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Puzzled: Solutions and Sources

Last month (Feb. 2012) we posted a trio of brainteasers concerning where sets meet, or Venn diagrams. Here, we offer solutions to two of them. How did you do?
  1. 1. Existence
  2. 2. Number of Crossing Points
  3. 3. Winkler's Conjecture
  4. Author

Solution. You were asked to prove that Venn diagrams of all orders exist. It is natural to construct them sequentially, adding one curve at a time, with many good ways to do it; you can even get all the curves to be convex if you prefer. My favorite way to prove existence is by observing that any Venn diagram of order n that is an extension of a Venn diagram of order n−1 can be extended to a Venn diagram of order n+1. Once you start with orders 1 and 2, you can then generate Venn diagrams of all orders.

To see how such an extension is done, let C be the new curve added to get a Venn diagram of order n. C must then cross through each previous face (connected region of the plane minus the curves) exactly once, because the points in that face must be divided into those that lie inside C and those that lie outside C. A new curve D can now be added by following closely along C and crossing C once inside each face of the original Venn diagram of order n–1. D will then split every face of the Venn diagram of order n containing C, ensuring we now have a Venn diagram of order n+1.

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2. Number of Crossing Points

Solution. You were asked to show that every Venn diagram of order n has the same number of crossing points and to compute that number. One way to do this computation is to think of the diagram as a planar graph, with crossing points as its vertices and pieces of curves as its edges. We know the graph has 2n faces (including the outer face) since each face corresponds to a subset of the curves. Moreover, it has exactly twice as many edges as vertices since each edge has two incident vertices, and each vertex (being a simple crossing) has four incident edges. Since the Euler characteristic (see, for example, http://en.wikipedia.org/wiki/Euler_characteristic) of the plane is 2, we have, assuming v is the number of vertices, v – 2v + 2n = 2; hence v = 2n − 2.

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3. Winkler’s Conjecture

Solution. Turns out, it was your correspondent who asked (in 1984) whether every Venn diagram of order n can be extended to a Venn diagram of order n+1. To my considerable surprise, the question remains open to this day. The argument in the solution to Problem 1 shows that the question of whether a Venn diagram can be extended is the same as the question of whether its dual graph is Hamiltonian. (The dual of a planar graph has a vertex for each face, with two vertices connected by an edge if the corresponding faces have a common piece of boundary. A graph is Hamiltonian if it contains a closed path that visits every vertex exactly once.)

If more than two curves are allowed to cross at a point, the answer turns out to be "yes," as was shown by Kiran B. Chilakamarri et al. in "Venn Diagrams and Planar Graphs" in Geometriae Dedicata 62, 1 (1996), 73–91.

For much more on Venn diagrams, including many references, see the Venn diagram survey at http://www.theory.cs.uvic.ca/~cos/venn/VennEJC.html by Frank Ruskey and Mark Weston of the University of Victoria, Victoria, BC, Canada.

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