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2048 puzzle

Sa967St

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If you made about 1.5 merges/turn then this game probably took you 700 turns? Does that sound reasonable?
1.5 merges/turn seems quite high. There are some turns that aren't merges at all, and I think they happen just as often as multiple merges. Now I'm tempted to try again to keep track. :p Gotta beat the Doge version first, though.


siva.shanmukh: You are insane.
 
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RicardoRix

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Given every turn you gain 1 tile, and every merge you lose a tile. This means you cannot have anything other than an average of 1 merge per turn.
Certainly an average of 1.5 merges per turn is not possible.
 

TDM

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Rocky0701

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When you press an arrow key on the keyboard let's call that a "turn". When two tiles combine together to form the next tile in the sequence let's call that a "merge".

Let's take a rough look at how many turns it takes to create a tile.

To create the \( N \) tile, where N is a power of 2 greater than 2, takes \( \frac{N}{2}-1 \) merges as long as these are the merges tracking toward your goal tile.

For example you can create a 16 tile in 7 merges:
2+2=4 -> 1 merge
2+2=4 -> 1 merge
2+2=4 -> 1 merge
2+2=4 -> 1 merge
4+4=8 -> 1 merge
4+4=8 -> 1 merge
8+8=16 -> 1 merge

However, practically when you are taking turns you may make multiple 4 tiles but you only need to merge four of those 4 tiles to be two 8 tiles to make the 16 tile. If you count the merges for the two 8 tiles that did eventually become part of your 16 tile, then the 16 tile took you 7 merges to create. It's just that not all the merges you make are towards your goal. Counting only the merges towards you goal the above result would count the number of merges necessary to make the \( N \) tile.

Also, you can perform multiple merges in one turn. So, to get a 512 tile would take at least 255 merges. To do that in 56 turns would mean you would have to be averaging about 4 merges per turn :p I would say that 1.1 merges per turn is a rough approximation for a beginner, like me, at this game. Also, I may only be creating tiles at 90% efficiency (not every merge I make will create a tile that will work towards my goal of the 512 tile). So I would estimate that I might be able to make the 512 tile in:

\( \frac{\left(\frac{512}{2}-1\right)}{1.1*0.90} \approx 258 \) turns.

I don't have an empirical result for my tile creating efficiency here so I guess about 90%. I also don't know if I really am doing about 1.1 merges per turn but I would say it's a rough guess. I think this style thinking would be an approximate guess for the number of turns required to create a certain tile.

As a side note, to those who created the 2048 tile, did it take you around:

\( \frac{\left(\frac{2048}{2}-1\right)}{1.1*0.90} \approx 1033 \) turns?

I think that tile creating efficiency will likely be even higher than 90% and I also think that the average number of merges per turn could be as high a 1.3 to 1.5

Another guess at number of turns to create the 2048 tile could be:

\( \frac{\left(\frac{2048}{2}-1\right)}{1.5*0.95} \approx 718 \) turns.

For those who beat the game, do these numbers sound reasonable for the number of turns it took you to beat the game?



Wow, I really want to try 2048 3D now!
Wow! Now i know that my random number was way off! You probably did more math than the creator did haha.
 

cmhardw

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siva.shanmukh you are amazing! I am really inspired by your success, and I hope you make it to the 8192 soon!

1.5 merges/turn seems quite high. There are some turns that aren't merges at all, and I think they happen just as often as multiple merges. Now I'm tempted to try again to keep track. :p
Given every turn you gain 1 tile, and every merge you lose a tile. This means you cannot have anything other than an average of 1 merge per turn.
Certainly an average of 1.5 merges per turn is not possible.
Yes, but you can't average more than one.
I just made 507 moves to a 1024 tile. Random, but I felt like counting after seeing chris's posts on movecount.
I've been thinking about this a lot and have found out the following:

RicardoRix is partially right, however the ratio (sum total number of merges) / (sum total number of turns) can be greater than 1 in the 2048 game. The reason why is that the game will sometimes give you a 4 tile after a turn.

However, there is a ratio that appears to always be less than or equal to 1.

