Hacker News new | ask | show | jobs
by anon4 4468 days ago
That's not a problem, because there are impossible games. Your task is just to form an impossible situation.

This game is MUCH easier than the original, at least for me. I never managed to beat the original, but on the other hand I've yet to be beaten by the AI.
1 comments
j2kun 4468 days ago
No, because the problem is deciding whether, for all possible moves you make, there is a move that the opponent can make, such that for all possible moves you make, there is a move the opponent can make, ... that will force a win or lose for one player.

This makes it smell like a PSPACE-hard problem (if you make the board size arbitrary).

[EDIT] Now I see what you mean, that you could start in a position where you can guarantee a win and so being in P doesn't matter (the computer would just be able to tell quickly that it cannot win if you play optimally). But this also isn't satisfying because it seems unlikely that a random starting position would put you in such a state (since it's so early in the game!).

link
anon4 4468 days ago
What I meant is that no matter how the computer plays, you can always serve it a sequence of 2s and 4s that will never reach 2048. Easiest is to make sure there's exactly one 2 on the board, then spawn 4s in the corners, always picking the corner farthest away from the other tiles. From time to time spawn a 2 if you can guarantee it won't align with the other 2 in a couple of moves. Near the endgame it is trivial to have three 2 tiles which you keep always separate with careful placement of 4s. At that point the game becomes literally impossible to win.

Edit: Try to play it the other way, too. Try dropping tiles in the best most convenient pattern for the AI. Notice the amount of clutter that inevitably results when reaching higher numbers. An interesting question would be, if you control both the tile placement and movement, what is the highest number you can reach? With the constraint you can place only 2s.

link
j2kun 4468 days ago
When you say things like, "you need a careful placement of 4s" it makes me think, "well that's not a proof."

So let me be formal. I'm posing the following decision problem: given a target score n and some initial board configuration of size (k * k), can one player force a score of at least n? I conjecture this problem is PSPACE-hard.

Your claim is that if n >= 2048 and k = 4 then the answer is always no (though I don't buy your justification). My question is more general, and it's clear that there is not a constant answer (e.g. if n is 2 the answer is always yes, but for some sufficiently large n the answer is always no)

link