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This page gives a few examples of Venn diagrams for 4 sets. Some examples:
alt text alt text
Thinking about it for a little, it is impossible to partition the plane into the $16$ segments required for a complete $4$-set Venn diagram using only circles as we could do for $<4$ sets. Yet it is doable with ellipses or rectangles, so we don't require non-convex shapes as Edwards uses.

So what properties of a shape determine its suitability for $n$-set Venn diagrams? Specifically, why are circles not good enough for the case $n=4$?

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    $\begingroup$ Related: 6 sets is possible with triangles, but not 7. Also, a paper I haven't finished reading, but looks like it could be summarized to at least partially answer this question. $\endgroup$ – Larry Wang Aug 3 '10 at 22:11
  • $\begingroup$ Is it actually doable with squares, or only with non-square rectangles? $\endgroup$ – Isaac Aug 3 '10 at 22:13
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    $\begingroup$ I wondered because the rectangles and ellipses are essentially the same diagram and perhaps the symmetry of squares would get in the way like the symmetry of circles does. $\endgroup$ – Isaac Aug 3 '10 at 22:47
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    $\begingroup$ @Isaac: It is possible to create a Venn diagram with four squares. I sketched a quick example: a.imageshack.us/img230/3009/venn4squares.jpg $\endgroup$ – e.James Aug 4 '10 at 10:24
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    $\begingroup$ You will probably find interesting the following online paper: "A survey of Venn diagrams", by Frank Ruskey and Mark Weston, it is one of the dynamic surveys of the Electronic Journal of Combinatorics, available at: combinatorics.org/Surveys/ds5/VennEJC.html $\endgroup$ – Andrés E. Caicedo May 10 '11 at 2:59
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The short answer, from a paper by Frank Ruskey, Carla D. Savage, and Stan Wagon is as follows:

... it is impossible to draw a Venn diagram with circles that will represent all the possible intersections of four (or more) sets. This is a simple consequence of the fact that circles can finitely intersect in at most two points and Euler’s relation F − E + V = 2 for the number of faces, edges, and vertices in a plane graph.

The same paper goes on in quite some detail about the process of creating Venn diagrams for higher values of n, especially for simple diagrams with rotational symmetry.

For a simple summary, the best answer I could find was on WikiAnswers:

Two circles intersect in at most two points, and each intersection creates one new region. (Going clockwise around the circle, the curve from each intersection to the next divides an existing region into two.)

Since the fourth circle intersects the first three in at most 6 places, it creates at most 6 new regions; that's 14 total, but you need 2^4 = 16 regions to represent all possible relationships between four sets.

But you can create a Venn diagram for four sets with four ellipses, because two ellipses can intersect in more than two points.

Both of these sources indicate that the critical property of a shape that would make it suitable or unsuitable for higher-order Venn diagrams is the number of possible intersections (and therefore, sub-regions) that can be made using two of the same shape.

To illustrate further, consider some of the complex shapes used for n=5, n=7 and n=11 (from Wolfram Mathworld):

Venn diagrams for n=5, 7 and 11

The structure of these shapes is chosen such that they can intersect with each-other in as many different ways as required to produce the number of unique regions required for a given n.

See also: Are Venn Diagrams Limited to Three or Fewer Sets?

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To our surprise, we found that the standard proof that a rotationally symmetric $n$-Venn diagram is impossible when $n$ is not prime is incorrect. So Peter Webb and I found and published a correct proof that addresses the error. The details are all discussed at the paper

Stan Wagon and Peter Webb, Venn symmetry and prime numbers: A seductive proof revisited, American Mathematical Monthly, August 2008, pp 645-648.

We discovered all this after the long paper with Savage et al. cited in another answer.

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  • $\begingroup$ I hope you don't mind that I included a direct link to your paper... $\endgroup$ – J. M. is a poor mathematician Feb 9 '12 at 0:13
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    $\begingroup$ Given that it is an MAA journal (nonprofit organization) I don't mind and I doubt anyone would. Still they do own the copyright, so this it not a trivial question. But I won't worry about it. $\endgroup$ – stan wagon Feb 9 '12 at 2:37
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Theorem:

An n-set Venn diagram cannot be created with circles for $n \geq 4$.

