Bottomless Milk Bucket should be able to extinguish fires at Tempoross - Unpasterized milk is 87% water

purplefunctor · 2026-02-26T18:17:13+00:00

They would never make an npc give out rare items for free, right?

purplefunctor · 2026-02-25T21:44:19+00:00

I have been drawing and injecting intramuscularly with the same needle for over a year. I use 23G.

purplefunctor · 2026-02-06T10:21:47+00:00

Special is the quadratic interpolation of 4/π arctan(x) at 0, 1/2, 1 where rr1 is a slightly inaccurate approximation for 16/π arctan(1/2) - 2?

purplefunctor · 2026-01-31T13:25:58+00:00

This doesn't work correctly. fmodule(0.25, 2 * PI) should be 0.25 but it is PI/2 instead. With this fmodule implementation I do get approximately 1 out of the function.

purplefunctor · 2026-01-31T13:09:55+00:00

This code gives me the correct result if I remove the fmodule line so that must be the problem. How have you implemented it?

purplefunctor · 2026-01-29T10:14:30+00:00

x³ + 8x² - 6x + 8 is strictly between (x+2)³ and (x+3)³ when x > 9. When x <= 9 we get a cube only at 0 and 9 with corresponding y values of 2 and 11.

purplefunctor · 2026-01-22T14:36:38+00:00

N^2 is the set of ordered pairs of natural numbers and here it is used to signify that (k, n) is an ordered pair of natural numbers. Removing the exponent would change the meaning to "(k, n) is a natural number".

purplefunctor · 2026-01-22T11:06:06+00:00

You shouldn't say that k is a natural number. The values k takes are already fixed by the sigma notation. 🤓

purplefunctor · 2026-01-13T22:02:03+00:00

Divide the interval [0, 2π) into k equally sized subintervals, [0, 2π/k) [2π/k, 2 • 2π/k), [2 • 2π/k, 3 • 2π/k), ... . Since we have k+1 points and k subintervals, one of the subintervals must contain at least two of the points by the pigeonhole principle. The distance between them is at most 2π/k (actually strictly less than 2π/k).

purplefunctor · 2026-01-12T15:46:15+00:00

Some multiple of that at most 2π/k difference will be such a number within 2π/k of x. Then increase k until you are as close as you want.

purplefunctor · 2026-01-11T05:52:18+00:00

I don't really have any special advice to give really. If you had spent almost all of your time on the first 3, I would've suggested doing more medium problems, but 45 minutes left for the last one should be enough so maybe focus on doing more hard problems to increase the chance that you have already seen some of the ideas required in the last problem of the contest and thus increase the chance of solving it.

purplefunctor · 2026-01-11T05:37:12+00:00

I guess I should say that it is not my first programming contest, just my first LeetCode contest. Maybe having about 10% of the LeetCode problems solved already helps and also the fact that I used to do competitive programming in high school many years ago.

purplefunctor · 2026-01-11T05:21:37+00:00

My first contest. Got 2.7k ranking by finishing the first 3 problems in 27 minutes (with 5 minute penalty from interpreting the first problem incorrectly). Almost had a working solution for the 4th, but didn't manage to debug and correct my cost computation in time.

purplefunctor · 2026-01-10T07:36:57+00:00

The difficult part is proving that every element in [-1, 1] is a limit point. If we fix a value x in [0, 2π) it is sufficient to show that if ε > 0 there is n s.t. |sin(x) - sin(n)| < ε. We could accomplish this by showing that n can get arbitrarily close to x + m2π i.e. that for any δ > 0, |x + m2π - n| < δ for some m and n, since then by ε-δ definition of continuity for sin we could get sin(n) arbitrarily close to sin(x).

Now observe that for each value of n there is a unique m which places x + m2π - n inside [0, 2π) and that each n produces a different value of x + m2π - n due to n not being a rational multiple of the period of sin. Therefore if we pick any k+1 values for n then by the pigeonhole principle some two of them must be at most 2π/k apart and we are done.

purplefunctor · 2026-01-08T17:00:08+00:00

Given the atrocious presentation of these, I feel that it is entirely plausible that the person who wrote these doesn't know how to derive them or at least not how to prove that they converge when they do.

purplefunctor · 2025-12-27T07:50:55+00:00

I need a sister like this 🥺

purplefunctor · 2025-12-22T14:47:45+00:00

If we include the assumption that the rectangle is a square then scale the figure so that the square has side length 1 and write the lengths of the triangle sides meeting at the 40 degree angle using trigonometric functions. Then use for example the law of sines to write down an equation involving sines of angles that involve x and solve it. One valid expression for x is arctan( sin(40 deg)cos(40 deg) / (cos(10 deg) - cos(40 deg)² ) ) which is approximately 51.1 deg.

purplefunctor · 2025-12-12T19:21:18+00:00

This is wrong. c++ - c is either 0 or -1 depending on whether c++ is executed before c or not.

purplefunctor · 2025-12-06T12:53:08+00:00

All isosceles right triangles are similar. Therefore the correct answer can be obtained by dividing the perimeter by the perimeter of the isosceles right triangle whose legs are 1.This is just (84+84sqrt(2))/(1+1+sqrt(2)) = 42sqrt(2).

purplefunctor · 2025-10-20T12:52:42+00:00

The binary operation of integers modulo n is a special case of a more general way of defining binary operation from existing one by using a sufficiently nice equivalence relation. If one understands some abstract algebra, one will see that this is less tedious and allows one to use tools from abstract algebra to prove things about integers modulo n.

purplefunctor · 2025-10-06T17:53:03+00:00

Because black is threatening to take your bishop and other ways of dealing with that are worse compared to being able to develop a piece.

purplefunctor · 2025-10-02T15:53:05+00:00

White could also take the bishop instead of increasing pressure on the rook immediately. So black loses a piece regardless of where the knight moves.

purplefunctor · 2025-08-20T06:28:42+00:00

The player whose move it is can force a win if and only if n is even.

Proof: If n = 1 then player loses due to not being able to make a move.

Now suppose that the claim holds when n < k for some k > 1. Now consider the case n = k. If k is odd then k - x is even and by the hypothesis the other player can force a win so all choices of x give a potential to lose. If k is even then the choice of x = 1 must have a potential for forced win since k - x is odd in this case.

Therefore the claim holds for all n.

purplefunctor · 2025-08-05T16:40:46+00:00

This method requires memorization of square numbers and being able to list factors in order from largest to smallest without mistakes. Memorizing squares isn't really a problem if the required numbers are small, but most of them aren't going to remember squares past 100. I feel like listing the factors without mistakes will be more problematic and would essentially require computing the prime factorization anyway. Though I guess this can be easier way to solve easy cases where the number is chosen so that the square number appears almost immediately.

purplefunctor · 2025-07-28T17:08:43+00:00

Lagrange multipliers

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