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
    Commented Aug 3, 2010 at 22:11
  • $\begingroup$ Is it actually doable with squares, or only with non-square rectangles? $\endgroup$
    – Isaac
    Commented Aug 3, 2010 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
    Commented Aug 3, 2010 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
    Commented Aug 4, 2010 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 $\endgroup$ Commented May 10, 2011 at 2:59

4 Answers 4


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?


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.

  • $\begingroup$ I hope you don't mind that I included a direct link to your paper... $\endgroup$ Commented Feb 9, 2012 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
    Commented Feb 9, 2012 at 2:37


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

Proof: Suppose it is possible. Then we may create a planar graph $G$ from the Venn diagram by placing vertices at each intersection, like this:


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

We will show that a graph created in such a way from a 4-set Venn diagram cannot be planar—a contradiction.

Claim: Every pair of circles intersect at two distinct points.

Suppose not. Then there is a pair of circles whose interiors do not intersect. This cannot be, for we require a Venn diagram graph to have a face for every pairwise intersection of sets.

Claim: Every vertex has degree four.

Suppose not. Then there are three circles $C_1, C_2, C_3$ which share an intersection. It can be readily checked by case analysis that not all intersections $\{C_1\cap C_2, C_1\cap C_3, C_2 \cap C_3, C_1\cap C_2 \cap C_3, \}$ are present as faces in this subgraph.

These two claims entail there are exactly two unique vertices for each pair of circles, so $v = 2{n \choose 2}$. Also, by the edge sum formula we have $e=\frac{d_1+\cdots+d_v}{2}=2v=4{n\choose 2}$ 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. Counting 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: $$v-e+f = 2{n\choose2}-4{n\choose 2} + 2^n= 2^n- 2{n\choose 2}\leq2,$$ 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

  • $\begingroup$ 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." $\endgroup$ Commented Dec 31, 2014 at 21:14
  • $\begingroup$ Thanks for the suggestions @Steffen Schuler! Looks a little better, yes? $\endgroup$ Commented Dec 31, 2014 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.) $\endgroup$ Commented Dec 31, 2014 at 22:25
  • $\begingroup$ I am a bit confused. Don't we have $2 = F-E+N \le 2^n - 2\binom{n}{2}$ which for $n=4$ just gives $2\le 4$. We would require the opposite bounds for $N$ and $E$ to make this argument work. $\endgroup$ Commented Mar 6, 2021 at 19:35
  • $\begingroup$ Hi @user2860104 thank you for the catch. I think the fix is to remove estimation from the argument as I have now sketched. If you have a more elegant approach to repairing the proof, I am interested to hear it! $\endgroup$ Commented Mar 6, 2021 at 23:59

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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