EULER PATHS & CHINESE POSTMAN SOL: DM.2 CLASSWORK WORKSHEET HOMEWORK (DAY 59) WORKSHEET

EULER PATH A CONNECTED GRAPH IS A PATH THAT TRAVELS THROUGH ALL THE EDGES OF THE GRAPH. THE EDGES CAN ONLY BE TRAVELED ONCE. H?V=YCRUO-U6RT8

EXAMPLE EULER PATH DOES NOT START AND END AT THE SAME VERTEX.

The problem of finding an Eulerian path for a graph is sometimes referred to as the “highway inspector’s problem,” because we try to inspect every road once without unnecessarily traveling any road. One inspector cannot accomplish this task if there are more than two odd vertices; but several inspectors might be able to succeed without duplication any of the work. In fact, having a different inspector for each stretch of the road would certainly succeed; but then more people would be hired than are really needed. Edge-disjoint paths are paths that have no edges in common with each other.

We now turn our attention to the question: “How many inspectors really are needed?” That is, given a connected graph with more than two odd vertices, what is the smallest number of edge-disjoint paths by which the graph can be covered? For example, the graph has four odd vertices (B, C, D and G) so it has no Eulerian path.

1. Find the least number of edge-disjoint paths needed to cover all the edges of the graph(s). 2. Highlight each multigraph that has an Euler circuit.

CHINESE POSTMAN A postman starting at the post office wishes to walk along each street exactly once and return to the post office a the end of the route. What is the least number of edges that will need to be repeated?