Re: GIN improvements part2: fast scan
От | Heikki Linnakangas |
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Тема | Re: GIN improvements part2: fast scan |
Дата | |
Msg-id | 51C170A8.9060509@vmware.com обсуждение исходный текст |
Ответ на | Re: GIN improvements part2: fast scan (Alexander Korotkov <aekorotkov@gmail.com>) |
Ответы |
Re: GIN improvements part2: fast scan
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Список | pgsql-hackers |
On 19.06.2013 11:30, Alexander Korotkov wrote: > On Wed, Jun 19, 2013 at 11:48 AM, Heikki Linnakangas< > hlinnakangas@vmware.com> wrote: > >> On 18.06.2013 23:59, Alexander Korotkov wrote: >> >>> I would like to illustrate that on example. Imagine you have fulltext >>> query >>> "rare_term& frequent_term". Frequent term has large posting tree while >>> >>> rare term has only small posting list containing iptr1, iptr2 and iptr3. >>> At >>> first we get iptr1 from posting list of rare term, then we would like to >>> check whether we have to scan part of frequent term posting tree where >>> iptr >>> < iptr1. So we call pre_consistent([false, true]), because we know that >>> rare term is not present for iptr< iptr2. pre_consistent returns false. >>> So >>> we can start scanning frequent term posting tree from iptr1. Similarly we >>> can skip lags between iptr1 and iptr2, iptr2 and iptr3, from iptr3 to >>> maximum possible pointer. >>> >> >> Thanks, now I understand the rare-term& frequent-term problem. Couldn't >> you do that with the existing consistent function? I don't see why you need >> the new pre-consistent function for this. > > In the case of two entries I can. But in the case of n entries things > becomes more complicated. Imagine you have "term_1& term_2& ...& term_n" > query. When you get some item pointer from term_1 you can skip all the > lesser item pointers from term_2, term_3 ... term_n. But if all you have > for it is consistent function you have to call it with following check > arguments: > 1) [false, false, false, ... , false] > 2) [false, true, false, ... , false] > 3) [false, false, true, ... , false] > 4) [false, true, true, ..., false] > ...... > i.e. you have to call it 2^(n-1) times. But if you know the query specific > (i.e. in opclass) it's typically easy to calculate exactly what we need in > single pass. That's why I introduced pre_consistent. Hmm. So how does that work with the pre-consistent function? Don't you need to call that 2^(n-1)-1 times as well? - Heikki
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