2048 and algebra (Part 7)

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 yesterday’s post, we 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:

2t + 4f = \displaystyle \sum_{n=2}^{17} 2^n.

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

In this post and tomorrow’s post, I consider how the two sums in the above equations can be obtained without directly adding the terms.

The first sum is certainly the easiest to handle, as it requires the sum of a finite geometric series:

a + ar + ar^2 + \dots + a r^{n-1} = \displaystyle \sum_{i=1}^n a r^{i-1} = \displaystyle \frac{a(1-r^n)}{1-r}

For the geometric series

\displaystyle \sum_{n=2}^{17} 2^n,

there are 16 terms (after all, there are 16 tiles on the board). The first term is 4, and the common ratio is 2. Therefore,

\displaystyle \sum_{n=2}^{17} 2^n = \displaystyle \frac{4(1-2^{16})}{1-2}

\displaystyle \sum_{n=2}^{17} 2^n = 4(2^{15} - 1)

\displaystyle \sum_{n=2}^{17} 2^n = 262,140

We’ll consider the more complicated sum in tomorrow’s post.


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