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1597 or 2584 is a variation of the game 2048 where pairs of tiles numbered with consecutive values in the Fibonacci sequence are merged into a new tile with the next value in the sequence.

An important difference is that new tiles may only have a value of either 1 or 2 (whereas in 2048 new tiles have a value of 2 or 4).

What is the largest value for a single tile that can be achieved in this game?

Here are some playable links: 1597, 2048, FIB.

There are also a few variations which are not of relevance for this puzzle, like this one where new tiles may have a value of 1, 2 or 3.

See also: What is the largest tile possible in 2048?

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  • btw If there is any 2048 fan (like me) who wants to try 1597, don't bother about it. I just kept alternating between the down and left arrow keys for 90% of the game, and managed to win it in 3 attempts. – ghosts_in_the_code Jan 02 '16 at 10:08
  • Well, that's because 1597 is way too low a goal. The "equivalent difficulty" of winning 2048 would be getting a 6765 tile. Making a 1597 tile is only as difficult as making a 512 and a 256 tile at the same time in 2048. – Zerris Jan 03 '16 at 00:42

1 Answers1

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The answer is:

3,524,578

Because:

The best you can do is spawning a 2 in the 16th spot, when everything else is lined up ready to be merged. This means your tiles are:

2 > 3 [makes 5] > 8 [makes 13] > 21... etc. The pattern here is fairly simple, we get to skip every other number, which means our tiles are f3, f4, f6, f8... ...f32. These all combine to create f33 = 3524578.

Proof that this situation is possible: if your tiles always spawn adjacent to the smallest previous tile, then we can unwrap the board into a single 16-long string of ordered tiles / empty spaces. Creating an f32 tile is identical to creating an f33 tile, except everything is one less in the series. This means that we have f31... ... f5, f3 = 2, f2 = 1, [blank]. Spawning a 1 in the blank spot clearly lets us combine all the way up, so we can create our f32. Creating an f30 uses identical tiles, except it doesn't need the initial f31 tile, so we can leave that slot filled by our recently created f32 and apply the same logic. This cascades all the way down to creating our f4, which is: ...f6, f3 = 2, [blank] - in which case, a spawned 1 will allow us to create our f4 = 3 without trouble. Thus there exists a sequence of tiles that will allow us to create every tile in f3, f4, f6... ...f32, which means that their combination (f33) is possible.

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