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From:   Kalesh Singh <kaleshsingh@google.com>
Date:   Tue, 26 Oct 2021 18:09:22 -0700
Message-ID: <CAC_TJve-mKSojaXtukdFeQKvPz-8TQtS=pgGD0Z18Wt6yJi7dg@mail.gmail.com>
Subject: Re: [PATCH v4 6/8] tracing/histogram: Optimize division by a power of 2
To:     Steven Rostedt <rostedt@goodmis.org>
Cc:     surenb@google.com, hridya@google.com, namhyung@kernel.org,
        kernel-team@android.com, Jonathan Corbet <corbet@lwn.net>,
        Ingo Molnar <mingo@redhat.com>, Shuah Khan <shuah@kernel.org>,
        Masami Hiramatsu <mhiramat@kernel.org>,
        Tom Zanussi <zanussi@kernel.org>, linux-doc@vger.kernel.org,
        linux-kernel@vger.kernel.org, linux-kselftest@vger.kernel.org
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On Tue, Oct 26, 2021 at 5:18 PM Steven Rostedt <rostedt@goodmis.org> wrote:
>
> On Tue, 26 Oct 2021 16:39:13 -0700
> Kalesh Singh <kaleshsingh@google.com> wrote:
>
> > >         // 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.
>
> mult = (2^20 + div - 1) / div;
>
> The "div - 1" is to round up.
>
> Basically, it's doing:  X / div  = X * (2^20 / div) / 2^20
>
> If div is constant, the 2^20 / div is constant, and the "2^20" is the
> same as a shift.
>
> So multiplier is 2^20 / div, and the shift is 20.
>
> But because there's rounding errors it is only accurate up to the
> difference of:
>
>   delta = mult * div / 2^20
>
> That is if mult is a power of two, then there would be no rounding
> errors, and the delta is zero, making the max infinite:
>
>   max = 2^20 / delta as delta goes to zero.
>
> > But doesn't this add 2 extra divisions? What am I missing here?
>
> The above is only done at parsing not during the trace, where we care
> about.

Hi Steve,

Thanks for the explanation, this cleared it up for me.

- 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;
>
> This is done at the time of recording.
>
> Actually, it would be:
>
>         if (val1 < max)
>                 return (val1 * mult) >> shift;
>         else
>                 return val1 / div;
>
> -- Steve