2048 with only 64 bits of state

https://github.com/izabera/bitwise-challenge-2048

by todsacerdoti • 1 year ago

173 43 1 year ago

github.com

meta-level• 1 year ago

Wow,I have a bash script that just starts a docker container with some fancy arguments roughly the same LOC...

JoeOfTexas• 1 year ago

I too am impressed with his bash script. Simple and concise.

kinduff• 1 year ago

The LOC are really impressive and the implemention pretty smart. Thanks for sharing!

lotyrin• 1 year ago

I was thinking "wow, how do you do that?" but then I forgot you're not dealing with any possible value between 1 and 2048 in a cell, only the powers of 2, so significantly fewer possible board states. Very cool.

I feel people often miss opportunities to map between (potentially complex, high entropy) state spaces into simple linear sequences of possible states and then either use those sequences to store the complex state data as a simple number or use the state of a system to encode data of some kind.

Like, if you use a limited 7-bit character set encoding for text and map that to a number in a sequence of possible orderings of a deck of 52 cards, you can store 32 characters (conveniently sized for passwords you might not want people to know are passwords).

carra• 1 year ago

It is worth mentioning that in the original game you can choose to keep going after you reach 2048. This game had to remove that option to achieve a 64bit state. Still, a clever implementation.

enriquto2• 1 year ago

keep going after you reach 2048. This game had to remove that option to achieve a 64bit state.

Since 15^16 < 2^64, you can still use a 64 bit state to reach 2^15=32768, which does not seem to be reachable in practice (the previous state would fill the whole board!)

ToValueFunfetti• 1 year ago

I'm not sure how the math works out, but some googling suggests that 32768 is achievable in practice while 65536 and 131072 are theoretically possible but have only been achieved with undos.

edit: Explanation of the 131072 cap: https://puzzling.stackexchange.com/questions/48/what-is-the-...

Also, in one of the answers there someone claims to have achieved 65536 in practice

sltkr3• 1 year ago

You need 16^16 for 32768 because 0 is used to represent an empty tile (2^0 = 1 is not used) which is exactly equal to 2^64.

The state in this implementation also stores a random seed (between 0 and 99, exclusive), so using 16^16 for the state would leave nothing for the random seed.

nneonneo• 1 year ago

Cool! I used a similar trick in my 2048 AI: https://github.com/nneonneo/2048-ai. In my case I allow the tiles to go all the way up to 32768, so I just have one tile per nybble.

Using this representation is great: it lets me pack a whole gameboard into a single machine register, which makes it super efficient. In this case I see that you’re able to pack in a seed value to enable replayability - neat!

kccqzy2• 1 year ago

Love the fact that you refer to your program as AI. We need more people to use AI to refer to non-LLM technologies lest the term be redefined.

enriquto2• 1 year ago

Total number of states = 11^16, assuming you don't reach 2048, just like you don't actually kill the king in chess.

Now log_2(11^16)=55.34, thus 56 bits are sufficient. That's tight!

nine_k2• 1 year ago

Encode them in CJK and combining Unicode characters, to keep the visual representation as short as possible.

eurleif2• 1 year ago

To me, the impressive part is implementing it in under 200 lines of bash script, not implementing it with only 64 bits of state. Nothing clever is required for 64 bits of state: a cell has 12 possible states (2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, and empty), which can be expressed in 4 bits, and 4*4*4 = 64.

charcircuit2• 1 year ago

a cell has 12 possible states

There is 18 states. The final possible board state is 16 increasing power of 2s starting at 4, since it's possible for a tile to spawn in as 4. Then you also need states for 2 and empty making for a total of 18.

Lerc• 1 year ago

That's a special case. You could have a single bit for special case mode and then very few bits to say what happened.

Usually you can encode special states in a more compact form because the type of specialness is additional information meaning the single special bit is all you need to grow by.

