[llvm-dev] [RFC] Interprocedural MIR-level outlining pass

[Daniel Berlin via llvm-dev]

On Wed, Aug 31, 2016 at 5:36 PM, Hal Finkel <hfinkel at anl.gov> wrote:

>
> ------------------------------
>
> *From: *"Daniel Berlin" <dberlin at dberlin.org>
> *To: *"Hal Finkel" <hfinkel at anl.gov>
> *Cc: *"Andrey Bokhanko" <andreybokhanko at gmail.com>, "llvm-dev" <
> llvm-dev at lists.llvm.org>
> *Sent: *Wednesday, August 31, 2016 7:02:57 PM
> *Subject: *Re: [llvm-dev] [RFC] Interprocedural MIR-level outlining pass
>
> (and in particular, the definition of equivalence used by code folding to
> make the dags is STH like  "two VNDAG expressions are equivalent if their
> operands come from VNDAG expressions with the same opcode")
>
> Thus,
>
> VN2 = VN0 + VN1
> VN3 = VN1 + VN2
>
> is considered equivalent to
>
> VN2 = VN0 + VN5
> VN3 = VN1 + VN2
>
> Despite the fact that this is completely illegal for straight redundancy
> elimination.
>
>
> But again, as I said if y'all want to make a pass that basically generates
> a new type of expression DAG, have fun :)
>
> I don't think that anyone wants to do anything unnecessary, or re-invent
> any wheels. I'm just trying to understand what you're saying.
>
> Regarding you example above, I certainly see how you might do this simply
> by defining an equivalence class over all "external" inputs. I don't think
> this is sufficient, however, for what is needed here. The problem is that
> we need to recognize maximal structurally-similar subgraphs, and I don't
> see how what you're proposing does that (even if you have some scheme where
> you don't hash every part of each instruction). Maybe if you had some
> stream-based similarity-preserving hash function, but that's another matter.
>
>

>


> Now, what I want to recognize is that the latter two instructions in each
> group are structurally similar. Even if I define an equivalence class over
> external inputs, in this case, E = { a, b, c, d, e, f, g, h}, then I have:
>
> q = E + E
> r = E + E
> s = q + r
>
t = q - r
>
> Doing literally nothing but building a standard VN dag (IE with normal
equivalence),
The dag here will contain:

V1 = VN Expression of E + E
V2 = V1
s = VN expression of V1 + V2 (you can substitute V1 if you like or not)
x = VN expression of V1 - V2 (ditto)

and:
>
> u = E + E
> v = E - E
> w = u + v
> x = u - v
>


The dag here would contain
V1 = E + E
V2 = E - E
w = VN expression of V1 + V2
x = VN expression of V1 - V2


So what's the issue you are worried about here, precisely?

If you extend this example to be longer, the answer does not change.

If you make it so one example is longer than the other, you can still
discover this by normal graph isomorphism algorithms.
If you literally only care about the operation being performed, you can
make that work too.



but then r and v are still different, so how to we find that {s, t} can be
> abstracted into a function, because we've found the isomorphism { s -> w, t
> -> x }, and then thus reuse that function to evaluate {w, x}?
>

So let's back up. In your scheme, how do you think you do that now,
precisely?
What is the algorithm you use to discover the property you are trying to
find?
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