Don't pick up your pencil!

The question is actually asking: "which of the 4 graphs are traversable?" . To tell whether or not a graph is traversable, we must first count the order of each vertex (the number of lines touching each dot).

From counting the order of each vertex, we can then sort each graph into 3 types: Eurlerian (all vertices are of even order), semi-Eurlerian (two of the vertices are of odd order) and non-Eurlerian (more than two of the vertices are of odd order).

We can see from the graphs above that graph A , B and D have no odd vertices, so they are Eurlerian graphs; however, graph C has more than two odd vertices so it is a non-Eurlerian graph.

From the definitions of each type of graph, a non-Eurlerian graph cannot create a Eulerian trail (a trail which visits every edge exactly once), because each edge adds two to the net order on all vertices (so you could go in, but not out of a vertex, sort of speak), so the answer is:

Graph C.

Although, if graph C had only two odd vertices (a semi-Eurlerian graph), it would be possible for a Eurlerian trail, although the trail would have to start and finish at two distinct vertices.

If you're interested in this area of maths, have a look at graph theory !

The figure in C cannot be traced according to the rules given. Any graph to be traced in this fashion must have at most two vertices connected by an odd number of paths. In figure C, we see that all four vertices are connected by an odd number of paths, so it cannot be traced.

Challenge: Which of the edges in figure C could be removed to make it traceable?

The question and answers are ambiguous. I can trace any of those shapes without picking up a pencil. If I leave a pencil pressed down on the topmost point of every figure, then use a pen to trace each shape, lifting and placing the pen as I go, I have still met the requirements. By this logic, none of the answers are correct.

To be clear, I cannot assume that the pencil is my only writing utensil, since nothing has been explicity stated, save that the pencil cannot be lifted and no line can be retraced.

I arrived at the correct response by practical experimentation and commonsense logic. Had I the Euclidean knowledge so eloquently espoused by Adam Blakey, I would feel more comfortable about my numeracy proficiency. Obviously, every conceivable mathematical dna has some logical and theoretical basis and all phenomena can be explained. I have a long way to go, but I'm willing to learn.

This question is based on Euler 's contradiction of Königsberg Problem. He stated if all points in the diagram have even no. Of lines joining at each point then there is a closed path possible. Even if there are points with odd no. of lines there mist be at mmax only 2 odd numbered line points. In C all 5 points were odd numbered lined. Hence we cannot have a closed path possible

To draw a figure of this type without lifting up your pencil is to find an euler circuit on the graph that these shapes form. If every vertice has an even degree, than the euler circuit exists. Since C has vertices with odd degree it doesn't have an euler circuit and, therefore, cannot be traced without lifting up your pencil.

By the way, C is the graph form of the Bridges of Könisberg problem.

If you take the math out and actually grab a pencil and trace it without lifting your pencil and only tracing over a single edge two times and one time on every other as the problem explains....all four can be traced by those rules. By the math yes C is correct, but I also traced it out and it can be done.