[llvm-dev] Scalar Evolution Expressions Involving Sibling Loops

[Bardia Mahjour via llvm-dev]

> I'm not following your example.  If you have two sibling loops with the
same parent, one will frequently, but not always dominate the other.  Can
you give a specific example of when forming a recurrence between two
siblings (without one dominating the other), is useful?

The situation can happen with guarded loops or with a user guard like
below:

if (c) {
  for (i = 0; i < n; i++)
    ...
}
for (j = 0; j < n; j++)
  ...


The specific example that we ran into is described in
https://reviews.llvm.org/D75628. Basically we have two triangular loops
that are siblings and we'd like to run Banerjee MIV tests on the memory
accesses in those loops. The loop looks like:

void foo(int *restrict A, int n1, int n2, int n3) {
  for (int i1 = 0; i1 < n1; i1++) {
    for (int i2 = 2; i2 < n2; i2++) {
      for (int i3 = i2 + 1; i3 < n3; i3++) {
        A[i2 + i3*n2] = 11;
      }
    }
    for (int i4 = 2; i4 < n3; i4++) {
      for (int i5 = 1; i5 < i4 - 1; i5++) {
        A[i5] = 22;
      }
    }
  }
}

To check the bounds of the dependence function we need to create a symbolic
expression that involves AddRecs for i2 and i4.

Bardia Mahjour





From:	Philip Reames <listmail at philipreames.com>
To:	Bardia Mahjour <bmahjour at ca.ibm.com>, LLVM Development List
            <llvm-dev at lists.llvm.org>
Date:	2020/03/30 02:50 PM
Subject:	[EXTERNAL] Re: [llvm-dev] Scalar Evolution Expressions
            Involving Sibling Loops






On 3/30/20 11:27 AM, Bardia Mahjour via llvm-dev wrote:


      Forwarding to the dev list, in case others ran into similar issues
      and/or have input on this topic.

      Bardia Mahjour

      ----- Forwarded by Bardia Mahjour/Toronto/IBM on 2020/03/30 02:25 PM
      -----

      From: Bardia Mahjour/Toronto/IBM
      To: listmail at philipreames.com
      Cc: "Michael Kruse" <llvm at meinersbur.de>
      Date: 2020/03/26 11:47 AM
      Subject: Scalar Evolution Expressions Involving Sibling Loops




      Hi Philip,

      I hope you are doing well.

      We've recently run into an issue with SCEV in the context of
      dependence analysis, and would like your opinion on it. Background
      discussion can be found here
      https://reviews.llvm.org/D75628#inline-689527.

      Basically in this case, the dependence equation requires us to
      symbolically create an expression involving two or more recurrences
      that recur with non-dominating loops (sibling loops).
I'm not following your example.  If you have two sibling loops with the
same parent, one will frequently, but not always dominate the other.  Can
you give a specific example of when forming a recurrence between two
siblings (without one dominating the other), is useful?
      Currently creating such a SCEV expression trips asserts in `
      CompareSCEVComplexity()` and `isKnownViaInduction()` saying that a
      given SCEV expression cannot be composed of recurrences that have no
      dominance relationship between them.

      Michael tried explaining to me why there is this restriction about
      dominance, and I'm beginning to understand why such restriction may
      be necessary when evaluating or expanding SCEV expressions in outer
      scopes (eg. `getSCEVAtScope(nullptr)`) but I still don't understand
      why this restriction is imposed at construction. Shouldn't this
      restriction be asserted on when calling getSCEVAtScope instead of
      when creating AddRecs, given that simplification steps may remove
      identical terms involving non-dominating loops?
Well, SCEV construction is generally done to parallel IR.  SSA requires
dominance, so having the SCEV operands require dominance would seem like a
reasonable thing.  If you want to change this, you'll need to motivate the
change.  (i.e. see above question)

      We would appreciate any other insight you might have about this
      issue.

      Regards,

      Bardia Mahjour
      Compiler Optimizations
      IBM Toronto Software Lab
      bmahjour at ca.ibm.com


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