What is the smallest digraph whose reflexive, symmetric, transitive closures (in all combinations) are distinct?

Question

For any given directed graph, we may consider the various closures of it with respect to reflexivity, symmetry, and transitivity, in any combination, like this:

For the particular graph shown above, this process results in eight distinct graphs, including the original graph. This graph is not the smallest instance with this feature, however, since if we delete the source point at right, we will still have eight distinct graphs, like this:

Question. What is the smallest directed graph such that these various closures are all distinct and distinct from the original?

The second example gets it down to five vertices and four edges.

The question arose in a reply of Bryan Bischof to my recent tweet https://twitter.com/JDHamkins/status/1318447368732397569. The first image is drawn from the chapter on Functions and Relations in my book, Proof and the Art of the Mathematics, available from MIT Press: https://mitpress.mit.edu/books/proof-and-art-mathematics.

Answer 1

The $4$-vertex digraph

a ---> b ---> c    d

is the smallest example possible.

To have the reflexive symmetric transitive closure be different from the symmetric transitive closure, we need an isolated vertex. (If a vertex $v$ has an edge to or from it, then in the symmetric transitive closure, we get the edge $v \to v$.) That isolated vertex will make all the reflexive closures different from the non-reflexive ones, but can't help us with anything else.

For the digraph a ---> b ---> c we can check that symmetric, transitive, and symmetric transitive closures are all different. If we want to beat this, we need the same thing to happen on a $2$-vertex digraph.

If the $2$-vertex digraph has edges $a \to b$ and $b \to a$, then its symmetric closure will not change anything. However, if the $2$-vertex digraph does not have both of those edges, then its transitive closure will not change anything. So either way, we need $3$ vertices.

Answer 2

The graph whose incidence matrix is

0   0   1
0   0   0
0   1   0

has all eight closures distinct. (Or my code has a bug...)

[And given Misha's answer, my code probably has a bug...]

The bug is obvious: the symmetric-transitive closure that Misha and OP were thinking about isn't just the symmetric closure of the transitive closure. You have to apply the two closures alternately until the graph stops changing. D'oh!

But if you interpret "symmetric transitive closure" as the "symmetric closure of the transitive closure" (and similarly for the other terms) then in fact all eight of the named closures are distinct for this graph --- they simply don't account for all possible "closures" (because swapping orders, or repeating things, like STST..., may lead to new ones).