# 2048 puzzle



## rj (Mar 12, 2014)

this is a really fun puzzle, and it's kinda like 15 puzzle:

http://gabrielecirulli.github.io/2048/


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## ajayd (Mar 13, 2014)

Goddang rj, ive wasted a full hour of my life on this game.


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## notfeliks (Mar 13, 2014)

As soon as I realised what this was, I knew I was going to waste so much time on it.

Damn you rj.


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## brian724080 (Mar 13, 2014)

Damnit, why did you have to show this.
On the other hand, this gives me something to do during boring lectures.


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## DAoliHVAR (Mar 13, 2014)

the best "strat" I found is that you just stick your big numbers on one side and then do all the combining with the other three keys
this way you stay alive for a while,but i always mess up when i accidentaly press some other key and all the big numbers get with the rest of em,so i lose in 5 turns


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## brian724080 (Mar 14, 2014)

DAoliHVAR said:


> the best "strat" I found is that you just stick your big numbers on one side and then do all the combining with the other three keys
> this way you stay alive for a while,but i always mess up when i accidentaly press some other key and all the big numbers get with the rest of em,so i lose in 5 turns



That doesn't yield much flexibility though

Edit: My class - 40% listening, 20% playing this, 40% doing something else


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## DAoliHVAR (Mar 14, 2014)

brian724080 said:


> That doesn't yield much flexibility though
> 
> Edit: My class - 40% listening, 20% playing this, 40% doing something else



i found that i can consistently do atleast 512 and a 256 with this,pb is 1024 + 512
please share if you know a better method


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## brian724080 (Mar 14, 2014)

DAoliHVAR said:


> i found that i can consistently do atleast 512 and a 256 with this,pb is 1024 + 512
> please share if you know a better method



By doing that, you limit the availability on the sides, and you're basically screwed if it forms a perfect rectangle. Just like in speedsolving, I never really stick to rules like that, I just do whatever seems fit, like blockbuilding. What is your actual score?


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## DAoliHVAR (Mar 14, 2014)

brian724080 said:


> By doing that, you limit the availability on the sides, and you're basically screwed if it forms a perfect rectangle. Just like in speedsolving, I never really stick to rules like that, I just do whatever seems fit, like blockbuilding. What is your actual score?



i just told you,are you saying i am lying?If you try to get a good score with the method, im sure you could.
did you pass the game?You are right about the rectangle thing though,i am used to having a method of doing something
and tried to make one up,i didn't say hey this is THE BEST METHOD USE THIS
i said its the best method i've found
btw pb in points is 11760


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## Mikel (Mar 14, 2014)

PB is 6268 so far


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## Jaycee (Mar 14, 2014)

First try: got a 512, a 256, and a bunch of 64s


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## kcl (Mar 14, 2014)

ugh I got to 1024 in math today.. The I started paying attention on the lesson too much and screwed up -_-


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## Jaycee (Mar 15, 2014)

I got so close... All the pieces were there, just in positions that made it difficult for me to get them all together. 



Got the score to 17012. If only the 256s had been closer ;_;


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## ajayd (Mar 15, 2014)

Yeah, the best strategy for me is just to pick a forbidden key, and never use it. High score is 15244. Only problem with it is that if you have to use the forbidden key, a 2 will get into the area, and then you're screwed.


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## TDM (Mar 15, 2014)

ajayd said:


> Yeah, the best strategy for me is just to pick a forbidden key, and never use it. High score is 15244. Only problem with it is that if you have to use the forbidden key, a 2 will get into the area, and then you're screwed.


I thought I could risk using the forbidden key once.





Score

btw, this was first try using this method. I'm definitely going to use this more and see how far I can get.


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## collins (Mar 15, 2014)

Finally got it, after several close attempts...
I got lucky at the end, I was forced to use the "forbidden key" but I could still gather the big pieces 





What is nice is that you can go on and try to get the 4096... but I could only get an additional 512.



Spoiler


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## Bh13 (Mar 15, 2014)

Win! Final score of 23784 and a 254 tile. Screenshots soon


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## TDM (Mar 15, 2014)

Bh13 said:


> Win! Final score of 23784 and a 254 tile. Screenshots soon


Nice... 1024 is so easy to get every time but getting 2048 is so hard


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## Jaycee (Mar 16, 2014)

We have another winner! I eventually got the score to 26248.


