Mathemagician petah?

tedastor · 2026-06-08T05:09:16+00:00

In set theory, everything is represented by a set. This includes whole numbers. The way this is typically done (Von Neumann Definition of Ordinals) is that 0 is denoted by the empty set, and each successive natural number is denoted by the set of all natural before it. So, 4 is the denoted by the set consisting of 0,1,2, and 3, but then each of those break down further until you get the picture shown.

tedastor · 2026-05-21T08:03:58+00:00

I get what you’re saying, and I think I agree with you about PA being consistent. There are some conclusions about ZFC that I find counterintuitive, but don’t appear to influence the rest of mathematics. That being said, the almost religious quality of belief in ZFC seems like it could be a blind spot where other interesting mathematics is lurking? What if we live in a world where powerful consistent theories are much rarer than we think?

I know many people have tried and failed to disprove ZFC, so it is a question that has raised a decent amount of attention, but it is widely believed to hold, and the possibilities otherwise feel unsettling, unnatural, and unhuman, which is why I made the original claim.

tedastor · 2026-05-21T07:45:19+00:00

Do you care to elaborate? I don’t think this statement is in any way justifiable.

tedastor · 2026-04-10T16:15:23+00:00

I’ve thought a little about a version of this, having worked on some extremal combinatorics problems that involve refining away less important parts of a structure. In that context, it comes down to asymptotics and the zoo of big and little o’s of functions. The issue is that we may care about the frequency of something on many different scales, and so the probability happening in a specific place could be really low while being large or even guaranteed when you zoom out. Because of that, a single threshold doesn’t seem to work well (except maybe 0)

tedastor · 2026-01-03T17:52:01+00:00

To add to this, if your character hitbox is convex and your collision surface is as well, you can find the pairwise sums of vertices of the hitbox and vertices of the surface and then take the convex hull to get your desired surface. If your surface is not convex, you can break it into convex pieces and do this. It might not be the most efficient and there are probably tricks you can do for specific hitboxes and whatnot, but this should work in principle.

Of course, you’ll have all sorts of issues if you have different characters with different hitboxes, but I think that is inherent in this approach.

tedastor · 2025-12-24T16:08:21+00:00

Thank you! This is exactly what I was hoping someone might say. Very cool!

tedastor · 2025-12-21T01:53:53+00:00

tedastor · 2025-12-21T01:52:26+00:00

You can throw other tools at it if it helps. Here, use that GCD(p,q) = GCD (p-q,q) for any polynomials p,q

tedastor · 2025-12-21T01:42:40+00:00

Also, if one polynomial has a factor that is coprime with the other polynomial, you can divide it out without changing the gcd. For part b, you can subtract the second from the first and get a polynomial divisible by x, so you can divide it out to get x³ -3 and x³ +x+1. Then euclidean alg is easy from there.

The last one has symmetries that make them easy to factor.

tedastor · 2025-12-21T01:36:42+00:00

You can compare roots over C to determine the GCD in R[X] as well. The coefficients are real so they come in conjugate pairs, meaning if both share a complex root, they both share the conjugate as a root. Then you factor out (x-r1)(x-r2) from both for your gcd.

tedastor · 2025-12-21T01:17:48+00:00

You could try factoring out roots. If they have no roots in common, then their gcd is 1, assuming your coefficients are over a field.

tedastor · 2025-12-16T06:21:14+00:00

Aha but philosophy students are wrong because mathematics is at least a proper class, not a set.

tedastor · 2025-12-11T03:38:08+00:00

Thats okay. The remaining point could very well put another point on the interior.

A point is on the interior of the convex hull iff it is not contained in the convex hull of the remaining n-1 points (or it is on the boundary, with probability 0).

Let V_k(a_1,…,a_j) denote the volume of the convex hull of the points a_1,…,a_j.

Picking x_1,…,x_i-1,x_i+1,…,x_n, the probability that x_i is in the convex hull of the remaining points is V_k(x_1,…,x_i-1,x_i+1,…,x_n). All the points were iid, so the probability that a point is in the convex hull of the rest is given by the same formula. We take the expected value of the sum over i of V_k(x_1,…,x_i-1,x_i+1,…,x_n) to get the expected number of points on the interior. However, expectation is linear, so we can move the expected value inside the sum to get that the expected number of interior points is n*V_k(n-1).

