On Wed, Jun 19, 2013 at 12:30 PM, Alexander Korotkov <aekorot...@gmail.com>wrote:
> On Wed, Jun 19, 2013 at 11:48 AM, Heikki Linnakangas < > hlinnakan...@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. > To be precise you don't need the first check argument I listed. So, it's 2^(n-1)-1 calls. ------ With best regards, Alexander Korotkov.
