Much more elegant: consider every circle that fits inside the unit circle, and we will work backward to find combinations of points. We only need consider centers on the x axis by symmetry, so these are parameterized by circle center at (0,c) and radius r with 0<c<1 and 0<r<1-c. Each circle contributes (2 pi r)^3 volume of triples of points, and this double integral easily works out to 2 pi^3/5 which is the answer (after dividing by the volume of point triples in the unit circle, pi^3)
int 0..1 2 pi c int 0..(1-c) (2 pi r)^3 dr dc / pi^3
int 0..1 2 pi c int 0..(1-c) (2 r)^3 dr dc
int 0..1 2 pi c 2 (1-c)^4 dc
-4 pi int 0..1 (1-g) g^4 dg
4 pi (1/6 - 1/5)
4 pi / 30
2 pi/ 15
(The article had already covered the r^3 in another part of the calculation.)
I'm trying to figure out an intuitive explanation as to why the work with the inner Jacobian is needed or an argument as to why it isn't.
Anyone want to simulate this accurately enough to distinguish between 40% and 41.9% probability? 5000 samples should be more than enough.
You've done
∫₀¹ ∫₀¹⁻ᶜ (2π r)³ dr dc
However,
- the `dr` integral is assuming that the radii are uniformly likely in [0, r]
- the `dc` integral is assuming that the centers are uniformly likely in [0, c]
You need to wait these integrals by the conditional probability distributions.
Doesn’t it have half its circle centers have 0 < c < ½, while that covers only a quarter of the area of the unit circle?
Thanks for editing your answer though. The thug got you in the end, but you saved me in the process.
Really enjoyed this keep writing!
The idea is to consider the set A of all circles that intersect the unit circle.
If you pick 3 random points inside the unit circle the probability that circle c ∈ A is the circle determined by those points should be proportional the length of the intersection of c's circumference with the unit circle.
The constant of proportionality should be such that the integral over all the circles is 1.
Then consider the set of all circles that are contained entirely in the unit circle. Integrate their circumferences times the aforementioned constant over all of these contained circles.
The ratio of these two integrals should I think be the desired probability.
And the probability distribution is simple: a given (x,y,r) is as likely as its circumference in the unit circle.
Reasoning: Let C:(x,y,r) a given circle. We want to know how likely is it that the circle on 3 random points are close to it, closer than a given value d. (A d wide ball or cube around C in (x,y,r) space. Different shapes lead to diffferent constants but same for every circle.) The set of good 3 points is more or less the same as the set of 3 points from the point set C(d): make C's circumference d thick, and pick the 3 points from this set. Now not any 3 points will suffice, but we can hope that the error goes to 0 as d goes to 0 and there is no systematic error.
Then we just have to integrate.
ChatGPT got me the result 2/3, so it's incorrect. I guess the circumference must not be the right distribution.
Although there is small error regarding the neutron number calculation. I assume 3/4 of the neutrons are lost and then the author can multiply by 1/4 to get the result that the naturally occurring uranium is safe (as its neutron number is less then 1)
A much better way to analyze it geometrically. The 6D problem has 2D trivial symmetries, and is parameterized by 4 polar coordinates: circumcenter distance from origin, circumradius, and 2 internal angles of the triangle. Then the solution is just [expected value of the area of a triangle on unit circle] times the radius integral in the OP article, divided by the π³ volume of the triangle space in rectangular coordinates.
Consider:
Does "random" mean
1. uniform distribution on x and y coordinates with some sort of capping at the circle boundary? Or perhaps uniform across all possible x,y pairs inside (on the edge also?) of the circle? what about a normal distribution?
2. a choice of an angle and a length?
3. A point using 1 or 2, and then a random walk for 2 and 3?
I could go on. The worked solution is for random = uniform distribution across all possible reals inside the boundary, I think.
Now it does say
> Three points are chosen independently and uniformly at random from the interior of a unit circle.
which sounded OK to me at the time but I understand there could have been some ambiguity. Especially around the "uniform on area" part.
Also, I think that with rejection sampling you could get the same with 1) [0], 2) would work (provided correct scaling) [1]. No idea about 3) or the normal distribution thing you mentioned - I figured the problem was hairy enough already!
[0] https://blog.szczepan.org/blog/monte-carlo/#sampling-uniform... [1] https://blog.szczepan.org/blog/monte-carlo/
It's been a lonnnng time since my geometry university courses, but my vague memory is there are some tricky differential geometry historical problems that founder on this precise imprecision.
Fun site, thank you for the write up. I skimmed each and every matrix and assumed you did a great job.
> Three points are chosen independently and uniformly at random from the interior of a unit circle
Has it been edited in the last 15 minutes to address your objection or something?
That has always been the statement (i.e. I've not updated it since adding the post). I do agree that the "uniform on area" bit should have been made more clear!
The distribution is under specified
Is it "uniformly" over area, even though it's not an area problem? That is, is it independent random coordinates (x, y) in rectangular coordinate space, or (r, theta) polar space, or in some other parameterization?
Short answer: yes it’s uniform in area. In the absence of the specificity you want, area makes the most sense, right? Uniformly sampling independent Cartesian variables yields uniform sampling in area, unlike polar where a uniform sampling of the independent variables gives you a non-uniform sampling of area.
I don’t understand what you mean about it not being an area problem, but I guess at some level this actually is an area problem. I’ll speculate wildly there might be a way to transform the question/setup into a different but equivalent problem that can be directly visualized as solving for area, and perhaps have a more intuitive solution that involves fewer determinants of Jacobians. Maybe, maybe not, I dunno.