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August 10th, 2017, 03:39 AM  #1 
Newbie Joined: Aug 2017 From: Zimbabwe Posts: 7 Thanks: 0  Thread: Graph Grammar Function For Two Exotic Graphs Question please.
I will start my question by describing a particular graph. 1. There are a finite, fixed, number of nodes. labeled {A,B....Z} (so for this example there are 24 nodes). 2. Each node consists of another graph (tree) with exactly 3 nodes in 3 depths. That means we have a graph where the nodes themselves are graphs. 3. We will take the labels for each of these new lower graphs from the set (a,b,....z), where we can use any small letter from the alphabet any number of times. 4. Since there are 72 lower letter nodes in total derived from the set of 24 letters of the alphabet, the nodes (A,B....Z) share nodes. E.g. A={a,d,u} B={b,u,p} C={c,b,a}; the nodes "A" and "B" share the lower level node labelled "u". In fact, a crucial point is that this is the exact same node. Additionally the rule for populating the nodes {A,B,....Z} with smaller nodes is, take the exact same letter as the Original node and label the root of the smaller graph with it in small letters, so for "A" it's "a", then choose any two other letters at random from the set {a,b....z}. E.g. B={b,g,t} Once we have done this for all {A,B...Z}, we save this configuration and use it from now on. So the topology of this graph takes on a new form. We could not simply have taken the smaller nodes and connected them and had a normal graph, because in some sense, the nodes in the small node normal graph are grouped in three's, and this grouping has an order . And the real interest in all of this is that we want to connect the larger nodes,( A,B...Z) with rules derived from considering the underlying graph with the smaller nodes. note: We have introduced a random element in that we arranged the paths between "a,b...z" randomly (while populating the graph), but we are going to start with a problem framed in terms of nodes (A,B...Z), and generate a solution by iteratively adjusting the (a...z). path arrangement till it is no longer random. It is clear that the nodes (A,B..Z) are connected. Also the paths between any two Nodes "A" to "N" becomes highly abstract. If we take the path in a line from say "A" to "N", the path that the underlying small letter nodes (a,....x) , will have will not itself be a straight line. It will consist of a 3dimensional lattice that is unique to the exact path taken from "A" to "N". Another point; If we select a path with the nodes (A,B...Z) , and the representation of the underlying graph (a,b...z) was initialized somehow, then each path A"" to "N" will have associated with it a graph with nodes (a to z). Also, if we take any sub graph from the graph (A,B...Z) the underlying graph will be highly complex and highly fractal, even if its initialization was random. Now for the problem I've been leading to: if we select a node "A" and form the root of the overlying graph, {A,B,...Z} and then take another instance of this overlying graph and raise it to a select B as its root, what is the most optimal (simplest) function we can have that can transform the graph with A as its root into the one with B as its root? Because they contain exactly the same material, there will be many solutions, so we stick with the simplest. Another question would be how could we adjust the arrangement we chose at random for the underlying layer of small letter nodes to make this function simpler. Last edited by skipjack; August 10th, 2017 at 04:15 AM. 
August 10th, 2017, 05:15 AM  #2 
Newbie Joined: Aug 2017 From: Zimbabwe Posts: 7 Thanks: 0 
If we have 24 nodes with 24 nodes instead of 3 nodes and we initialise it with an order (simple way of permuting the letters), i.e. we take one extreme... A={a,b.......z} B={b,c......z,a} ...... Z={z,a.....y} then the underlying layer is a perfectly homogeneous evenly spaced out symmetric lattice.(spiral?) The path between nodes A and N will also be rotation of that lattice in some sense, I'm just thinking as I type... I am not that knowledgeable in this topic, but perhaps someone here could describe this rotation somehow in terms of a function of the underlying latice...if a rotation is the right word. It also means that the nodes A to Z would exist in the same "plane" here in the rulebased initialised version, while in the random initialization example in the OP every node exists in some sense in its own plane and these planes intersect. What type of mathematics would there be to deal with a geometry where the dimensions don't go up in integer amounts (or at right angles)? When taking A as root and then operating on that instance with a function that converts it to its form with B as root, we would need to move all the nodes A...Z through this geometry with skewed dimensions. That's my limit in thinking about this. Last edited by skipjack; August 11th, 2017 at 12:35 AM. 
August 10th, 2017, 11:21 PM  #3 
Newbie Joined: Aug 2017 From: Zimbabwe Posts: 7 Thanks: 0 
These nodes A to Z do not exist in different dimensions, They form the vertices of some geometric object. Each of these represent a point of starting from when navigating the lattice with smaller (a...z) (26 Letter) nodes. The shape of this geometric object depends on our choice for how we arrange the smaller letter nodes. If we simply shift each letter one position left in each A,B...Z, and use all 26 letters , just to take it to one extreme (so its simpler and easier to understand) e.g. A={a,b,c....z} B={b,c,d....z,a) ..... Z={z,a....x,y} Then the shape of this geometric object becomes a sphere, meaning the question of moving from one node (i.e. A as the root) to any other (e.g. B as root) becomes a smooth function. A randomly innitialised underlying layer would give us a path that has descrete edges, or a discontinuos function from A to B. Last edited by moyo; August 10th, 2017 at 11:25 PM. 
August 11th, 2017, 12:13 AM  #4 
Newbie Joined: Aug 2017 From: Zimbabwe Posts: 7 Thanks: 0 
Here is one application of this. The axiom of identity (A=A & A!=!A) contains parts. That is the source of the whole problem in Münchhausen's trilemma. Because having parts imply that each of those parts , i.e. each A is different in at least one respect. That respect is evident in that relations, are *ordered* pairs. So the relation A=A places those two A's in some relation with each other, i.e. in the ordered pair (A,A) ...they have unique positions. It doesn't matter that equivalence relations are symmetric, because symmetry is also expressed using ordered pairs. If you still have doubts consider that anything without parts cannot contain any information. If the axiom of identity *does* contain the information (parts) then those parts are not equivalent, otherwise they would be the same part and the axiom would contain no information. ==== The mathematical solution above is the equivalent of expressing the axiom in a way where it both has parts and has no parts. A is then related to A because it is related to B and C......Z i.e. A is there because what's not A is there more importantly A is made of the same material as B. (so this system expresses "has no parts" But the function from moving from A to B represents the separatedness of A from B (so the system expresses "has parts" because A=A now has no parts but still contains information. Last edited by skipjack; August 11th, 2017 at 12:30 AM. 

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