Re: Statistics and selectivity estimation for ranges

On 01.03.2013 16:22, Alexander Korotkov wrote:
> On Tue, Mar 12, 2013 at 8:03 PM, Heikki Linnakangas<hlinnakangas@vmware.com
>> wrote:
>
>> So, after some hacking, I ended up with this version. I find it more
>> readable, I hope I didn't miss anything. This seems to produce results that
>> are close, but not identical, to the original patch. I'm not sure where the
>> discrepancy is coming from, or which patch is more correct in that respect.
>> I'll continue from this tomorrow, but if you have the time, please take a
>> look and let me know what you think.
>
> I've read your explanation and version of patch. In general it seems
> correct to me.
> There is one point why I have breaked up abstraction in some functions is
> infinities. For example, in calc_length_hist_frac one or both of length1
> and length2 can be infinity. In the line
>> frac = area / (length2 - length1);
> you can get NaN result. I've especially adjusted the code to get more of
> less correct result in this case.

Hmm, good point. I think I managed to fix those cases in the attached
version. Is there any other corner case that I missed?

> Another minor note about this line
>> bin_width *= get_position(typcache, lower,&hist_lower[i],
>>   &hist_lower[i + 1]);
> ITSM it sould looks like
>> bin_width -= 1.0 - get_position(typcache, lower,&hist_lower[i],
>>    &hist_lower[i + 1]);
> Imagine lower and upper bounds fall into same histogram bin. In this case
> we should subtract lengths of both parts which were cut in the left and in
> the right.

Yes, true. There's one negation too many above, though; should be:

bin_width -= get_position(typcache, lower,&hist_lower[i],
    &hist_lower[i + 1]);

Fixed that. Barring any more issues, I'll read through this once more
tomorrow and commit.

- Heikki