use set equality when checking for expression equality of commutative terms (and potentially generally use sets for commutative terms)

[Krastanov]

we already have

julia> Set([X1, X2]) == Set([X1, X2])
true

julia> Set([X1, X2]) == Set([X1, Z1])
false

Set uses hashing for quick (potentially false positive) equality check in inclusion (log(n)), followed by isequal check to avoid false positive. Thus set equalities have cost of n log(n) instead of n^2 without the use of a smart datastructure.

Initially there was a worry that Set uses ==, which would have broken its use with Symbolics (where == is used to create a symbolic equation), but that does not seem to be the case.

To avoid creating a new Set at every equality check, we should probably have it as part of the structure. E.g.

@withmetadata struct SAdd{T<:QObj} <: Symbolic{T}
    dict
    _set_precomputed
    _arguments_precomputed
end

It will take a bit more memory (not much given that the objects are not copied, only referenced), but it should speed up a lot of other code.

pinging @apkille as we had been discussing this in #40

[apkille]

Could you show me your source that explains how Set checks equality? I'm getting the following result:

julia> Set([dagger(X1)]) == Set([dagger(X1)])
false

julia> dagger(X1) == dagger(X1)
false

julia> isequal(dagger(X1), dagger(X1))
true

julia> X1 == X1
true

[apkille]

Ah, the hash changes each time dagger is called:

julia> hash(X1)
0x00671145608bd477

julia> hash(X1)
0x00671145608bd477

julia> hash(dagger(X1))
0x44e2b6a69704df53

julia> hash(dagger(X1))
0xb846fec1d33c803b

[Krastanov]

I think a hash is not defined for these objects so it defaults to memory id as hash. Defining a hash function should take care of this (and they have a pretty simple standard definition in this case along the lines of hash(arguments(expr)) (but it should be the two-argument hash)).