[HN Gopher] Dark Alley Mathematics
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Dark Alley Mathematics
Author : quibono
Score : 119 points
Date : 2026-02-02 15:58 UTC (5 days ago)
(HTM) web link (blog.szczepan.org)
(TXT) w3m dump (blog.szczepan.org)
| derelicta wrote:
| 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.
| del_operator wrote:
| We are not so differential
| layman51 wrote:
| 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.
| del_operator wrote:
| Aye
| dooglius wrote:
| 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)
| ccvannorman wrote:
| took me a few reads but this is indeed correct (lol)
| clutter55561 wrote:
| 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.
| wedog6 wrote:
| 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
|
| Genuinely not sure if this is wrong or if TFA is.
| wedog6 wrote:
| 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.
| quibono wrote:
| There's actually a second post on exactly that [0]
|
| https://blog.szczepan.org/blog/monte-carlo/
| Someone wrote:
| 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 < 1/2, while
| that covers only a quarter of the area of the unit circle?
| carstimon wrote:
| You pointed out one error in your own answer, here's another. I
| haven't figured out the adjustments, but presumably they
| cancel.
|
| You've done
|
| [?]01 [?]01- (2p 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.
| carstimon wrote:
| Sorry, this was incorrect. In this integral we're just
| calculating the geometrical question, what is is the 5d
| volume of the collection of 3 points whose circumcenter is on
| {(x, 0) : x in [0, 1]}. So the "overall" probability of the
| problem has nothing to do with it.
| mehulashah wrote:
| So, I'm left wondering why he did it the hard way.
| fancyswimtime wrote:
| I'd prefer a world like this; higher levels of whimsy accompanied
| with greater danger
| fosco wrote:
| The intro strongly reminded me of
| https://existentialcomics.com/comic/604
|
| Really enjoyed this keep writing!
| cyberax wrote:
| Or maybe XKCD: https://xkcd.com/123/
| it4rb wrote:
| We were told a (kind of) similar story in high school:
| https://medium.com/intuition/explain-this-or-i-will-shoot-yo...
| tzs wrote:
| 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.
| bmacho wrote:
| 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.
| tzs wrote:
| Maybe 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.
| elcapitan wrote:
| 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.
| del_operator wrote:
| Galois pistols loaded like hold my coffee
| hansvm wrote:
| I'm not sure Galois losing that duel proves your point.
| jb1991 wrote:
| It's funny because it's true.
| del_operator wrote:
| 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...
| alkyon wrote:
| 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)
| lupire wrote:
| "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 wrote:
| it seems to me that the answer is clear: you take the uniform
| probability measure on the unit disc.
| masfuerte wrote:
| 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.
| dahart wrote:
| 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.
| voidmain wrote:
| 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.
| bmacho wrote:
| "Uniform" means uniform in area in these contexts. It's
| precise.
| analog8374 wrote:
| I would just take a billion random samples and derive my
| probability from that. But I'm bad at math.
| vessenes wrote:
| 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.
| voidmain wrote:
| The 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 wrote:
| Hey, 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 wrote:
| 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!
|
| [0] https://blog.szczepan.org/blog/monte-carlo/#sampling-
| uniform... [1] https://blog.szczepan.org/blog/monte-carlo/
| vessenes wrote:
| Totally agreed! Conceptually I think of the radian/length
| distribution as having uniformly increasing density closer to
| the origin - you could imagine a whole bunch of discs
| concentrically stacked ending in the circumference of the
| circle - each of them - if a constant "radius length" - will
| have the same "number" of points but spread out over a larger
| total area.
|
| 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.
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