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## TheNextFeliks (Mar 16, 2014)

Jaycee said:


> We have another winner! I eventually got the score to 26248.
> 
> 
> 
> Spoiler


Nice. I've gotten to 512 once in 5 tries. 

Is that chrome OS


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## Kit Clement (Mar 16, 2014)

Lackluster effort after the 2048, but hell, after getting 1024+512+256 three straight times and losing each time, I'll take it.



Spoiler


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## bryson azzopard (Mar 17, 2014)

im so addicted to this fricking game its insane! I just want to win that all I ask for


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## JTW2007 (Mar 17, 2014)

At last I can go do other things.



Spoiler











Tip: Sometimes it's possible to get into a spiral pattern where your highest tile is in top left and the top row decreases in value from left to right. You build up the (4, 2) tile and then pair with (4, 1), and then the whole thing cascades leftward as long as you have consecutive powers of two in the top row. This has an added benefit which is that since you don't normally need the whole rest of the board to build, biggish powers tend to accumulate in the middle left side of the board, allowing you to sometimes pair upward into the top row from there.


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## mande (Mar 17, 2014)

Spoiler











Has anybody here ever gotten 4096?


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## Jaycee (Mar 17, 2014)

JTW: The cascading effect feels really satisfying when it happens :3

mande: if anyone has i'd scream at them for being too awesome


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## RicardoRix (Mar 17, 2014)

Just hit yahoo news:

http://uk.news.yahoo.com/one-19-old-won-internet-addictive-puzzler-2048-083752788.html


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## rj (Mar 17, 2014)

TheNextFeliks said:


> [/spoiler]
> Nice. I've gotten to 512 once in 5 tries.
> 
> Is that chrome OS



I got 2 1024s, then the board filled up. It's Chrome OS.


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## TheNextFeliks (Mar 17, 2014)

I got 1024 with a 256+128. I switched to change the song on my music. And the Internet stopped responding.  I could've won it.


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## JTW2007 (Mar 17, 2014)

I messed up, but I was well on the way to 4096. 


Spoiler











I think I figured out a method that works fairly well:



Spoiler



Notation: Up/left/right/down arrows are U/L/R/D. Squares are indexed like pixels, in pairs of (x, y) starting at (1, 1) in the top left.

Use U and L to build as large a block as possible in (1, 1). Avoid shifting this block to the right at all costs. If necessary, burn off turns by mashing U until the top row fills. When the top row is static (no horizontal combinations or open spaces) you can make R moves without fear of displacing your large block from (1, 1). You should have inadvertently built up some large numbers toward the left of the middle two rows. Now that you can use R moves, move these over and combine them into the top when possible, and then collapse the top row as far as possible to the left. This should ensure your largest term always stays in (1, 1). Your goal is to have the top row occupied by consecutive powers decreasing rightward. 

Eventually you'll end up with a pretty big number in (4, 1). Now you begin a new phase where the goal is just to grow (1, 4) so you can collapse left again. Specifically, you want to increase the value of (4, 2) as quickly as possible so you can combine up into (1, 4) (I'll try my best to describe how I do this but it's a little unclear to me still). You can separate out the new spawns in rows 3 and 4 using U and R, so that you're sending larger numbers toward (4, 2) from both the left and the bottom at the same time. You should be able to get to a point where no matter what the new block is, you can combine it in one direction or the other. It is very likely that one of these chains will make it all the way to your build square. Occasionally, the zugzwang will force you to do an L. If the new block does not spawn in (4, 2), go back immediately and continue. If the new block does spawn where you were building, you'll have to start over with the building-up process and look for an opportunity to combine your old (4, 2) tile up into the top row.

Eventually, new spawns in your build square will have forced you to move a lot of high-value tiles to the upper left. When this happens and you can't combine them with R/L/U, you'll run out of space to build and need to drop down a row. Now you can build in (4, 3) using the bottom row and left two columns to maneuver in the same way you did when you were building (4, 1).

One thing I noticed in this method is that it's very hard to get stuck with an uncombinable rectangle on your turn. Since you're using the 2s and 4s to build both horizontally and vertically, you'll almost always have an asymmetrical board, and when you do have a rectangular board, you'll be almost guaranteed to have either U or R/L as a legal move.

