EDIT: ok this was nagging at me for a while as something being off, I think this is actually wrong (in some way that must cancel out to accidentally get the right answer) because I need to multiply by 2 pi c to consider all rotations of centers around (0,0) at a given radius, but then my integral no longer works. Ah well, that's what I get for trying to method act and solve quickly, I guess the hooligan stabs me. I think at least this approach done properly could save some dimensions out of the Jacobian we need to calculate. Original post below:
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)
I think it's fairly straightforward to adapt your method. Given circle center c you just need to multiply by 2 pi c to get all the circles.
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
One can discuss what “choosing three points independently and uniformly at random from the interior of a unit circle” means, but whatever you pick, I don’t think that method is doing it.
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?
When I first read the title, I thought it was gonna be about a book similar to one I heard about called “Street Fighting Mathematics” and it would be about like heuristics, estimation, etc. but this one seems to be about a specific problem.
I've got an idea for a simpler approach, but I've forgotten too much math to be able to actually try it.
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.
I like this reasoning. Define a probability distribution on all circles of (x,y,r>0) based on how likely a given circle is. Then we can just sum the good circles and all the circles.
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)
This is a very unenlightening exposition that could discourage people from studying this beautiful problem.
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.
It's improperly formed as a question - the ruffian can shoot whenever he likes;
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.
Author here: when calculating this I _did_ assume a uniform (area) distribution on the unit disk.
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!
"Three points are chosen independently and uniformly at random from the interior of a unit circle. "
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?
The rest of the article answers that question. The followup article answers it more directly, and compares polar to rectangular. https://blog.szczepan.org/blog/monte-carlo/
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.
I think this is reasonably precise. "Uniformly" means that all points within the unit circle are equally likely. You can sample this distribution by picking independent rectangular coordinates and rejecting points outside the unit circle. I'm sure you can sample it in polar space by using an appropriate nonuniform distribution for radius (because a uniform radius would not result in a uniform distribution over points in the unit circle). If you want to sample directly in some really weird parameterization I guess markov chain monte carlo methods are available.
If you choose uniformly from a set then all possible selections are equally likely, by definition. The set is the interior of a circle, which is an area. There's no ambiguity.
I would calculate that the probability of a mathematician doing anything practical like operating a gun is even lower than the probability that I could solve the riddle (even with pen, paper, wikipedia and a liter of coffee on a good day), and choose to sprint off.
dooglius|24 days ago
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)
wedog6|24 days ago
carstimon|23 days ago
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.
Someone|24 days ago
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?
clutter55561|24 days ago
Thanks for editing your answer though. The thug got you in the end, but you saved me in the process.
ccvannorman|24 days ago
fosco|24 days ago
Really enjoyed this keep writing!
cyberax|24 days ago
layman51|24 days ago
del_operator|24 days ago
tzs|24 days ago
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.
bmacho|24 days ago
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.
it4rb|24 days ago
alkyon|24 days ago
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)
alisonkisk|23 days ago
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.
vessenes|23 days ago
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.
szczepan1|23 days ago
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/
voidmain|23 days ago
> 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?
derelicta|24 days ago
del_operator|24 days ago
lupire|24 days ago
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?
dahart|23 days ago
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.
voidmain|23 days ago
masfuerte|23 days ago
robotpepi|23 days ago
bmacho|23 days ago
elcapitan|24 days ago
del_operator|24 days ago
del_operator|24 days ago
mehulashah|24 days ago
analog8374|24 days ago
fancyswimtime|24 days ago
jb1991|24 days ago
sukilot|15 days ago
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