Dne 2018-01-22 14:29, Jakub Jelinek napsal:
On Mon, Jan 22, 2018 at 01:52:11PM +0100, Jan Hubicka wrote:
In new code bb4 is reached by first_prob + (1-first_prob)*second_prob
which should be equal to prob.
Say your choosen constant is to obtain first_prob=c*prob is c=0.99 and
you
want to know c2 to obtain second_prob=c2*prob
You want the combined probability of branch
(c*prob)+(1-c*prob)*(prob*c2) to
be the same prob
This should give c2=(1-c)/(1-c*prob) so
(c*prob)+(1-c*prob)*prob*(1-c)/(1-c*prob) cancels out to
c*prob+(1-c)*prob
that is
profile_probability cprob =
profile_probability::guessed_always ().apply_scale (1, 100);
first_prob = prob * cprob;
Yes, if we want it to be done with arbitrary cprob, we can do it the
above
way and move the latter into the helper.
prob = cprob.inverse () / (first_prob).inverse ();
But this looks incorrect, because that computes only the above c2 in
prob, not
second_prob. It needs to be
prob = cprob.invert () * prob / first_prob.invert ();
or
prob *= cprob.invert () / first_prob.invert ();
or that
prob = (prob - first_prob) / first_prob.invert ();
I had in the patch. The last one looked to me like cheapest to
compute.
Aha, I see. Ok that makes sense to me now :)
void profile_probability::split (profile_probability cprob,
profile_probability *first_prob, profile_probability *second_prob)
This doesn't take prob as an argument, or should it be a method and
take
prob as *this? Couldn't it simply return first_prob and adjust itself,
so
profile_probability split (profile_probability cprob)
{
profile_probability ret = *this * cprob;
*this = (*this - prob) / ret.invert ();
// or *this = *this * cprob.invert () / ret.invert ();
return ret;
}
?
Yep, that is fine.
I would bet we already have something like that but don't see it.
Other cases seems to choose c as 1/2 that simplifies the formulate
bit it
is also easy to get misunderstand such as in:
profile_probability false_prob = prob.invert ();
profile_probability op0_false_prob =
false_prob.apply_scale (1,
2);
profile_probability op1_false_prob =
false_prob.apply_scale (1,
2)
/ op0_false_prob.invert ();
Or profile_probability op1_false_prob = op0_false_prob /
op0_false_prob.invert ();
I would use the splitting function here so once someone decides to
change the weights the
code will not produce corrupt profile.
Thanks,
Honza
Jakub
