Re: [RFC PATCH] clk: fractional-divider: Correct max_{m,n} handed over to rational_best_approximation()

[Andy Shevchenko <andriy.shevchenko@linux.intel.com>]

On Wed, Jul 14, 2021 at 01:38:22PM +0300, Andy Shevchenko wrote:
> On Wed, Jul 14, 2021 at 06:10:46PM +0800, Liu Ying wrote:
> > On Wed, 2021-07-14 at 12:12 +0300, Andy Shevchenko wrote:
> > > On Wed, Jul 14, 2021 at 02:41:29PM +0800, Liu Ying wrote:
> 
> ...
> 
> > > > /*
> > > >  * Get rate closer to *parent_rate to guarantee there is no overflow
> > > >  * for m and n. In the result it will be the nearest rate left shifted
> > > >  * by (scale - fd->nwidth) bits.
> > > >  */
> > > 
> > > I don't know how to rephrase above comment better.
> > > 
> > > > scale = fls_long(*parent_rate / rate - 1);
> > > > if (scale > fd->nwidth)
> > > > 	rate <<= scale - fd->nwidth;
> > > 
> > > This takes an advantage of the numbers be in a form of
> > > 
> > > 	n = k * 2^m, (1)
> > > 
> > > where m will be scale in the snippet above. Thus, if n can be represented by
> > > (1), we opportunistically reduce amount of bits needed for it by shifting right
> > > by m bits.

> > > Does it make sense?
> > 
> > Thanks for your explaination.
> > But, sorry, Jacky and I still don't understand this.


It seems I poorly chose the letters for (1). Let's rewrite above as

This takes an advantage of the numbers be in a form of

	a = k * 2^b, (1)

where b will be scale in the snippet above. Thus, if a can be represented by
(1), we opportunistically reduce amount of bits needed for it by shifting right
by b bits.

Also note, "shifting right" here is about the result of n (see below).

> Okay, We have two values in question:
>  r_o (original rate of the parent clock)
>  r_u (the rate user asked for)
> 
> We have a pre-scaler block which asks for
>  m (denominator)
>  n (nominator)
> values to be provided to satisfy the (2)
> 
> 	r_u ~= r_o * m / n, (2)
> 
> where we try our best to make it "=" instead of "~=".
> 
> Now, m and n have the limitation by a range, e.g.
> 
> 	n >= 1, n < N_lim, where N_lim = 2^nlim. (3)
> 
> Hence, from (2) and (3), assuming the worst case m = 1,
> 
> 	ln2(r_o / r_u) <= nlim. (4)
> 
> The above code tries to satisfy (4).
> 
> Have you got it now?
> 
> > > The code looks good to me, btw, although I dunno if you need to call the newly
> > > introduced function before or after the above mentioned snippet.
> > 
> > Assuming that snippet is fully orthogonal to this patch, then it
> > doesn't matter if it's before or after.
> 
> Please, double check this. Because you play with limits, while we expect them
> to satisfy (3).
> 
> -- 
> With Best Regards,
> Andy Shevchenko
> 
> 

-- 
With Best Regards,
Andy Shevchenko