[Nauty] lexicografic ordered canonical form
Angela Jo Pignotti
pignotti at email.arc.nasa.gov
Sat Aug 16 07:06:02 EST 2003
Hi Bill,
Now, I'm trying to decipher the message I received from the
author of nauty. It sounds as if I can compare graphs by looking
at the orbits (which node labels change to produce an isomorphism).
I haven't tried to code this algorithm, but have been trying another
approach. I'd like to try this approach with 2 graphs that are
isomorphic to each other and see if the code can identify this.
-Angela
>From: Brendan McKay <bdm at cs.anu.edu.au>
>To: nauty-list at cs.anu.edu.au
>Subject: Re: [Nauty] lexicografic ordered canonical form
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>Date: Thu, 7 Aug 2003 17:48:07 +1000
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>* Joost Winne <Joost.Winne at UGent.be> [030807 16:24]:
>>
>>
>> > for each row i+1 compatible with D[i]
>> > form D[i+1]
>> > apply nauty to find the orbits of points
>> > and a canonical order of the points
>> > reject D[i+1] if point i+1 is not in the same orbit
>> > as the point which is last in the canonical order
>>
>> This would mean (after calling nauty for D[i+1]) compare orbits[i+1] with
>> orbits[lab[i+1]] (with getcanon true) ?
>
>Actually orbits[i] and orbits[lab[i]].
>
>Lets say that I(D) is the set of all designs isomorphic to design D.
>
>The basic explanation: you want to make each isomorphism type I(D[i+1])
>exactly once, by adding a row to some D[i]. Because we are thinking
>about isomorphism types and not labelled designs, the rows are not in a
>special order. So I(D[i+1]) could in fact be made from i+1 different
>smaller designs (though some of them may be the same due to rows of
>D[i+1] being equivalent). The main step in isomorph rejection is to
>choose which I(D[i]) you want to make I(D[i+1]) from and reject it if
>you make it in any other way. The algorithm above says that D[i+1] must
>be made from D[i+1]-row[j], where j is the row that nauty labels last
>(i.e. j = lab[i]. The use of nauty ensures that if you meet another
>design D isomorphic to D[i+1] then the smaller design you want to make
>D from is isomorphic to the smaller design you want to make D[i+1] from.
>
>So when you make D[i+1] you reject it if the row you just added is not
>row j, or, more correctly, if the row you just added is not equivalent
>to row j.
>
>This eliminates the possibility that members of the same I(D[i+1])
>are made (and accepted) from more than one D[i]. This leaves only the
>possibility that several different members of I(D[i+1]) are made and
>accepted from the same D[i]. It is not hard to see that this can only
>happen due to automorphisms of D[i].
>
>Cheers,
>Brendan.
>
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Angela Pignotti
Graduate Student Intern
Mission Critical Technologies
located at NASA-Ames Research Ctr.
Bldg. 258 Rm. 201
ph: 415-604-4313
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