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)
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?
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
This result is out from the article by a factor of pi/3. This is the multiplicative difference between his inner integral with all the sins 24pi^2 and the GP's observation that 3 points on the chosen circle have density (2 pi r)^3 = 8pi^3 r^3.
(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.
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)
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.
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.
"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?
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.
28 comments:
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)
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?
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.
Genuinely not sure if this is wrong or if TFA is.This result is out from the article by a factor of pi/3. This is the multiplicative difference between his inner integral with all the sins 24pi^2 and the GP's observation that 3 points on the chosen circle have density (2 pi r)^3 = 8pi^3 r^3.
(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.
There's actually a second post on exactly that [0]
https://blog.szczepan.org/blog/monte-carlo/
Damn! I read your answer before bed and actually had trouble sleeping trying to understand it!
Thanks for editing your answer though. The thug got you in the end, but you saved me in the process.
took me a few reads but this is indeed correct (lol)
I also enjoyed: https://blog.szczepan.org/blog/los-alamos-primer/
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)
The intro strongly reminded me of https://existentialcomics.com/comic/604
Really enjoyed this keep writing!
Or maybe XKCD: https://xkcd.com/123/
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.
Aye
We were told a (kind of) similar story in high school: https://medium.com/intuition/explain-this-or-i-will-shoot-yo...
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.
I would just take a billion random samples and derive my probability from that. But I'm bad at math.
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.
Galois pistols loaded like hold my coffee
I'm not sure Galois losing that duel proves your point.
"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?
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.
it seems to me that the answer is clear: you take the uniform probability measure on the unit disc.
Ah, 24, reminds me of ole days the lattice of those math alleys had a monstrous moonshine leeching into reality stranger than we’d care to code…
So, I’m left wondering why he did it the hard way.
It’s funny because it’s true.
I'd prefer a world like this; higher levels of whimsy accompanied with greater danger
What's even scarier than such encounter, is that I personally know some people who would survive it. Unfortunately, I'm not one of them.
We are not so differential