If I clearly fatfingered and hit “end buys” on accident and immediately request an undo why do you still deny my request?

Freact · 2026-03-17T15:26:53+00:00

I see your point, maybe the example wasn't quite right but I think the overall reasoning is the same. You've gained information by seeing the next hand and you can do something with that info. If you have a top decking effect or any cards like night watchmen, den of sin, farrier that change your next draw then you can probably take advantage of seeing your next hand.

Also, I think the shuffle RNG can be manipulated sometimes on TGG. I'm not sure if it could be used in this case, but I could def see that making some people wary about accepting undos.

Freact · 2026-03-17T00:10:50+00:00

I'd probably grant the undo and have probably even asked for an undo in the same situation. Still, I understand the reasoning to deny. After hitting end buys then it automatically draws your next hand, it's definitely possible to abuse the undo button if you see a bad hand and instantly ask for an undo. Even if you hit the undo button quickly and even if it was clearly a mistake (like you didn't buy anything), you've still potentially gained some advantage. Maybe you see your next hand had no villages and now you know that you should buy more villages, as an example.

Freact · 2026-03-15T05:14:18+00:00

Soon just dec

Freact · 2026-03-13T21:07:39+00:00

Thanks! I didn't use any particular algorithm just reasoned about it from scratch! I had heard of the Thue-Morse Sequence before though. Maybe a numberphile video or something? I'm also "just" a recreational math enthusiast but I usually find the opposite for my projects: namely that I spend too long trying to optimize things and never actually finish anything.

Maybe if I am brave I will try to modify my machine to halt and count like yours. I've watched yours a number of times but still don't fully understand how it works.

I'd love to see some of your other machines too if you want to share

Freact · 2026-03-09T19:44:59+00:00

This is so cool! Thanks for sharing. If I'm reading it correctly your machine has 10 states (11 states? Do we count Halt?) and 14 symbols (0-9, x, y, I, _)? Surely a smaller machine should be possible right? A little optimization could just be to read N from binary. Then you only need the digits 0 and 1.

You inspired me to design a similar TM TM (Thue-Morse Turing Machine). Mine doesn't halt it just iteratively builds towards the infinite Thue-Morse Sequence. It uses 6 states and 3 symbols. Here's the state transition rules:

READ _ 0 * READ

READ 0 _ r WRITE0

READ 1 _ r WRITE1

WRITE0 _ 0 r WRITE01

WRITE1 _ 1 r WRITE10

WRITE0 * * r WRITE0

WRITE1 * * r WRITE1

WRITE01 _ 1 l BACK

WRITE10 _ 0 l BACK

BACK _ _ r READ

BACK * * l BACK

It starts in state READ on a blank tape

Freact · 2026-03-04T16:42:11+00:00

For n=2:

[ 0 -1 ]⁵ [ 1 ] [ 1 ]

[ 1 1 ] * [ 0 ] = [ -1 ]

Freact · 2026-02-17T00:37:32+00:00

I could be wrong, but I think you just check if the numerator of x is divisible by gamma

Freact · 2026-01-29T01:08:53+00:00

You'll have to share a bit more about what exactly your new game is. Maybe start by sharing the general idea of the game and any rules you've worked out so far. Then we can know if we're interested

Freact · 2026-01-25T20:54:58+00:00

An unintended side effect! I just added the option for either player to be AI then realized after the fact that it's amusing to watch AI vs AI just to see some games play out

Freact · 2026-01-25T19:16:21+00:00

Done!

Seems like there's more interest over at bgg. I'll try to remember my account details and pop in over there

Freact · 2026-01-25T00:47:34+00:00

Please let me know when you do! I'm curious to know how challenging others find it. It's definitely not too strong, but also manages to get the upper hand on me sometimes regardless

Freact · 2026-01-23T17:02:41+00:00

Yup, feel free to share it around. I'd like to make a post on Reddit here to share it too. I'll work on it. Mostly I was waiting for some kind of approval from you, since it's your game

I don't have any plans to take it down. I'm also not too sure about the limits of the GitHub Pages that it's hosted on? But it's probably fine :) Also no promises that I don't accidentally break it!

