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Date: Mon, 8 Aug 2016 22:45:52 +0000 (UTC)
From: Spyros Kapnissis <skapni@yahoo.com.INVALID>
Reply-To: Spyros Kapnissis <skapni@yahoo.com>
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Subject: Re: BooleanQuery rewrite optimization
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Hm, I hadn't really thought about the minShouldMatch part, I thought it' d =
be covered but I see your point being semantically different if you keep it=
 as is.
However.. Running your edge case example on an actual local index I get the=
 following:
"(X X Y #X)" w/minshouldmatch=3D2 vs. (+X X Y) w/minshouldmatch=3D2 =3D> sa=
me top score, less results in second case."(X X Y #X)" w/minshouldmatch=3D2=
 vs. (+X X Y) w/minshouldmatch=3D1 =3D> same top score, same number of resu=
lts"(X X X Y #X)" w/minshouldmatch=3D3 vs. (+X X X Y) w/minshouldmatch=3D2 =
=3D> same top score, same number of results
But still not really convinced myself if decrementing minshouldmatch by 1 w=
ill do the trick.. I'll have to verify - maybe I'll try more examples to se=
e if it holds as a general case.. Nice exercise either way :)

=20

    On Tuesday, August 9, 2016 12:40 AM, Chris Hostetter <hossman_lucene@fu=
cit.org> wrote:
=20

=20
Off the top of my head, i think any optimiation like that would also need=
=20
to account for minNrShouldMatch, wouldn't it?

if your query is "(X Y Z #X)" w/minshouldmatch=3D2, and you rewrite that=20
query to "(+X Y Z)" w/minshouldmatch=3D2 you now have a semantically diff=
=20
query that won't match as many documents as the original.

in that example, you could decrement minshouldmatch (=3D1) ... but i'm not=
=20
sure off that holds as a general rule for all possible permutations/values=
=20
... i'd have to think about it.

An interesting edge case to think about is "(X X Y #X)" w/minshouldmatch=3D=
2=20
... pretty sure that would give you very diff scores if you rewrote it to=
=20
"(+X X Y)" (or "(+X Y)") w/minshouldmatch=3D1


: Hello all, I noticed while debugging a query that BooleanQuery will=20
: rewrite itself to remove FILTER clauses that are also MUST as an=20
: optimization/simplification, which makes total sense. So (+f:x #f:x)=20
: will become (+f:x). However, shouldn't there also be another=20
: optimization to remove FILTER clauses that are also SHOULD, while=20
: converting them to MUST? So, for eg. query (f:x #f:x) will become=20
: (+f:x). I did an initial simple implementation and the tests seem to=20
: pass. Are there any cases where this does not hold?=C2=A0
:=20
:=20

-Hoss
http://www.lucidworks.com/


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