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

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Date:   Tue, 26 Oct 2021 20:18:46 -0400
From:   Steven Rostedt <rostedt@...dmis.org>
To:     Kalesh Singh <kaleshsingh@...gle.com>
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, 26 Oct 2021 16:39:13 -0700
Kalesh Singh <kaleshsingh@...gle.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.

> > 
> >
> > 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