In network therory, how can tou tell if a network can be crossed without retracing paths?

Like a shongo network. i want to see how too make shongo networks and i wondered what network therory says about them and the rules to determine if they're solveable. I know its something like an even or odd number of connections or nodes. Thanks!

In network therory, how can tou tell if a network can be crossed without retracing paths?
Suppose the network is represented by a graph with edges and vertices. Do you mean you want to take a path through the graph that passes through every edge without travelling on an edge more than once ? This is an easy problem in graph theory, and such a path is called an Euler path. An Euler path through a connected graph exists if and only if the graph contains at most two vertices with odd degree (the number of edges the vertex has). This is a trivial theorem.


Suppose a graph is connected, which means there exists some path between any pair of vertices. For each vertex, there is some number of edges connected to it. Suppose the number of edges is even for all the vertices. Then there is no problem drawing an Euler path through these edges. Explain why (Examine the problem at each vertex).


From the above, the only possible problems must come from vertices with odd degree. Without loss of generality, we proceed by adding vertices of odd degree to the nice graph described above. Adding one vertex is no problem, we start at that vertex, travel along the one additional edge and then we're in a nice graph that already contains an Euler path, since the vertex we left at the start now has even degree, for all intents and purposes.


If we add two vertices of odd degree, we are alright as well, with the exception that any path must start at one of these vertices and end at the other. Show this by labelling each odd degree vertex with little "in" and "out" signs on the edges.


It should now be obvious that we can't add any more vertices with odd degree for the simple reason that a path that doesn't double back on itself can only have 2 endpoints. Adding more vertices doesn't change the fact that you put your pencil down at one odd vertex and you're required to lift it back up at one other odd vertex. After that, the game is up.
Reply:Euler worked it out, I think. If you have 2 odd nodes, you can start at one and end at the other, but if you have more than 2, not solveable.
Reply:In geometry, we learned that there must be either 0 or 2 odd nodes (nodes with an odd number of paths intersecting at them). I think if there are 2 odd nodes, you must begin at one of the odd ones, but it doesn't matter if there are no odd nodes.