2018-04-10 This post was discussed on Hacker News.

So far in this series on the mathematics of 2048, we’ve used Markov chains to learn that it takes at least 938.8 moves on average to win, and we’ve explored the number of possible board configurations in the game using combinatorics and then exhaustive enumeration.

In this post, we’ll use a mathematical framework called a Markov Decision Process to find provably optimal strategies for 2048 when played on the 2x2 and 3x3 boards, and also on the 4x4 board up to the 64 tile. For example, here is an optimal player for the 2x2 game to the 32 tile:

The random seed determines the random sequence of tiles that the game adds to the board. The ‘strategy’ the player follows is defined by a table, called a policy, that tells it which direction it should swipe in every possible board configuration. In this post, we’ll see how to construct a policy that is optimal, in the sense that it maximizes the player’s chances of reaching the target 32 tile.

It turns out that the 2x2 game to the 32 tile is very hard to win — even when playing optimally, the player only wins about 8% of the time, which probably does not make for a very fun game. The 2x2 games are qualitatively quite different to the 4x4 games, but they’ll still be useful to introduce the key ideas.

Ideally we’d be able to find an optimal policy for the full game on the 4x4 board to the 2048 tile, but as we saw in the previous post, the number of possible board configurations is very large. This makes it infeasible to construct a complete optimal policy for the full game, at least with the methods used here.

We will however be able to find an optimal policy for the shortened 4x4 game to the 64 tile, and fortunately we’ll see that optimal play on the 3x3 boards looks qualitatively similar, in an admittedly hand-wavy way, to some successful strategies for the full game.

## Markov Decision Processes for 2048

Markov Decision Processes (MDPs) are a mathematical framework for modeling and solving problems in which we need to make a sequence of related decisions in the presence of uncertainty. Such problems are all around us, and MDPs find many applications in economics, finance, and artificial intelligence. For 2048, the sequence of decisions is the direction to swipe in each turn, and the uncertainty arises because the game adds new tiles to the board at random.

To set up the game of 2048 as an MDP, we will need to write it down in a specific way. This will involve six main concepts: states, actions and transition probabilities will encode the game’s dynamics; rewards, values and policies will be used to capture what the player is trying to accomplish and how they should do so. To develop these six concepts, we will take as an example the smallest non-trivial 2048-like game, which is played on the 2x2 board only up to the 8 tile. Let’s start with the first three.

### States, Actions and Transition Probabilities

A state captures the configuration of the board at a given point in the game by specifying the value of the tile, if any, in each of the board’s cells. For example, is a possible state in a game on a 2x2 board. An action is swiping left, right, up or down. Each time the player takes an action, the process transitions to a new state.

The transition probabilities encode the game’s dynamics by determining which states are likely to come next, in view of the current state and the player’s action. Fortunately, we can find out exactly how 2048 works by reading its source code. Most important 1 is the process the game uses to place a random tile on the board, which is always the same: pick an available cell uniformly at random, then add a new tile either with value 2, with probability 0.9, or value 4, with probability 0.1.

At the start of each game, two random tiles are added using this process. For example, one of these possible start states is . For each of the possible actions in this state, namely Left, Rright, Up and Down, the possible next states and the corresponding transition probabilities are 2:

In this diagram, there is an arrow for each possible transition to a successor state, on the right hand side. The weight of the arrow and the label indicate the corresponding transition probability. For example, if the player swipes right (R), both 2 tiles go to the right edge, leaving two available cells on the left. The new tile will be a 4 with probability 0.1, and it can either go into the top left or the bottom left cell, so the probability of the state is $$0.1 \times 0.5 = 0.05$$.

From each of those successor states, we can continue this process of enumerating their allowed actions and successor states, recursively. For , the possible successors are:

Here swiping right is not allowed, because none of the tiles can move right. Moreover, if the player reaches the successor state , highlighted in red, they have lost, because there are no allowed actions from that state. This would happen if the player were to swipe left, and the game were to place a 4 tile, which it would do with probability 0.1; this suggests that swiping left may not be the best action in this state.

