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Comment by dooglius

21 days ago

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)

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dooglius

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wedog6  20 days ago

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  20 days ago

    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.

carstimon  20 days ago

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⁻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.

  • carstimon  20 days ago

    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.

Someone  20 days ago

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?

clutter55561  20 days ago

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.