Fixed, thanks. The proof was right, and the original claim was trivial.
Im not intending to go back to thinking about this anymore, but diffractor is the person who was thinking /writing about it. https://www.lesswrong.com/posts/SgkaXQn3xqJkGQ2D8/cooperative-oracles
Update: we will be using a private gather town rather than Walled Garden.
Also, hear is a hint on puzzle 1, you could read first:
I believe the above answer is optimal. It only passes through natural numbers it its intermediate computations. I believe it is impossible to match the score of the above solution if you limit yourself to natural numbers below a googol.
Typing >! at the start of a line makes a spoiler
I conjecture you can get every positive integer with cost 3 and a single floor function, but it also seems very hard to prove.
Oh, yeah, oops, I didn’t read it well enough. Anyway, it is right now, so ill leave my comment endorsing it.
This comment contains the best I can do on puzzle 2. Read it only if you want to be spoiled.
This comment is the best I can do on puzzle 3. Read it only if you want to be spoiled.
This comment is the best I can do on puzzle 1. Read it only if you want to be spoiled.
Yeah, I changed it, thanks
Puzzle 2 does not allow primes over 2021
Puzzle 1 does not allow the use of 1 (also, I changed the scoring to the product rather than the sum)
I changed the scoring, so now √24!−2−33 scores 144.
I changed the scoring, so now 211−33 scores 198.
You can right now (I believe until Jan 5, 2021) get all five of the tier 1 games in this list on Steam for a total of $34.30.
Note that in my experience, Recursed works easily on MacOS 10.15, in spite of the warning, but Braid does not.
I want and appreciate nits. Fixed.
I don’t think you lose much by focusing on finite Cartesian frames. I have mostly only been imagining finite cases.
I think there is some potential for later extending the theory to encompass game theory and probabilistic strategies, and then we might want to think of the infinite space of mixed strategies as the agent, but it wouldn’t surprise me if in doing this, we also put continuity into the system and want to assume compactness.
I think the h‘s are correct. Any morphism in which either component is the identity must be homotopic to the identity, since homotopic is symmetric. In this proof, checking the h’s is easier.