1
$\begingroup$

There are several questions about the 2048 game, looking at strategy, worst case, max tile, etc.

This question is about the sum of a subset of tiles mid-game.

Consider the following boards from real games:

enter image description here enter image description here

In each case, the top row is filled and the rest of the board (tiles on rows 2 to 4) sums to 6.

In a real game, is this sum minimal, or is it possible to reduce the sum to a number less than 6? Note that the top row must be filled.

(Please supply a proof in each answer. There's no need for spoiler tags.)

EDIT: As @Deusovi and @Trenin have shown, this can be done almost trivially.

So for something a little meatier, we will only look at the sum (of the lower 3 rows) when the top row is in strict descending order, with all top-row tiles greater than 2. We will assume that newly-spawned tiles are all '2's, and you can nominate where each tile is spawned. Once spawned, though, the tiles must move and merge according to the rules of the game. You may start with either of the above boards, or start with the standard 2-tile opening, but from there, you'll have to explain how you got to the final layout, or explain why no sequence of moves can produce a sum less than 6.

Improve this question
$\endgroup$
5
  • $\begingroup$ As I recall, you can never reduce the value of a tile or set of tiles in 2048, you can only condense the same value into fewer tiles. In the second instance, you might be able to claim that your reduced the 'rest of the board' to 2, but that's only by sliding right to turn the twos to a four, then sliding up to merge that four with the other to form an 8. $\endgroup$
    – Sconibulus
    Dec 7, 2016 at 15:09
  • $\begingroup$ @Sconibulus Yes, the idea is to merge the tiles with / into the top row (normal rules of play), and look at the tiles in the remaining rows. $\endgroup$
    – Lawrence
    Dec 7, 2016 at 15:11
  • $\begingroup$ @Sconibulus By reduce the sum to a number less than 6, I meant find a board / sequence of moves yielding with a smaller sum than 6. $\endgroup$
    – Lawrence
    Dec 7, 2016 at 15:18
  • 1
    $\begingroup$ I got i.stack.imgur.com/eFUhL.png today while playing 2048. $\endgroup$
    – mathlander
    Nov 16, 2023 at 0:07
  • 1
    $\begingroup$ @mathlander Nice! $\endgroup$
    – Lawrence
    Nov 19, 2023 at 14:24

2 Answers 2

Reset to default
5
$\begingroup$

Too big for a comment.

If new tiles must appear in the bottom three rows and the top row must be in descending order, then the following works. For convenience, the new tile always appears in the bottom row.

....   4...   8...   8...   84..   88..   884.   884.
.... u .... u .... u 4... u 4... u 4... u 4... u 8... u
....   ....   ....   ....   ....   ....   ....   ....
4...   4...   4...   .4..   .4..   ..4.   4...   ...2

The minimum here is 2 if the new tile that appears is a 2.

It is very unlikely considering the number of consecutive "4"s. If we only had "2"s appear (much more likely, but requires pretty precise spawn locations), then the following works.

....   2...   4...   4...   4...   8...   8...   8...   82..
.... u .... u .... u 2... u 4... u 2... u 4... u 4... u 4...
....   ....   ....   ....   ....   ....   ....   2...   2...
2...   2...   2...   2...   2...   2...   2...   .2..   .2..

84..   84..   842.   8422   8422   8422
4... u 42.. u 42.. 2 42.. u 42.. u 84..  
2...   2...   2...   2...   4...   ....
.2..   ..2.   ...2   2...   .2..   ..2.

Again, the minimum is 2 if a 2 appears.

The latest constraint is minimum til in the top must be a 4. It is easy to see how we can get the following by doing the above. Now we will make this into 32-16-8-4.

16-8-4-2    16-8-4-2    16-8-4-2    16-8-4-2    16-8-4-2    16-8-4-2
 . . . .     2 . . .     4 . . .     4 . . .     8 . . .     8 . . .
 . . . .     . . . .     . . . .     2 . . .     . . . .     2 . . .
 2 . . .     2 . . .     2 . . .     2 . . .     2 . . .     2 . . .

16-8-4-2    16-8-4-2    16-8-4-2    16-8-4-2    16-8-4-2    16-8-4-2
 8 . . .     8 2 . .     8 4 . .     8 4 . .     8 4 2 .     8 4 2 .
 4 . . .     4 . . .     4 . . .     4 2 . .     4 2 . .     4 2 . .
 . 2 . .     . 2 . .     . 2 . .     . . 2 .     . . . .     . . . 2

16-8-4-2    16-8-4-2    16-8-4-2    16-8-4-2    32-16-8-4 
 8 4 2 .     8 4 2 .     8 4 2 .    16 8 4 .     .  . . .
 4 2 . .     4 2 . .     8 4 . .     . . . .     .  . . .
 2 . . 2     4 2 . .     . . 2 .     . . . 2     .  . . 2
Improve this answer
$\endgroup$
1
  • $\begingroup$ +1 also. We're discussing how to get past the 'trivial' stage, but I'm having some difficulty defining something non-trivial but not too contrived. $\endgroup$
    – Lawrence
    Dec 7, 2016 at 15:50
4
$\begingroup$

If the first row goes through these states:

...2
2..4
24.2
2424

then it's possible to fill up that row without anything being in the bottom three rows. (Each time, the new tile spawns in the top right, and the key pressed is left.)

If new tiles must appear in the bottom three rows:

....
....
....
2...
2...
....
....
.4..
24..
....
....
..2.
242.
....
....
...4
2424
....
.2..
....

is the best you can do (since the last move must involve filling an empty square in the top row, and that will spawn a new tile).

Improve this answer
$\endgroup$
10
  • $\begingroup$ +1 That's true. Is it true in a non-trivial situation as well? $\endgroup$
    – Lawrence
    Dec 7, 2016 at 15:09
  • $\begingroup$ @Lawrence: Define nontrivial. $\endgroup$
    – Deusovi ♦
    Dec 7, 2016 at 15:10
  • $\begingroup$ Let's try this: new tiles must appear in the bottom 3 rows. (I hope this doesn't over-constrain the problem.) $\endgroup$
    – Lawrence
    Dec 7, 2016 at 15:15
  • $\begingroup$ Haha. I was just thinking of that scenario! $\endgroup$
    – Lawrence
    Dec 7, 2016 at 15:20
  • $\begingroup$ Hmm, can you help me with the non-trivial constraint, please? I've been trying to get a sum less than 6, but haven't yet succeeded - the question is genuine, though I have an inkling of an answer. Perhaps I need to specify that we need to start with the top row already filled (with more than just 2s and 4s). $\endgroup$
    – Lawrence
    Dec 7, 2016 at 15:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.