Six Houses and Three Utilities on A Coffee Mug

Abstract

In a recent Monthly article [1], Hammack and Kainen displayed drawings of the complete bipartite graphs K3,3 and K4,4 on the torus without crossed edges. Notice that if K3,n can be so drawn on the torus, then by the Euler characteristic, 0 = V − E + F = (3 + n) − 3n + F . Each face has at least four edges, and each edge is shared by two faces, so F ≤ E/2 = 3n/2. Therefore, 0 ≤ 3 − 2n + 3n/2, so n ≤ 6. It is known that this necessary condition is also sufficient [1]. The graph K3,3 has been posed as a classical puzzle about connecting three houses with three utilities without crossing utility lines. This is impossible on the sphere (or plane) but possible on a torus, or equivalently, on a coffee mug. In fact, such a puzzle on a coffee mug has been popularized by an entertaining star-studded video [2]. Therefore, it would make a more challenging puzzle to put on a coffee mug six houses a, b, c, d, e, and f , and three utilities u, v, and w! Two solutions are shown below, by gluing the opposite sides of the square or hexagon to make a torus. In the less familiar hexagonal identification [3, pp. 4–5], we glue the top and bottom sides first to get a cylinder, then we glue the two end-circles of the cylinder with a 180◦ twist to obtain a torus.