billionaire's gates; make C & E adjacent

[Ghost Post]

As with the solution to the first edition of this problem, we can see that there is no path or asteroid configuration that will allow us to pass through each door only once. Using some graph theory, construct a graph representing this shed in the following way:

Each room is a vertex (label it using the room's name) and let's add a vertex for "outside labeled "O".
Add an edge for each door, the edge connects whichever vertices that the door allow passage between.

Note that this graph is connected and that the vertices O, C, and E are of odd degree.

The solution we are looking for is whether this graph has a Eulerian trail. A Eulerian Trail is a walk on the edges of a graph which uses each edge (door) exactly once. A Connected Graph has an Eulerian trail Iff it has at most two Vertices of Odd Degree.

Therefore there is no Eulerian Trai in the orignal configuration.

Topology changes the graph (basically, in anyway we want). Keep in mind all vetices connect to O because they have exterior doors (that lead outside) but we can now create topolgies that allow us to make rooms E & C adjacent (or any room for that matter), so these now become even vertices.