The relations $R$ in abstract graphs (with genuinely propertyless vertices) cannot be *defined* because there is nothing the relations can base on: they have to be presupposed.


But consider derived relations $\Phi(x,y)$ between vertices of a graph which can be defined in terms of the base relation $R$. I don't want to fix a language, but as an example I have in mind relations of the form $\phi(d(x),d(y))$, with $d(x)$ the degree of $x$ (with respect to $R$) and $\phi(n,m)$ a relation between natural numbers.


**Definition:** $\Phi$ is a *self-fulfilling property* (SFP) w.r.t. $G$ iff $\Phi(x,y) \equiv Rxy$ for all $x,y \in G$.


The other way round: $G$ is a *self-defining structure* w.r.t. $\Phi$ iff $\Phi(x,y) \equiv Rxy$ for all $x,y \in G$. 


The only SFPs I found so far are $d(x) = d(y)$, which is self-fulfilling exactly w.r.t. complete graphs, and $d(x) \neq d(y)$, which is self-fulfilling exactly w.r.t. empty edgeless graphs. Can anyone come up with more intriguing examples?


What else can be said about SFPs?



> 
> * Can we decide whether there is a graph $G$ for which a given
> $\Phi$ is an SFP? Can we construct
> such a graph?
> * Can we decide whether there is an SFP $\Phi$ for a given graph
> $G$?
> * How can graphs with an SFP be characterized?
> 
> 
> 



Self-defining structures
I've found a necessary and sufficient characterization for
when a relation $\Phi$ is nontrivially self-fulfilling, in
the theorem below.


(As Aaron pointed out, every $\Phi$ is realized trivially
in the graph with no vertices and also in the graph with
one vertex, so by *nontrivially* self-fullfilling, let us
insist that the graph have at least two vertices. Let us
speak of undirected graphs with no loops, so that we adopt
Aaron's correction that a binary relation $\Phi$ on
$\mathbb{N}$ is self-fulfilling in graph $\langle
V,R\rangle$ just in case $x\mathrel{R} y\iff
\Phi(d(x),d(y))$ and $x\neq y$.)


**Theorem.** A symmetric binary
relation $\Phi$ on $\mathbb{N}$ is nontrivially
self-fulfilling in some graph if and only if
$\neg\Phi(0,0)$ or $\Phi$ is not a subrelation of
$\{0\}\times\mathbb{N}\cup\mathbb{N}\times\{0\}$; that
is, if and only if $\neg\Phi(0,0)$ or it has $\Phi(n,k)$
for some $n,k\neq 0$.


Proof. If $\Phi(n,n)$ holds for some $n\neq 0$, then $\Phi$
is self-fulfilling in the complete graph on $n+1$ vertices.
All vertices have degree $n$, and so only $d(x)=n$ arises
in this graph, and so only $\Phi(n,n)$ is relevant when
checking the self-fulfilling property.


Otherwise, if $\Phi(n,k)$ holds for some $n\neq 0\neq k$,
but $\Phi(n,n)$ and $\Phi(k,k)$ both fail, then $\Phi$ is
self-fulfilling in the bipartite graph $B(n,k)$, having $n$
nodes on the left connected to each of $k$ nodes on the
right. In this graph, every vertex has degree either $k$ or $n$, and all such vertices are connected by edges, fulfilling $\Phi$. 


If $\neg\Phi(0,0)$, then $\Phi$ is self-fulling in the
graph on any number of vertices, but with no edges. Again,
every vertex in this graph has $d(x)=0$, and so the only
relevant part of $\Phi$ is $\Phi(0,0)$, which fails, and
none of them are connected, as required.


Otherwise, $\Phi$ only relates numbers to $0$ and
$\Phi(0,0)$ holds. No such $\Phi$ can be self-fullfilling
in a graph with at least two nodes $x\neq y$, since if
there are no edges in the graph, then $d(x)=d(y)=0$, and so
they would have to be connected because of $\Phi(0,0)$,
contradiction. And if there is an edge in the graph between
some $x$ and $y$, then $d(x)$ and $d(y)$ have some value
for which $\Phi(d(x),d(y))$, but by assumption one of those
values must be $0$, contradicting the fact that there is an
edge. QED


Note that the graphs arising in the proof have vertices only of very few degrees, which makes the self-fulfilling property easier.




---


Here is the example I gave earlier, which does have vertices of every non-zero degree. 


The relation $|d(y)-d(x)|=1$ seems to be self-fulfilling in the following graph, where each node is labeled with its degree. One can construct the graph in levels, where at each level, the degree increases by $1$, and the number of nodes on the next level is determined by the self-fulfilling requirement. Each node on each level is connected to all nodes on any prior or next level.



```
 
  .......            etc.

 11 11 11 11 11   (each 11 is connected to five 10s and six 12s)

 10 10 10 10 10   (each 10 is connected to five 9s and five 11s)

 9  9  9  9  9   (each 9 is connected to four 8s and five 10s)

 8  8  8  8   (each 8 is connected to three 7s and five 9s)

  7  7  7     (each 7 is connected to three 6s and four 8s)

  6  6  6     (each 6 is connected to three 5s and three 7s)

   5 5 5      (each 5 is connected to two 4s and three 6s)
   |//\\|
   4   4
    \ /
     3
     |
     2
     |
     1

```

The sizes of the levels grows in the pattern: three of the same size, then one level with one more, then three of the same size one step larger, etc.


It seems that this idea can be generalized to make many more examples where the degrees increase in levels.


