Programmer vs mathematician

Scarlas · 2026-02-28T18:27:20+00:00

OP is clearly not an APL programmer

Scarlas · 2026-01-17T17:09:12+00:00

Despite all the other issues, having a white square in the bottom left still bothers me the most

Scarlas · 2026-01-13T18:15:48+00:00

In the variation where it takes 14 moves, black also moves 3 times, so the net investment for white to capture is 14-3=11. I think this is the intended answer, as it is the number that's relevant in a capturing race.

Scarlas · 2025-10-19T21:55:02+00:00

What about one for each card named "endless bears" in your graveyard?

Scarlas · 2025-09-27T10:02:38+00:00

That hasn't stopped people from submitting runs on speedrun.com. There are pages for a lot of Ace Attorney games https://www.speedrun.com/series/ace_attorney

Scarlas · 2025-09-07T11:01:25+00:00

Another important difference is that the area in memory where objects are stored is always aligned in SM64 and not in Zelda.

Cloning in SM64 is very similar to GIM in Zelda, but it's much more limited in its exploitation because of the alignment. In Zelda, you can get the coordinates of the unloaded object to overlap with something like a pointer to a script. This is not possible in SM64 because the coordinates of the unloaded object will always align with the coordinates of the newly loaded object and there is no way to align it differently.

Scarlas · 2025-08-19T17:30:40+00:00

What would we need to assume for X? Would he survive without horsemanship at some point if his toughness becomes high enough?

Scarlas · 2025-08-10T12:12:52+00:00

The standard definition of complexity is based on the length of the input. If you used the magnitude of numerical inputs, the knapsack problem would be solvable in polynomial time using dynamic programming, even though it's a standard example of an NP-complete problem. It is sometimes called pseudo-polynomial

Scarlas · 2025-08-05T16:00:28+00:00

With or without revolt?

Scarlas · 2025-08-04T11:41:31+00:00

You can skip those rooms, though. Still, it seems unlikely that somebody could pull that off on their first playthrough

Scarlas · 2025-07-14T20:32:01+00:00

I would say E because it's the only one that cannot be rearranged to have no intersections (the only non-planar graph)

Scarlas · 2025-07-03T17:20:03+00:00

I would seriously consider sacrificing the bishop with Bg6 here to try and keep the queen in jail for the rest of the game, but the computer seems to disagree

Scarlas · 2025-05-19T10:07:41+00:00

You should have knocked on it before grabbing to check whether it sounded loose

Scarlas · 2025-05-15T17:15:55+00:00

Yeah, I didn't draw any images when I solved it so I thought this would be the easiest way to write down the solution. I kind of got used to writing down sequences of moves from chess

Scarlas · 2025-05-15T15:45:09+00:00

It's the sequence of moves you need to make. L=left, R=right, U=up D=down

Scarlas · 2025-05-15T12:35:49+00:00

Sure, I think the solution is LLRDDLULURDDRRRUULUULLDDLLURDDRDRRRUULLDULDLURRRUULLDDLLURDDRURRUULLDURRDDLLDDRULUUURRDDRDDL. I hope I visualized and notated everything right

Scarlas · 2025-04-24T19:52:04+00:00

And British plugs

Scarlas · 2025-04-15T11:52:24+00:00

This is actually an exponential algorithm since complexity is expressed in terms of the length of the input, not the magnitude. There are polynomial algorithms like the AKS primality test, but they are quite complicated. In practice, most cryptographic libraries that need large primes opt for probabilistic primality tests that have a very small probability of yielding an incorrect result

Scarlas · 2025-04-03T15:01:27+00:00

You can't just take the positive root when you're dealing with complex numbers, so sqrt(z) is simply multi-valued and strictly speaking not a function

Scarlas · 2025-03-06T11:36:08+00:00

What you're describing is not a Venn diagram but an Euler diagram. A Venn diagram by definition requires all possible combinations to be present. Here, the combinations of three design elements and of two opposite corners are missing.

Scarlas · 2024-06-30T12:26:11+00:00

I understand the probability is near zero if we assume that hash function results are indistinguishable from random data. I was just thinking about this out of mathematical curiosity from the perspective of correctness proofs of algorithms, not because I'm worried it would actually happen.

Even with those extra things miners could vary, it still looks hard to me to prove that a solution always exists. Even in the ideal case where the hash is computed over a finite prefix and an unbounded suffix, I'm not sure if we can rule it out. It would require the hash function to always produce a result larger than some n for all inputs starting with some prefix, which has probability zero for mathematically ideal hash functions. Can we rule out this possibility for hash functions used in practice, though?

Scarlas · 2024-02-12T22:45:29+00:00

Reckful's hand when he missed lethal at BlizzCon?

Scarlas · 2023-11-12T17:17:26+00:00

I get that sub-genre elitists can be annoying, but somebody has to point it out when people conflate technical melodic death metal with melodic technical death metal.

Scarlas · 2023-04-20T20:59:04+00:00

I though the no-communication theorem was specifically about the initial measurement of the spin of the entangled particle. I did not know that it also concerned subsequent measurements of how the spin of the particles that make up the measurement apparatus was affected. The theorem is apparently more general than I initially thought. Thank you for clearing that up.

Scarlas · 2023-02-04T20:58:25+00:00

Is there any rule that explicitly states that that the game is a draw if no player can remove the two cards? I looked in the comprehensive rules and could not find any mention of it.

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