25

I barely managed to reach 64 in the 3×3 board when it was game over. 4×4 is still doable, but 3×3 is pretty impossible to me.

0
47

That's mathematically not possible. Let's assume that you are very-very lucky and you only get number 4 tiles as the new ones.

That is 2^2, and we have 9 tiles, each of them twice as big as the previous one. That adds up to:

2^2, 2^3, 2^4, 2^5, 2^6, 2^7, 2^8, 2^9, 2^10

As you can see, 2^10 is the maximum possible we can have, which is just a 1024

So to answer your question: it's not possible to reach 2048 tile in a 3x3 version of the game(assuming that the rules are the same as with 4x4: 1 new tile after each move, which is either 2 or 4)

10
  • 7
    @EricDuminil, not a stupid question. To reach a 2048, at one point the sum of all pieces must have been 2044 or 2046, i.e. binary 1111111110 or 1111111100. The only way to get this with at most 9 pieces is to have the pieces which are shown in the answer. Unfortunately, having these 9 pieces means that the position is locked.
    – Carsten S
    Jan 10 at 14:55
  • 1
    @EricDuminil to put it another way, to get the next tile, you must merge two copies of the previous tile, right? so if you're missing a tile, there will be only one copy of some tile after the previous merge (= no next merge)
    – somebody
    Jan 10 at 18:06
  • 5
    @Dave incorrect, note that the sequence given is 9 tiles long - it is game over, there is no way to free up a space to let the other 4 spawn in
    – somebody
    Jan 10 at 18:09
  • 2
    ah because it needs a second 4 to merge with, yes. Quite right
    – Dave
    Jan 10 at 18:10
  • 3
    @Dave: In order to reach a situation with two 1024 tiles, one must have been in a situation with one 1024 tile and two 512 tiles. In order to reach such a situation, one must have previously been in a situation with either four three 512 tiles and two 256 tiles, or else one 1024 tile, one 512 tile and two 256 tiles. In other words, a situation with two 256 tiles and at least two higher tiles. To reach that, one must have been in a situation with two 128 tiles and at least three higher tiles, two 64 tiles and four higher tiles; also 2x32 and five; 2x16 and six; 2x8 and seven; 2x4 and eight.
    – supercat
    Jan 11 at 17:13

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