[LLVMdev] MergeFunctions: reduce complexity to O(log(N))

[Duncan P. N. Exon Smith]

On 2014 Jan 22, at 07:35, Stepan Dyatkovskiy <stpworld at narod.ru> wrote:

> Hi Raul and Duncan,
> 
> Duncan,
> Thank you for review. I hope to present fixed patch tomorrow.
> 
> First, I would like to show few performance results:
> 
> command: "time opt -mergefunc <test>"
> 
> File: tramp3d-v4.ll, 12963 functions
>                  Current : Patched
> Time:              9.8 s  : 2.0 secs  ; >400%!!
> Functions merged:  3503   : 3507
> 
> File: spirit.ll, 2885 functions
>                  Current : Patched
> Time:              2.2 s  : 0.5 s
> Functions merged:  1503   : 1505
> 
> File: k.ll, 2788 functions
>                  Current : Patched
> Time:              1.5 s  : 0.7 s
> Functions merged:  1626   : 1626
> 
> File: sqlite3.ll, 1057 functions
>                  Current : Patched
> Time:              0.4 s  : 0.4 s
> Functions merged:  7      : 7
> 
> All the tests were token from test-suite object directory. In average it shows >200% performance improvement. 

These results look promising!  When you say 200% average, do you mean
200% average for all files in test-suite, or just these four?  Are
there any negative results?

> You could also see that patched version detects a bit more functions for merging, since it has improved constant-equivalence detection.

1. I’d prefer to see the improvement to constant-equivalence detection
   as a separate initial patch (to the old code).  Is that possible?
   In particular, it would be reassuring to see your patch *not* change
   the number of functions merged.

2. I’m also still concerned about correctness.  Previously, when the
   comparison result was unknown, returning “false” would conservatively
   consider the functions distinct, so we had no correctness issues.

   However, with your patch, *everything* must be accurately compared,
   right?  Are there any cases where you can’t accurately order
   functions?  (Is FunctionComparator::cmpEnumerate an example?)

   If these are problems, maybe you can use the previous ==-style
   comparison to confirm.

> Raul,
> 
> Yes. you right it is O(N*log(N)). And yes, I have some ideas about hashing.
> Though in the end of hashing improvement I still see the tree. I'll explain.
> 
> Once you calculated hash (even really good one), you still have probability of N>1 functions with the same hash. Of course it could be really low. But then, you have to spend some time just being creating such a complex hash (definitely, you have to scan whole function body).
> 
> I think there should be sequence of short hash numbers. In another words we can try to represent function as a chain of hash codes:
> F0: hashF0-0, hashF0-1, hashF0-2, ...
> F1: hashF1-0, hashF1-1, ...
> 
> In this case you can create kind of hash-tree. Consider we have F0, F1 and F2.
> 
> Imagine, hashF0-0 == hashF1-0 == hashF2-0, hashF0-1 == hashF1-1:
> 
> Then we can build the next tree (requires fixed-width fonts):
> 
>             [hashF0-0]
>              /      \
>     [hashF0-1]      [hashF2-1]
>      /      \             \
> [hashF0-2]  [hashF1-2]   [hashF2-2]
> 
> In this case as you can see, whole process would be of complexity O(N*n), where:
> 
> N - number of functions
> n - function size.
> 
> Note, hashF[i]-[j] could be generated on demand. It is not obvious to generate whole hash at once.
> 
> I have also implemented this approach. Though it is looks more complex, and not so fast, as I hoped. Perhaps I can make it simplier. But it could not be simpler that just to replace "bool" with -1,0,1. Actually I really wanted to propose idea with hash-trees, since it combines advantages of hashing and tree-based lookup routines. But then I discovered idea I presented finally, and saw that it shows almost same improvement and looks much simpler.
> 
> -Stepan

I think you get diminishing returns for each level of the tree.  I’d be
interested in seeing numbers on a simpler 2-stage hash.

- Leave the first hash as it is now in the code.
- Add a secondary hash that’s built from the quantities that the equals
  comparison (or your new less-than comparison) is based on.
- Fall back on O(n^2) for the third stage when there are collisions.