A bloated form for the final state would be a bit to indicate specialness, 7 bits for specialness mode. End state mode is encoded as the turn before plus direction. So adding 10 bits in all, there's almost certainly enough in the knowledge that it is an end state to encode the board in 10 bits fewer, eliminating all expansion except for the specialness bit.

pixelpoet• 1 year ago

There are*

vikingerik• 1 year ago

This is correct (if cells higher than 2048 are allowed, which varies by implementation.) But the game can still be represented in fewer than 16×18 bits, because not every board state is reachable. There can't be more than one cell with the maximum value. And if that is present, then there can't be more than one cell with the next-highest value, and so on. So you could devise some scheme to enumerate all reachable states and skip unreachable ones, and it would take fewer than 16×18 bits to indicate which one. The upper bound is at least bounded by ignoring representing the maximum value and then using 4 bits to indicate its necessarily singular position. There are also some other unreachable configurations, like if many copies of the same value are touching each other, since at least one pair of them would have been combined on the previous move.

nine_k2• 1 year ago

13 possible states, because the empty state is important, too. But they fit nicely into 4 bits.

It would be mildly interesting to super-package the state into the theoretically smallest possible amount of bits: 13 * 16 = 665416609183179841 possible states, which is approximately 2 * 59.2, so 60 bits would be enough, with some margin. A whole hex digit shorter!

eurleif2• 1 year ago

I included empty in my count.

sltkr3• 1 year ago

That's what the code already does. It packs 16 fields into a base 12 representation that takes up 57.36 bits and uses the remaining 6.64 bits to store a random seed between 0 and 99 (exclusive).

dmurray2• 1 year ago

Is that really the theoretically smallest?

I think you can prove that some states representable by your proposal are unreachable. E.g. a board with 16 2s. So there might be another bit to spare.

kccqzy2• 1 year ago

The board rotation trick is very nice to achieve the short code length!

        w) squish ;;
        a) rot; squish; rot; rot; rot ;;
        s) rot; rot; squish; rot; rot ;;
        d) rot; rot; rot; squish; rot ;;

izabera2• 1 year ago

thank you i was so proud of it lmao

mark_undoio• 1 year ago

Always fun to see what izabera has come up with - every single time I'm somewhere between delighted and terrified to see what she's made the computer do this time!

Normal_gaussian2• 1 year ago

YOLO install command

  curl -fsSL https://raw.githubusercontent.com/izabera/bitwise-challenge-2048/develop/2048.bash -o "$HOME/bin/2048" && chmod +x "$HOME/bin/2048" && 2048

teaearlgraycold• 1 year ago

You could pin to a commit for safety

Normal_gaussian2• 1 year ago

Thats a fair point

    curl -fsSL https://raw.githubusercontent.com/izabera/bitwise-challenge-2048/413abf35be0f47c4947a5ebd6d581355383041e1/2048.bash -o "$HOME/bin/2048" && chmod +x "$HOME/bin/2048" && 2048

hofrogs• 1 year ago

Why not

    $HOME/.local/bin

bvanderveen• 1 year ago

Like others I found the concise implementation to be impressive! I have noticed a bug though. Using the "drive into the corner" strategy (I keep the high tile in the bottom left) sometimes the top left tile randomly gets a smaller value (e.g., goes from 16 -> 4) when I slide to the left.

chadrs• 1 year ago

Yeah, the first few times I thought maybe I was just imagining it; but I went from STATE=7280398215952, tried to merge my 64s into the left corner and instead ended up in STATE=7280398025476; my 64s have become 4s.

Can't reproduce it exactly as it happened, but if I run `STATE=7280398215952 bash 2048.bash` and press a 8 times, the 128 always becomes a 2.

shawabawa3• 1 year ago

Doesn't happen for me, i wonder if it's only on certain architectures or certain versions of bash or something

izabera2• 1 year ago

thanks, i fixed it. it was only happening on bash 3.2 and i don't use a mac so i didn't see it. https://github.com/izabera/bitwise-challenge-2048/commit/73f...

rossant• 1 year ago

I'd love to see a minimal C program or similar, implemented as a single function `int64 run_2048(int64 state, int input)`, which takes the current state and an input (left, right, up, down), and returns the next state. Could be an interesting code golf exercise.

2048 with only 64 bits of state

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