I have my first conjecture of the 2048 game:

Hardwick's 2048 game Conjecture #1:

\( \frac{-F+ \sum_k m_k}{T} \leq 1 \)

where
F = the number of 4 tiles, so far, that the game has given you (either after taking taking a turn or from the start)
T = the total number of turns taken in the game so far
\( \sum_k m_k \) = the sum total for all tiles of the minimum # of merges required to create that tile (k is the number of tiles on the board)

Note that the minimum number of merges to create a tile with value N is:
\( \frac{N}{2}-1 \)
(and this does mean that it takes 0 merges to create a 2 tile)

Let's generalize this equation a bit to become:

\( S_e = \frac{-F+ \sum_k m_k}{T} \)

where \( S_e \) is an empirical rating of the efficiency of whatever strategy you happen to be employing. Let's call \( S_e \) your "adjusted merges/turn efficiency".

Using the above result we know that:
\( S_e \leq 1 \)

Let's do an example:

I just played a game where I was able to create the 128 tile in 83 turns. On my board I have the following:

one 128-tile
one 16-tile
three 8-tiles
three 4-tiles
four 2-tiles

The minimum required merges to create each tile is:
63 minimum merges to create the 128 tile
7 minimum merges to create the 16-tile
9 minimum merges to create the three 8 tiles (each tile requires 3 merges * three 8-tiles)
3 minimum merges to create the three 4-tiles (each tile requires 1 merge * three 4-tiles)
0 minimum merges to create the four 2-tiles (each tile requires 0 merges * four 2-tiles)

The sum total of minimum merges to create all tiles is:
\( \sum_k m_k=63+7+9+3+0 \)
\( \sum_k m_k=82 \)

During this game I received ten 4-tiles after taking a turn. Now let's put it all together:

\( S_e=\frac{-F + \sum_k m_k}{T} \)
\( S_e=\frac{-10 + 82}{83}=\frac{72}{83}\approx. 0.8675 \) which is less than or equal to 1.

I'm still working on trying to generalize this to an even more accurate prediction method for the number of turns to create your first N tile on the board. I have some ideas, but this is still a work in progress.
 

Stefan

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RicardoRix is partially right, however the ratio (sum total number of merges) / (sum total number of turns) can be greater than 1 in the 2048 game. The reason why is that the game will sometimes give you a 4 tile after a turn.
No, that ratio really can't be greater than 1.

Or am I misinterpreting your "sum total number" (which does sound very oddly bloated to me, does it differ from "total number" or "number"?)? How are those appearing 4-tiles relevant here?

I suspect you haven't quite understood Ricardo's argument. I'll rephrase it:
You start with two tiles and after T turns and M merges you have 2+T-M tiles which is less than two if M/T>1, but you can never have less than two.
 
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cmhardw

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No, that ratio really can't be greater than 1.
Jaycee created a 1024 tile in 507 turns. This would be at least 511 merges in 507 turns. That's \( \frac{511+\sum_{k-1} m_k}{507}=\frac{511}{507}+\frac{\sum_{k-1} m_k}{507} \)

where \( \sum_{k-1} m_k \) means the minimum required merges of every other tile on Jaycee's board (without counting the 1024 tile).

For this ratio to be less than or equal to 1 then \( \frac{\sum_{k-1} m_k}{507} \) must be negative. However, every tile has a positive merge number so this didn't make sense to me. It was then that I realized that the 4-tiles the game gives you do have a negative merge number since they are effectively counted in the merge number of the larger tiles you create using these "free" 4s.

So let's use the formula that I believe to be always less than or equal to 1 and take another look at this.

\( \frac{-F+\sum_k m_k}{T} \)

Based on my previous trial where I was given ten 4-tiles (that's 10 "free" merges that I never did) in 83 turns then I would guess that Jaycee received somewhere in the neighborhood of sixty one 4-tiles during a game of 507 turns. Using that information we get:

\( \frac{-61+\sum_k m_k}{507} \leq 1 \)

\( -61 + \sum_k m_k \leq 507 \)

One of the \( m_k \) tiles is the 1024 tile that Jaycee created. It's minimum number of merges required to be created is 511.