I'm still trying to figure out a way to protect the top-row block order from being disrupted. I sometimes get forced to make R moves when I don't want to, and if the new block spawns in (1, 1), it's pretty much gg. Anyone have any ideas for improvements to this system?


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## rj (Mar 18, 2014)

YES!!!!!!


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## Kit Clement (Mar 18, 2014)

Aaaand that's the last time I play this game.



Spoiler











EDIT: jk, had to take care of this too



Spoiler


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## IQubic (Mar 18, 2014)

What is your elusive strategy Kit? Come you can tell us, We won't abuse it.

-IQubic


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## Kit Clement (Mar 18, 2014)

It's nothing special, really. It's mostly what JTW does, but I avoid building in (1, 4) and instead build my 4th largest number in (2, 3). Corners are just too hard to build with. The next two largest I try to have as (2, 2) and (2, 1), but given the added flexibility of working with the middle of the board, this can change depending on the situation. I use (1, 4) and (2, 4) as filler, and build them up so I have extra pieces to both block off my snake pattern and have some pieces that can often bail me out of trouble.


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## Rocky0701 (Mar 18, 2014)

Did anyone else notice that Kevin Hays said that he was obsessed with this game him and Feliks' live show?


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## TDM (Mar 18, 2014)

Kit Clement said:


> Spoiler


That's impressive... how do you do it?
Also trying the doge one now. For some reason I find it easier to not know what everything is...


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## Jaycee (Mar 18, 2014)

^My friend says he likes the doge one because "it requires less thinking". >__> 

Just beat 2048 for the second time! I'll edit with score when I finish the game. :3 EDIT: Lost shortly after; I got to a point where I had to move my 2048 tile from the corner and it went downhill from there. The biggest thing I got afterwards was a 256. Final score was in the 22000s, doesn't beat the score from my other win which was 26k.

I think I caused this game to spread like a disease through the honors and AP population of my school. A classmate saw me playing while we were watching a movie clip in English, he started playing and showed it to a friend of his, and that friend showed it to more people and it spread from there. My AP Government had a 5 minute conversation about the game when we had some free time, and half the class (15 kids/30) were discussing the game. Yay me! Lol.


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## uvafan (Mar 18, 2014)

First time getting 2048, was also very close to 2,048+1,024.


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## TDM (Mar 18, 2014)

Spoiler













Spoiler



http://ov3y.github.io/2048-AI/


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## Rocky0701 (Mar 18, 2014)

We could also have little competitions on the lowest number of moves to get to certain numbers. Such as 56 moves to get to 512.


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## TDM (Mar 19, 2014)

Rocky0701 said:


> We could also have little competitions on the lowest number of moves to get to certain numbers. Such as 56 moves to get to 512.


Did you really get to 512 in 56 moves? Is that even possible?
20 for 32
37 for 64
82 for 128
155 for 256
196 for 256+128
230 for 256+128+64+32+16+8+4+2
270 for 256+128+128
311 for 512


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## rj (Mar 19, 2014)

Nice. I'm just happy I have half the community addicted to this. I'M NOT ALONE!


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## ThomasJE (Mar 19, 2014)

TDM said:


> Did you really get to 512 in 56 moves? Is that even possible?
> 20 for 32
> 37 for 64
> 82 for 128
> ...



It sounded to me like 56 was just a random number.

Also, is anyone else having problems with the mobile site? It's not working for me.


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## brian724080 (Mar 20, 2014)

Btw, 3D, 4D, and 5D

2048-3D is surprisingly easy:


Also:


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## Rocky0701 (Mar 20, 2014)

ThomasJE said:


> It sounded to me like 56 was just a random number.
> 
> Also, is anyone else having problems with the mobile site? It's not working for me.



Yes, 56 was just a random number, sorry for not clarifying. Or we could even do time competitions.


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## Jaycee (Mar 20, 2014)

Sometimes I like to time how fast I can get a 128 tile. I've only done it under 10 times though; my best is 28.xx


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## TDM (Mar 20, 2014)

Jaycee said:


> Sometimes I like to time how fast I can get a 128 tile. I've only done it under 10 times though; my best is 28.xx


How do you time? I always lose a few seconds going from qqTimer to 2048 and back again...
E: incredibly lucky 31.50. Lost 5 seconds almost accidentally deleting a time clicking on the window...


