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The standard answer I've seen presented for "What is the largest tile possible in 2048?" is 2^17, produced by collapsing a 16 tile long chain into one large tile in the corner.

The answer then generally states that it would be impossible to create a larger tile because it would take a 17 tile chain to create, which is space that we no longer have.

I find this answer compelling, but I have a friend that disagrees and posits that because it takes 15 moves to collapse down to the 2^17th, 15 more tiles are generated, providing the opportunity for them to collapse, giving you a "head start" so to speak toward 2^18.

To be clear: I do not agree with my friend, and don't believe the extra tiles make a difference, but is it possible that it could? If not, is there a simple way to explain why?

Dr_StrangeKill
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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 enough the easiest way to prove it's not solveable would be to prove that you can't make a 2^17 tile in 15 spaces, or a 2^16 in 14 spaces working down to it's impossible to make a 2^3 tile in a single space.

enter image description here

Evan Capstick
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Here's a rigorous proof:

At every point in the game, there must be at least one 2 or 4. For a 2^18 tile to be made, the total of all tiles must surpass 2^18. At some point, the total must either be exactly 2^18 - 2 or 2^18. Having a total of 2^18 - 2 would require 17 tiles, which is impossible. Having a total of 2^18 and a 2 or 4 on the board also requires 17 tiles, which is impossible. Therefore, we cannot make 2^18.

mathlander
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