On Thu, 6 Feb 2025, Michael Niedermayer wrote:
On Thu, Feb 06, 2025 at 02:38:48PM +0200, Martin Storsjö wrote:
On Thu, 6 Feb 2025, Michael Niedermayer wrote:
+ // If the timer resolution is high, and we get the same timer
+ // value multiple times, use variances in the number of repeats
+ // of each timer value as entropy. If the number of repeats
changed,
+ // proceed to the next index.
Does it still work if you check against the last 2 ?
or does this become too slow ?
What iam thinking of is this
7,8,7,8,8,7,8,7,8,8,7,8,7,8,8,7,8,7,8,8,... and a 9 or 6 or further distant
would trigger it
I assume both the CPU clock and the wall time are quite precisse so if we
just compare them the entropy could be low even with 2 alternating values
Yes, that still works for making it terminate in a reasonable amount of
time. I updated the patch to keep track of 3 numbers of repeats, and we
consider that we got valid entropy once the new number of repeats is
different from the last two.
So in the sequence above, e.g. for 7,8,7,8,8,7, at the point of the last
one, we have old repeats 8 and 8, and the new repeat count 7, which in that
context looks unique.
I was thinking that in 7,8,8 that 7 and 8 be the 2 least recent used
values not 8,8
Sure, that's probably doable too.
that is, something like:
if (old2 == new) {
FFSWAP(old,old2);
I don't see why we'd need to check this if clause at all, it seems to me
that it's enough to have the "if (old != new)" case. If we have old2 ==
new, we'd just end up with old2 = old, and old = (previous old2 value)
anyway.
} else if (old != new) {
old2 = old;
old = new;
}
but again, iam not sure this will work or just need too much time to gather
enough entropy
It still executes in reasonable amount of time; my patch now looks like
this:
if (t == last_t) {
repeats[0]++;
} else {
// If we got a new unique number of repeats, update the history.
// (We don't need to check repeats[2]; if it is equal to the new
// value we'll end up keeping the same two values as before, in
// opposite order.
if (repeats[0] != repeats[1]) {
repeats[2] = repeats[1];
repeats[1] = repeats[0];
}
repeats[0] = 0;
}
// Martin
_______________________________________________
ffmpeg-devel mailing list
ffmpeg-devel@ffmpeg.org
https://ffmpeg.org/mailman/listinfo/ffmpeg-devel
To unsubscribe, visit link above, or email
ffmpeg-devel-requ...@ffmpeg.org with subject "unsubscribe".