For one final example, if the player moves up instead of left, one of the possible successor states is , and if we enumerate the allowed actions and and successor states from that state, we can see that swiping left or right will then result in an 8 tile, which means the game is won (highlighted in green):

If we repeat this process for all of the possible start states, and all of their possible successor states, and so on recursively until win or lose states are reached, we can build up a full model with all of the possible states, actions and their transition probabilities. For the 2x2 game to the 8 tile, that model looks like this (click to enlarge; you may then need to scroll down):

To make the diagram smaller, all of the losing states have been collapsed into a single lose state, shown as a red oval, and all of the winning states have been similarly collapsed into a single win state, shown as a green star. This is because we don’t particularly care how the player won or lost, only that they did.

Play proceeds roughly from left to right in the diagram, because the states have been organized into ‘layers’ by the sum of their tiles. A useful property of the game is that after each action the sum of the tiles on the board increases by either 2 or 4. This is because merging tiles does not change the sum of the tiles on the board, and the game always adds either a 2 tile or a 4 tile. The possible start states, which are in the layers with sum 4, 6 and 8, are drawn in blue.

Even for this smallest example, there are 70 states and 530 transitions in the model. It is possible significantly reduce those numbers, however, by observing that many of the states we’ve enumerated above are trivially related by rotations and reflections, as described in Appendix A. This observation is important in practice for reducing the size of the models so that they can be solved efficiently, and it makes for more legible diagrams, but it is not essential for us to move on to our second set of MDP concepts.

### Rewards, Values and Policies

To complete our specification of the model, we need to somehow encode the fact that the player’s objective is to reach the win state 3. We do this by defining rewards. In general, each time an MDP enters a state, the player receives a reward that depends on the state. Here we’ll set the reward for entering the win state to 1, and the reward for entering all other states to 0. That is, the one and only way to earn a reward is to reach the win state. 4

Now that we have an MDP model of the game in terms of states, actions, transition probabilities and rewards, we are ready to solve it. A solution for an MDP is called a policy. It is basically a table that lists for every possible state which action to take in that state. To solve an MDP is to find an optimal policy, which is one that allows the player to collect as much reward as possible over time.

To make this precise, we will need our final MDP concept: the value of a state according to a given policy is the expected, discounted reward the player will collect if they start from that state and follow the policy thereafter. To explain what that means will require some notation.

Let $$S$$ be the set of states, and for each state $$s \in S$$, let $$A_s$$ be the set of actions that are allowed in state $$s$$. Let $$\Pr(s’ | s, a)$$ denote the probability of transitioning to each successor state $$s’ \in S$$, given that the process is in state $$s \in S$$ and the player takes action $$a \in A_s$$. Let $$R(s)$$ denote the reward for entering state $$s$$. Finally, let $$\pi$$ denote a policy and $$\pi(s) \in A_s$$ denote the action to take in state $$s$$ when following policy $$\pi$$.

For a given policy $$\pi$$ and state $$s$$, the value of state $$s$$ according to $$\pi$$ is $V^\pi(s) = R(s) + \gamma \sum_{s’} \Pr(s’ | s, \pi(s)) V^\pi(s’)$ where the first term is the immediate reward, and the summation gives the expected value of the successor states, assuming the player continues to follow the policy.

The factor $$\gamma$$ is a discount factor that trades off the value of the immediate reward against the value of the expected future rewards. In other words, it accounts for the time value of money: a reward now is typically worth more than the same reward later. If $$\gamma$$ is close to 1, it means that the player is very patient: they don’t mind waiting for future rewards; likewise, smaller values of $$\gamma$$ mean that the player is less patient. For now, we’ll set the discount factor $$\gamma$$ to 1, which matches our assumption that the player cares only about winning, not about how long it takes to win 5.

