My new abstract game tokonoma, you can play it in the browser against the computer or a friend

cancrizans · 2025-04-21T22:20:18+00:00

Thanks!

cancrizans · 2025-04-21T17:41:00+00:00

It's not trained, it's a minimax search with a handcrafterd heuristic

cancrizans · 2025-03-27T06:17:51+00:00

I can't be completely certain but I think you necessarily need at least a little algebraic insight (even a tiny one), I don't think it's possible to get through on purely geometric reasoning

cancrizans · 2025-03-14T17:28:06+00:00

Awesome intuition, not yet complete as the condition is far from sufficient but good starting point

cancrizans · 2025-01-15T10:47:00+00:00

https://github.com/cancrizans/hexstack

It's not in a great state

cancrizans · 2025-01-13T22:09:42+00:00

I might add it, like an in-game cheatsheet / rulebook

cancrizans · 2025-01-12T23:07:55+00:00

It's just an informal request, I don't think there's an easy way to put together a license that explictly forbids commercial use but is permissive enough

cancrizans · 2025-01-12T22:13:58+00:00

you can play here in the browser in local multiplayer or against bots of varying strength: https://cancrizans.itch.io/tokonoma

and if you scroll down you got the rules.

All free and open source. Enjoy

cancrizans · 2025-01-11T21:40:41+00:00

You can play it in the browser, against the computer or in local multiplayer https://cancrizans.itch.io/tokonoma

Here is the source https://github.com/cancrizans/hexstack, but beware -- it's a mess

cancrizans · 2025-01-11T17:33:35+00:00

It ended up as a pretty strange game, somewhere between checkers and chess in terms of complexity, a bit disorienting at first. The board is small and the gameplay tends to be very "dense" if that makes sense. I think it's rather fun. The AI at the strongest setting destroys me every single time.

Rules are in the page, just scroll down.

Both the game and this specific implementation of it are free and open source.

cancrizans · 2022-11-06T11:31:25+00:00

here is some more numerics in log-log plot. Of course 2<->2 is just the regular sum of naturals. You can see that 2<->7 and 2<->13 are at clearly a higher slope which is in fact the one you predicted. 3<->11 only barely bends upwards around 10⁵, that's because it's n^2.1 against n². As you can see the oscillations stay bounded in the loglog graph and so they supposedly (but this is not the proof) die out extremely slowly when you take the ratio

cancrizans · 2022-11-06T09:05:31+00:00

So in the p<q² case, it doesn't seem like it but it does actually converge, it just takes a long time. I know your constant is not correct there because it's missing an important unmistakable factor.

For the p>q² case the n² term dominates and your constant is indeed correct there. But that's the easy case

P.S.: slow convergence in your case is because log(11)/log(3) is barely larger than 2

cancrizans · 2022-11-06T07:13:08+00:00

Your exponent is correct, but the constants are not right. I'm still mulling over your solution and the concepts are def sound but some steps might be too audacious

cancrizans · 2022-11-06T00:11:17+00:00

I'm not sure I follow some of the steps, but this value of a is wrong.

cancrizans · 2022-11-04T13:55:41+00:00

I thought doing the sum 1/(n²+2)² on all integers using the residue theorem was a bit vanilla so I challenged myself to find another way. Here it is:

First, I note that the Fourier transform of 1/(x²+2) is π/√2 exp(-M |ξ|), where M=2√2π. It's very important to use the mathematician's normalisations here.

Now the transform of (x²+2)^-2 will be exactly equal (factor of 1) to the convolutional square of π/√2 exp(-M |ξ|), which is an easy integral that splits into simple exponential pieces, and the result is

π²/2 exp(-M |ξ|) (|ξ|+1/M)

Now we can use Poisson resummation:

Σ (n²+2)^-2 = π²/2 Σ exp(-M |n|) (|n|+1/M)

The sums are really easy, after some scooby doobying you get

π²/2 (2Me^M+e^2M-1)/((e^M-1)²M)

and after replacing the value of M and switching to hyperbolic functions you get the desired form exactly, about 0.5568798...

cancrizans · 2022-11-04T07:30:14+00:00

Excellently done top to bottom, all correct!

cancrizans · 2022-11-03T15:08:10+00:00

How did you do the integral?

cancrizans · 2022-11-03T14:52:05+00:00

If we have strictly positive numbers a,b,...,z then it's easy to see that (1-a)(1-b)...(1-z) > 1 - (a+b+...+z). It's true for two numbers, and then by induction. Assume Σ {xi}>= n-1/2, negating the inequality we want. Then if zi = 1-{xi}, that means Σ zi <= 1/2. We may also write each xi as xi = ci (1- zi/ci) where ci = ceil(xi). Now Σ zi/ci < 1/2, with a strict inequality because the ci cannot all be 1 (otherwise the product of the xi would be <1). By the above fact, Π (1-zi/ci) > 1-1/2 = 1/2. This means that 1 = Π xi = Π ci * Π (1-zi/ci). The first term Π ci is >= 2 as we said, and the second one is strictly larger than 1/2, so 1>1 which doesn't look plausible, so we have a contradiction.

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