\( -61 + \left(511 + \sum_{k-1} m_k\right) \leq 507 \)
\( 450 + \sum_{k-1} m_k \leq 507 \)
\( \sum_{k-1} m_k \leq 57 \)

What this tells me is that the sum of all the minimum merges for all the other tiles on the board other than the 1024 tile is less than or equal to 57. One possible distribution of this could be:
one 64 tile (31 merges)
one 32-tile (15 merges)
one 16-tile (7 merges)
one 8-tile (3 merges)
one 4-tile (1 merge)

Now it makes more sense why Jaycee's ratio of merges/turns is over 1. In a sense the units didn't match. The # of turns (507) that Jaycee reported is the number of global turns for the whole game. The number of merges to create a 1024 tile (511 merges) is only the number of merges for the one 1024-tile. The ratio of (511 merges / 507 turns) being larger than 1 confused me for some time, until I realized that the turns counted for the whole board, and the 511 merges only counted for the one tile. Not only that, but without taking into account the free 4-tiles given to you the ratio will go over 1.


Or am I misinterpreting your "sum total number" (which does sound very oddly bloated to me, does it differ from "total number" or "number"?)? How are those appearing 4-tiles relevant here?
If you begin the game with two 2-tiles in line to be merged and you merge them, your board now has a 4-tile and you've taken 1 turn. Now let's say that the game gives you a 4-tile for free and that it is in line with the 4 tile you created. You take your 2nd turn to merge those two 4-tiles into an 8 tile, and then the game gives you a 4-tile for free anywhere on the board. Your board now consists of an 8-tile and a 4-tile. You have taken two turns. If you did not account for the number of free 4 tiles given to you, then you would say that the ratio of merges/turns for the whole board is:

\( \frac{3+1}{2}=2 \)

So now your global ratio of merges/turn is 2, which breaks our understanding of this ratio needing to be less than or equal to 1. I fixed this by accounting for the number of 4-tiles given to you for free. Since you are looking at the face value of tiles when calculating \( \sum_k m_k \) then you are effectively counting all the 4 tiles that were used to create each of your tiles as having taken 1 merge. Some of those 4s did not take one merge, you got them for free.

Now let's use the version of the formula that takes into account how many "free" 4s you received.

\( \frac{-F+\sum_k m_k}{T} \)

\( \frac{-2+(3+1)}{2} \leq 1 \)

And now this ratio really is less than or equal to 1 again and there is no problem.

I suspect you haven't quite understood Ricardo's argument. I'll rephrase it:
You start with two tiles and after T turns and M merges you have 2+T-M tiles which is less than two if M/T>1, but you can never have less than two.
How do you count how many merges have been done?

I choose to count it by looking at the face value of each tile and knowing that a tile of value N needs \( \frac{N}{2}-1 \) merges to be created. This formula accounts for every merge required (including down to all the 4-tiles you had to create to create each of the larger tiles to get to N). If you've been given some number of 4 tiles for free, but you have counted every free 4-tile as having required 1 merge to be created, then you are over counting your merge number when you look at tile face values. This problem can be remedied by tracking the number of 4-tiles the game gives you, since each one of these is a "free" merge that you never had to do by taking a turn to merge two 2-tiles to create that 4-tile.
 

Stefan

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If you begin the game with two 2-tiles in line to be merged and you merge them, your board now has a 4-tile and you've taken 1 turn. Now let's say that the game gives you a 4-tile for free and that it is in line with the 4 tile you created. You take your 2nd turn to merge those two 4-tiles into an 8 tile, and then the game gives you a 4-tile for free anywhere on the board. Your board now consists of an 8-tile and a 4-tile. You have taken two turns. If you did not account for the number of free 4 tiles given to you, then you would say that the ratio of merges/turns for the whole board is:

\( \frac{3+1}{2}=2 \)
You did two merges (I highlighted them in your text), that's exactly one merge per turn.
 

cmhardw

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You did two merges (I highlighted them in your text), that's exactly one merge per turn.
So we have resolved the issue. You choose to count the number of merges in this example by counting them by hand. I choose to count the merges by looking at the face values of the tiles on the board. I choose my method so that I can extend it to boards with tiles with very large face values.
 

Methuselah96

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So we have resolved the issue. You choose to count the number of merges in this example by counting them by hand. I choose to count the merges by looking at the face values of the tiles on the board. I choose my method so that I can extend it to boards with tiles with very large face values.
But isn't a merge: "When two tiles combine together to form the next tile in the sequence?"
Your calculations seem to disagree with this definition.
 
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