unblocked - What is the optimal algorithm for the game 2048?

mcts 2048 (10)

I have recently stumbled upon the game 2048. You merge similar tiles by moving them in any of the four directions to make "bigger" tiles. After each move, a new tile appears at random empty position with a value of either 2 or 4. The game terminates when all the boxes are filled and there are no moves that can merge tiles, or you create a tile with a value of 2048.

One, I need to follow a well-defined strategy to reach the goal. So, I thought of writing a program for it.

My current algorithm:

while (!game_over) {
    for each possible move:
        count_no_of_merges_for_2-tiles and 4-tiles
    choose the move with a large number of merges

What I am doing is at any point, I will try to merge the tiles with values 2 and 4, that is, I try to have 2 and 4 tiles, as minimum as possible. If I try it this way, all other tiles were automatically getting merged and the strategy seems good.

But, when I actually use this algorithm, I only get around 4000 points before the game terminates. Maximum points AFAIK is slightly more than 20,000 points which is way larger than my current score. Is there a better algorithm than the above?


    for each possible move:
        evaluate next state

    choose the maximum evaluation


Evaluation =
    128 (Constant)
    + (Number of Spaces x 128)
    + Sum of faces adjacent to a space { (1/face) x 4096 }
    + Sum of other faces { log(face) x 4 }
    + (Number of possible next moves x 256)
    + (Number of aligned values x 2)

Evaluation Details

128 (Constant)

This is a constant, used as a base-line and for other uses like testing.

+ (Number of Spaces x 128)

More spaces makes the state more flexible, we multiply by 128 (which is the median) since a grid filled with 128 faces is an optimal impossible state.

+ Sum of faces adjacent to a space { (1/face) x 4096 }

Here we evaluate faces that have the possibility to getting to merge, by evaluating them backwardly, tile 2 become of value 2048, while tile 2048 is evaluated 2.

+ Sum of other faces { log(face) x 4 }

In here we still need to check for stacked values, but in a lesser way that doesn't interrupt the flexibility parameters, so we have the sum of { x in [4,44] }.

+ (Number of possible next moves x 256)

A state is more flexible if it has more freedom of possible transitions.

+ (Number of aligned values x 2)

This is a simplified check of the possibility of having merges within that state, without making a look-ahead.

Note: The constants can be tweaked..

EDIT: This is a naive algorithm, modelling human conscious thought process, and gets very weak results compared to AI that search all possibilities since it only looks one tile ahead. It was submitted early in the response timeline.

I have refined the algorithm and beaten the game! It may fail due to simple bad luck close to the end (you are forced to move down, which you should never do, and a tile appears where your highest should be. Just try to keep the top row filled, so moving left does not break the pattern), but basically you end up having a fixed part and a mobile part to play with. This is your objective:

This is the model I chose by default.

1024 512 256 128
  8   16  32  64
  4   2   x   x
  x   x   x   x

The chosen corner is arbitrary, you basically never press one key (the forbidden move), and if you do, you press the contrary again and try to fix it. For future tiles the model always expects the next random tile to be a 2 and appear on the opposite side to the current model (while the first row is incomplete, on the bottom right corner, once the first row is completed, on the bottom left corner).

Here goes the algorithm. Around 80% wins (it seems it is always possible to win with more "professional" AI techniques, I am not sure about this, though.)


    checkCornerChosen(); // Unimplemented, but it might be an improvement to change the reference point

    for each 3 possible move:
    execute move with best score
    if no move is available, execute forbidden move and undo, recalculateModel()

 evaluateResult() {
     calculates distance to chosen model
     stores result

 calculateBestCurrentModel() {
      (according to the current highest tile acheived and their distribution)

A few pointers on the missing steps. Here:

The model has changed due to the luck of being closer to the expected model. The model the AI is trying to achieve is

 512 256 128  x
  X   X   x   x
  X   X   x   x
  x   x   x   x

And the chain to get there has become:

 512 256  64  O
  8   16  32  O
  4   x   x   x
  x   x   x   x

The O represent forbidden spaces...

