5 ms·
Dark Alley Mathematics
- derelicta 8mo agoWhat's even scarier than such encounter, is that I personally know some people who would survive it. Unfortunately, I'm not one of them.
- del_operator 8mo agoWe are not so differential
- layman51 8mo agoWhen 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.
- del_operator 8mo agoAye
- dooglius 8mo agoEDIT: 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)
- ccvannorman 8mo agotook me a few reads but this is indeed correct (lol)
- clutter55561 8mo agoDamn! 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.
- wedog6 8mo agoI 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 Genuinely not sure if this is wrong or if TFA is.
- wedog6 8mo agoThis 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.
- quibono 8mo agoThere's actually a second post on exactly that [0] https://blog.szczepan.org/blog/monte-carlo/ https://blog.szczepan.org/blog/monte-carlo/
- Someone 8mo agoOne 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?
- mehulashah 8mo agoSo, I’m left wondering why he did it the hard way.
- fancyswimtime 8mo agoI'd prefer a world like this; higher levels of whimsy accompanied with greater danger
- fosco 8mo agoThe intro strongly reminded me of https://existentialcomics.com/comic/604 https://existentialcomics.com/comic/604 Really enjoyed this keep writing!
- cyberax 8mo agoOr maybe XKCD: https://xkcd.com/123/ https://xkcd.com/123/
- it4rb 8mo agoWe were told a (kind of) similar story in high school: https://medium.com/intuition/explain-this-or-i-will-shoot-you-mathematics-saved-a-physicists-life-568c3d9f5bc5 https://medium.com/intuition/explain-this-or-i-will-shoot-yo...
- tzs 8mo agoI'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.
- bmacho 8mo agoI 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.
- tzs 8mo agoMaybe it should be the cube of the amount of circumference inside the unit circle, because 3 random points have to land on the circle for it to be chosen.
- bmacho 8mo agoI think that's correct. But that wouldn't work either. The problem is that there is an infinite volume of circles with ~2 as the circumference inside the unit circle. (Volume in x,y,r space, and, consequently in the proposed probability distribution as well.) So this proposed probability distribution gives infinity, when summed up.
- elcapitan 8mo agoI 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.
- del_operator 8mo agoGalois pistols loaded like hold my coffee
- hansvm 8mo agoI'm not sure Galois losing that duel proves your point.
- jb1991 8mo agoIt’s funny because it’s true.
- del_operator 8mo agoAh, 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…
- alkyon 8mo agoI also enjoyed: https://blog.szczepan.org/blog/los-alamos-primer/ 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)
- lupire 8mo ago"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?
- robotpepi 8mo agoit seems to me that the answer is clear: you take the uniform probability measure on the unit disc.
- masfuerte 8mo agoIf 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.
- dahart 8mo agoThe 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/ 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.
- voidmain 8mo agoI 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.
- analog8374 8mo agoI would just take a billion random samples and derive my probability from that. But I'm bad at math.
- vessenes 8mo agoIt'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.
- voidmain 8mo agoThe article currently says > 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?
- szczepan1 8mo agoHey, author here :) 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!
- szczepan1 8mo agoAuthor 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! [0] https://blog.szczepan.org/blog/monte-carlo/#sampling-uniformly-from-a-disk https://blog.szczepan.org/blog/monte-carlo/#sampling-uniform... [1] https://blog.szczepan.org/blog/monte-carlo/ https://blog.szczepan.org/blog/monte-carlo/
- vessenes 8mo ago
- alisonkisk 8mo agoThis 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.
- sukilot 8mo ago[dead]