Will someone do my homework?
12
150Ṁ269
resolved Dec 31
Resolved
NO

Resolves YES if someone posts answers with proof of work, resolves NO if nobody does it by close.

Definition:

S_n(q) is the number of permutations that avoids the pattern 'q'. There might be multiple patterns to avoid like in a)

Here it is:

a)

b)

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You’re “someone”, if you do your homework and post it below, then someone has done it and the market resolves “yes”.

@JohnSmithb9be I don't count.

@levifinkelstein This explains your trouble with math. Hard to pass a math class if you don’t count…

predictedNO

Answers:

(a) 2n

(b) Follows immediately due to Pythagorean theorem

Proof of work: $ echo -n Yff2d799f213f3670bf88081e3c095a5b647fab297ebc95c6e2d29712f63934a9POW | sha256sum

000001afa5c4bd8a20d1ccb3399e979758425b24745dd15c81ec750d99e9b57b -

The market didn't specify that they had to be correct answers, nor that the proof of work needed to be related to the questions...

(a) is actually really close to the correct answer, like 3 chars off

predictedYES

@jacksonpolack 3 chars off doesn't sound as good when the answer only has 2 chars in it 🤣

wait, (a) was fun, but (b) is a troll, right?

your formula for S goes negative for even n, and the number of permutations isn't negative

i noticed that after 30 seconds

@jacksonpolack (b) is not a troll, why would you think so?

because (7n^2-3n-2)/2(-1)^(n-1), which appears to be Sn(1342), is negative for large and even n, and the count of permutations can't be negative? good one

@jacksonpolack idk, I guess it's some extended definition? Please help me man, I really need these points

i have access to gpt-5, i bet it knows the answer, so i'll ask it. the api is expensive though, so i'll need you to chip in a bit...

@jacksonpolack I thought you were good at math man, come just solve it. I promise it's not a troll question.

whoa, whoa, settle down with the flattery! I'm not moved THAT easily.

Tell you what, let's get lunch together and chat. Maybe we can work out something that'll work for both of us.

also, are you having trouble with this one, are you running out of time, just don't want to do it, or have you already solved it but thought it'd be an interesting question

why not bounty

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