So it will press right, then right again, then (right or top depending on where the 4 has created) then will proceed to complete the chain until it gets:

So now the model and chain are back to:

 512 256 128  64
  4   8  16   32
  X   X   x   x
  x   x   x   x

Second pointer, it has had bad luck and its main spot has been taken. It is likely that it will fail, but it can still achieve it:

Here the model and chain is:

  O 1024 512 256
  O   O   O  128
  8  16   32  64
  4   x   x   x

When it manages to reach the 128 it gains a whole row is gained again:

  O 1024 512 256
  x   x  128 128
  x   x   x   x
  x   x   x   x

I became interested in the idea of an AI for this game containing no hard-coded intelligence (i.e no heuristics, scoring functions etc). The AI should "know" only the game rules, and "figure out" the game play. This is in contrast to most AIs (like the ones in this thread) where the game play is essentially brute force steered by a scoring function representing human understanding of the game.

AI Algorithm

I found a simple yet surprisingly good playing algorithm: To determine the next move for a given board, the AI plays the game in memory using random moves until the game is over. This is done several times while keeping track of the end game score. Then the average end score per starting move is calculated. The starting move with the highest average end score is chosen as the next move.

With just 100 runs (i.e in memory games) per move, the AI achieves the 2048 tile 80% of the times and the 4096 tile 50% of the times. Using 10000 runs gets the 2048 tile 100%, 70% for 4096 tile, and about 1% for the 8192 tile.

See it in action

The best achieved score is shown here:

An interesting fact about this algorithm is that while the random-play games are unsurprisingly quite bad, choosing the best (or least bad) move leads to very good game play: A typical AI game can reach 70000 points and last 3000 moves, yet the in-memory random play games from any given position yield an average of 340 additional points in about 40 extra moves before dying. (You can see this for yourself by running the AI and opening the debug console.)

This graph illustrates this point: The blue line shows the board score after each move. The red line shows the algorithm's best random-run end game score from that position. In essence, the red values are "pulling" the blue values upwards towards them, as they are the algorithm's best guess. It's interesting to see the red line is just a tiny bit above the blue line at each point, yet the blue line continues to increase more and more.

I find it quite surprising that the algorithm doesn't need to actually foresee good game play in order to chose the moves that produce it.

Searching later I found this algorithm might be classified as a Pure Monte Carlo Tree Search algorithm.

Implementation and Links

First I created a JavaScript version which can be seen in action here. This version can run 100's of runs in decent time. Open the console for extra info. (source)

Later, in order to play around some more I used @nneonneo highly optimized infrastructure and implemented my version in C++. This version allows for up to 100000 runs per move and even 1000000 if you have the patience. Building instructions provided. It runs in the console and also has a remote-control to play the web version. (source)


Surprisingly, increasing the number of runs does not drastically improve the game play. There seems to be a limit to this strategy at around 80000 points with the 4096 tile and all the smaller ones, very close to the achieving the 8192 tile. Increasing the number of runs from 100 to 100000 increases the odds of getting to this score limit (from 5% to 40%) but not breaking through it.

Running 10000 runs with a temporary increase to 1000000 near critical positions managed to break this barrier less than 1% of the times achieving a max score of 129892 and the 8192 tile.


After implementing this algorithm I tried many improvements including using the min or max scores, or a combination of min,max,and avg. I also tried using depth: Instead of trying K runs per move, I tried K moves per move list of a given length ("up,up,left" for example) and selecting the first move of the best scoring move list.

Later I implemented a scoring tree that took into account the conditional probability of being able to play a move after a given move list.

However, none of these ideas showed any real advantage over the simple first idea. I left the code for these ideas commented out in the C++ code.

I did add a "Deep Search" mechanism that increased the run number temporarily to 1000000 when any of the runs managed to accidentally reach the next highest tile. This offered a time improvement.

I'd be interested to hear if anyone has other improvement ideas that maintain the domain-independence of the AI.

2048 Variants and Clones

Just for fun, I've also implemented the AI as a bookmarklet, hooking into the game's controls. This allows the AI to work with the original game and many of its variants.

This is possible due to domain-independent nature of the AI. Some of the variants are quite distinct, such as the Hexagonal clone.

I copy here the content of a post on my blog

The solution I propose is very simple and easy to implement. Although, it has reached the score of 131040. Several benchmarks of the algorithm performances are presented.