So, how do we find the policy? For each state, we want to choose the action that maximizes the expected future value:

$\pi(s) = \mathop{\mathrm{argmax}}\limits_{a \in A_s} \left\{ \sum_{s’} \Pr(s’ | s, a) V^\pi(s’) \right\}$

So, this gives us two linked equations, and we can solve them iteratively. That is, pick an initial policy, which might be very simple, compute the value of every state under that simple policy, and then find a new policy based on that value function, and so on. Perhaps remarkably, under very modest technical conditions, such an iterative process is guaranteed to converge to an optimal policy, $$\pi^*$$, and an optimal value function $$V^{\pi^*}$$ with respect to that optimal policy.

This standard iterative approach works well for the MDP models for games on the 2x2 board, but it breaks down for the 3x3 and 4x4 game models, which have many more states and therefore take much more memory and compute power. Fortunately, it turns out that we can exploit the particular structure of our 2048 models to solve these equations much more efficiently, as described in Appendix B.

## Optimal Play on the 2x2 Board

We’re now ready to see some optimal policies in action! If you leave the random seed at 42 and press the Start button below, you’ll see it reach the 8 tile in 5 moves. The random seed determines the sequence of new tiles that the game will place; if you choose a different random seed by clicking the ⟲ button, you will (usually) see a different game unfold.

For the 2x2 game to the 8 tile, there is not actually much to see. If the player follows the optimal policy, they will always win. (As we saw above, when building the transition probabilities, if the player does not play optimally, it is possible to lose.) This is reflected in the fact that the value of the state remains at 1.00 for the whole game — when playing optimally, there is at least one action in every reachable state that leads to a win.

If we instead ask the player to play to the 16 tile, a win is no longer assured even when playing optimally. In this case, picking a new random seed should lead to a win 96% of the time for the game to the 16 tile, so I’ve set the initial seed to one of the rare seeds that leads to a loss.

As a result of setting the discount factor, $$\gamma$$, to 1, the value of each state also conveniently tells us the probability of winning from that state. Here the value starts at 0.96 and then eventually drops to 0.90, because the outcome hinges on the next tile being a 2 tile. Unfortunately, the game delivers a 4 tile, so the player loses, despite playing optimally.

Finally, we’ve previously established that the largest reachable tile on the 2x2 board is the 32 tile, so let’s see the corresponding optimal policy. Here the probability of winning drops to only 8%. (This is the same game and the same policy used in the introduction, but now it’s interactive.)

It’s worth remarking that each of the policies above is an optimal policy for the corresponding game, but there is no guarantee of uniqueness. There may be many optimal policies that are equivalent, but we can say with certainty that none of them are strictly better.

If you’d like to explore these models for the 2x2 game in more depth, Appendix A provides some diagrams that show all the possible paths through the game.

# Optimal Play on the 3x3 Board

On the 3x3 board, it is possible to play up to the 1024 tile, and that game has some 25 million states. Drawing an MDP diagram like we did for the 2x2 games is therefore clearly out of the question, but we can still watch an optimal policy in action 6:

Much like the 2x2 game to 32, the 3x3 game to 1024 is very hard to win — if playing optimally, the probability of winning is only about 1%. For some less frustrating entertainment, here also is the game to 512, for which the probability of winning if playing optimally is much higher, at about 74%:

At the risk of anthropomorphizing a large table of states and actions, which is what a policy is, I see here elements of strategies that I use when I play 2048 on the 4x4 board 7. We can see the policy pinning the high value tiles to the edges and usually corners (though in the game to 1024, it often puts the 512 tile in the middle of an edge). We can also see it being ‘lazy’ — even when it has two high value tiles lined up to merge, it will continue merging lower value tiles. Particularly within the tight constraints of the 3x3 board, it makes sense that it will take the opportunity to increase the sum of its tiles at no risk of (immediately) losing — if it gets stuck merging smaller tiles, it can always merge the larger ones, which opens up the board.

