Each of the first two figures in the diagram below has the property that if one places a pencil at a certain point, the figure can be drawn over without lifting the pencil from the paper or retracing over any line. (Cross-overs are allowed.) Try it!

We say these figures can be drawn unicursally. None of the next three figures in the diagram has this property.

If you spend a little time trying to draw over these figures, you may discover that there is a very simple condition which needs to be satisfied if a figure is to be drawn unicursally. What is it? Now, can you tell easily whether or not the final complicated figure below can be drawn over unicursally?

Hint 1

Look closely at what happens at each vertex (where the edges meet).

Hint 2

Do you notice anything about the ‘parity’ (evenness and oddness) of the vertices?

Solution

Each of the figures is made up of straight or curved ‘edges’ meeting at ‘vertices’. We notice that there can be an even number of edges meeting at a vertex (call this an even vertex), or an odd number of edges meeting at a vertex (an odd vertex).

Let us first ignore the initial and end points of the curve, as we draw it. When the drawing of the figure is in progress, each time the pencil approaches a vertex along an edge, it must also leave the vertex via another edge. This leaves two less edges through this vertex which have to be traversed. Thus in the process of drawing the figure, any such even vertex remains even and any such odd vertex remains odd. If the vertex is even, we expect finally to get 0 edges left. If the vertex is odd, then finally there will be 1 edge left. In this last case, we will not have drawn over all the edges.

If we begin and end the curve at vertices, then there are two exceptional cases. At the initial vertex, we start out along one edge. After this the above argument continues to hold for this vertex. Thus, an initial vertex may be odd, and similarly a final vertex may be odd, but all other vertices must be even.

So the test is: if there are more than two odd vertices, the figure cannot be drawn unicursally.

We deduce that since the figure of Diagram 3 has more than two odd vertices, it cannot be drawn unicursally.

Extensions

Suppose we insist that the curve be drawn unicursally, and that the end-point should be the same as the starting point. What simple condition must now be placed on the curve? Do any of the given curves satisfy this condition?

Hint 1

Hint 2

Solution

Extensions