Heuristic scoring algorithm

The assumption on which my algorithm is based is rather simple: if you want to achieve higher score, the board must be kept as tidy as possible. In particular, the optimal setup is given by a linear and monotonic decreasing order of the tile values. This intuition will give you also the upper bound for a tile value: where n is the number of tile on the board.

(There's a possibility to reach the 131072 tile if the 4-tile is randomly generated instead of the 2-tile when needed)

Two possible ways of organizing the board are shown in the following images:

To enforce the ordination of the tiles in a monotonic decreasing order, the score si computed as the sum of the linearized values on the board multiplied by the values of a geometric sequence with common ratio r<1 .

Several linear path could be evaluated at once, the final score will be the maximum score of any path.

Decision rule

The decision rule implemented is not quite smart, the code in Python is presented here:

def nextMove(board,recursion_depth=3):
    m,s = AI.nextMoveRecur(board,recursion_depth,recursion_depth)
    return m

def nextMoveRecur(board,depth,maxDepth,base=0.9):
    bestScore = -1.
    bestMove = 0
    for m in range(1,5):
            newBoard = copy.deepcopy(board)

            score = AI.evaluate(newBoard)
            if depth != 0:
                my_m,my_s = AI.nextMoveRecur(newBoard,depth-1,maxDepth)
                score += my_s*pow(base,maxDepth-depth+1)

            if(score > bestScore):
                bestMove = m
                bestScore = score
    return (bestMove,bestScore);

An implementation of the minmax or the Expectiminimax will surely improve the algorithm. Obviously a more sophisticated decision rule will slow down the algorithm and it will require some time to be implemented.I will try a minimax implementation in the near future. (stay tuned)


  • T1 - 121 tests - 8 different paths - r=0.125
  • T2 - 122 tests - 8-different paths - r=0.25
  • T3 - 132 tests - 8-different paths - r=0.5
  • T4 - 211 tests - 2-different paths - r=0.125
  • T5 - 274 tests - 2-different paths - r=0.25
  • T6 - 211 tests - 2-different paths - r=0.5

In case of T2, four tests in ten generate the 4096 tile with an average score of 42000


The code can be found on GiHub at the following link: https://github.com/Nicola17/term2048-AI It is based on term2048 and it's written in Python. I will implement a more efficient version in C++ as soon as possible.

I think I found an algorithm which works quite well, as I often reach scores over 10000, my personal best being around 16000. My solution does not aim at keeping biggest numbers in a corner, but to keep it in the top row.

Please see the code below:

while( !game_over ) {
    if( !move_is_possible(up) ) {
        if( move_is_possible(right) && move_is_possible(left) ){
            if( number_of_empty_cells_after_moves(left,up) > number_of_empty_cells_after_moves(right,up) ) 
                move_direction = left;
                move_direction = right;
        } else if ( move_is_possible(left) ){
            move_direction = left;
        } else if ( move_is_possible(right) ){
            move_direction = right;
        } else {
            move_direction = down;

I wrote a 2048 solver in Haskell, mainly because I'm learning this language right now.

My implementation of the game slightly differs from the actual game, in that a new tile is always a '2' (rather than 90% 2 and 10% 4). And that the new tile is not random, but always the first available one from the top left. This variant is also known as Det 2048.

As a consequence, this solver is deterministic.

I used an exhaustive algorithm that favours empty tiles. It performs pretty quickly for depth 1-4, but on depth 5 it gets rather slow at a around 1 second per move.

Below is the code implementing the solving algorithm. The grid is represented as a 16-length array of Integers. And scoring is done simply by counting the number of empty squares.

bestMove :: Int -> [Int] -> Int
bestMove depth grid = maxTuple [ (gridValue depth (takeTurn x grid), x) | x <- [0..3], takeTurn x grid /= [] ]

gridValue :: Int -> [Int] -> Int
gridValue _ [] = -1
gridValue 0 grid = length $ filter (==0) grid  -- <= SCORING
gridValue depth grid = maxInList [ gridValue (depth-1) (takeTurn x grid) | x <- [0..3] ]

I thinks it's quite successful for its simplicity. The result it reaches when starting with an empty grid and solving at depth 5 is:

Move 4006

Game Over

Source code can be found here: https://github.com/popovitsj/2048-haskell

Many of the other answers use AI with computationally expensive searching of possible futures, heuristics, learning and the such. These are impressive and probably the correct way forward, but I wish to contribute another idea.

