The Maximum Score in the Game “Entanglement” is 9080
Entanglement is a browser-based game that has gained a fair bit of popularity lately due to its recent inclusion in Google’s Chrome Web Store and Chrome 9. The way the game works is probably best understood by actually playing it, but here is my brief attempt:
- You are given a hexagonal tile with six paths printed on it, with two path ends touching each side of the hexagon. One such tile is as follows:
- You may rotate, but not move the hexagon that has been provided to you.
- Once you have selected an orientation of the hexagon, a path is traced along that hexagon, and you are provided a new hexagon that you may rotate at the end of your current path.
- The goal of the game is to create the longest path possible without running into either the centre hexagon or the outer edge of the game board.
To make things a bit more interesting, the game was updated in November 2010 to include a new scoring system that gives you 1 + 2 + 3 + … + n (the nth triangular number) points on a turn if you extend the length of your path by n on that turn. This encourages clever moves that significantly extend the length of the path all at once. The question that I am going to answer today is what the maximum score in Entanglement is under this scoring system (inspired by this reddit thread).
On a Standard-Size Game Board
The standard Entanglement game board is made up of a hexagonal ring of 6 hexagons, surrounded by a hexagonal ring of 12 hexagons, surrounded by a hexagonal ring of 18 hexagons, for a total of 36 hexagons. In order to maximize our score, we want to maximize how much we increase the length of our path on our final move. Thus, we want to just extend our path by a length of one on each of our first 35 moves, and then score big on the 36th move.
Well, each hexagon that we lay has six paths on it, for a total of 6*36 = 216 paths on the board. 35 of those paths will be used up by our first 35 moves. It is not possible to use all of the remaining 181 paths, however, because many of them lead into the edge of the game board or the central hexagon, and connecting to such a path immediately ends the game. Because there are 12 path ends that touch the central hexagon and 84 path ends that touch the outer border, there must be at least (12+84)/2 – 1 = 47 unused paths on the game board (we divided by 2 because each unused path takes up two path ends and we subtracted 1 because one of the paths will be used by us).
Thus we can add a length of at most 181 – 47 = 134 to our path on the 36th and final move of the game, giving a total score of at most 35 (from the first 35 moves of the game) + 1 + 2 + 3 + … + 134 = 35 + 9045 = 9080. Not only is this an upper bound of the possible scores, but it is actually attainable, as demonstrated by the following optimal game board:
Paths in red are unused, the green line depicts the portion of the path laid by the first 35 moves of the game, and the blue line depicts the portion of the path (of length 134) gained on the 36th move. One fun property of the above game board is that it is actually completely “unentangled” – no paths cross over any other paths.
On a Larger or Smaller Game Board
Other than being a good size for playability purposes, there is no reason why we couldn’t play Entanglement on a game board of larger or smaller radius (by radius I mean the number of rings of hexagons around the central hexagon – the standard game board has a radius of 3). We will compute the maximum score simply by mimicking our previous analysis for the standard game board. If the board has radius n, then there are 6 + 12 + 18 + … + 6n = 3n(n+1) hexagons, each of which contains 6 paths. Thus there are 18n(n+1) lengths of path, 3n(n+1)-1 of which are used in the first 3n(n+1)-1 moves of the game, and we want to add as many as possible of the remaining 15n(n+1)+1 lengths of path in the final move of the game. There are 12 path ends that touch the central hexagon and 12 + 24n path ends that touch the outer edge of the game board. Thus there are at least (12 + 12 + 24n)/2 – 1 = 11 + 12n unused paths on the game board.
Tallying the numbers up, we see that on the final move, we can add at most 15n(n+1)+1 – (11 + 12n) = 15n2 + 3n – 10 lengths of path. If T(n) = n(n+1)/2 is the nth triangular number, then we see that it’s not possible to obtain more than 3n(n+1)-1 + T(15n2 + 3n – 10) = (225/2)n4 + 45n3 – 135n2 – (51/2)n + 44 points. In fact, this score is obtainable via the exact same construction as the optimal board in the n = 3 case – just extend the (counter)clockwise rotation of the path in the obvious way. Thus, the maximum score for a game of Entanglement on a board of radius n for n = 1, 2, 3, … is given by the sequence 41, 1613, 9080, 29462, 72479, … (A180667 in the OEIS).