It’s important to note that we did not teach the policy about these strategies or give it any other hints about how to play. All of its behaviors and any apparent intelligence emerges solely from solving an optimization problem with respect to the transition probabilities and reward function we supplied.

# Optimal Play on the 4x4 Board

As established in the last post, the game to the 2048 tile on the 4x4 board has at least trillions of states, and so far it has not been possible to even enumerate all the states, let alone solve the resulting MDP for an optimal policy.

We can, however, complete the enumeration and the solve for the 4x4 game up to the 64 tile — that model has “only” about 40 billion states. Like the 2x2 game to 8 above, it is impossible to lose when playing optimally. This does not make for very interesting viewing, because in many cases there are several good actions, and the choice between them is arbitrary.

However, if we reduce the discount factor, $$\gamma$$, that makes the player slightly impatient, so that they prefer to win sooner rather than later. It then looks a bit more directed. Here is an optimal player for $$\gamma = 0.99$$:

The value starts around 0.72; the exact initial value reflects the expected number of moves it will take to win from the randomly selected start state. It gradually increases with each move, as the reward for reaching the win state gets closer. Again the policy shows good use of the edges and corners to build sequences of tiles in an order that’s convenient to merge.

## Conclusion

We’ve seen how to represent the game of 2048 as a Markov Decision Process and obtained provably optimal policies for the smaller games on the 2x2 and 3x3 boards and a partial game on the 4x4 board.

The methods used here require us to enumerate all of the states in the model in order to solve it. Using efficient strategies for enumerating the states and efficient strategies for solving the model makes this feasible for models with up to 40 billion states, which was the number for the 4x4 game to 64. The calculations for that model took roughly one week on an OVH HG-120 instance with 32 cores at 3.1GHz and 120GB RAM. The next-largest 4x4 game, played up to the 128 tile, is likely to contain many times that number of states and would require many times the computing power. Calculating a provably optimal policy for the full game to the 2048 tile will likely require different methods.

It is common to find that MDPs are too large to solve in practice, so there are a range of proven techniques for finding approximate solutions. These typically involve storing the value function and/or policy approximately, for example by training a (possibly deep) neural network. They can also be trained on simulation data, rather than requiring enumeration of the full state space, using reinforcement learning methods. The availability of provably optimal policies for smaller games may make 2048 a useful test bed for such methods — that would be an interesting future research topic.

## Appendix A: Canonicalization

As we’ve seen with the full model for the 2x2 game to the 8 tile, the number of states and transitions grows quickly, and even games on the 2x2 board become hard to draw in this form.

To help keep the size of the model under control, we can reuse an observation from the previous post about enumerating states: many of the successor states are just rotations or reflections of each other. For example, the states

and

are just mirror images — they are reflections through the vertical axis. If the best action in the first state was to swipe left, the best action in the second state would necessarily be to swipe right. So, from the perspective of deciding which action to take, it suffices to pick one of the states as the canonical state and determine the best action to take from the canonical state. A state’s canonical state is obtained by finding all of its possible rotations and reflections, writing each one as a number in base 12, and picking the state with the smallest number — see the previous post for the details. The important point here is that each canonical state stands in for a class of equivalent states that are related to it by rotation and reflection, so we don’t have to deal with them all individually. By replacing the successor states above with their canonical states, we obtain a much more compact diagram:

It’s somewhat unfortunate that the diagram appears to imply that swiping up (U) from somehow leads to . However, the paradox is resolved if you read the arrows to also include a rotation or reflection as required to find the actual successor’s canonical state.

### Equivalent Actions

We can also observe that in the diagram above it does not actually matter which direction the player swipes from the state — the canonical successor states and their transition probabilities are the same for all of the actions. To see why, it helps to look at the ‘intermediate state’ after the player has swiped to move the tiles but before the game has added a new random tile:

The intermediate states are drawn with dashed lines. The key observation is that they are all related by 90° rotations. These rotations don’t matter when we eventually canonicalize the successor states.

