# The largest tile in 2048, groups of 3 variant?

Question closed on SO for being too mathematical, I've re-asked it here.

Following the rules of the, as http://arxiv.org/pdf/1501.03837v1.pdf puts it, "slide and merge" game 2048, canonically implemented here https://gabrielecirulli.github.io/2048, but for dropping 3 and 9 tiles each turn instead of 2 and 4, and merging and summing three tiles together instead of two, to create this semi-common variant of the game.

Question:

What is the highest tile value for the above described, groups of 3 variant of the game 2048?

The vanilla 2048 answer is 2^17. However in this crowded groups of three variant of the game, the answer seems potentially to be a lot less straightforward.

Given the, as I understand it conclusions of the above mentioned paper, I wonder if a search (hard/exponential time) algorithm would be required to solve this problem.

Remember, in the real game (link above), and implementations of this variant, unlike in the majority of the referenced paper (link above), you start with an empty board.

As we are looking at perfect games here, 3^1 tiles need never be dropped and we can assume all tiles are added, once per turn, as the rarer 3^2 tile.

An example of a turn in the groups of three variant follows:

3 9 3 3
3 3 3 9
3 . . 9
. . . 9


Pressing Down:

. . . .
. . . .
. 9 3 3
9 3 3 27


Hypothetical boards:

The numbers in the graphics below represent $3^n$:

From the answers:

2 2 2
3 3 4 4
6 6 5 5
7 7 8 8


However it's probably not possible to get here, and it becomes:

3
3 3 4 4
6 6 5 5
7 7 8 8


And the three $3^3$s cannot collapse around the corner.

I have worked out that this board ought to be possible, but not if higher valued tiles are also possible:

2 2 3 3
2 2 3 3
5 5 4 4
5 5 4 4


In this case, then so far $3^5 = 243$ could be the largest possible tile.

• If you want to run parallel to the $2048$ game you will either need to drop $3$ and $6$ tiles, and add the values together - in which case the maximum value will be $3072$. Or drop $3$ and $9$ tiles and multiply tiles together when they clash - in which case it will be $3^{17}$. – Mathmo123 Aug 10 '15 at 10:08
• Neither, three tiles are combined in a row in these variants. – alan2here Aug 10 '15 at 11:05
• When a $3$ hits another $3$ what new tile do you get? – Mathmo123 Aug 10 '15 at 11:07
• Thanks :) Nothing, they cannot combine, I've added a diagram in the question to show this as it was not clear. – alan2here Aug 10 '15 at 11:14

## 2 Answers

Let's just imagine $9=3^2$ tiles are dropped all the time.

You need three tiles to get a $27=3^3$.

You need five to get $3^4$ because the step before you have two $3^3$s and three 9's.

You need seven to get $3^5$ because at some point before that you'll need two $3^4$'s and two $3^3$'s and three $3^2$'s.

So nine for $3^6$, $11=2.7-3$ for $3^7$, $13=2.8-3$ for $3^8$, and then 15 for $3^9$. So how about $3^9$ for an answer. What did I miss?

• Well worked out :) But see new information at the end of the question about this. – alan2here Aug 10 '15 at 11:45
Where x represents 3^x:

1 1 1 7
1 4 2 7
3 3 2 6
5 5 4 6

Move right:

2 7
1 4 2 7
3 3 2 6
5 5 4 6

Down:

7
1 4   7
3 3 3 6
5 5 4 6

Right:

7
1 4 7
4 6
5 5 4 6

Down:

7
7
1   6
5 5 5 6

Right:

7
7
1 6
6 6

Up:

1 7
6 7
6
6

Left:

1 7
6 7
6
6

Up:

1 7
7 7

Phew. Done.


And you can probably shift these around to get an 8 because of the new tiles.

• How did you get to the first diagram here from the start of the game? And no need to use any 3^1 tiles as in these perfect games we can presume our luck always gives us 3^2 tiles. – alan2here Oct 11 '15 at 9:48