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strokes, forming the star, will bring you back to A; then one stroke round the circle to B, one stroke to C, one round the circle to D, and one final stroke to E—twelve in all. Of course, in practice the second circular stroke will be over the first one; it is separated in the diagram, and the points of the star not joined to the circle, to make the solution clear to the eye.

242.—THE TUBE INSPECTOR'S PUZZLE.

The inspector need only travel nineteen miles if he starts at B and takes the following route: BADGDEFIFCBEHKLIHGJK. Thus the only portions of line travelled over twice are the two sections D to G and F to I. Of course, the route may be varied, but it cannot be shortened.

243.—VISITING THE TOWNS.

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.

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.

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 difierent. 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.

There are thirty edges, of which eighteen were visible in the original illustration, represented