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Re: RDF-Entailment: Remove duplicate anonymous resources - looking for an algorithm

From: Joshua Tauberer <tauberer@for.net>
Date: Tue, 22 Nov 2005 08:49:01 -0500
Message-ID: <438321CD.5010302@for.net>
To: Reto Bachmann-Gmür <reto@gmuer.ch>
CC: Andreas Andreakis <andreas.andreakis@gmx.de>, semantic-web at W3C <semantic-web@w3c.org>

Reto Bachmann-Gmür wrote:
> How do I check the 2^N subgraphs when making b lean?

So just to go back a step (to the beginning, really)-  When checking if, 
for g=a+b, a entails b, there must be a mapping M from blank nodes only 
in b (not in a) to nodes such that M 'applied' to b is a subset of a. 
Let's call those blank nodes variables.  Further, no statement in b can 
have no variables in it.  (If there was such a statement s in b, M(b) 
would still contain s.  a and b don't overlap, so s couldn't be in a, so 
M(b) couldn't be a subset of a.)

 From this, we can characterize every b by its variables.  If n is a 
variable, then every statement mentioning n is in b.  And b contains 
only those statements in g that mention a variable.  Rather than 
choosing b, we can choose a set of variables and from that generate b.

So rather than considering all 2^N subgraphs of g where N is the number 
of statements (each statement can be in or out of the subgraph), one can 
just consider the 2^N possible choices of variables where N is the 
number of blank nodes (each blank node can be in or out of the choice of 
variables).  Then for each of the 2^N permutations of blank nodes, 
choose the subgraph b that has just the statements that mention a variable.

I think that answers your question, but I'll continue rambling a proof 
about how MSGs get involved:

We can make a further restriction on b that it can't be split into two 
parts b1 and b2 where each variable is in either b1 or b2 but not both. 
  (This is like a MSG where you can't cut off a piece that contains all 
of the instances of some variables, without taking the whole MSG.)  If b 
can be split that way, then a entails b just when a entails b1 and b2 
separately, so we don't need to check b if we check b1 and b2.

This gives us the result that if b is a MSG (in which case every blank 
node in b is a variable and does not appear outside of b), then no 
superset of b needs to be checked.  So the largest chunks that need to 
be checked are MSGs.

We still (so far) need to look at subgraphs of MSGs, which means taking 
an MSG and looking at its 2^N subgraphs where N is the number of blank 
nodes within that MSG.  But we still only want to look at 'unsplittable' 
subgraphs.  If node x is connected to node z only through node y, and if 
we take the variables to be x and z but not y, then b can be split into 
a part containing x and a part containing z, so this was a bad choice of 
variables.  There might be a way to quickly find the unsplittable parts 
of a MSG using a connectivity map or something... but that needs more 
thinking.

-- 
- Joshua Tauberer

http://taubz.for.net

** Nothing Unreal Exists **
Received on Tuesday, 22 November 2005 13:51:06 UTC

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