53

In the original game, sometimes the new tile that pops up is a 4. Let's ignore that possibility for the sake of simplicity. We've 16 squares in total and considering the new tile popped up at the position we chose, to make it to 4096, we need minimum these tiles: 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 2 These can be combined from right to left ...


18

BlueFlame gives an essentially correct answer, but I wanted to add a rigorous proof (for anyone looking for it). [EDIT: user2357112's comment points out an annoying hole in the proof, under the "proof that 65536 is theoretically possible" section. While I believe the hole can be filled, I don't have time to do it right now.] I will prove that 65536 is the ...


16

No. Both 2s and 4s can spawn on a turn. So it's possible to have a board with a single 4 and a 2 at the start of the game, or from starting with two 2s, merging them, and having another 2 spawn in. If only 2s spawn, then yes. Look at one of the largest pieces, $2^n$: it could only have come from two pieces of size $2^{n-1}$, and it must have yielded $2^n$ ...


12

If you were unlucky enough and were not trying...You could in theory end up with a square full of alternating 2s and 4s. It should even be possible to do this without matching, which means a score of 0.


12

Try to line up numbers in descending order on a row of your choice (I prefer the top row), and avoid messing this up at all costs (especially the cell with the biggest number, which should be in a corner). Example: $$ \begin{array}{|r|r|} \hline 1024 & \phantom{0}512 & \phantom{00}16 & \phantom{000}2 \\ \hline 2 & 2 & 4 & 2 \\ \end{...


10

Okay, so I actually made it this far (no photoshop or anything, just a secret trick and play endlessly without any hassle. It works for long time also on normal mode, until you make a mistake). No more possibilities... EDIT: So someone did ask me for the "trick", so here it is: First, always corner your highest tile. This is very important, as this gives ...


7

No, it is impossible to make a 2^18 tile in a 4x4 2048. You could make a 2^17 tile and a 2^16 tile, but you run into the issue of needing to make a 2nd 2^16 tile to merge into that. I've attached a picture of the largest possible board state that you could get, even with a 4 (2^2) tile spawning in the last remaining slot. If the picture isn't proof ...


5

When playing deterministic 2048, where a new tile is placed in the first available spot, you can get a pattern like this using left-down-left-down: The worst I have been able to get in regular 2048 is this: It is possible to get that worst-case scenario if your tile generator and human are working together to do so.


4

Too big for a comment. If new tiles must appear in the bottom three rows and the top row must be in descending order, then the following works. For convenience, the new tile always appears in the bottom row. .... 4... 8... 8... 84.. 88.. 884. 884. .... u .... u .... u 4... u 4... u 4... u 4... u 8... u .... .... .... .... .... .... ...


4

If the first row goes through these states: ...2 2..4 24.2 2424 then it's possible to fill up that row without anything being in the bottom three rows. (Each time, the new tile spawns in the top right, and the key pressed is left.) If new tiles must appear in the bottom three rows: .... .... .... 2... 2... .... .... .4.. 24.. .... .... ..2. 242. .......


4

The answer is: Because:


4

A pattern that I have had success with is as follows: Alternately swipe up ($\uparrow$) and left ($\leftarrow$) (or whichever two directions you choose) as quickly as you can. When you get stuck (or when you see an opportunity$^1$), swipe right ($\rightarrow$), then up ($\uparrow$) a few times$^2$, then continue with step 1. Keep this up until you get the ...


3

I think


3

My brother used to play this game for a long time. He also used to play in practice mode (so he could undo his last move). When I saw this question, I asked him if he could send me a screenshot of his best result. This is what he sent me (no cheating or photoshop):


3

We can simplify 2048 by discarding the layout and focusing on a multiset (or bag) of tiles. We can do this, if we assume that a new tile will always appear in a position that is most advantageous to us. We can then define two rules to make a new bag out of the current one (we also ignore the possibility of getting 4s for simplicity's sake): Combine one or ...


3

The number one most important tip in my opinion is to keep your highest tile in one corner, and your next highest tiles adjacent to it. You should only move your top tile from the corner when absolutely necessary.


2

According to the programs written on Stack Overflow, you can get the 8192 tile always, after which it depends on luck. Note that they are not 'perfect', and use alpha-beta pruning. If you are asking about the probability of greater tiles, try the math SE.


2

Up until 4096, @Doorknob's solution seems optimum, using a zig zag along one row then back along the next. Once you have a 4096 in one corner, the number of free tiles limits your options if you go for a single direction, though, so the technique I use once I have my bottom row filled (I like my top number at bottom left) is to add to both ends of the next ...


1

The answer is $2^{n^2+1}$. This is because $n^2$ is the number of squares in an $n\times n$ grid, and the $+1$ comes from requiring that the last square spawned be a $4 = 2^2$.


1

There is no single anawer to this question because the game randomly adds a 2 or 4 tile. But it is possible to give a range. This answer shows the maximum possible score. The sum of all tiles is: $2^{18}-4=262140$ To reach this result you need a 4 as a new tile at least at 16 specific times: first time, when the rest of the board is filled with values ...


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