More generally, if two or more actions have the same canonical ‘intermediate state’, then those actions must have identical canonical successor states and transition probabilities and therefore are equivalent. In the example above, the canonical intermediate state happens to be the last one, .

We can therefore simplify the diagram for further if we just collapse all of the equivalent actions together:

Of course, states for which all actions are equivalent in this way are relatively rare. Considering another potential start state, , we see that swiping left and right are equivalent, but swiping up is distinct; swiping down is not allowed, because the tiles are already on the bottom.

### MDP Model Diagrams for 2x2 Games

Using canonicalization, we can shrink the models enough to just about draw out the full MDPs for some small games. Let’s again start with the smallest non-trivial model: the 2x2 game played just up to the 8 tile (click to enlarge):

Compared to the figure without canonicalization, this is much more compact. We can see that the shortest possible game comprises only a single move: if we are lucky enough to start in the state with two adjacent 4 tiles, which happens in only one game in 150, we just need to merge them together to reach an 8 tile and the win state. On the other hand, we can see that it is still possible to lose: if from state the player swipes up, they reach with probability 0.9, and then if they swipe up again, they reach with probability 0.9, at which point the game is lost.

We can further simplify the diagram if we know the optimal policy. Specifying the policy induces a Markov chain from the MDP model, because every state has a single action, or group of equivalent actions, identified by the policy. The induced chain for the 2x2 game to the 8 tile is:

We can see that the lose state no longer has any edges leading into it, because it is impossible to lose when playing optimally in the 2x2 game to the 8 tile. Each state is also now labelled with its value, which in this case is always 1.000. Because we’ve set the discount factor $$\gamma$$ to 1, the value of a state is in fact the probability of winning from that state when playing optimally.

They get a bit messier, but can build similar models for the 2x2 game to the 16 tile:

If we look at the optimal policy for the game to the 16 tile, we see that the start states (in blue) all have values less than one, and that there are paths to the lose state, in particular from two states that have tile sum 14:

That is, even if we play this game to the 16 tile optimally, we can still lose, depending on the particular sequence of 2 and 4 tiles that the game deals us. In most cases, we will win, however — the values of the start states are all around 0.96, so we’d expect to win roughly 96 games out of a hundred.

Our prospects in the game to the 32 tile on the 2x2 board, however, are much worse. Here is the full model:

We can see a lot of edges leading to the lose state, which is a bad sign. This is confirmed when we look at the diagram restricted to optimal play:

The average start state value is around 0.08, so we’d expect to win only about 8 games out of a hundred. The main reason becomes clear if we look at the right hand edge of the chain: once we reach a state with tile sum 28, the only way to win is to get a 4 tile in order to reach the state . If we get a 2 tile, which happens 90% of the time, we lose. It’s probably not a very fun game.

## Appendix B: Solution Methods

To efficiently solve an MDP model like the ones we’ve constructed here for 2048, we can exploit several important properties of its structure:

1. The transition model is a directed acyclic graph (DAG). The sum of the tiles must increase, namely by either 2 or 4, with each move, so it is never possible to go back to a state you have already visited.

2. Moreover, the states can be organized into ‘layers’ by the sums of their tiles, as we did in the first MDP model figure, and all transitions will be from the current layer with sum $$s$$ to either the next layer, with sum $$s+2$$, or the one after, with sum $$s+4$$.

3. All states in the layer with the largest sum will transition to either a lose or win state, which have a known value, namely 0 or 1, respectively.

Property (3) means that we can loop through all of the states in the last layer, in which all successor values are known, to generate the value function for that last layer. Then, using property (2), we know that the states in the second last layer must transition to either states in the last layer, for which we have just calculated values, or to a win or lose state, which have known values. In this way we can work backward, layer by layer, always knowing the values of states in the next two layers; this allows us to build both the value function and the optimal policy for the current layer. In general, this approach to solving MDPs is called backward induction, which is a particular type of dynamic programming.

