What are the possible numbers of regions that 4 planes can create? We know that the minimum number is 5 and the maximum number is 15. (http://mathworld.wolfram.com/SpaceDivisionbyPlanes.html)

Is it possible to make a generalization based on the ways the planes could intersect?

  • $\begingroup$ If I take two parallel planes and then take a bunch of orthogonal planes to them, wont that produce a lot more than 15? What am I missing? $\endgroup$ – B. Bischof Sep 10 '10 at 4:56
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    $\begingroup$ I think he said 4 planes. $\endgroup$ – John Jiang Sep 10 '10 at 5:05
  • $\begingroup$ In dimension $n,$ the maximum number of regions created by $n+1$ hyperplanes will be $2^{n+1} - 1,$ caused by extending the sides of a ( regular if you like ) simplex to complete hyperplanes. $\endgroup$ – Will Jagy Sep 10 '10 at 5:10
  • $\begingroup$ It is not clear to me what kind of generalization is sought here. Hyperplane arrangements have been extensively studied; in particular, the face lattice of an arrangement records a lot more information than just the number of the components in the hyperplane complement ("regions"). $\endgroup$ – Victor Protsak Sep 10 '10 at 5:42
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    $\begingroup$ Are you limiting the answer to $\mathbb{R}^3$ or not? This question is rather vague and almost seems like a first or second week homework question, doesn't it? It is an interesting question, but there's no motivation listed, no domain listed, no mention of any attempts at solving it... And the original author of the question has not edited the question to clarify anything. $\endgroup$ – sleepless in beantown Sep 10 '10 at 14:58

Actually this seems like an interesting question to me. One can easily calculate the maximum number of regions obtained by n hyperplanes:

For lines in $\mathbb{R}^2$, by induction, the maximum number of regions achievable with $n$ lines is $1+1+2+ \ldots + n$. For planes in $\mathbb{R}^3$, denote the maximum regions by $N_n$. Then one sees that $N_{n+1} - N_n = $ the maximum number of regions in $\mathbb{R}^2$ achievable by $n$ lines, hence equals $1+2+ \ldots + n$. Thus $N_n = 1+ n + (n-1) + 2(n-2) + 3(n-3) + \ldots (n-1)$.

The in-between numbers seem much more elusive. Even the version of the problem for lines in $\mathbb{R}^2$ seems hard. I found by experimenting that 5 is not achievable by any number of lines in $\mathbb{R}^2$ less than 4. So a natural question could be what number $n$ has the property of not being achievable by any number of lines less than $n-1$.

For the special case of 4 planes in $\mathbb{R}^3$. I think the correct answer is: 5, 8, 9, 10, 11, 12, 14, 15. It's clear 6,7 aren't constructible. 13 is not constructible by brute force checking all constructible numbers with 3 planes and seeing that it's impossible to add another plane to get 13.

  • $\begingroup$ This was exactly what I got. I just determined how 4 lines can intersect. Lines can represent planes, so any number of regions that 4 lines can form is possible. Also, 2 times any number of regions that 3 lines can form. Correct me if I'm wrong $\endgroup$ – user9107 Sep 10 '10 at 7:33
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    $\begingroup$ Eliminating 13: if three planes meet in a line the create 6 regions; the last at most doubles that. If two planes are parallel they make 3 regions and the others at most quadruple it. If three planes meet pairwise in three parallel lines they create 7 regions. A plane not parallel to these lines doubles the number of regions, a plane parallel to them creates at four more regions. Thus all triples of planes meet in one point. If all planes have a point in common, the arrangement is symmetric about it and the number of regions is even. Otherwise we are in general position and get 15 regions. $\endgroup$ – Robin Chapman Sep 10 '10 at 9:19
  • $\begingroup$ @unknown(google): I think you are right except the ends: 5 and 15. But I feel it's not generalizable to more than 4 planes in $\mathbb{R}^3$. $\endgroup$ – John Jiang Sep 10 '10 at 19:12

Computing the maximum number of pieces into which $r$ hyperplanes disconnect the space $\mathbb{R}^n$ is an old little problem, easily solved by induction, and the answer is $${r \choose 0}+{r \choose 1}+\dots+{r \choose n},$$ and it is achieved if they are in generic position. So in dimension $n$ the number of pieces as $r$ increases is asymptotically $r^n/n!$. Note also that for $r\leq n$ it's $2^r$ and for $r=n+1$ it's $2^r-1$ (so in particular in dimension $3$ the number of pieces with $0,1,2,3,4$ generic planes is resp. $1,2,4,8$ and then $15$ (not $16$ as one would guess, so this may be used as an example of fallacious argument based on analogy).


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