Verified dynamic programming with Σ-types in Lean
tannerduve.github.io
tannerduve.github.io
https://lean-lang.org/doc/reference/latest/Basic-Types/Subty...
A crucial difference between type theory (as its known in Lean) and set theory is that an inhabitant/element is of exactly one type.
Object-oriented programming languages are not that different: The types induced by classes can easily be viewed as sets: A child class is a specialized version of its parent's class, hence a subtype/subset thereof if you define all the sets by declaring `instanceof` to be their predicate function.
Technically speaking the elements in the supertype are all distinct from the elements in the subtype and viceversa. They are not a subset of the other, hence why it's improper to consider one a subtype of the other.
Emphasis on "technically". The embedding is trivial. The Lean docs linked by the GP suggest to put those technicalities aside:
> Even though they are pairs syntactically, Subtype should really be thought of as elements of the base type with associated proof obligations.
https://bertrandmeyer.com/2015/07/06/new-paper-theory-of-pro...
1. build a table tab[n]
2. check that for every i, tab[i] == maxDollars_spec[i]
if you take the latter approach i proposed (bottom up) there is nothing to check the totality of.
> For an arbitrary n, compute the table full of values and their proofs, and just pull out the nth proof
if you thought harder about it you'd realize what you're suggesting is impossible
For example, in a theorem prover, you can write an inductive proof that x^2 + x is even for all x. And you can write this via a computation that demonstrates that it's true for zero, and if it's true for x, then it's true for x + 1. However, you don't need to run this computation in order to prove that it's true for large x. That would be computationally intractable, but that's okay. You just have to typecheck to get a proof.
my friend you should either read the article more closely or think harder. he's not proving that the recurrence relation is correct (that would be meaningless - a recurrence relation is just given), he's proving that DP/memoization computes the same values as the recurrence relation.
the obvious indicator is that no property of any numbers is checked here - just that one function agrees with another:
theorem maxDollars_spec_correct : ∀ n, maxDollars n = maxDollars_spec n
Hmmm.
> no property of any numbers is checked here - just that one function agrees with another:
theorem maxDollars_spec_correct : ∀ n, maxDollars n = maxDollars_spec n
> https://en.wikipedia.org/wiki/Richardson%27s_theorem
Given that both `maxDollars n` and `maxDollars_spec n` are defined to be natural numbers, I'm not sure why Richardson's theorem is supposed to be relevant.
But even if it was, the structure of the proof is to produce the fact `maxDollars_spec n = maxDollars n` algebraically from a definition, and then apply the fact that equality is symmetric to conclude that `maxDollars n = maxDollars_spec n`. And once again I'm not sure how you could possibly fail to conclude that two quantities are equal after being given the fact that they're equal.
Did you know that the naturals are a subset of the reals? If Richardson's doesn't convince you there's also
https://en.m.wikipedia.org/wiki/Rice%27s_theorem
> Examples
> Is P equivalent to a given program Q?
Irrespective of where you're convinced it's 100% true that equality of two functions is undecidable in general.
The Lean program in the article, adds `maxDollars_spec n` as a type on `helper`, with strong induction actually proves for all N possible that the implementation of the dynamic program is correct.
You can go further. Write the iterative form of a dynamic program (which uses array to store values, instead of hash, and uses a for loop instead of recursive memoized call) and prove it is computing the recursive maxDollars_spec.
Similar things were done with Z3 prover for other functions. Bit tricks, you want to go from one subset repr to the next. Subset {1, 3} is encoded as 101. Subset {1, 3, 7} as 1010001. You want to go to the next lexicographically greater subset of size 3. You can do that with efficient bit tricks, or you can write a recursive spec. You can use Z3 prover to prove for bitset of size N, that your algorithm that uses efficient tricks is equivalent to the recursive spec.
If Z3 prover actually had to go through all pairs (x,y) to prove that f(x)=y, you'd never get the proof in time.