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## Rocky0701 (Mar 20, 2014)

Just tried this and did an average of 5 timing to 128, and got 16.33 This was my first time trying this so i am pretty happy. I mostly just hit random keys, but there was still some strategy.

16.95 15.13 (14.02) 16.92 (18.27)


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## Sajwo (Mar 20, 2014)

128 tile

Single: 5.61

Avg5: 7.57

(10.59), 8.11, 7.33, 7.27, (5.61)

Avg12: 8.28

8.11, 7.33, 7.27, (5.61), (10.04), 9.60, 8.35, 8.64, 8.40, 9.35, 6.82, 8.92

Easy. Just hit the buttons very fast and try to match big numbers


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## Rocky0701 (Mar 20, 2014)

256 tile average of 5: 25.99

(22.14) 26.74 (DNF: I got a game over) 26.69 24.54

New 128 tile average of 5: 11.33

11.59 (DNF) 9.45 12.96 8.32


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## Sajwo (Mar 20, 2014)

Rocky0701 said:


> 256 tile average of 5: 25.99
> 
> (22.14) 26.74 (DNF: I got a game over) 26.69 24.54




This is impressive  I have gotten like 8 DNF's in a row. But then I got 16.04 single


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## TDM (Mar 20, 2014)

256: 22.58, 17.68, 19.13, DNF, 14.03 = 19.80


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## Rocky0701 (Mar 20, 2014)

64: 5.49 (7.49) 5.21 4.96 (4.06) average: 5.22

512: PB 52.32


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## Sa967St (Mar 21, 2014)

Yay. 



Spoiler


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## Jaycee (Mar 21, 2014)

Would've been my 4th time beating it. All I had to do was connect the 32's and it would have the cascading effect from there. 



Edit 2 hours later: I eventually got it. 4th win. Yay. I still haven't beat my original high score of 26xxx.

ANother edit 2 hours later: 5th win. This is getting good :3


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## TheNextFeliks (Mar 21, 2014)

Ugh. 2nd 1024. Had 256 and 128 also. Changed forbidden keys halfway through on accident.


Edit:

YES!!!!!



Listening to the song Happy when this happened. Lol. Now I have stuff I can do.


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## cmhardw (Mar 21, 2014)

Rocky0701 said:


> We could also have little competitions on the lowest number of moves to get to certain numbers. Such as 56 moves to get to 512.



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?



brian724080 said:


> Btw, 3D, 4D, and 5D
> 
> 2048-3D is surprisingly easy:
> View attachment 3774



Wow, I really want to try 2048 3D now!


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## cmhardw (Mar 21, 2014)

Sa967St said:


> Yay.
> 
> 
> 
> Spoiler



Sarah, congrats! I haven't gotten the 2048 tile yet, but I really enjoy this game.

I would guess that you took approximately 1050 merges to beat this game. I got this by adding up the required number of merges for each tile on your board. Do you know about how many moves this solution took you? I'm trying to come up with an empirical result for the number of merges per turn on average that a person makes when playing this game.

If you made about 1.5 merges/turn then this game probably took you 700 turns? Does that sound reasonable?


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## Jaycee (Mar 21, 2014)

Aaand here we go! Fantastic continuation on my 5th win. This is what the board looked like before my Game Over move.

View attachment 3784

So I actually got quite close to the 4096 tile! Most of the game, the 2048 was in the corner, but I had to hit the up key at one point. I still lasted quite a while after that, though! (I think I made the entire 512 with the 2048 out of the corner) I also smashed my high score, went from 26k to 34788.


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## TDM (Mar 21, 2014)

cmhardw said:


> I haven't gotten the 2048 tile yet, but I really enjoy this game.


Same... it's addictive. 1024 is so easy, but 2048 is impossible. I'll get there eventually!


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## JTW2007 (Mar 21, 2014)

cmhardw said:


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



I haven't counted my turns yet, but at a guess I'd say these are underestimates for the method I've been using. I waste a lot of moves intentionally to avoid being stuck in a bad position on my turn, which means that, especially in the early stages of the game, I take lots of merge-free turns. I'd say due to this I probably average a little less than one merge per turn. I also end up with lots of high-value remnant tiles at the end of the game (Sarah: Your board has none of these; how did you avoid them?), so a non-negligible percentage of my merges don't go towards the 2048 tile. Other people's solutions may be much more efficient though, so perhaps ~1.1 merges/turn is a good overall estimate.