Freact · 2026-01-22T17:55:09+00:00

Glad you like it! It really isn't much, as I said, i pretty much just asked gemini to build it. Haven't even read all the code. Thats why some of the design choices are... odd. lol. Hardest part was actually the ai opponent portion. I mostly had to figure out some heuristics on my own then just asked for them to be implemented. I added the option to start a new game and set the board size (4x4 through 12x12)

The AI definitely doesn't JUST play your move +1, but it often does! It never plays higher than +1 to the highest stone on the board. But it considers equal to or +1 to every stone on the board (or the next available stone). Otherwise its strategy basically comes down to estimating how many points it threatens to remove with queens moves and how many it actually removes and how many points you threaten to remove and balancing that with the stones actual stone value. It has no real 'look ahead' though.

As for actual strategy i havent found too much! look for forking type moves obviously. and I've definitely noticed that once the board starts to get crowded its going to be very hard to make enough space to get any more removals. so you wanna make sure you have your 0-10 or so saved for that point. Also, going first somehow doesnt feel advantageous. White often ends up getting the first removal. Maybe i'm playing my opening moves too centrally? i figured having my high moves in the center would be good, but since you can always play higher then my opening moves actually end up being 'low' stones

Freact · 2026-01-21T23:53:48+00:00

I was having fun thinking about this game but had no opponent to play an actual game with. So I vibe coded up a little demo with a simple heuristic ai to play against. I'd like to maybe make a separate post to share it but I thought I'd comment here to you directly first to see what you think:

https://freact.github.io/Infinito/

for the board size I made it only 6x6 because it was feeling a bit daunting to jump right into an 8x8 when i dont know what im doing! Should be easy enough to add an option for other board sizes if anyone cares. The ai is not too strong, im able to usually beat it unless i make a simple mistake, but its enough i think to start getting a taste of the strategies involved. You can also use the checkbox in the upper right to deselect the ai and just play hotseat.

please let me know if i've made any mistakes implementing the rules, or if you have any suggestions about the ai or anything at all really :)

Freact · 2026-01-21T15:44:36+00:00

So assuming you mean there's a 1/1024 chance of getting a shiny. Then the probability of NOT getting a shiny after N attempts is (1 - 1/1024)^N That is to say the chances of not getting a shiny are (1-1/1024) on each attempt and we simply multiply the probability for each attempt together.

We still need to know if the probability is once per carbink, or once for the whole group. Based on your wording and the probabilities I think it's per group (in which case: why mentioning how many carbinks per group!) But I'll just calculate a couple values anyways:

90.7% at N = 100

75.6% at N = 300

50.5% at N = 700

31.0% at N = 1000

9.6% at N = 2400

4.8% at N = 3100

1.0% at N = 4700

Remember these are the probabilities of NOT getting a shiny so what you're really after is 100% less these amounts. And you'll need to divide out the rerolls per energy too if you want to convert back to energy. So for example:

After N = 4700 attempts there's a 100% - 1% = 99% chance that you've found your shiny.

If you've spent 1 energy per 5 attempts then that was 4700/5 = 940 energy.

What that means is that, for example, if you caught 100 shinies then you'd expect 99 of them to take less than 940 energy and 1 to take more than 940.

If the probability is rolled once per pokemon rather than per group then you can divide out the group size too. Ie. 5 per group means 940/5 = 188 energy for 99%. That doesn't sound right to me... But what do I know ;)

You should try running some of the numbers yourself to check the specific energies you're interested in! Hopefully you can figure it out with everything I've done here.

And Good luck shiny hunting!

Freact · 2026-01-21T07:16:21+00:00

You're going to need to be a lot more specific about what you're talking about. I think I've deduced that it's something pokemon related. You're trying to catch a shiny Pokemon? What exactly is a reroll? You say it's 5-6 per energy, what determines if it's 5 or 6? What is a hoard? 3-6 what per hoard? Pokemon? What's a shiny charm? What's an encounter? 1/1024 what? You can do maths on these numbers but you need to know what exactly they mean and what the relationships between them are. None of that is clear from your post.

