Is it possible to win $2048$ in a $3 \times 3$ grid? I was playing the famous $2048$ game with the non-standard $3 \times 3$ grid. I quickly realized it was awfully difficult to even get up to $256$, with the best possible algorithm(of course in my head, not using a computer).
So my question is - Is it possible to win $2048$ in $3 \times 3$ grid? If yes, can it be shown mathematically? 
 A: Let's prove by induction that with $n$ squares whose initial values are all below $2^n$ you can get at most $1$ square with $2^n$ and all other squares will have less.
Base: $n=1$. You can get a $2$ and that's it, which equals $2^1$
Step: Suppose we have $n+1$ squares, and let the inductive hypothesis be that with $n$ squares you can at most get $1$ square with $2^n$ and all other squares have less than that
Well, before we can get to $2^{n+1}$ on the $n+1$ squares board you need to get two squares with $2^n$. Now, once you get one square with $2^n$, you need to use the remaining $n$ squares to get another $2^n$, because the square with the $2^n$ will be useless for that endeavor. And since you just got your first $2^n$ square, that means that all those squares have values below $2^n$, and thus we can apply the inductive hypothesis. 
So, by inductive hypothesis, you might be able to get one square with $2^n$ with those remaining $n$ squares, but all other squares will have less than that. So, even if you can get another $2^n$ square, once you combine the two squares with $2^n$ to get a square with $2^{n+1}$, you are left with $n$ squares all with less than $2^n$, and by inductive hypothesis the most you can get from those squares is just $1$ more square with $2^n$, and the rest will have less (again, the square with $2^n$ will not help in getting any bigger number, since it cannot be combined with any of the other $n$ squares while they all have values of less than that, which is why we can apply the inductive hypothesis). 
So, you cannot get a second square with $2^{n+1}$ and so you cannot get a square with $2^{n+2}$ either. Hence, you can at best get $1$ square with $2^{n+1}$ and all other squares will have less.
Applying this general result to your case: on a $3 \times 3$ grid, we can at most get $2^9=512$
A: It would be possible!! The turns right before the creation of the 2048 block would fill the 3x3 with $$1024,\ 512,\ 256$$
$$32\ \ ,\ \ 64\ ,\ 128$$ $$16\ ,\ \ 8\ \ ,\ \ 4$$
This is possible because in the standard game, about 10% of the generations will be 4s. In general, we can show that the max possible value of a tile can be described as $2^{n+2}$ where $n$ is the number of squares in the puzzle.
A: the largest tile you can get is a 1024, you need a lucky 4 spawn, (about 10%)
and if you were using the "snake" method the board would look like
4  ,4  ,8
16 ,32 ,64
128,256,512
however, this is nearly impossible to do, a more consistent method is what we call
"DPDF" or Double Perimeter Defense Formation which is when you arrange the board like this, and feed in the from corners.
8  ,2  ,x
64 ,16 ,4
256,128,4
the 8 2 and 4s are where you combine.
(i have gotten 3 1024s, but reached that last 4 around 35 times.)
best of luck!
