2048 and algebra (Part 3)

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?

To study this question, here’s a graphic showing the first nine moves in a typical game of 2048. I’ve included black circles to highlight the new 2-tiles and 4-tiles that are placed with each successive move, and I’ve added dark red ovals to indicate when two tiles are about to be combined in the next move.



In yesterday’s post, I raised one key insight about this game: we can calculate how many points were added for making each tile on the final board.

In today’s post, I raise a second insight. The final board has three 2-tiles, one 4-tile, and one 16-tile. So the sum of the tiles is 6 + 4 + 16, or 26. Also, during the course of the game, nine 2-tiles and two 4-tiles were introduced by the game. The sum of these tiles is 18 + 8, which is also 26. In other words, the sum of the tiles on the final board must equal the sum of the tiles that are introduced during the successive moves of the game.

With these two insights, we will (in tomorrow’s post) set up a system of two equations in two unknowns that will allow us to solve for the number of 2-tiles and 4-tiles that were introduced during the game using only the information on the final board.


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