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. Gotta beat the Doge version first, though.If you made about 1.5 merges/turn then this game probably took you 700 turns? Does that sound reasonable?
waatNot going to give up till 8192,
My PB so farView attachment 3785
You can merge more than one at a time.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.
You can merge more than one at a time.
Yes, but you can't average more than one.
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 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!
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.
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.
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.
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:
\( \frac{3+1}{2}=2 \)
Here's a version I made up real quick that counts your turns and merges: http://fantasy.cubing.net/2048.htm
You should change the "NOTE: This site is the official version of 2048..." text.
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.
Here's a version I made up real quick that counts your turns and merges: http://fantasy.cubing.net/2048.htm
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