We will use the thread safe version for implementing persistent hash maps.
The hash maps will be used to implement decision procedures and refactor
the congruence closure and ematching modules.
The persistent hash maps based on thread safe parrays are performant
when most of the time there is a single thread updating them.
We use a small hack to make sure we don't have any overhead for
parray<T, false>
i.e., the thread unsafe version used in the VM.
There were two performance bottlenecks in the recursive equation
compiler. Both bottlenecks were due to conversion checking.
1- We allow patterns such as (x+1) in the left-hand-side of a
recursive equation. This is kind of pattern has to be reduced
since it is not a constructor. Moreover, when we are trying to
compile using structural recursion, we need to find an element
that is structurally smaller in recursive applications.
Again, we need to use reduction since the pattern may be (x+2),
and in the recursive application we have (x+1). Now, consider
the following equation
f (x+1) (y+1) := f complex_term y
It will first check whether complex_term is structurally smaller
than (x+1), and the compiler will timeout trying to reduce
complex_term.
This commit adds the following workaround. The structural
recursion module from now on will only unfold reducible constants
and constants marked as patterns. This is not a complete
solution. It will timeout in the following equation:
f (x+1) (y+1) := f (x+1000000000000) y
For this one, we need to add a whnf "fuel" option to type_context
2- Equational lemma generation was producing lemmas that are too
expensive to check. Suppose we the following two definitions
| f x 0 := 1
| f x (y+1) := f complex_term y
and
| g 0 y := 1
| g (x+1) y := g x complex_term
Before this commit, we would generate the following proofs for
the second equation of each definition:
eq.refl (f complex_term y)
eq.refl (g x complex_term)
This proof triggers the following definitionally equality test:
f x (y+1) =?= f complex_term y
g (x+1) y =?= g x complex_term
Since, we have f/g on both sides, the type checker will try
first to unify the arguments, and may timeout trying to solve
x =?= complex_term
y =?= complex_term
since it may take a long time to reduce `complex_term`.
We workaround this problem by creating a slightly different
proof.
eq.refl (unfold_of(f x (y+1)))
eq.refl (unfold_of(g (x+1) y))
where unfold_of(t) is the result of applying one delta reduction
step.
