[There's still the strategy to go. A suitably robust argument that establishes what is statistically the best strategy will be accepted.]

Here's my description of the game:

There's a $4\times 4$ grid with some random, numbered cards on. The numbers are either one, two, or multiples of three. Using up, down, left, and right moves, you add the numbers on adjacent cards to make a new card like so: $$\begin{align}\color{blue}1+\color{red}2&=3\tag{1} \\ n+n&=2n\end{align}$$ for $n=2^k3\ge 3$, where $k\in\{0, 1, . . . , 10\}$, so the highest a card can be is $2^{11}3$. But at each move the "free" cards move too and a random new card appears at a random point along the edge you slide away from. Everything is kept on the grid. The card for the next move is indicated by colour at the top of the screen: blue for $1$, red for $2$, and white for $n\ge 3$ (such that $n$ is attainable using the above process). The white $2^\ell 3$-numbered cards are worth $3^{\ell+1}$ points; the rest give no points. Once there are no more available moves, the points on the remaining cards are summed to give your score for the game.

Here's another description I've found; it's the least promotional. It has the following gif.

enter image description here


What's the best strategy for the game? What's the highest possible score?


We could model this using some operations on $4\times 4$ matrices over $\mathbb{N}$. A new card would be the addition of $\alpha E_{ij}$ for some appropriate $\alpha$ and standard basis vector $E_{ij}$. That's all I've got . . .

NB: If this is a version of some other game, please let me know so I can avoid giving undue attention to this version :)

The number on each card can be written $n=2^k3^{\varepsilon_k}$, where $$\varepsilon_k=\cases{\color{blue}0\text{ or }1 &: $k=0$ \\ \color{red}0\text{ or }1 &: $k=1$ \\ 1 &: $k\ge 2$;}$$that is, $\varepsilon_k=\cases{0 &:$n<3$ \\ 1 &:$n\ge 3$}$. So we can write $(k, \varepsilon_k)$ instead under $$(k, \varepsilon_k)+(\ell, \varepsilon_\ell)\stackrel{(1)}{=}\cases{(k+1, 1)&: $\varepsilon_k, \varepsilon_\ell, k=\ell > 0$ \\ (0, 1)&: $\color{blue}k=\color{blue}{\varepsilon_k}=\color{red}{\varepsilon_\ell}=0, \color{red}\ell=1$ \\ (0, 1)&: $\color{blue}\ell=\color{red}{\varepsilon_k}=\color{blue}{\varepsilon_\ell}=0, \color{red}k=1$.}$$

Looking at a $2\times 2$ version might help: the moves from different starting positions show up if we work systematically. It fills up quickly.

It'd help to be more precise about what a good strategy might look like. The best strategy might be one that, from an arbitrary $4\times 4$ grid $G_0$ and with the least number of moves, gives the highest score attainable with $G_0$, subject to the random nature of the game. That's still a little vague though . . .

  • $\begingroup$ Sorry I've typed so much . . . $\endgroup$
    – Shaun
    Mar 18 '14 at 17:58
  • $\begingroup$ Note that I've corrected the scoring system :) $\endgroup$
    – Shaun
    Mar 19 '14 at 8:49
  • 3
    $\begingroup$ I doubt the use of matrices would be useful. Just because something is a grid of numbers doesn't mean it should be called a matrix, matrices specifically represent linear maps (except when they don't, but even then you're still at least multiplying them or taking $\det$s). $\endgroup$
    – Jack M
    Mar 19 '14 at 8:53
  • $\begingroup$ Is the value of the new tile chosen at random? $\endgroup$
    – naslundx
    Mar 19 '14 at 9:02
  • 3
    $\begingroup$ Great question! There's related programming question for a very similar game, 2048 (intial tiles are 2 and 4; only rule is $n + n= 2n $): stackoverflow.com/questions/22342854/… which includes an answer from multiple people who wrote algorithms which solve it. A graph theoretical approach is mentioned in the top answer. $\endgroup$
    – Roland
    Mar 24 '14 at 10:23

The strategy I employ is simply to make the move that leaves the most available moves, and to disregard score entirely. As a natural consequence of playing more moves, the score of the board will increase simply because the only way to continue playing is to make combinations, and combinations generate higher scores.

At the beginning of the game, the most important thing to consider is the placement of $1$s and $2$s. They are unique in that nothing can be combined adjacent to them to make a valid combination, they will only combine with their complement, which can only be achieved by board translation (verses, say a $12$, which can be adjacent to a $6$, which combines with another $6$ and then the $12$ can subsequently combine with that $12$. There's no way to make a $1$ or a $2$, it must simply be moved around the board).