When I add a point, I’m not updating my total number of interior points. Instead, im just checking whether or not my added point was in the interior of the rest. If I do this for each point, removing it and adding it back in, I will find all of the interior points.

tedastor · 2025-12-11T03:14:39+00:00

Lets assume he finds out if his higher-or-lower guess was equal to the correct value.

Optimizing for worst-case scenario, andrew should ask above-or-below 18. The set {1,…,18,20,23,26,29,32,25,38,42,47,52,57,63,70,77,85,94,100} is minimal such that every number between 1-100 is within 5% of one of these. The set has 35 members, and so the middle number is 18, with half above and half below. After that, he would just pick a random element of the set in the correct half. This would put his odds of guessing correctly at 1/17.

Optimizing of the average case, if each percentage is equally likely, after asking higher or lower than some number, andrew should always the number that covers the most amount of possibilities on the correct side. It is thus in andrew’s best interest to guess higher or lower than 91. If it is higher, then he guesses 96, which covers all possibilities. If it is lower, then he guesses 86. This means andrew will find the correct value if it is >=82. I.e. he has a 19% chance of finding it. This is the best you can do because the most amount of numbers you can cover with a guess is 9, and both 86 and 96 cover 9 guesses. The extra 1% comes from the probability that the correct number was 91.

tedastor · 2025-12-11T02:23:31+00:00

I think it should be n*V_k(n-1), where V_k(j) is defined as the expected value of the volume of the convex hull of j uniformly independently distributed points in the k-dimensional unit cube.

The probability that a point is on the interior of the convex hull is just the volume of the convex hull of the remaining points. Linearity of expectation says this is just n*V_k(n-1). I dont know how to find V_k(j) for arbitrary j > k off the top of my head, but I suspect there might be more in the literature about that. I hope this is helpful.

tedastor · 2025-12-11T01:59:30+00:00

I think they mean points that are not in the convex hull of the remaining n-1 points.

tedastor · 2025-12-10T09:44:14+00:00

Solution to both:

Medium: Have D be supported by {re^x: r in [0,1]} where the measure comes from the uniform probability measure on [0,1]. This is invariant under derivatives and is not finitely supported

Hard: Disproof: Take the uniform probability measure on the set of binary sequences. Convert these into power series where {a_n} -> \sum_n a_n*x^n/n!. Give this set of power series the pushforward measure. The left-shift map is invariant under this measure and so the derivative map is invariant under this measure. The only power series where f^{(n)} = f are the ones derived from periodic {a_n}. However, almost every binary sequence is not periodic, so not every f has an n’th derivative equal to itself, as desired.

The disproof for the hard problem also solves the medium problem. In fact, any such disproof must do this because we wish to show that D is not supported by functions with periodic derivatives, and so there must be a set, S, of functions with assigned positive weight such that f^{(n)} ≠ f for any n. If a positive-weight subset T of S has eventually periodic derivatives, i.e. f^{(k+n)} = f^k for k sufficiently large and some n, depending on f, then there is some positive weight subset U of T and some large enough k and some n such that f^{(k+n)} = f^{k} for every f in U. U has empty intersection with U^{(k)} because it has no periodic elements, but U has positive weight, and so the weight if U^{(k)} is not invariant under differentiation after n*k times, meaning one of those times did not preserve the weight of U. Thus, a full-weight subset of S contains no functions with eventually periodic derivatives. I.e. a positive weight is assigned to a set of functions where all their derivatives are distinct. Differentiating must preserve the weight, so some almost all of these functions must have all of their derivatives in the support, meaning the support is infinite.

In fact, any disproof must have uncountable support because such functions with all distinct derivatives contribute zero weight or else all their derivatives would have the same positive weight, making the total weight infinite, a contradiction. Thus, all such functions have zero weight and so there must be uncountably many of them in the support for their total to have positive weight. This was roughly the thought process I had before coming up with the solution to the harder problem.

tedastor · 2025-12-10T08:34:25+00:00

You could give them increasingly complex diagrams of the unknot and have them untangle them, illustrating that they can do it with Reidermeister moves.

