# Question about a quoted expression: {x,y} matches {x,x} ?

Posted 10 days ago
107 Views
|
0 Replies
|
0 Total Likes
|
 Hello everyone,I find that I have gotten myself stuck on what I think was a Stephen Wolfram quote, the specifics of which escape me, but which can be roughly paraphrased as: "..and remember, {x,y} in the rule will match an {x,x} in the hypergraph.." I can't locate exactly where I might have read that (I've looked}, but I seem to remember that it made little sense to me at the time, and makes even less sense to me now.If an {x,y} in the rule will match an {x,x} in the hypergraph, then surely an {x,y,z} in the rule should match an {x,x,x} in the hypergraph. And by extension, also a {y,y,y}. And even a {z,z,y}.I mean, doesn't that imply, if not explicitly state, that any string of vertices in the rule will match any string of vertices in the hypergraph? As in, WTF? (excuse my French)So at the time, I dismissed it as a "Wolfram compatibility" issue, and continued investigating Wolfram Physics Model rules, hypergraphs, and multi-way graphs. But the problem cropped up again.Specifically, there is a rule that Stephen Wolfram (et al) mentions in Section 4.1 "Recognizable Geometry", of his white paper entitled, "A Class of Models with the Potential to Represent Fundamental Physics".As rules go, it's a beaut. When requisitely iterated, it is claimed to resemble a spatial "orbifold" - just what the doctor ordered for someone wanting to build a spatial Universe.Only thing is, as far as I can tell, the rule doesn't seem to work. It dead ends before the second iteration.The rule is: {{x,x,y},{x,z,u}}->{{u,u,v},{v,u,y},{z,y,v}}If we start with the left-hand side as the Initial Condition, as is the custom, then the first iteration will produce the right-hand side (of course) - but only the right-hand side. So far, so good.But then, where is the {{x,x,y},{x,z,u}} in {{u,u,v},{v,u,y},{z,y,v}} ??There isn't one. It dead-ends.If {{u,u,v},{v,u,y},{z,y,v}} is broken out into its component edges, then we can see that:{u,u} is matched by the {x,x} edge in the rule, and{u,v} is matched by the {x,y} edge in the rule, and{v,u} is not matched, and{u,y} is matched by the {x,z} edge in the rule, and{z,y} is not matched, and{y,v} is NOT matched by the {z,u} edge in the rule. - because the v is NOT unique.There doesn't appear to be any other possible permutations.So I guess my question is: If it is indeed true that the Wolfram Physics Model considers any {x,x} in the hypergraph to be "matched" by any {x,y} in the rule, then doesn't that essentially negate using unique symbols for vertices at all?If not, then what exactly are the limitations of this particular.. methodology?Any response would be appreciated.BTW, just so you know, I don't currently own a copy of Mathematica. All my current investigations are done using C++ and T-SQL.Thanks in advance.