3 ms·

If someone could implement "512 in 2D", that'd be great. I might actually win a game.

by nmeofthestate 13y ago

killertypo 13y ago
"128" or even "64" is more my cup of tea :)
- RBerenguel 13y ago
  Let's make it 4 and get back to work ;) (only joking: I only got to 1024 myself, but I prefer Threes, the split between 1 and 2 is neat, makes me think about re-reading The Gods Themselves)
solox3 13y ago
Did you know that using only three directions drastically increases the chance of you winning, because it alters the way numbers are being stacked?
thaumaturgy 13y ago
I've beaten 2048 (i.e., developed a general algorithm that consistently wins). Do you end up with game boards with more than one 128 tile, 64 tile, etc.? The game has a natural tendency to spawn multiple high-value tiles, and preventing it from doing that is the nut of the game. Pick the bottom row and make that row your "accumulator". Pick a bottom corner; this bottom corner will store your largest value. Now, keep your bottom row sorted from low value tile to high value tile at all costs: 4 8 16 128 Let's say the bottom two rows look like this: 2 8 4 2 4 8 16 128 You drop the 8 down: 2 4 2 4 16 16 128 Now, merge left or right? To the right of course, you want to keep that bottom row sorted: 2 4 2 4 32 128 Now drop down (or continue merging to the right until you can drop a tile down into the empty space in the bottom row): 8 4 2 2 4 32 128 This "locks" the bottom row into position, allowing you to freely move the upper three rows left and right without disturbing the ordering of the bottom row. Row-locking is more important than clearing the board. Focus only on the bottom two rows for most of the game (until/unless most of the board fills up). Accumulate values in the bottom row, and build matching values in the row above it. What if the game plays dirty and you end up with: 4 32 128 2 Reverse the ordering of your bottom row. You now want to gradually shift that "128" tile to the left, so build on the "4" and "32" tiles until you get something like: 64 64 128 4 Now shift left and left again: 256 4 And drop new tiles into the bottom row: 256 4 2 2 And your bottom row is ordered once again. What if a higher value tile drops into the bottom row out of order? 32 4 16 128 Ignore it. Build those "4" and "16" tiles: 32 32 32 128 And then merge right and right again: 32 64 128 And drop a new tile into the empty position: 2 32 64 128 This is the situation you want to avoid: 16 64 128 64 This is a natural result of the rules of the game, and this is where it becomes difficult for the player to win, especially when those values start to look more like: 32 128 512 128 Now you suddenly have to do a lot of work in just three rows of space to double one of those 128 values. This approach makes the game pretty easy.
- superlucy 13y ago
  thx for the strategy... I developed something similar while playing, but this explains it much more explicitly. :)
- ajuc 13y ago
  I developed similar strategy, just in the top row (usually top 2 rows, because it's hard to keep hivh value tiles just the first row). The only problem is when you have no valid moves that keep the bottom row in place and you need to move up. In such case mova back down as fast as possible, and deal with that one "2" that was most probably spawned on the wrong side of accumulator row later.
- nmeofthestate 13y ago
  Thanks for that - I have now won my first game of 2048. I can die happy :)
lmm 13y ago
This version is actually much easier than the original.
- jaxn 13y ago
  I agree.
- acjohnson55 13y ago
  I think you're right. What makes the original so difficult is that if you're not careful, you can easily end up in a situation where the only move you're allowed is potentially disastrous. With 6 possible moves, that's significantly less likely. I just won on my first shot.
deleted 13y ago
[deleted]

If someone could implement "512 in 2D", that'd be great. I might actually win a game.

killertypo 13y ago

"128" or even "64" is more my cup of tea :)

RBerenguel 13y ago
Let's make it 4 and get back to work ;) (only joking: I only got to 1024 myself, but I prefer Threes, the split between 1 and 2 is neat, makes me think about re-reading The Gods Themselves)

solox3 13y ago

Did you know that using only three directions drastically increases the chance of you winning, because it alters the way numbers are being stacked?

thaumaturgy 13y ago

I've beaten 2048 (i.e., developed a general algorithm that consistently wins). Do you end up with game boards with more than one 128 tile, 64 tile, etc.? The game has a natural tendency to spawn multiple high-value tiles, and preventing it from doing that is the nut of the game. Pick the bottom row and make that row your "accumulator". Pick a bottom corner; this bottom corner will store your largest value. Now, keep your bottom row sorted from low value tile to high value tile at all costs: 4 8 16 128 Let's say the bottom two rows look like this: 2 8 4 2 4 8 16 128 You drop the 8 down: 2 4 2 4 16 16 128 Now, merge left or right? To the right of course, you want to keep that bottom row sorted: 2 4 2 4 32 128 Now drop down (or continue merging to the right until you can drop a tile down into the empty space in the bottom row): 8 4 2 2 4 32 128 This "locks" the bottom row into position, allowing you to freely move the upper three rows left and right without disturbing the ordering of the bottom row. Row-locking is more important than clearing the board. Focus only on the bottom two rows for most of the game (until/unless most of the board fills up). Accumulate values in the bottom row, and build matching values in the row above it. What if the game plays dirty and you end up with: 4 32 128 2 Reverse the ordering of your bottom row. You now want to gradually shift that "128" tile to the left, so build on the "4" and "32" tiles until you get something like: 64 64 128 4 Now shift left and left again: 256 4 And drop new tiles into the bottom row: 256 4 2 2 And your bottom row is ordered once again. What if a higher value tile drops into the bottom row out of order? 32 4 16 128 Ignore it. Build those "4" and "16" tiles: 32 32 32 128 And then merge right and right again: 32 64 128 And drop a new tile into the empty position: 2 32 64 128 This is the situation you want to avoid: 16 64 128 64 This is a natural result of the rules of the game, and this is where it becomes difficult for the player to win, especially when those values start to look more like: 32 128 512 128 Now you suddenly have to do a lot of work in just three rows of space to double one of those 128 values. This approach makes the game pretty easy.

superlucy 13y ago
thx for the strategy... I developed something similar while playing, but this explains it much more explicitly. :)
ajuc 13y ago
I developed similar strategy, just in the top row (usually top 2 rows, because it's hard to keep hivh value tiles just the first row). The only problem is when you have no valid moves that keep the bottom row in place and you need to move up. In such case mova back down as fast as possible, and deal with that one "2" that was most probably spawned on the wrong side of accumulator row later.
nmeofthestate 13y ago
Thanks for that - I have now won my first game of 2048. I can die happy :)

lmm 13y ago

This version is actually much easier than the original.

jaxn 13y ago
I agree.
acjohnson55 13y ago
I think you're right. What makes the original so difficult is that if you're not careful, you can easily end up in a situation where the only move you're allowed is potentially disastrous. With 6 possible moves, that's significantly less likely. I just won on my first shot.

deleted 13y ago

[deleted]