Re: New algorithm for the isomorphism of graphs
- From: i.am.a.whiz@xxxxxxxxxxxxxxxx
- Date: Sat, 28 Jul 2007 00:08:35 -0000
mimouni gave another trivial example:
G=3D(12 ;24 ;34 ;36 ;45 ;56) and G'=3D(ab ;ae ;bc ;cf ;df ;ef)
In G, 1 is the only node of lowest order (1), so we are forced to
pick it first. Then 2 is the only node adjacent to 1, so it must be
picked next. Then 4 is the only node adjacent to what was picked
before, so it must be picked next. Next, 6 is unique as the only
remaining node not adjacent to any already picked, so it must be
picked next. Finally there's a symmetry between 3 and 5, so we can
pick either arbitrarily.
In G', d is the only node of lowest order (1), so we are forced to
pick it first. [Hence 1 must map to d.] Then f is the only node
adjacent to d, so it must be picked next. [Hence 2 must map to f.]
But then there are two nodes (c,e) each adjacent to f, and two
nodes (a,b) neither adjacent to f, so the two graphs aren't
isomorphic, because there's nothing like 4 that it can map to.
Nothing as complicated as your algorithm is needed to decide this example.
Please try to find a pair of graphs whereby my simpleminded
algorithm isn't sufficient (to avoid backtracking) and your more
complicated algorithm works better (avoids backtracking entirely,
or reduces the amount of backtracking compared to my algorithm).
.
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