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.
> 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;
}
?
>
> 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
();
Jakub