Proof: We represent a Venn diagram $D$ composed of circles as a graph by placing vertices at each intersection, like this:

venndiagram

$\small\text{(an example for a 3-set Venn diagram)}$

To prove the theorem, we will show that a 4-set (or greater) Venn diagram graph is not planar and thus cannot be drawn in two dimensions.

Each circle overlaps every other one, and it is clear that distinct overlapping circles create two intersections. Thus there are at most two vertices for each pair of circles, and we have $v \leq 2{n \choose 2}$ vertices for an $n$-set Venn diagram created with circles.

Because vertices are placed at intersections of two or more distinct circles, each vertex must have an even degree $d_i \geq 4$. By the degree sum formula, we have $e=\frac{d_1+\cdots+d_v}{2}\geq 2v=4{n\choose 2}$ edges. Because $D$ represents all possible intersections of $n$ sets, we have a face for each set alone, each pair of sets, each triplet of sets, and so on. If we count the outer face of the graph as ${n\choose 0}$, an application of the binomial theorem yields $$f= {n\choose 0} + {n\choose 1} + {n\choose 2} + \cdots + {n\choose n} = 2^n \text{ faces.}$$

We now substitute each of these results into Euler's formula: $$2{n\choose2}-4{n\choose 2} + 2^n= 2^n- 2{n\choose 2}\leq2,$$ which results in a clear contradiction for $n = 4$. To reach one for $n>4$, notice that if there existed such a Venn diagram for some $n > 4$ we could remove circles from the diagram to obtain one for $n=4$.


The following is a good resource for exploring the combinatorics of Venn Diagrams more deeply: http://www.combinatorics.org/files/Surveys/ds5/VennEJC.html

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  • $\begingroup$ You could correct your style of the indirect proof and check your estimation of $e$ for correctness. The estimation is a bit confused. You use $n$ for two distinct things. Also you should replace "one or more closed curves" by "two or more circles." $\endgroup$ – Steve Dec 31 '14 at 21:14
  • $\begingroup$ Thanks for the suggestions @Steffen Schuler! Looks a little better, yes? $\endgroup$ – JosephSlote Dec 31 '14 at 21:38
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    $\begingroup$ It's better now. Please, consider that two distinct closed curves can coincide on a segment but not two distinct circles. (Thus, the degree of a vertex at an intersection of two closed curves can be 3.) $\endgroup$ – Steve Dec 31 '14 at 22:25
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I'm quite late to the party, but I'd like to add a very simple but much less rigorous answer for variety.

Each region of a Venn diagram must represent some combination of true or false for each category. For instance, the two circle Venn diagram has 4 regions: the intersection, the outside, and the last two circle parts. The intersection contains everything true to both categories, the outside neither category, and the last two only one of the two categories. (You could list them like TT, TF, FT, FF)

In order to fully cover all cases of true/false combinations, we need $2^n$ sections. Each new category we add doubles the number of regions as it must split each existing region into 2 smaller regions. (One for the true case of the new category, the other for false, so the TT region is now split into TTT and TTF)

We can see how the third circle intersects all 4 regions, but there's no way to draw a 4th circle that intersects all 8 of the new regions, making it impossible to draw a 4 circle Venn diagram.

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  • $\begingroup$ Why is there no way to draw a 4th circle that intersects all 8 of the new regions? $\endgroup$ – JosephSlote Nov 21 '17 at 7:45
  • $\begingroup$ That's what the other answers for this question attempt to look at more closely. This is more of an informal outline for those looking for a very simple intuition assuming they accept that the 4th circle is impossible. $\endgroup$ – CosmoVibe Nov 21 '17 at 11:32

protected by J. M. is a poor mathematician Mar 11 '18 at 14:23

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