Model the sort of strategy that good players of the game use.

For example:

13 14 15 16
12 11 10  9
 5  6  7  8
 4  3  2  1

Read the squares in the order shown above until the next squares value is greater than the current one. This presents the problem of trying to merge another tile of the same value into this square.

To resolve this problem, their are 2 ways to move that aren't left or worse up and examining both possibilities may immediately reveal more problems, this forms a list of dependancies, each problem requiring another problem to be solved first. I think I have this chain or in some cases tree of dependancies internally when deciding my next move, particularly when stuck.

Tile needs merging with neighbour but is too small: Merge another neighbour with this one.

Larger tile in the way: Increase the value of a smaller surrounding tile.


The whole approach will likely be more complicated than this but not much more complicated. It could be this mechanical in feel lacking scores, weights, neurones and deep searches of possibilities. The tree of possibilities rairly even needs to be big enough to need any branching at all.

My attempt uses expectimax like others solutions above, but without bitboards. Nneonneo's solution can check 10millions of moves which is approximately a depth of 4 with 6 tiles left and 4 moves possible (2*6*4)4. In my case, this depth takes too long to explore, I adjust the depth of expectimax search according to the number of free tiles left:

depth = free > 7 ? 1 : (free > 4 ? 2 : 3)

The scores of the boards are computed with the weighted sum of the square of the number of free tiles and the dot product of the 2D grid with this:


which forces to organize tiles descendingly in a sort of snake from the top left tile.

Code below or jsbin:

var n = 4,
	M = new MatrixTransform(n);

var ai = {weights: [1, 1], depth: 1}; // depth=1 by default, but we adjust it on every prediction according to the number of free tiles

var snake= [[10,8,7,6.5],
snake=snake.map(function(a){return a.map(Math.exp)})


function run(ai) {
	var p;
	while ((p = predict(ai)) != null) {
		move(p, ai);
	//console.log(ai.grid , maxValue(ai.grid))
	ai.maxValue = maxValue(ai.grid)

function initialize(ai) {
	ai.grid = [];
	for (var i = 0; i < n; i++) {
		ai.grid[i] = []
		for (var j = 0; j < n; j++) {
			ai.grid[i][j] = 0;
	ai.steps = 0;

function move(p, ai) { //0:up, 1:right, 2:down, 3:left
	var newgrid = mv(p, ai.grid);
	if (!equal(newgrid, ai.grid)) {
		//console.log(stats(newgrid, ai.grid))
		ai.grid = newgrid;
		try {
		} catch (e) {
			console.log('no room', e)

function predict(ai) {
	var free = freeCells(ai.grid);
	ai.depth = free > 7 ? 1 : (free > 4 ? 2 : 3);
	var root = {path: [],prob: 1,grid: ai.grid,children: []};
	var x = expandMove(root, ai)
	//console.log("number of leaves", x)
	//console.log("number of leaves2", countLeaves(root))
	if (!root.children.length) return null
	var values = root.children.map(expectimax);
	var mx = max(values);
	return root.children[mx[1]].path[0]


function countLeaves(node) {
	var x = 0;
	if (!node.children.length) return 1;
	for (var n of node.children)
		x += countLeaves(n);
	return x;

function expectimax(node) {
	if (!node.children.length) {
		return node.score
	} else {
		var values = node.children.map(expectimax);
		if (node.prob) { //we are at a max node
			return Math.max.apply(null, values)
		} else { // we are at a random node
			var avg = 0;
			for (var i = 0; i < values.length; i++)
				avg += node.children[i].prob * values[i]
			return avg / (values.length / 2)

function expandRandom(node, ai) {
	var x = 0;
	for (var i = 0; i < node.grid.length; i++)
		for (var j = 0; j < node.grid.length; j++)
			if (!node.grid[i][j]) {
				var grid2 = M.copy(node.grid),
					grid4 = M.copy(node.grid);
				grid2[i][j] = 2;
				grid4[i][j] = 4;
				var child2 = {grid: grid2,prob: .9,path: node.path,children: []};
				var child4 = {grid: grid4,prob: .1,path: node.path,children: []}
				x += expandMove(child2, ai)
				x += expandMove(child4, ai)
	return x;

