2048 and algebra (Part 5)

In this series of posts, I consider how algebra can be used to answer a question about the 2048 game: From looking at a screenshot of the final board, can I figure out how many moves were needed to reach the final board? Can I calculate how many new 2-tiles and 4-tiles were introduced to the board throughout the course of this game?

(The above board is the only picture I currently have of reaching 4096 without using an undo — the version of the game that I had at the time permitted two undos per game. I have also reached 8192 in game mode — I was really lucky that day — but I sadly don’t have a screenshot to memorialize the occasion.)

In the previous post, I used two insights  to develop of a system of two equations in two unknowns. Let $t$ and $f$ denote the number of 2-tiles and 4-tiles, respectively, that were introduced by the game. The sum of the tiles on the final board must also be the sum of the 2-tiles and 4-tiles that were introduced during the course of the game. Therefore,

$2t + 4f = 3 \times 2 + 3 \times 4 + 2 \times 8 + 16 + 4096$

$2t + 4f = 4146$

Next, we consider how the total of 44,148 points was reached by looking at the final tiles.

1. Each 8-tile was formed by joining two 4-tiles (for 8 points). Some of those 4-tiles were added by the computer; others were formed by joining two 2-tiles (for 4 points each). So each 8-tile is worth $1 \times 8$ points, plus 4 times some undetermined number.
2. Each 16-tile was formed by joining two 8-tiles (for 16 points). Each of those two 8-tiles was formed by joining two 4-tiles (for 8 points for each 8-tile, or $2 \times 8 = 16$ more points). Again, some of those 4-tiles were added by the computer; others were formed by joining two 2-tiles (for 4 points each). So each 16-tile is worth $16 + 16 = 2 \times 16 = 32$ points, plus 4 times some undetermined number.
3. There are no 32-tiles on this board. However, to continue the pattern, any 32-tiles are formed by joining two 16-tiles (for 32 points). Each of those two 16-tiles was formed by joining two 8-tiles (for 16 points for each 16-tile, or $2 \times 16 = 32$ more points). Each of those four 8-tiles was formed by joining two 4-tiles (for 8 points for each 8-tile, or $4 \times 8 = 32$ more points). Again, some of those 4-tiles were added by the computer; others were formed by joining two 2-tiles (for 4 points each). So each 16-tile is worth $32 + 32 + 32 = 3 \times 32 = 96$ points, plus 4 times some undetermined number.
4. By now, the pattern should be clear. Any 64-tile on the board would contribute $4 \times 64 = 256$ points, plus 4 times some undetermined number (as a reminder, the number of 4-tiles formed by adding in the course of making the 64-tile).
5. Any 128-tile on the board would contribute $5 \times 128 = 640$ points, plus 4 times some undetermined number.
6. And, in general, a $2^n$-tile would contribute $(n-2) \times 2^n$ points, plus 4 times some undetermined number.
7. In particular, when $n = 12$, a 4096-tile would contribute $10 \times 2^{12} = 40,960$ points, plus 4 times some undetermined number.

In summary, for the above board, there are:

• Three 4-tiles (for 4 times some undetermined number),
• Two 8-tiles (for $2 \times 8 = 16$ points plus 4 times some undetermined number),
• One 16-tile (for $32$ points plus 4 times some undetermined number), and
• One 4096-tile (for $40,960$ points plus 4 times some undetermined number).

Adding, this board will have $16 + 32 + 40,960 = 41,008$, plus 4 times some undetermined number.

What is this undetermined number? It is the number of 4-tiles that are formed by combined two 2-tiles throughout the course of the game thus far. Since $t$ is the number of 2-tiles that have appeared and there are three 2-tiles on the board above, we conclude that $t-3$ 2-tiles have been combined into $(t-3)/2$ 4-tiles throughout the course of the game, resulting in $4 \times (t-3)/2 = 2(t-3)$ points. Therefore, we have the second equation

$41,008 + 2(t-3) = 44,148$

Let’s start solving:

$2(t-3) = 3,140$

$t -3 = 1570$

$t = 1573$

Substituting into the first equation:

$2 \times 1573 + 4f = 4146$

$3146 + 4f = 4146$

$4f = 1000$

$f = 250$

So we conclude that 1573 2-tiles and 250 4-tiles were introduced to the board. Stated another way, about 86.3% of the new tiles were 2-tiles, while about 13.7% of the new tiles were 4-tiles. Also, since two tiles were on the board before any moves were made, a total of $1573 + 250 - 2 = 1821$ moves were needed to reach the above board.

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1. 2048 and algebra: Index | Mean Green Math

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