Generalizations upright=0.5|K3,3 drawn with only one crossing.  Two important characterizations of planar graphs, Kuratowski's theorem that the planar graphs are exactly the graphs that contain neither K3,3 nor the complete graph K5 as a subdivision, and Wagner's theorem that the planar graphs are exactly the graphs that contain neither K3,3 nor K5 as a minor, make use of and generalize the non-planarity of K3,3. Solution to the three utilities problem on a torus. K3,3 is a toroidal graph, which means it can be embedded without crossings on a torus. In terms of the three cottage problem this means the problem can be solved by punching two holes through the plane (or the sphere) and connecting them with a tube. This changes the topological properties of the surface and using the tube allows the three cottages to be connected without crossing lines. An equivalent statement is that the graph genus of the utility graph is one, and therefore it cannot be embedded in a surface of genus less than one. A surface of genus one is equivalent to a torus. A toroidal embedding of K3,3 may be obtained by replacing the crossing by a tube, as described above, in which the two holes where the tube connects to the plane are placed along one of the crossing edges on either side of the crossing. Another way of changing the rules of the puzzle is to allow utility lines to pass through the cottages or utilities; this extra freedom allows the puzzle to be solved.
Generalizations upright=0.5|K3,3 drawn with only one crossing. Solution to the three utilities problem on a torus. Two important characterizations of planar graphs, Kuratowski's theorem that the planar graphs are exactly the graphs that contain neither K3,3 nor the complete graph K5 as a subdivision, and Wagner's theorem that the planar graphs are exactly the graphs that contain neither K3,3 nor K5 as a minor, make use of and generalize the non-planarity of K3,3. K3,3 is a toroidal graph, which means it can be embedded without crossings on a torus. In terms of the three cottage problem this means the problem can be solved by punching two holes through the plane (or the sphere) and connecting them with a tube. This changes the topological properties of the surface and using the tube allows the three cottages to be connected without crossing lines. An equivalent statement is that the graph genus of the utility graph is one, and therefore it cannot be embedded in a surface of genus less than one. A surface of genus one is equivalent to a torus. A toroidal embedding of K3,3 may be obtained by replacing the crossing by a tube, as described above, in which the two holes where the tube connects to the plane are placed along one of the crossing edges on either side of the crossing. Another way of changing the rules of the puzzle is to allow utility lines to pass through the cottages or utilities; this extra freedom allows the puzzle to be solved.
A
Solution to the three utilities problem on a torus.
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Solution to the three utilities problem on a torus.