What are the possible numbers of regions that 4 planes can divide space? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T06:33:32Z http://mathoverflow.net/feeds/question/38267 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38267/what-are-the-possible-numbers-of-regions-that-4-planes-can-divide-space What are the possible numbers of regions that 4 planes can divide space? unknown (google) 2010-09-10T04:46:37Z 2010-09-10T09:47:33Z <p>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)</p> <p>Is it possible to make a generalization based on the ways the planes could intersect? </p> http://mathoverflow.net/questions/38267/what-are-the-possible-numbers-of-regions-that-4-planes-can-divide-space/38276#38276 Answer by John Jiang for What are the possible numbers of regions that 4 planes can divide space? John Jiang 2010-09-10T06:19:00Z 2010-09-10T06:19:00Z <p>Actually this seems like an interesting question to me. One can easily calculate the maximum number of regions obtained by n hyperplanes: </p> <p>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 $N_n = 1+ n + (n-1) + 2(n-2) + 3(n-3) + \ldots (n-1)$. </p> <p>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 $n-1$. </p> <p>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. </p> http://mathoverflow.net/questions/38267/what-are-the-possible-numbers-of-regions-that-4-planes-can-divide-space/38287#38287 Answer by Pietro Majer for What are the possible numbers of regions that 4 planes can divide space? Pietro Majer 2010-09-10T09:40:45Z 2010-09-10T09:47:33Z <p>Computing the <em>maximum</em> 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). </p>