This is irrelevant.
> You can go further. Write the iterative form of a dynamic program (which uses array to store values, instead of hash, and uses a for loop instead of recursive memoized call) and prove it is computing the recursive maxDollars_spec.
Yes my original comment said exactly this.
The rest is irrelevant.
Reread my original comment again - or any of my follow-up comments - I didn't say the lean code doesn't prove equality, I said it proves it using exhaustion.
What is exhaustion for you? For everyone else in this thread, exhaustion means trying out all values and proving it works.
Exhaustive proof: pick 64-bits for a bit trick, run inefficient algorithm and get 64-bit outputs for all 2^64 bitsets, and then run the bit trick algorithm and exhaustively prove slow(n)==bittrick(n) for all n from 0 to 2^64-1.
Similar way you'd prove four color theorem.
Lean prover, in this case, does no such thing, it uses strong induction to prove correctness. Strong induction does not depend on the size of the input N at all, it depends on the size of the typed problem (which includes type annotations and how they relate).
Of course, proof that needs to deal with semantics of a hashmap is more complex than just dealing with lookup array, but it still proves that exponential recursive calculation can be done with a faster algorithm whose implementation is right there.
You have similar systems, where you can write a recursive quicksort and get average time complexity analysis for free. The system proves the average time complexity as O(n log n) directly from the implementation. (from memory, system would output C_n = 2 * (n + 1) * H_n - 4 * n, were C_n is the average number of comparisons of quicksort, H_n the harmonic number, average is calculated over all possible inputs of array of size N, it does not prove it exhaustively and finds the best approx for comparison counts, it proves it symbolically by computing directly on the representation of the average comparisons.)
My instinct was that you can still call it proof by exhaustion if you divide the values into classes and do the proof for each class.
> Similar way you'd prove four color theorem.
And this seems to support that idea? It's the same thing as proof by cases.
So the proof that n^2 + n is even for all integers is exhaustive: you do one proof for even integers, and another one for odd ones. But we wouldn't generally use the term "exhaustion" there because the vibes are wrong.
Lean is not powerful enough to do this, to somehow conclude that there's a finite set of cases on which you can run brute-force checks and prove a general case.
Well, all I can say here is that my intuition suggested to me that proofs about the complexity of a set are generally not extensible to much-less-complex subsets.
But OK. If you want an objection stated in terms of Richardson's theorem, where are we invoking the sine function?
> Irrespective of where you're convinced it's 100% true that equality of two functions is undecidable in general.
So what? We're not trying to decide the equality of two functions in general. We're deciding the equality of two functions in specific.
> If Richardson's doesn't convince you there's also https://en.m.wikipedia.org/wiki/Rice%27s_theorem
>> Is P equivalent to a given program Q?
Again, why is this supposed to matter?
Here are two programs in C:
int main(int argc, char* argv[]) {
return 0;
}
int main(int argc, char* argv[]) {
return 3 + 1 - 4;
}
Rice's theorem says there is no algorithm which will take two programs as input and return yes if they're equivalent while returning no if they aren't. But no attempt has been made to supply such an algorithm.
It is also true that no numerical reasoning is happening: the memoized version of any recursive relation will return the same result as the original function, assuming the function is not stateful and will return the same outputs given the same inputs.
However, it is not true to say that it does this by exhaustion, since there are infinitely many possible outputs and therefore it cannot be exhaustively checked by direct computation. The “n” for which we are “taking the proof” is symbolic, and hence symbolic justification and abstract invariants are used to provide the proof. It is the symbolic/abstract steps that are verified by the type checker, which involves only finite reasoning.
Of course, the symbolic steps somewhat mirror the concrete computations that would happen if you build the table, especially for a simpler proof like this. But it also shouldn’t be surprising that a program correctness proof would look at the steps that a program takes and reason about them in an abstract way to see that they are correct in all cases.