Is the code for the game available? It'd be cool to have the game track these statistics.


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## siva.shanmukh (Mar 21, 2014)

Not going to give up till 8192,  
My PB so far


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## Jaycee (Mar 21, 2014)

what sort of madman are you


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## Sa967St (Mar 21, 2014)

cmhardw said:


> 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.  Gotta beat the Doge version first, though.


siva.shanmukh: You are insane.


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## RicardoRix (Mar 21, 2014)

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.


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## TDM (Mar 21, 2014)

siva.shanmukh said:


> Not going to give up till 8192,
> My PB so farView attachment 3785


waat


RicardoRix said:


> 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.


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## RicardoRix (Mar 21, 2014)

TDM said:


> You can merge more than one at a time.



Yes, but you can't *average* more than one.


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## Jaycee (Mar 21, 2014)

I just made 507 moves to a 1024 tile. Random, but I felt like counting after seeing chris's posts on movecount.


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## uberCuber (Mar 21, 2014)

RicardoRix said:


> Yes, but you can't *average* more than one.



Well, slightly more than one since you start with 2 tiles and end with 16 

EDIT: nvm I'm dumb


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## Rocky0701 (Mar 22, 2014)

cmhardw said:


> 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.
> 
> ...



Wow! Now i know that my random number was way off! You probably did more math than the creator did haha.


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## cmhardw (Mar 22, 2014)

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



Sa967St said:


> 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.





RicardoRix said:


> 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.





RicardoRix said:


> Yes, but you can't *average* more than one.





Jaycee said:


> 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.


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## TP (Mar 22, 2014)

3-D version is really fun, I managed to get a nice pattern.

http://oi58.tinypic.com/zv3nmd.jpg


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## Stefan (Mar 22, 2014)

cmhardw said:


> 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 (Mar 22, 2014)

Stefan said:


> 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.




Stefan said:


> 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.



Stefan said:


> 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 (Mar 22, 2014)

cmhardw said:


> 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.


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## Methuselah96 (Mar 22, 2014)

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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## Stefan (Mar 22, 2014)

Methuselah96 said:


> 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.


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## Methuselah96 (Mar 22, 2014)

Stefan said:


> You should change the _"NOTE: This site is the official version of 2048..."_ text.



Good point. Fixed.


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## cmhardw (Mar 22, 2014)

Stefan said:


> 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.


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## Methuselah96 (Mar 22, 2014)

cmhardw said:


> 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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## Stefan (Mar 22, 2014)

Methuselah96 said:


> Here's a version I made up real quick that counts your turns and merges: http://fantasy.cubing.net/2048.htm



Here's my result (I don't think I've ever had such a clean finish before):
 (one 2-tile shy of being perfect)


943 turns
936 merges

Edit: Did one with timing, took 17 minutes and only 939 turns and 932 merges.


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## cmhardw (Mar 22, 2014)

Methuselah96 said:


> 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.



Please help me to see how. Stefan counts 2 merges by hand.

I count 3 merges necessary to create the 8 tile, 1 merge necessary to create the 4 tile then I subtract 2 merges (1 for each of the two free 4-tiles). Therefore I count (3+1)-2=2 merges.

I don't understand how my calculation disagrees with "When two tiles combine together to form the next tile in the sequence" ? Stefan and I both count 2 merges by this definition.


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## Stefan (Mar 22, 2014)

cmhardw said:


> Please help me to see how. Stefan counts 2 merges by hand.
> 
> I count 3 merges necessary to create the 8 tile, 1 merge necessary to create the 4 tile then I subtract 2 merges (1 for each of the two free 4-tiles). Therefore I count (3+1)-2=2 merges.
> 
> I don't understand how my calculation disagrees with "When two tiles combine together to form the next tile in the sequence" ? Stefan and I both count 2 merges by this definition.



So you do count two merges there? Then do you agree now that Ricardo was correct? That you can't average more than one merge per turn?


