# 2048 and algebra (Part 6)

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? In this post, we consider the event horizon of 2048, which I reached after about four weeks of intermittent doodling:

In the previous posts, we have developed a system of two equations in two unknowns to solve for $t$ and $f$, the number of 2-tiles and 4-tiles (respectively) that appeared throughout the course of the game.

The first equation,

$2t + 4f = \sum T_i$,

says that the sum of the tiles that were introduced has to be equal to the sum of the tiles $T_i$ that appear on the final board. Directly adding the sixteen tiles above yields

$2t + 4f = 262,140$.

This sum can also be calculated using a trick to be discussed in tomorrow’s post.

The second equation,

$2(t - t_0) + \displaystyle \sum_{T_i \ge 8} (\log_2 T_i -2) T_i = P$,

says that the total number of points $P$ may be divided into the contributions provided by the tiles on the final board. For example, the 16-tile was formed by joining two 8-tiles for 16 points. Each of those 8-tiles were formed by joining two 4-tiles for another $2 \times 8 = 16$ points. Added together, the 16-tiles results in $2 \times 16 = (\log_2 16 - 2) \times 16 = 32$ points. This analysis does not account for any 4-tiles that were created by adding two 2-tiles. The number of such 2-tiles is $t - t_0$, where $t_0$ is the number of 2-tiles that appear on the final board (in this case, $t_0 = 0$). These additions result in $(t - t_0)/2$ 2-tiles worth $4 \times (t-t_0)/2 = 2(t-t_0)$ points.

For the board above, this equation becomes

$2t + \displaystyle \sum_{T_i \ge 8} (\log_2 T_i - 2) T_i = 3,867,072$

and (for this particular board) the sum can be written more simply as

$2t + \displaystyle \sum_{n=1}^{15} n \cdot 2^{n+2} = 3,867,072$

Directly adding the sum $1 \cdot 8 + 2 \cdot 16 + 3 \cdot 32 + \dots + 15 \cdot 131,072$ — and being very careful to double-check the arithmetic — yields the second equation

$2t + 3,670,024 = 3,867,072$

This sum can also be calculated using a trick to be discussed in a future post.

Solving, we find

$2t = 197,048$

$t = 98,524$

Substituting into the first equation:

$2 \times 98,524 + 4f = 262,140$

$197,048 + 4f = 262,140$

$4f = 65,092$

$f = 16,273$

So we conclude that 98,524 2-tiles and 16,273 4-tiles were introduced to the board. Stated another way, about 85.8% of the new tiles were 2-tiles, while about 14.2% of the new tiles were 4-tiles. Also, since two tiles were on the board before any moves were made, a total of $98,524 + 16,273 - 2 = 114,795$ moves were needed to reach the above board.

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