function expandMove(node, ai) { // node={grid,path,score}
	var isLeaf = true,
		x = 0;
	if (node.path.length < ai.depth) {
		for (var move of[0, 1, 2, 3]) {
			var grid = mv(move, node.grid);
			if (!equal(grid, node.grid)) {
				isLeaf = false;
				var child = {grid: grid,path: node.path.concat([move]),children: []}
				x += expandRandom(child, ai)
	if (isLeaf) node.score = dot(ai.weights, stats(node.grid))
	return isLeaf ? 1 : x;

var cells = []
var table = document.querySelector("table");
for (var i = 0; i < n; i++) {
	var tr = document.createElement("tr");
	cells[i] = [];
	for (var j = 0; j < n; j++) {
		cells[i][j] = document.createElement("td");

function updateUI(ai) {
	cells.forEach(function(a, i) {
		a.forEach(function(el, j) {
			el.innerHTML = ai.grid[i][j] || ''

function runAI() {
	var p = predict(ai);
	if (p != null && ai.running) {
		move(p, ai)
runai.onclick = function() {
	if (!ai.running) {
		this.innerHTML = 'stop AI';
		ai.running = true;
	} else {
		this.innerHTML = 'run AI';
		ai.running = false;

hint.onclick = function() {
	hintvalue.innerHTML = ['up', 'right', 'down', 'left'][predict(ai)]
document.addEventListener("keydown", function(event) {
	if (event.which in map) {
		move(map[event.which], ai)
var map = {
	38: 0, // Up
	39: 1, // Right
	40: 2, // Down
	37: 3, // Left
init.onclick = function() {

function stats(grid, previousGrid) {

	var free = freeCells(grid);

	var c = dot2(grid, snake);

	return [c, free * free];

function dist2(a, b) { //squared 2D distance
	return Math.pow(a[0] - b[0], 2) + Math.pow(a[1] - b[1], 2)

function dot(a, b) {
	var r = 0;
	for (var i = 0; i < a.length; i++)
		r += a[i] * b[i];
	return r

function dot2(a, b) {
	var r = 0;
	for (var i = 0; i < a.length; i++)
		for (var j = 0; j < a[0].length; j++)
			r += a[i][j] * b[i][j]
	return r;

function product(a) {
	return a.reduce(function(v, x) {
		return v * x
	}, 1)

function maxValue(grid) {
	return Math.max.apply(null, grid.map(function(a) {
		return Math.max.apply(null, a)

function freeCells(grid) {
	return grid.reduce(function(v, a) {
		return v + a.reduce(function(t, x) {
			return t + (x == 0)
		}, 0)
	}, 0)

function max(arr) { // return [value, index] of the max
	var m = [-Infinity, null];
	for (var i = 0; i < arr.length; i++) {
		if (arr[i] > m[0]) m = [arr[i], i];
	return m

function min(arr) { // return [value, index] of the min
	var m = [Infinity, null];
	for (var i = 0; i < arr.length; i++) {
		if (arr[i] < m[0]) m = [arr[i], i];
	return m

function maxScore(nodes) {
	var min = {
		score: -Infinity,
		path: []
	for (var node of nodes) {
		if (node.score > min.score) min = node;
	return min;

function mv(k, grid) {
	var tgrid = M.itransform(k, grid);
	for (var i = 0; i < tgrid.length; i++) {
		var a = tgrid[i];
		for (var j = 0, jj = 0; j < a.length; j++)
			if (a[j]) a[jj++] = (j < a.length - 1 && a[j] == a[j + 1]) ? 2 * a[j++] : a[j]
		for (; jj < a.length; jj++)
			a[jj] = 0;
	return M.transform(k, tgrid);

function rand(grid) {
	var r = Math.floor(Math.random() * freeCells(grid)),
		_r = 0;
	for (var i = 0; i < grid.length; i++) {
		for (var j = 0; j < grid.length; j++) {
			if (!grid[i][j]) {
				if (_r == r) {
					grid[i][j] = Math.random() < .9 ? 2 : 4