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## Methuselah96 (Mar 22, 2014)

cmhardw said:


> Please help me to see how. Stefan counts 2 merges by hand.
> 
> I count 3 merges necessary to create the 8 tile, 1 merge necessary to create the 4 tile then I subtract 2 merges (1 for each of the two free 4-tiles). Therefore I count (3+1)-2=2 merges.
> 
> I don't understand how my calculation disagrees with "When two tiles combine together to form the next tile in the sequence" ? Stefan and I both count 2 merges by this definition.



Sorry, I read the quote from Stefan's message and did not keep on reading from your original post.  I'll read more carefully next time.


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## TDM (Mar 22, 2014)

So close  Am I the only one to have not yet completed it?


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## Sa967St (Mar 22, 2014)

Third 2048 success: 7:16.62

Fourth success: 6:20.97

Seventh success: 5:33.66

ninth success: 4:47.91


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## kcl (Mar 22, 2014)

Finally 2048 success! Didn't time it. Oh well.


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## Jaycee (Mar 22, 2014)

Methuselah96: Nice edit!

16 tile: 9 turns, 7 merges
32 tile: 17 turns, 13 merges
64 tile: 39 turns, 34 merges (got a bit inefficient for a brief period)
128 tile: 59 turns, 56 merges (aaand now I'm back on track)
256 tile: 120 turns, 114 merges
512 tile: 243 turns, 236 merges 
1024 tile: 490 turns, 480 merges
2048 tile: 971 turns, 958 merges

6th win, BTW. I'm positive that I'm much slower than Sarah (7 minutes?! )


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## Rocky0701 (Mar 22, 2014)

TDM said:


> So close  Am I the only one to have not yet completed it?


I haven't either, i got to where i had two 512's and one 1024, but i couldn't combine them.


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## Stefan (Mar 22, 2014)

cmhardw said:


> 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.



I did count it by hand in that tiny example. For larger examples I'd let the computer count. Or if that's not available, I'd count the number T of turns and the number E of tiles I end up with and then simply calculate the number of merges as M=2+T-E. I think that's far simpler than your method, both the final calculation as well as the counting (T is easier than F because you don't need to look at the new tile and decide whether to count it, plus you already have T in your conjecture anyway).



cmhardw said:


> Note that the minimum number of merges to create a tile with value N is:
> \( \frac{N}{2}-1 \)



Not true, for example for N=4 this would mean you need a merge. But you really don't. And you can build an 8 with just one merge, don't need three. And so on.

You're clearly using two different definitions of "merge", and that's no good.


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## cmhardw (Mar 22, 2014)

Stefan said:


> So you do count two merges there? Then do you agree now that Ricardo was correct? That you can't average more than one merge per turn?



Either you're still not reading my posts because you consider them "too long" or I'm not explaining myself well enough. I already covered that question. See the two quotations below:



cmhardw said:


> 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.





cmhardw said:


> Now let's use the version of the formula that takes into account how many "free" 4s you received.
> 
> \( \frac{-F+\sum_k m_k}{T} \)
> 
> ...



------------------



Stefan said:


> I did count it by hand in that tiny example. For larger examples I'd let the computer count. Or if that's not available, I'd count the number T of turns and the number E of tiles I end up with and then simply calculate the number of merges as M=2+T-E. I think that's far simpler than your method, both the final calculation as well as the counting (T is easier than F because you don't need to look at the new tile and decide whether to count it, plus you already have T in your conjecture anyway).



I think that's really elegant. I did not come up with the most elegant way to figure out how the merges/turns ratio during the game must be less than 1. You clearly have superior math skills and my method is clearly not the most elegant way to perform this calculation. Based on what I have seen so far my method still works, although it seems I was not clear in my definition of "merges" which I will attempt to correct at the end of my post below.

Stefan, I cannot stress enough how infuriating your posts can be to read sometimes. I write my posts keeping in mind how you are going to rip them to shreds when you read them. I have been doing this for years, as you have consistently, for years, ripped my math posts to shreds for either being "too long" or not being rigorous enough. I notice now that I get frustrated whenever I make a math post and see that you have responded to it.

Your solution is more elegant than the one I came up with. You win. You are the man. Is this what you are looking for with your posts? You have succeed in correcting the worthless dribble that I post, good job sir.



