==> geometry/K3,3.s <== The problem you describe is to draw a bipartite graph of 3 nodes connected in all ways to 3 nodes, all embedded in the plane. The graph is called K3,3. A famous theorem of Kuratowsky says that all graphs can be embedded in the plane, EXCEPT those containing a subgraph that is topologically equivalent to K3,3 or K5 (the complete graph on 5 vertices, i.e., the graph with 5 nodes and 10 edges). So your problem is a minimal example of a graph that cannot be embedded in the plane. The proofs that K5 and K3,3 are non-planar are really quite easy, and only depend on Euler's Theorem that F-E+V=2 for a planar graph. For K3,3 V is 6 and E is 9, so F would have to be 5. But each face has at least 4 edges, so E >= (F*4)/2 = 10, contradiction. For K5 V is 5 and E is 10, so F = 7. In this case each face has at least 3 edges, so E >= (F*3)/2 = 10.5, contradiction. The difficult part of Kuratowsky is the proof in the other direction! A quick, informal proof by contradiction without assuming Euler's Theorem: Using a map in which the houses are 1, 2, and 3 and the utilities are A, B, and C, there must be continuous lines that connect the buildings and divide the area into three sections bounded by the loops A-1-B-2-A, A-1-B-3-A, and A-2-B-3-A. (One of the areas is the infinite plane *around* whichever loop is the outer edge of the network.) C must be in one of these three areas; whichever area it is in, either 1, or 2, or 3, is *not* part of the loop that rings its area and hence is inaccessible to C. The usual quibble is to solve the puzzle by running one of the pipes underneath one of houses on its way to another house; the puzzle's instructions forbid crossing other *pipes*, but not crossing other *houses*.