Later, with higher scoring tiles on the board, the "bonus" tile (which shows up as a white tile with a $+$ in it at the top) becomes increasingly important, and the best strategy I've found is to attempt to place the bonus tile as near a mixed group of larger tiles as possible. The bonus tile will always be at least $6$, but never the same score as your highest scoring tile in play.

There is also the nature of the tile selection. It's been reverse engineered that the random generator uses a "bag" where $12$ tiles are shuffled. The original board layout uses this method, and $9$ tiles are placed into the board with $3$ remaining in the "bag". Once the bag is exhausted, the tiles are shuffled again. There are always $4$ of each: $1$, $2$, and $3$. Once you reach a high tile of $48$ a "bonus" tile is inserted with a potential value of greater than $3$. This changes the size of the "bag" to $13$ instead of $12$. So, keeping track of where you are in the "bag" and how many of each color you've seen can give you an advantage when looking at future moves.

Curiously, the possibility space for scoring is actually quite sparse. All scores will necessarily be multiples of $3$, but it turns out that only about $\frac38$ of the multiples of $3$ between $0$ and the max score are actually valid. There are a lot that are simply impossible to get, like a $19$ in cribbage.

The lowest one that isn't trivially small is still $39,363$, though, which seems well out of the range of the average player. The next lowest I found is $52,485$. There are lots of gaps at the high end, due to the fact that highest scoring tile is worth over $500$k by itself.

  • $\begingroup$ Thank you for this answer. You've clearly put effort into it. However, I'm not sure if it's the best strategy, so I'm reluctant to accept it. I wasn't aware of any bonus tile feature: are you sure it's not just the clue for the next move? $\endgroup$
    – Shaun
    Apr 22 '14 at 7:11
  • 1
    $\begingroup$ There's a lot of in depth analysis of the behavior of the stack over at touch arcade, where people have been building AIs to play the game. forums.toucharcade.com/showthread.php?p=3133680#post3133680 $\endgroup$
    – XaiaX
    Apr 26 '14 at 1:18
  • $\begingroup$ I don't think there is a simple mathematical solution, due to the nature of the randomness of the board, the board is very much not stateless, unlike something like chess or go, where all you need to know to calculate the next move is the current state of the board and whose turn it is. With the random bag the best you can do is what is statistically the best move. $\endgroup$
    – XaiaX
    Apr 26 '14 at 2:13
  • $\begingroup$ Well then, let's work with Statistics; I'm not exactly surprised :) $\endgroup$
    – Shaun
    Apr 26 '14 at 9:25

A partial answer:

The highest possible score is $16\times 3^{12}$.

If the game could start with no available moves, then just suppose it starts with $2^{11}3$ everywhere.

Alternatively, suppose you start with $2^{11}3$ in the top left and suppose every new card happens to be $2^{11}3$. Assume the cards all show up in the top left corner, which we can do in the following. Slide right until the top row is full. Slide down once. Repeat. This will eventually fill the grid; once it does, there'll be no more available moves so the game ends (with a score of $16\times 3^{11+1}$).

  • $\begingroup$ Why not have $2^{12}3$ be the spawning card? $\endgroup$ Apr 1 '14 at 23:11
  • $\begingroup$ @StellaBiderman Because $2^{12}3$ is not a card. The highest a card can be is $2^{11}3$, which contributes $3^{12}$ to the final score. $\endgroup$
    – Shaun
    Apr 1 '14 at 23:18
  • $\begingroup$ Oh, my bad. I missed that part. $\endgroup$ Apr 1 '14 at 23:21
  • $\begingroup$ . . . or it could just start with $2^{11}3$ in every place, but that's boring :) $\endgroup$
    – Shaun
    Apr 4 '14 at 15:32

If this game is similar to 2048, one strategy I know is to exclude one direction except for certain circumstances. This greatly simplifies the game and makes combining easier. Large differences of numbers next to each other should be avoided, so besides choosing a main direction a secondary direction is chosen to simplify more. Also, you want to create "chains" of numbers that can easily be pushed together.

  • $\begingroup$ That does seem like a good strategy. It somewhat describes my intuition and most successful attempts at the game, alongside the heuristic of keeping high cards in one corner (but that might be equivalent to using primary & secondary directions). But how would I know it's the best? I'm reluctant to accept it without some form of argument/evidence for why it is. $\endgroup$
    – Shaun
    Apr 18 '14 at 7:25
  • $\begingroup$ Perhaps it would help to classify those exceptional circumstances. I can sort of envision what you mean, like having to break things down into a sub-$3\times 3$ game for a while because of a rogue $1$ or something. $\endgroup$
    – Shaun
    Apr 18 '14 at 12:38

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