Let G be a weakly-connected directed graph on n nodes such that:
n is odd.
Every node has exactly 2 arrows going into it.
Every node has exactly 2 arrows going out of it.
Between two nodes there is an arrow in at most one direction.
Exactly 2 nodes, call them A and B, have precisely one self-referencing arrow i.e. one of A's arrows points to A. Same for B. These are the only self-references allowed.
Exactly 1 node, C, has both of its outgoing arrows going to the same node, D. Neither C or D can be A or B.
With the exception of nodes A, B, C, the outgoing arrows of a node go to different nodes.
With the exception of node A, B, D, the incoming arrows of a node comes from 2 different nodes.
Under these conditions, is it true that there is a path between any two nodes of G? If so, how does one prove it? If not, what is the simplest counter-example?
As I have no drawing capability, here is a table of 5 nodes representing a graph which I think meets the above criteria. Note that all nodes have 2 outgoing and incoming nodes. Two nodes A, B use one arrow to point to themselves. One node, C, uses 2 of its arrows to point to the same node D. Neither C or D are A or B. Except for A, B, C the outgoing arrows of a node go to 2 different nodes. Except for A, B, D, the incoming arrows of a node come from 2 different nodes.