function equal(grid1, grid2) {
	for (var i = 0; i < grid1.length; i++)
		for (var j = 0; j < grid1.length; j++)
			if (grid1[i][j] != grid2[i][j]) return false;
	return true;

function conv44valid(a, b) {
	var r = 0;
	for (var i = 0; i < 4; i++)
		for (var j = 0; j < 4; j++)
			r += a[i][j] * b[3 - i][3 - j]
	return r

function MatrixTransform(n) {
	var g = [],
		ig = [];
	for (var i = 0; i < n; i++) {
		g[i] = [];
		ig[i] = [];
		for (var j = 0; j < n; j++) {
			g[i][j] = [[j, i],[i, n-1-j],[j, n-1-i],[i, j]]; // transformation matrix in the 4 directions g[i][j] = [up, right, down, left]
			ig[i][j] = [[j, i],[i, n-1-j],[n-1-j, i],[i, j]]; // the inverse tranformations
	this.transform = function(k, grid) {
		return this.transformer(k, grid, g)
	this.itransform = function(k, grid) { // inverse transform
		return this.transformer(k, grid, ig)
	this.transformer = function(k, grid, mat) {
		var newgrid = [];
		for (var i = 0; i < grid.length; i++) {
			newgrid[i] = [];
			for (var j = 0; j < grid.length; j++)
				newgrid[i][j] = grid[mat[i][j][k][0]][mat[i][j][k][1]];
		return newgrid;
	this.copy = function(grid) {
		return this.transform(3, grid)
body {
	text-align: center
table, th, td {
    border: 1px solid black;
    margin: 5px auto;
td {
    width: 35px;
    height: 35px;
    text-align: center;
<button id=init>init</button><button id=runai>run AI</button><button id=hint>hint</button><span id=hintvalue></span>

This algorithm is not optimal for winning the game, but it is fairly optimal in terms of performance and amount of code needed:

  if(can move neither right, up or down)
    direction = left
      direction = random from (right, down, up)
    while(can not move in "direction")

This is not a direct answer to OP's question, this is more of the stuffs (experiments) I tried so far to solve the same problem and obtained some results and have some observations that I want to share, I am curious if we can have some further insights from this.

I just tried my minimax implementation with alpha-beta pruning with search-tree depth cutoff at 3 and 5. I was trying to solve the same problem for a 4x4 grid as a project assignment for the edX course ColumbiaX: CSMM.101x Artificial Intelligence (AI).

I applied convex combination (tried different heuristic weights) of couple of heuristic evaluation functions, mainly from intuition and from the ones discussed above:

  1. Monotonicity
  2. Free Space Available

In my case, the computer player is completely random, but still i assumed adversarial settings and implemented the AI player agent as the max player.

I have 4x4 grid for playing the game.


If I assign too much weights to the first heuristic function or the second heuristic function, both the cases the scores the AI player gets are low. I played with many possible weight assignments to the heuristic functions and take a convex combination, but very rarely the AI player is able to score 2048. Most of the times it either stops at 1024 or 512.

I also tried the corner heuristic, but for some reason it makes the results worse, any intuition why?

Also, I tried to increase the search depth cut-off from 3 to 5 (I can't increase it more since searching that space exceeds allowed time even with pruning) and added one more heuristic that looks at the values of adjacent tiles and gives more points if they are merge-able, but still I am not able to get 2048.

I think it will be better to use Expectimax instead of minimax, but still I want to solve this problem with minimax only and obtain high scores such as 2048 or 4096. I am not sure whether I am missing anything.

Below animation shows the last few steps of the game played by the AI agent with the computer player:

Any insights will be really very helpful, thanks in advance. (This is the link of my blog post for the article: https://sandipanweb.wordpress.com/2017/03/06/using-minimax-with-alpha-beta-pruning-and-heuristic-evaluation-to-solve-2048-game-with-computer/)

The following animation shows the last few steps of the game played where the AI player agent could get 2048 scores, this time adding the absolute value heuristic too:

The following figures show the game tree explored by the player AI agent assuming the computer as adversary for just a single step: