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There are $n$ random lines drawn in a circle, defined by endpoints being uniform on circle. I am trying to figure out the expected number of regions separated by $n$ lines. I know $f(0)=1$, $f(1)=2$ and $f(2)=\frac{10}{3}$. Though naive, I can work it out through enumeration. But when $n$ is large, is there a formula to do this?

user3800977
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  • Suppose there are $L$ lines, and they intersect at $P$ different points. (We assume no $3$ lines meet at the same point.) From various small examples it seems the number of regions would be $R= 1+L+P$. Can someone confirm this? It seems some smart induction proof might work, but I can't quite make it rigorous enough and that leaves me doubtful. Anyway, if true, this may provide a way to find $E[R]$ by finding $E[P]$. – antkam Oct 03 '19 at 19:41
  • It seems to me that it will matter how the lines are randomly drawn. Are they drawn through two random points on the circle? Are they drawn with a random point and angle? This is a well known statistical paradox. I'm not sure it will apply here since the problem is slightly different, but it seems like it will. – Charles Hudgins Oct 08 '19 at 20:18

2 Answers2

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Euler's formula states that, for any planar graph, $f - e + v = 2$, where $f$ is the number of faces (what you call regions, with an important caveat to be discussed at the end of this post), $e$ is the number of edges, and $v$ is the number of vertices.

Each line necessarily intersects the circle at exactly $2$ distinct points (the case of a tangent line happens with probability $0$, and the case of no intersections is not relevant). If, in addition, there are $P$ points of intersection, then $v = 2L + P$, where $L$ is the number of lines.

It happens with probability $0$ that any point $P$ lies at the intersection of more than $2$ lines or that any vertex on the edge of the circle is at the intersection of more than $1$ line with the circumference. Thus, there are exactly $4$ edges meeting at each of the $P$ intersection points and exactly $3$ edges ($2$ circular arcs and $1$ edge corresponding to a line) meeting at each of the $2L$ circumference points. Thus, we might guess $e = 3(2L) + 4(P)$. But this actually double counts, since each edge is connected to exactly $2$ different vertices. Thus, we must divide by $2$ to get the true number of edges $e = 3L + 2P$.

Putting this all together, we find $$ f - (3L + 2P) + (2L + P) = 2 $$ That is $$ f - L - P = 2 $$ Or $$ f = 2 + P + L $$ You might wonder why we have $2$ instead of $1$ in this last expression. That is because the exterior of the graph counts as a face for Euler's formula. That is, we have overcounted by $1$. Therefore, the number of regions is given by $$ R = 1 + P + L $$ as you suspected.

The only remaining question is how to assign a probability to $P$. As I mentioned in my first comment, my guess is that this will be impossible without more precisely defining how you will be randomizing your lines.

Charles Hudgins
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  • +1 nice rigorous justification of $R=1+P+L$ via the observation that it depends on exactly $4$ edges meeting at an internal intersection and exactly $3$ meeting at an intersection with the circle, both happening with probability $1$. – antkam Oct 09 '19 at 21:29
  • re: how the lines are randomized, the OP did say "endpoints being uniform on circle" (at least, as of 2019/10/9). what the OP didn't explicitly say is that the two endpoints of a line are independent, but i thought that was meant implicitly - just as endpoints of different lines are clearly meant to be independent. so IMHO Bertrand's paradox didnt apply because the OP specified the distribution was variation no.1 of the wikipedia article. hence the permutation argument applies and any two lines intersect with prob $1/3$. – antkam Oct 09 '19 at 21:36
  • I've actually never seen a proof that the probability of intersection is $\frac{1}{3}$, but now I have a fun little homework assignment. Assuming the probability is $\frac{1}{3}$, it still isn't completely clear to me that $E(P(L)) = \frac{C^2_L}{3}$, but probability has never been my strong suit. – Charles Hudgins Oct 10 '19 at 00:58
  • since you said homework, i wont give it away. if you want an explanation at a future date, just @antkam msg me again! :) – antkam Oct 10 '19 at 01:20
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Following the hint from antkam, With $L$ lines intersecting $P$ points, we have number of regions at $R=1+L+P$. $E(P(L))$ can be calculated by linearity of expectation, which means for any two lines, there is $\frac{1}{3}$ probability that they intersect and we have $C^2_L$ combinations. So, $E(R(L))=1+L+\frac{C^2_L}{3}$.

user3800977
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