Stefan said:


> Not true, for example for N=4 this would mean you need a merge. But you really don't. And you can build an 8 with just one merge, don't need three. And so on.
> 
> You're clearly using two different definitions of "merge", and that's no good.



Thank you for correcting the discrepancy in my terminology. Let me try to explain my thoughts.

Without having said it, I have been approaching the 2048 game as a variation on a "pure" form version of this game where after each turn the user is given a 2-tile, with no exceptions.

In this "pure" form version of the game, a 4-tile would _always_ require 1 merge to be created. It would be created by the player merging two 2-tiles. In this "pure" form version of the game, a tile with the value N would always be created using \( \frac{N}{2}-1 \) merges by the user. It is possible for the user to perform multiple merges during one turn.

The "real" version of 2048 will give the user 4-tiles after a turn sometimes. I then decided to use the formula from the "pure" version of the game, namely that the N-tile can be created in \( \frac{N}{2}-1 \) merges and adjust it for the fact that sometimes a 4-tile has been given for free. This is less efficient than your method, but from what I have seen so far my formula will still calculate the number of merges initiated by the user just as accurately as your formula. I called it a conjecture because I don't have the math knowledge yet to prove my formula for every case. You have more math schooling and knowledge than I do, so if you can disprove my formula rigorously then clearly my formula was not correct. As far as I can tell you have pointed out the error in my explaining my formula so far, but no errors in how the formula calculates the number of merges made by the user.

I may have made even more errors in this very post, and if you point them out I will do my best to either try to correct them if they can be corrected, or admit that my formula is not true if you disprove it.

Stefan, I really respect you as a person and as a cuber. I like spending time with you in person at competitions. I like you enough to tell you that your forum personality can be very infuriating to talk to.


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## Stefan (Mar 22, 2014)

I think I have more to say, but I'll re-read your posts more carefully first. For now let me just say I'm sorry. You're among the people I like and respect and care about most, and I'm really unhappy having frustrated and infuriated you. I'm aware I'm somewhat confrontational, but I didn't realize it could make you feel that way that much. Sorry.


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## uyneb2000 (Mar 23, 2014)

My first success!
What strategies do you guys use?
Here's a video of my ending.


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## brian724080 (Mar 23, 2014)

uyneb2000 said:


> My first success!
> What strategies do you guys use?
> Here's a video of my ending.



Nice, but you could've won much more easily. You can fill up the three left-most columns with useless blocks, making sure that they don't merge with an up/down move, and then shift the rightmost 128 up by one block to win.


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## Lucas Garron (Mar 23, 2014)

Methuselah96 said:


> Here's a version I made up real quick that counts your turns and merges: http://fantasy.cubing.net/2048.htm



Feature request: Record all the moves (and the location/value of new tiles) and/or support playbacks!

(Preferably store all my history of solves in localStorage. Don't really care how sharing works.)


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## Jaycee (Mar 23, 2014)

I fell into a trance while making the first 2048, and it just went smoothly from there.



Game over:


Spoiler


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## Sa967St (Mar 23, 2014)

Methuselah96 said:


> Here's a version I made up real quick that counts your turns and merges: http://fantasy.cubing.net/2048.htm



Awesome, thanks!


Here's a (somewhat) interesting observation: \( \#\ of\ turns\ =\ \#\ of\ merges\ +\ \#\ of\ active\ tiles\ -\ \#\ of\ active\ tiles\ on\ turn\ 0 \).

Assuming this is true, in order for \( \frac{\#\ of\ merges}{\#\ of\ turns} \) to equal 1, there must be exactly two tiles on the board, which (other than on turn 0) is possible on your first turn, if you start with two 2s that are aligned (and also on the following second turn if the next block is 4 and is aligned with the previous 4). In order for that ratio to be greater than one, there must be exactly one tile on the board, which is impossible. So, the ratio is always \( <=1 \).


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## Stefan (Mar 23, 2014)

Sa967St said:


> Here's a (somewhat) interesting observation: \( \#\ of\ turns\ =\ \#\ of\ merges\ +\ \#\ of\ active\ tiles\ -\ \#\ of\ active\ tiles\ on\ turn\ 0 \).
> 
> Assuming this is true



It is indeed true, I [post=963309]ended up[/post] with the same, though it's really just [post=963076]Ricardo's insight[/post] fleshed out a little.


