Re: [RFC PATCH 2/3] clk: sunxi-ng: Implement precalculated NKM rate selection

[Julian Calaby]

Hi Frank,

On Sun, May 28, 2023 at 8:10 PM Frank Oltmanns <frank@xxxxxxxxxxxx> wrote:
>
> Hi Julian,
>
> On 2023-05-28 at 09:19:36 +1000, Julian Calaby <julian.calaby@xxxxxxxxx> wrote:
> > Hi Frank,
> >
> > On Sat, May 27, 2023 at 11:37 PM Frank Oltmanns <frank@xxxxxxxxxxxx> wrote:
> >>
> >> Add a new precalculation method for NKM clock rate selection in the
> >> sunxi-ng clock driver. Introduce ccu_nkm_find_best_precalc which uses a
> >> precalculated table of valid NKM combinations (struct clk_nkm_table and
> >> struct clk_nkm_combo) to find the best rate. This approach provides
> >> faster rate selection by searching a table of valid combinations rather
> >> than calculating for all possible combinations.
> >>
> >> The table of NKM combinations needs to be initialized with meaningful
> >> combinations only, i.e. removing redundant combinations that result in
> >> the same rate.
> >>
> >> Keep the existing ccu_nkm_find_best function in place and use it as a
> >> fallback if no precalculated table is provided.
> >>
> >> Signed-off-by: Frank Oltmanns <frank@xxxxxxxxxxxx>
> >> ---
> >>  drivers/clk/sunxi-ng/ccu_nkm.c | 84 +++++++++++++++++++++++++++-------
> >>  drivers/clk/sunxi-ng/ccu_nkm.h | 26 +++++++++++
> >>  2 files changed, 94 insertions(+), 16 deletions(-)
> >>
> >> diff --git a/drivers/clk/sunxi-ng/ccu_nkm.c b/drivers/clk/sunxi-ng/ccu_nkm.c
> >> index 94d2a83992b2..9652f6df17bd 100644
> >> --- a/drivers/clk/sunxi-ng/ccu_nkm.c
> >> +++ b/drivers/clk/sunxi-ng/ccu_nkm.c
> >> @@ -54,6 +54,49 @@ static unsigned long ccu_nkm_find_best(unsigned long parent, unsigned long rate,
> >>         return best_rate;
> >>  }
> >>
> >> +static unsigned long ccu_nkm_find_best_precalc(unsigned long parent,
> >> +                                              unsigned long rate,
> >> +                                              struct _ccu_nkm *nkm,
> >> +                                              struct clk_nkm_table *table)
> >> +{
> >> +       unsigned long best_rate = 0, best_diff = ULONG_MAX;
> >> +       unsigned long best_n = 0, best_k = 0, best_m = 0;
> >> +       int start = 0, end = table->num - 1, mid;
> >> +
> >> +       while (start <= end) {
> >> +               unsigned long tmp_rate;
> >> +               unsigned long tmp_diff;
> >> +
> >> +               mid = (start + end) / 2;
> >> +
> >> +               tmp_rate = parent * table->combos[mid].n * table->combos[mid].k /
> >> +                          table->combos[mid].m;
> >> +
> >> +               tmp_diff = abs(rate - tmp_rate);
> >> +
> >> +               if (tmp_diff < best_diff) {
> >> +                       best_rate = tmp_rate;
> >> +                       best_diff = tmp_diff;
> >> +                       best_n = table->combos[mid].n;
> >> +                       best_k = table->combos[mid].k;
> >> +                       best_m = table->combos[mid].m;
> >> +                       if (best_diff == 0)
> >> +                               goto out;
> >> +               }
> >
>
> Thank you for your feedback!
>
> In my proposal, the code performs a binary search by
>  1. taking the element in the middle (mid)
>  2. calculating the rate of the element (tmp_rate)
>  3. calculating the difference to the requested rate (tmp_diff)
>  4. if the diff is better than the best_diff making it the new best
>     n-k-m-combo (the if block)

I'm so sorry, I thought that this was still doing a linear search as
it's so close to the original code.

>
> > If the table was sorted by n * k / m, this could just be a process of
>
> Please note, the table already has to be sorted for the function to
> work, as is the nature of a binary search. I should definitely add
> comments. I'm sorry, the code was intended more as a basis to discuss
> the general idea that I described in the cover letter. I should have
> made that clearer.
>
> > searching through until we either:
> > - find that the first rate in the table is too high
>
> I could see that I could add two steps in the beginning, before the loop:
>  - Take the first element and see if its rate is greater than the
>    requested rate, if so immediatly return it
>  - Take the last element and see if its rate is less than the requested
>    rate, if so immediatly return it
>
> Is that what you mean? I'd have to run some simulations to see, if this
> is a real improvement, because we would need two additional rate
> calculations. Worst case would therefore be 2+log(n) calculations
> instead of log(n) and the code would be slightly more complicated in my
> opinion. But if we run this function with all possible parents rate (as
> suggested in the end of my cover letter) these two special cases could
> very well be often applicable. Thanks!
>
> > - find an exact rate
>
> What do you mean by "exact rate"? Do you mean a rate that matches the
> requested rate exactly. This is what the code is already trying to do.
> But, as this is not always possible, in cases where it does not find an
> exact match, it takes the closest match instead.
>
> > - go above the requested rate, then there's only two to compare: our
> > current rate and the previous one
>
> Sorry, you've lost me here. How would I go above the requested rate? You
> would have to do the binary search to find that rate, but then why not
> search the closest rate directly (as the code does) instead of searching
> the closest rate above the requested (as you proposed). I feel like
> either one of us is missing something. :)

What we're missing is that I'm not explaining this well.

Let's take a very simple table: (value = parent * n * k / m)

0. 100
1. 200
2. 300
3. 400

If we search for 50, our closest is the first rate, so index 0: this
is the "find that the first rate in the table is too high" case.

If we search for 300, we'll converge on index 2: this is the "exact
rate" situation.

If we search for 275, then we'll converge on either 200 or 300: this
is the "two to compare" situation: if we converge until we get to the
lowest rate above our target, we only need to check the rate
immediately before it in the table and the one we converged on to find
the closest.

So in pseudo-code, we'd end up with something like this:

--------

start = 0;

cur_rate = parent * table[start].n * table[start].k / table[start].m;

if (cur_rate >= target)
    return table[start];

while (start <= end) {
    mid = (start + end) / 2;

    cur_rate = parent * table[mid].n * table[mid].k / table[mid].m;

    if (cur_rate == target)
        return table[mid];

   if (target < cur_rate)
       end = mid - 1;
   else
       start = mid + 1;
}

prev_rate = parent * table[mid - 1].n * table[mid - 1].k / table[mid - 1].m;

if (abs(target - prev_rate) < abs(target - cur_rate))
    return table[mid - 1];

return table[mid];

--------

Which seems simpler to my eye and moves all the difference
calculations out of the loop so they only have to be done once,
effectively trading a difference calculation on each checked rate for
a rate calculation, and dropping some variables in the process.

Thanks,

-- 
Julian Calaby

Email: julian.calaby@xxxxxxxxx
Profile: http://www.google.com/profiles/julian.calaby/