Note that there are six towns, from which only two roads issue.  Thus 1 must lie between 9 and 12 in the circular route.  Mark these two roads as settled.  Similarly mark 9, 5, 14, and 4, 8, 14, and 10, 6, 15, and 10, 2, 13, and 3, 7, 13.  All these roads must be taken.  Then you will find that he must go from 4 to 15, as 13 is closed, and that he is compelled to take 3, 11, 16, and also 16, 12.  Thus, there is only one route, as follows:  1, 9, 5, 14, 8, 4, 15, 6, 10, 2, 13, 7, 3, 11, 16, 12, 1, or its reverse—­reading the line the other way.  Seven roads are not used.

244.—­THE FIFTEEN TURNINGS.

[Illustration]

It will be seen from the illustration (where the roads not used are omitted) that the traveller can go as far as seventy miles in fifteen turnings.  The turnings are all numbered in the order in which they are taken.  It will be seen that he never visits nineteen of the towns.  He might visit them all in fifteen turnings, never entering any town twice, and end at the black town from which he starts (see “The Rook’s Tour,” No. 320), but such a tour would only take him sixty-four miles.

245.—­THE FLY ON THE OCTAHEDRON.

[Illustration]

Though we cannot really see all the sides of the octahedron at once, we can make a projection of it that suits our purpose just as well.  In the diagram the six points represent the six angles of the octahedron, and four lines proceed from every point under exactly the same conditions as the twelve edges of the solid.  Therefore if we start at the point A and go over all the lines once, we must always end our route at A. And the number of different routes is just 1,488, counting the reverse way of any route as different.  It would take too much space to show how I make the count.  It can be done in about five minutes, but an explanation of the method is difficult.  The reader is therefore asked to accept my answer as correct.

246.—­THE ICOSAHEDRON PUZZLE.

[Illustration]

There are thirty edges, of which eighteen were visible in the original illustration, represented in the following diagram by the hexagon NAESGD.  By this projection of the solid we get an imaginary view of the remaining twelve edges, and are able to see at once their direction and the twelve points at which all the edges meet.  The difference in the length of the lines is of no importance; all we want is to present their direction in a graphic manner.  But in case the novice should be puzzled at only finding nineteen triangles instead of the required twenty, I will point out that the apparently missing triangle is the outline HIK.

In this case there are twelve odd nodes; therefore six distinct and disconnected routes will be needful if we are not to go over any lines twice.  Let us therefore find the greatest distance that we may so travel in one route.