No, this is what would happen _if you ran the proof_, but proofs are not meant to be ran in the first place! The usual goal is proving their correctness, and for that it's enough for them to _typecheck_.
> For an arbitrary n, compute the table full of values and their proofs, and just pull out the nth proof
if you thought harder about it you'd realize what you're suggesting is impossible
That's what the intermediate function is doing, but _it doesn't need to be executed_ for the final proof to be valid.
> if you thought harder about it you'd realize what you're suggesting is impossible
No, it is perfectly possible. This is not a program running all cases and `assert`ing that they have the expected result, this is a proof encoded as a program that when executed will surely produce a proof that for the given n the theorem holds. But it surely produces a proof when executes you don't need to execute it to know that a proof exists and thus the theorem holds!
This is exactly what mathematicians do when they prove theorems. Do you think they go on and check for every number n that some theorem holds? That's impossible to do for every possible n, and this is exactly what you're claiming you can do in every other language.
And just to be extra clear, a signature like this:
> theorem maxDollars_spec_correct : ∀ n, maxDollars n = maxDollars_spec n
Does not mean that for every n the function returns a boolean indicating whether `maxDollars n` is equal to `maxDollars_spec n` or not. Instead, it _surely_ returns a proof that they are equal. And with _surely_ I mean it cannot return errors or exceptions or whatever. It is guaranteed that the function will terminate with a proof for that proposition, which I want to reiterate again, it's the same as knowing that the proposition is true.
Well, you at least need dependent types just to state the theorem, which eliminates nearly all other languages.
let rec helperMemo : Nat → HashMap Nat Nat → Nat × HashMap Nat Nat
def MemoMap := HashMap Nat Nat
def MemoResult := Nat × MemoMap
let rec helperMemo : Nat → MemoMap → MemoResultTupples don't really indicate what I can expect from the members.
let rec helperMemo (n : Nat) (map : HashMap Nat Nat) : Nat × HashMap Nat Nat
Defn. f: X → Y is flat ⇔ O_{Y,f(x)} → O_{X,x} flat ∀ x.
Induction is often taught as something you do with natural numbers. But it's actually something you do with sets that are defined inductively. Any time you have a set that is defined like so:
1. x is in the set.
2. for all y in the set, f(y) is in the set
If you have multiple elements that you guarantee will be in the set, each of them must be dealt with as a base case, and if you have multiple rules generating new elements from existing ones, each of them must be dealt with as another inductive step.
For the case of the natural numbers, x is 0, and f is the successor function.
If you wanted to apply this model to the Fibonacci sequence, you could say that the tuple (0, 1, 1) is in the set [representing the idea "F_1 is 1"] and provide the generator f((a, b, c)) = (b, a+b, c+1). Then since (0, 1, 1) is in the set, so is (1, 1, 2) [or "F_2 is 1"], and then (1, 2, 3) ["F_3 is 2"], and so on.
(Or you could say that the two tuples (1, 1) and (2, 1) are both in the set, and provide the generator f( (i, x), (i+1, y) ) = (i+2, x+y). Now your elements are simpler, your generator is more complex, and your set has exactly the same structure as before.)
The approach taken by the author's "improved solution" is to define a set consisting of the elements of the memoization table, with the generator being the function chain that adds elements to the table. He annotates the type of the elements to note that they must be correct (this annotation is administrative, just making it easy to describe what it is that we want to prove), and then does a proof over the addition operation that this correctness is preserved (the inductive step!).
https://live.lean-lang.org/#codez=JYWwDg9gTgLgBAZRgEwHQBECGN...
Doing the proof inside the algorithm (i.e. doing inline induction over the algorithm recursion structure) has the advantage that the branching structure doesn't have to be duplicated as in an external proof like the one I did.
In my proof I didn't struggle so much with induction, but much more with basic Lean stuff, such as not getting lost in the amount of variables, dealing with r.fst/r.snd vs r=(fst, snd) and the confusing difference of .get? k and [k]?.