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2048 is played in a 4*4 space. At the start there are 2 tiles spawned.

Move - a swipe followed by a new tile (either 2 or 4) spawning.

Position - any configuration of the various tiles and blanks on the board.

End Position - a position where no moves are possible (must have either 2 or 4 on one of outer 12 blocks to be obtainable)

Valid Position - obtainable through play.

Invalid Position - unobtainable through play. (all 2's)

For a given position p and natural number n, let $n(p)$ be the number of possible positions n moves earlier. Let $n^+(p)$ be the sum of all the different ways of each position from n(p) to reach position p.

Note that $n^+(p) \geq n(p)$; there are two ways figure 1 could transpose to figure 2. enter image description here

figure1______________________________figure2

Example where $1(p) = 1^+(p)$: enter image description here

Is it possible to backtrack from a valid position and get an invalid position? If so, can n(p) not increase when add 1 to n?

How to tell valid positions from invalid?

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Xiangyu Chen
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  • You forgot to explain a main element of the rules, that a move involves swiping a number onto an adjacent equal number, resulting in a tile with the double value. – joriki Jun 30 '18 at 07:16
  • It's very inopportune to have $n$ and $\mathbf n$ denote two different things -- there are so many letters available, why reuse the same one? – joriki Jun 30 '18 at 07:17
  • I don't understand your first question. Obviously there's an algorithmic way to calculate these functions; just check all possible ancestor positions. Are you looking for any particular properties in the algorithm? – joriki Jun 30 '18 at 07:21

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