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Once one has mastered winning 2048 in general, it's often an interesting exercise to try to do it as fast as possible.

What strategies are there for speeding up 2048-solving?
For example, how can you pass the boring beginning game quickly (ex. mashing up and left),
and how do you know when to stop so you don't mess up your middlegame?

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  • $\begingroup$ I'm voting to close this question as "primarily opinion -based" (though in retrospect, "too broad" may be more applicable). I don't think there's a definitive way to do this, so the best answers we'll get for this question are going to be "this is how I optimize it visually." $\endgroup$ – Aza May 14 '14 at 23:22
  • $\begingroup$ I think I'd have to vote to close as well. My technique is purely speeding up the reflexes on whether I go down/right or down/left/right (I use a different corner to you, obviously) - so I think this is entirely subjective $\endgroup$ – Rory Alsop May 14 '14 at 23:35
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A pattern that I have had success with is as follows:

  1. Alternately swipe up ($\uparrow$) and left ($\leftarrow$) (or whichever two directions you choose) as quickly as you can.
  2. When you get stuck (or when you see an opportunity$^1$), swipe right ($\rightarrow$), then up ($\uparrow$) a few times$^2$, then continue with step 1.
  3. Keep this up until you get the 512 tile in the top-left corner.

Once you get to 512, you need to be more careful, because you can easily fill your entire board without the opportunity to collapse anything.

At this point, I usually adopt an alternating right($\rightarrow$)/up($\uparrow$) strategy, but making sure that I only swipe right ($\rightarrow$) when the top row is full. When you get two matching tiles in the top row, swipe left ($\leftarrow$) until you've collapsed all pairs, and then go back to the first strategy above (left($\leftarrow$)/up($\uparrow$)) for a bit, until the numbers on the top row get too big. Then continue with right($\rightarrow$)/up($\uparrow$), etc.

Of course, as with any strategy, you need to use your knowledge of the game and judgement to determine when to deviate from the "speed" tactic and pay closer attention to what you're doing, but as a general strategy, I have found this to work quite well.

With this method, I can fairly consistently get to 2048 in under 15 minutes. (Often under 10 minutes.)


1Look for situations like this:

$$\begin{array}{|c|c|c|c|} \hline 128 & 64 & 16 & 4\\ \hline 64 & 16 & 4\\ \hline 32 & 16 & 2\\ \hline 4 & 2\\ \hline \end{array}$$

Even though you can continue swiping up and left, it makes sense here to stop and swipe right, because the 2nd row matches up so nicely with the top row. You need to watch for these opportunities, because if you focus too heavily on the up/right strategy, you will clog up the whole board and leave yourself no moves.


2After you do the one swipe up, you may still have several large-ish numbers in rows 2 and 3. If you immediately continue with the up/left tactic, you will probably trap a small number on the left side of row 2, which will upset the entire strategy. Often, the numbers will line up quite nicely, and you'll be able to collapse many of your larger numbers into the top row. You have to use your judgement to determine how many up-swipes you need before you go back to the up/left tactic.

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You can't speed it up by much. To get to the 2048 tile, you need 1024 2 tiles, 512 4 tiles or a combination, and that many moves is your basic minimum. There is no way to cross those limits. Because 4 shows up 1/4 times, it will take about 850 moves minimum to finish, regardless of skill or strategy.

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  • $\begingroup$ I'm not talking about minimal moves, I'm talking about time taken. $\endgroup$ – Doorknob May 15 '14 at 0:26
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    $\begingroup$ Then it is merely about how quickly you can do the moves on your keyboard. If you disable animations, they take no time, and it all depends on how quickly you can react smartly. $\endgroup$ – JohnJPershing May 15 '14 at 0:27
  • $\begingroup$ @Greenturtle3141 thanks, I had it set to the wrong language, so it came out as a comma. :/ $\endgroup$ – Mithrandir Dec 19 '16 at 20:56

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