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

Hi Frank,

On Sun, May 28, 2023 at 8:10 PM Frank Oltmanns [off-list ref] wrote:

Hi Julian,

On 2023-05-28 at 09:19:36 +1000, Julian Calaby [off-list ref] wrote:

quoted

Hi Frank,

On Sat, May 27, 2023 at 11:37 PM Frank Oltmanns [off-list ref] wrote:

quoted

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 <redacted>
---
 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.

quoted

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.

quoted

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!

quoted

- 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.

quoted

- 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@gmail.com
Profile: http://www.google.com/profiles/julian.calaby/

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