linux-kernel - Re: [PATCH v4 6/8] tracing/histogram: Optimize division by a power of 2

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Message-ID: <CAC_TJveHgsPZw7p7BWOgQw6h8GNU_Pv_WUjNmw3AUq+wnSzk6Q@mail.gmail.com>
Date:   Tue, 26 Oct 2021 16:39:13 -0700
From:   Kalesh Singh <kaleshsingh@...gle.com>
To:     Steven Rostedt <rostedt@...dmis.org>
Cc:     surenb@...gle.com, hridya@...gle.com, namhyung@...nel.org,
        kernel-team@...roid.com, Jonathan Corbet <corbet@....net>,
        Ingo Molnar <mingo@...hat.com>, Shuah Khan <shuah@...nel.org>,
        Masami Hiramatsu <mhiramat@...nel.org>,
        Tom Zanussi <zanussi@...nel.org>, linux-doc@...r.kernel.org,
        linux-kernel@...r.kernel.org, linux-kselftest@...r.kernel.org
Subject: Re: [PATCH v4 6/8] tracing/histogram: Optimize division by a power of 2

On Tue, Oct 26, 2021 at 12:14 PM Steven Rostedt <rostedt@...dmis.org> wrote:
>
> On Mon, 25 Oct 2021 13:08:38 -0700
> Kalesh Singh <kaleshsingh@...gle.com> wrote:
>
> > == Results ==
> >
> > Divisor is a power of 2 (divisor == 32):
> >
> >    test_hist_field_div_not_optimized  | 8,717,091 cpu-cycles
> >    test_hist_field_div_optimized      | 1,643,137 cpu-cycles
> >
> > If the divisor is a power of 2, the optimized version is ~5.3x faster.
> >
> > Divisor is not a power of 2 (divisor == 33):
> >
> >    test_hist_field_div_not_optimized  | 4,444,324 cpu-cycles
> >    test_hist_field_div_optimized      | 5,497,958 cpu-cycles
>
> To optimize this even more, if the divisor is constant, we could make a
> separate function to not do the branch, and just shift or divide.

Ack. I can update to use separate functions for the constant divisors.

>
> And even if it is not a power of 2, for constants, we could implement a
> multiplication and shift, and guarantee an accuracy up to a defined max.
>
>
> If div is a constant, then we can calculate the mult and shift, and max
> dividend. Let's use 20 for shift.
>
>         // This works best for small divisors
>         if (div > max_div) {
>                 // only do a real division
>                 return;
>         }
>         shift = 20;
>         mult = ((1 << shift) + div - 1) / div;
>         delta = mult * div - (1 << shift);
>         if (!delta) {
>                 /* div is a power of 2 */
>                 max = -1;
>                 return;
>         }
>         max = (1 << shift) / delta;

I'm still trying to digest the above algorithm. But doesn't this add 2
extra divisions? What am I missing here?

Thanks,
Kalesh

>
> We would of course need to use 64 bit operations (maybe only do this for 64
> bit machines). And perhaps even use bigger shift values to get a bigger max.
>
> Then we could do:
>
>         if (val1 < max)
>                 return (val1 * mult) >> shift;
>
> -- Steve