In the previous post we worked forward from the start states to enumerate all of the states, layer by layer, using a map-reduce approach to parallelize the work within each layer. For the solve, we can use the output of that enumeration, which is a large list of states, to work backward, again using a map-reduce approach to parallelize the work within each layer. And like last time we can further break up the layers into ‘parts’ by their maximum tile value, with some additional book keeping.

The main solver implementation is still fairly memory-intensive, because it has to keep the value functions for up to four parts in memory at once in order to process a given part in one pass. This can be reduced to one part in memory at a time if we calculate what is usually called $$Q^\pi(s,a)$$, the value for each possible state-action pair according to policy $$\pi$$, rather than $$V^\pi(s)$$, but that solver implementation proved to be much slower, so all of the results presented here use the main $$V^\pi$$ solver on a machine with lots of RAM.

With both solvers, the layered structure of the model allows us to build and solve a model in just one forward pass and one backward pass, which is a substantial improvement on the usual iterative solution method, and one of the reasons that we’re able to solve these fairly large MDP models with billions of states. The canonicalization methods in Appendix A, which reduce the number of states we need to consider, and the low level efficiency gains from the previous post are also important reasons.

If you’ve read this far, perhaps you should follow me on twitter, or even apply to work at Overleaf. :)

# Footnotes

1. There is also some nuance in how tiles are merged: if you have four 2 tiles in a row, for example, and you swipe to merge them, the result is two 4 tiles, not a single 8 tile. That is, you can’t merge newly merged tiles in a single swipe. The original code for merging tiles is here, and the simplified but equivalent code I used to merge a line (row or column) of tiles is here with tests here

2. The graph diagrams here come from the excellent dot tool in graphviz

3. There are several other possible objectives. For example, in the first post in this series, I tried to reach the target 2048 tile in the smallest possible number of moves; and many people I’ve talked to play to reach the largest possible tile, which is also what the game’s points system encourages. These different objectives could also be captured by setting up the model and its rewards appropriately. For example, a simple reward of 1 per move until the player loses would represent the objective of playing as long as possible, which would I think be equivalent to trying to reach the largest possible tile.

4. Technically, we need one more special state, in addition to the win and lose states, to make this reward system work as described. The equations that we develop for $$\pi$$ and $$V^\pi$$ assume that all states have at least one allowed action and successor state, so we can’t just stop the process at the lose and win states. Instead, we can add an absorbing state, end, with a trivial action that just brings the process back into the end state with probability one. Then we can add a trivial action to both the lose and win states to transition to the absorbing end state with probability 1. So long as the end state attracts zero reward, it will not change the outcome. It’s also worth mentioning that there are more general ways of defining an MDP that would provide other ways of working around this technicality, for example by making the rewards depend on the whole transition rather than just the state and by making policies stochastic which, as a side effect, means that we can handle states with no allowed actions, but they require more notation.

5. In addition to being well founded in economic theory, the discount factor is often required technically in order to ensure that the value function converges. If the process runs forever and continues to accumulate additive rewards, it could accumulate infinite value. The geometric discounting ensures that the infinite sum still converges even if this happens. For the processes with the reward structure we’re considering here, we can safely set the discount factor to 1, because the process is constructed so that there are no loops with nonzero reward, and therefore all rewards are bounded.

6. There is a caveat for the 3x3 and 4x4 policies: the full optimal policies for every state are too large to ship to the browser (without unduly imposing on GitHub’s generosity in hosting this website). The player therefore only has access to the policy for the states that have a probability of at least $$10^{-7}$$ of actually occurring when playing according to the optimal policy. This means that, unfortunately, roughly one in hundred readers will choose a random seed that takes the process to a state that is not included in the data available on the client, in which case it will stop with an error. These states are selected by calculating the transient probabilities for the absorbing Markov chain induced by the optimal policy. The mathematics are essentially the same as those in the first post about Markov chains for 2048

7. Of course, I’m not claiming here to be great at 2048. The data in my first post suggest otherwise!