid addresses \\(v'\\)".
|
||||||
|
Before creating the inference rules for this evaluation relation, let's define the effect of evaluating a single
|
||||||
|
instruction, using notation \\((c, a) \rightarrow_i (c', a')\\). An addition instruction changes the accumulator,
|
||||||
|
and increases the program counter by 1.
|
||||||
|
|
||||||
|
{{< latex >}}
|
||||||
|
\frac{}
|
||||||
|
{(c, a) \rightarrow_{\texttt{add} \; n} (c+1, a+n)}
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
A no-op instruction does even less. All it does is increment the program counter.
|
||||||
|
|
||||||
|
{{< latex >}}
|
||||||
|
\frac{}
|
||||||
|
{(c, a) \rightarrow_{\texttt{nop} \; n} (c+1, a)}
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
Finally, a jump instruction leaves the accumulator intact, but adds a number to the program counter itself!
|
||||||
|
|
||||||
|
{{< latex >}}
|
||||||
|
\frac{}
|
||||||
|
{(c, a) \rightarrow_{\texttt{jmp} \; n} (c+n, a)}
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
None of these rules have any premises, and they really are quite simple. Now, let's define the rules
|
||||||
|
for evaluating a program. First of all, a program starting in a state that is not considered "valid"
|
||||||
|
is done evaluating, and is in a "failed" state.
|
||||||
|
|
||||||
|
{{< latex >}}
|
||||||
|
\frac{c \not \in v \quad c \not= \text{length}(p)}
|
||||||
|
{(c, a, v) \Rightarrow_{p} (c, a, v)}
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
We use \\(\\text{length}(p)\\) to represent the number of instructions in \\(p\\). Note the second premise:
|
||||||
|
even if our program counter \\(c\\) is not included in the valid set, if it's "past the end of the program",
|
||||||
|
the program terminates in an "ok" state. Here's a rule for terminating in the "ok" state:
|
||||||
|
|
||||||
|
{{< latex >}}
|
||||||
|
\frac{c = \text{length}(p)}
|
||||||
|
{(c, a, v) \Rightarrow_{p} (c, a, v)}
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
When our program counter reaches the end of the program, we are also done evaluating it. Even though
|
||||||
|
both rules {{< sidenote "right" "redundant-note" "lead to the same conclusion," >}}
|
||||||
|
In fact, if the end of the program is never included in the valid set, the second rule is completely redundant.
|
||||||
|
{{< /sidenote >}}
|
||||||
|
it helps to distinguish the two possible outcomes. Finally, if neither of the termination conditions are met,
|
||||||
|
our program can take a step, and continue evaluating from there.
|
||||||
|
|
||||||
|
{{< latex >}}
|
||||||
|
\frac{c \in v \quad p[c] = i \quad (c, a) \rightarrow_i (c', a') \quad (c', a', v - \{c\}) \Rightarrow_p (c'', a'', v'')}
|
||||||
|
{(c, a, v) \Rightarrow_{p} (c'', a'', v'')}
|
||||||
|
{{< /latex >}}
|
||||||
|
|
||||||
|
This is quite a rule. A lot of things need to work out for a program to evauate from a state that isn't
|
||||||
|
currently the final state:
|
||||||
|
|
||||||
|
* The current program counter \\(c\\) must be valid. That is, it must be an element of \\(v\\).
|
||||||
|
* This program counter must correspond to an instruction \\(i\\) in \\(p\\), which we write as \\(p[c] = i\\).
|
||||||
|
* This instruction must be executed, changing our program counter from \\(c\\) to \\(c'\\) and our
|
||||||
|
accumulator from \\(a\\) to \\(a'\\). The set of valid instructions will no longer include \\(c\\),
|
||||||
|
and will become \\(v - \\{c\\}\\).
|
||||||
|
* Our program must then finish executing, starting at state
|
||||||
|
\\((c', a', v - \\{c\\})\\), and ending in some (unknown) state \\((c'', a'', v'')\\).
|
||||||
|
|
||||||
|
If all of these conditions are met, our program, starting at \\((c, a, v)\\), will terminate in the state \\((c'', a'', v'')\\). This third rule completes our semantics; a program being executed will keep running instructions using the third rule, until it finally
|
||||||
|
hits an invalid program counter (terminating with the first rule) or gets to the end of the program (terminating with the second rule).
|
||||||
Loading…
Reference in New Issue
Block a user