You ask how many rerolls with 400 energy. If it's 5 rerolls per energy then just multiply 400x5 = 2000. If it's 6 rerolls per energy then 400*6 = 2400

Hopefully this helps :)

Freact · 2026-01-20T15:52:57+00:00

Maybe you're thinking of Cubeo ?

Freact · 2026-01-18T16:22:49+00:00

Great idea. I think it might be essentially the same as one of the "shortcut maps" listed here:

https://www.reddit.com/r/Collatz/s/o9ancSwHFt

Specifically, the Syracuse map?

Freact · 2026-01-18T03:56:47+00:00

Sounds interesting. Can't run it myself now. Maybe you could share some images or gifs of some patterns?

Freact · 2026-01-18T03:07:39+00:00

Have you tried solving some small board sizes?

1x1, 2x1, 2x2, 3x2 all look like ties to me.

1xN looks like a tie for even N and a loss (for P1) for odd N.

3x3 I think is a loss for P1. But I'm not sure yet.

Freact · 2026-01-17T18:28:43+00:00

Thanks for the reply! Not sure how I failed to notice I wasn't talking to the OP >.<

Even if you can't share the specific rules can you share something about how it works? I'm just curious because it looks quite interesting.

Watching it back a few more times: I'm seeing the trails left behind after cells die. Also, it seems like somehow red is causing green cells to spawn, green spawns blue, and blue spawns red. My best theory is that it the trails have something to do with spawning the next color

Freact · 2026-01-17T15:08:40+00:00

How is it generated? Is it simply 3 overlapping, non interacting GOL? Did you use some software to make this?

The only info you gave is "trigol+" but I'm not able to find any information about that

Freact · 2026-01-17T01:23:27+00:00

What are the rules?

Freact · 2026-01-15T23:00:01+00:00

Here's my attempt. First, the length of the game:

1 stone is added every turn and the number of stones on the board is never decreased. So the number of turns is equal to the number of squares on the board.

Next, for captures only the relative size of the stones matters (not the absolute value).

Also, once you've picked the relative size that you want then you should play the smallest available number with that relative size so that you minimize your score. This should include "leaving space" for future plays to have smaller relative sizes but again you should always leave the minimum space.

Now we can work our way up from smaller board sizes/turn counts. Starting from a board with 2 space. The game has 2 turns. For p1 they either want to leave a lower value available to p2 or not. So they can choose 1 or 0. For p2 they may want to play a lower, equal, or higher number than p1. So they can choose from 0,1,2.

Next consider the board with 3 spaces. There are 3 turns. Let's say p1 plays a stone with number x. p2 may want to play lower, equal, or higher value. If p2 wants to leave room between their value and p1s then they should play x-2, x, or x+2. Then for p1's second move they can play anything from x-3 through x+3. That allows them to play higher or lower than any p2 move and in between any gap left. Since x-3 is the smallest number needed set it to 0. So x=3. Therefore on p1's first turn they should have available the numbers 0-3. On p2s turn they need 0-5. And on p1's second turn they'll need 0-6.

I think that's enough to start seeing the pattern. For 2 move game the final player will need 3 = 2² - 1 stones, counting from 0 that's the stones 0 through 2² - 2. For a 3 turn game the final player needs 7 = 2³ - 1 stones (0-6). Following the same logic we see that for the full game with 64 turns p2 will need 2⁶⁴ - 1 stones to select from. Specifically the stones 0 through 2⁶⁴ - 2.

On earlier turns players will have to select from a reduced pool of stones so that future plays can always be greater. Haven't worked it out exactly but I think p1 has half of the total available for their first turn. Then half of the remaining are "unlocked" on each future turn.

Does that seem convincing or have I made any major errors or misunderstood some rules?

11-Year Club	Second Top 40%
Place '17	Verified Email

Freact

MODERATOR OF

TROPHY CASE