1
$\begingroup$

Let two graphs with vertices $G_1 = \{A, B, C\}$, $G_2 = \{A, D, E\}$ be unioned $\cup$.

Normally a union of these would result in the graph $G_1 \cup G_2 = \{A, B, C, D, E\}$.

However, given that all vertices hold distinct data both vertices identified by $A$ should be in the unioned graph as one of them can simply not be thrown away.

What is the most common way of dealing with this? For instance, are the identifiers changed to e.g. $G_1A$ and $G_2A$? Or is one of them discarded by some default rule? Or do you combine the data in some way?

Keep in mind, that I am looking for the most common way of solving this: What would the common mathematician do?

1 Answers 1

0

Common mathematician would separate the data of vertices from the data about edges. Then vertice $A$ contains data only about this vertice, independent from the graph (eg. name, id, shoe size...) and edges contains all the data that determines the connection in graph.

For example: $$V_1=\{A,B,C\}, \, E_1=\{(AB), (AC), (CA)\}$$ $$V_2=\{A,D,E\}, \, E_2=\{(AD), (DE)\}$$ $$G_1=(V_1,E_1),\, G_2=(V_2,E_2)$$ Union of this two graphs is determined by union of vertices and union of edges: $$G_1\cup G_2 = (V_1 \cup V_2, E_1\cup E_2)=\\ (\{A,B,C,D,E\},\{(AB), (AC), (CA),(AD), (DE)\})$$

On the other hand, if vertices $A$ in $G_1$ and $A$ in $G_2$ are different objects, then they should be marked as a different objects (the best way to do so is naming them differently). For example consider two families with relation of parenting:

Schmidts=({[A]lice, [B]etty, [M]ark, [J]ohn}, {([A][B]),([A][J]),([M][B]),([M][J])})
Kowalscy=({[A]lice, [C]lark, [W]inston}, {([A][W]),([C][W])})

But Alice Schmidt is not the same person as Alice Kowalski, so if we want to consider union of these two graphs, we have to name them dofferently:
Schmidts and Kowalscy =
({[A]lice[S]chmidt, [B]etty, [M]ark, [J]ohn, [A]lice[K]owalski,[C]lark,[W]inston},
{([AS][B]),([AS][J]),([M][B]),([M][J]),([AK][W]),([C][W])})

  • 0
    Thanks for clarifying this. This does, however, not solve the proposed problem. In your example, notice how when you union the two sets of vertices $V_1$ and $V_2$, you discard one of the vertices identified by A. What if these vertices had the variable p which was 2 and 5, respectively? Which one would the common mathematician choose to discard and how did he come to that conclusion? This is what my question's point is. Sorry, if that was unclear. – 2017-02-20
  • 0
    @JensBurla If vertce $A$ in graph $G_1$ had property $p=2$ and $A$ in graph $G_2$ had property $p=5$, then these vertices were completeley different objects and shouldn't be both entitled as $A$. These vertices are then $A_1$ and $A_2$ (or $A$ and $F$, or "Mark" and "Betty" or...) and union of vertices is $V_1\cup V_2=\{A_1, B, C, A_2, D, E\}$ (see, that now these vertices are not skipped ). In the other words - you have to mark, that these two vertices are different objects, for example by entitling them with two different names. – 2017-02-20
  • 0
    Just to be completely sure. If two vertices entitled A are in $G_1$ and $G_2$ respectively, then one can assume that these are the same object, because if they in fact are different objects, then they should be named differently, e.g. $A_1$ and $A_2$? Is that correct? – 2017-02-20
  • 0
    @JensBurla Yes. I've edited my answer. – 2017-02-20
  • 0
    Thank you very much for you help. – 2017-02-21