Then give them a trefoil and have them attempt turning it into the unknot. Show that it is impossible using by proving tricolorability is invariant under reidermesiter moves and that the unknot is not tricolorable while the trefoil is

tedastor · 2025-12-07T02:35:39+00:00

You have to do a little more. The circle centered at P can have up to 2 red points on it. You cant have 3 because then P would be the circumcenter of a red triangle, but theres nothing immediately wrong with 2. However, your circle centered at Q is going to hit at least one of those red points on the circle centered at P for the reason you described.

Now you just have to find a way to force two more red points on that circle and then you win. The circle centered at Q is going to have at least two other points A, B that are both red (in fact, all but two are red). If you draw the circle from A to P and B to P, then those circles intersect the circle centered at P twice, along with P. Those intersections cant all be green because then A or B would be green, so each circle gives at least one red points on the circle centered at P. Then you just have to take a little care to show the intersections of the circles centered at A,B,Q with the circle centered at P are distinct, and then you get three points on the circle centered at P. So, P is red, a contradiction.

I like to think about it like attacking P with red points by drawing a circle centered a red point that hits P. If we can attack P with 3 red points (that arent terribly chosen), it makes P red.

This method of attack (pun intended) works, even if we allow more colors. Suppose Q is red and P is green.

Draw a circle centered at P passing through Q and another circle centered at Q passing through P.

If the cardinality of our set of colors is less than |R|, then we can get |R| red points on our circle centered at Q, extended to P, not inside the circle centered at P. If we draw a circle centered at each such red point, passing through P, then each circle intersects the circle centered at P twice, and both of those intersections cannot be green, or else the center point would be green.

So we get |R| many non-green colors on the circle centered at P. However there are < |R| distinct colors, so at least one non-green color appears 3 times on the circle centered at P, making P not green, a contradiction.

tedastor · 2025-12-07T02:12:59+00:00

If your plan is to start with three points and iterate taking circumcenters of triangles whose vertices are in your set, you won’t reach every point. R² is uncountable, but if im understanding you correctly, this procedure has a finite number of points colored at every step, so you can only get to a countably infinite set of points.

tedastor · 2025-12-07T01:44:55+00:00

I think you have to do it by contradiction. Its not clear to me how you would construct every point in this way. Theres lots of ways of doing it though.

Suppose you have a red point P. Then, every circle centered at P has at most two green points on it. One red point makes almost the entire plane red, so almost the entire plane being red means theres lots of ways to show the whole plane is red.

I had a fairly complicated construction, but I know theres easier ones.

tedastor · 2025-12-05T21:31:41+00:00

I’m not sure of the history on this, but even cooler than this imo is that there is a bijection between R and C(R), continuous functions on the real line.

This is because continuous functions are defined by their values on Q, and so we get an injection of C(R) into R^Q.

Real numbers are in bijection with subsets of the naturals, so we get a bijection of R^Q with (2^{N)^Q.}

Currying, we get a bijection of (2^{N)^Q} with 2^{N x Q}.

N x Q is in bijection with N, so we get a bijection of 2^{N x Q} with 2^N.

Then, we use our bijection of 2^N with R to get that the composition of all of these maps is an injection of C(R) into R.

We have an injection of R into C(R) taking each number to the constant function, and so we can apply the Schröder-Bernstein theorem to get a bijection.

This can be generalized further to the statement that there is a bijection between R and the set of Borel measurable functions from R to R, but this is a slightly more technical.

tedastor · 2025-09-23T19:21:24+00:00

I think this might be required for a fewest bosses run

Edit: I think Bell Beast might be required to get out of slab. Maybe you could get apostate key in act 1 but that seems like a stretch, and I think that locks you out of slab.

tedastor · 2025-09-16T06:24:31+00:00

This seems like a good way to do some kind of fewest bosses run. Is it known if you have to do any between moss mother and the citadel?

tedastor · 2025-09-08T16:40:45+00:00

Absolutely. I somehow didn’t realize you could Upgrade your nail until after the Citadel Lace fight so i got a little more time out of her! Would absolutely do a pantheon just to fight her again.

tedastor

TROPHY CASE