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## Sa967St (Mar 23, 2014)

Stefan said:


> It is indeed true, I [post=963309]ended up[/post] with the same, though it's really just [post=963076]Ricardo's insight[/post] fleshed out a little.


Ah, I should've read more of the thread. Can you think of a way to prove it without just observing arbitrarily many cases? Also, would it still be true if you start with three or four tiles instead of two? I assume so, but it's strictly from observation.


Another (more) interesting observation, which I can actually prove: 

Let the nth tile have value t(n). (t(0)=2, t(1)=4, t(2)=8, etc.) Then if you receive only a 2 on every turn (no 4s), the your score is \( \sum_s nt(n) \), for all active tiles s with degree n and value t(n).

Proof:
Using the scoring system, each active tile with degree n, when first merged, accumulates 2p(n-1)+t(n) points, where p(n-1) are the points accumulated from the (n-1)th tile.
Since p(0)=0, we get the following when we receive only 2s:

p(1) = (2 x 0) + 4 = 4
p(2) = (2 x 4) + 8 = 16
p(3) = (2 x 16) + 16 = 48
p(4) = (2 x 48) + 32 = 128
p(5) = (2 x 128) + 64 = 320
p(6) = (2 x 320) + 128 = 768
p(7) = (2 x 768) + 256 = 1792
p(8) = (2 x 1792) + 512 =4096
p(9) = (2 x 4096) + 1024 = 9216
p(10) = (2 x 9216) + 2048 = 20480

Note the following:

p(0) = 0 = 0 x 2
p(1) = 4 = 1 x 4
p(2) = 16 = 2 x 8
p(3) = 48 = 3 x 16
p(4) = 128 = 4 x 32
p(5) = 320 = 5 x 64
p(6) = 768 = 6 x 128
p(7) = 1792 = 7 x 256
p(8) = 4096 = 8 x 512
p(9) = 9216 = 9 x 1024
p(10) = 20480 = 10 x 2048

This is exactly p(n)=nt(n) (since t(n)=2^(n+1)). [very lazy way of finding an explicit formula for a recursive equation, I know]

So each active tile of degree n and value t(n) accumulates nt(n) points, and the total score is the sum of all points of each active tile, which is precisely \( \sum_s nt(n) \), for all active tiles s with degree n and value t(n).

Q.E.D.
________________________________________________________________________________________________________________________________


Also, since it's unrealistic to only get 2s and no 4s, it follows that your score is actually \( \sum_s nt(n)-4f \), for all active tiles s with degree n and value t(n), where f is the total number of 4s given. 


Question: Does anyone know the probability of a getting a 4 instead of a 2?


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## Stefan (Mar 23, 2014)

Sa967St said:


> Can you think of a way to prove it without just observing arbitrarily many cases? Also, would it still be true if you start with three or four tiles instead of two?



I consider Ricardo's first sentence the proof. Is it not clear enough?



Sa967St said:


> Question: Does anyone know the probability of a getting a 4 instead of a 2?



Looks like 10%:


```
// Adds a tile in a random position
GameManager.prototype.addRandomTile = function () {
  if (this.grid.cellsAvailable()) {
    var value = Math.random() < 0.9 ? 2 : 4;
    var tile = new Tile(this.grid.randomAvailableCell(), value);

    this.grid.insertTile(tile);
  }
};
```
From http://gabrielecirulli.github.io/2048/js/game_manager.js


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## vcuber13 (Mar 23, 2014)

Sa967St said:


> Question: Does anyone know the probability of a getting a 4 instead of a 2?



1/10, assuming javascript has an equal distribution, which I think it does.


```
// Adds a tile in a random position
GameManager.prototype.addRandomTile = function () {
  if (this.grid.cellsAvailable()) {
    var value = [B]Math.random() < 0.9[/B] ? 2 : 4;
    var tile = new Tile(this.grid.randomAvailableCell(), value);

    this.grid.insertTile(tile);
  }
};
```


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## Sa967St (Mar 23, 2014)

Stefan said:


> Sa967St said:
> 
> 
> > Can you think of a way to prove it without just observing arbitrarily many cases? Also, would it still be true if you start with three or four tiles instead of two?
> ...



I was referring to proving the formula itself, not the ratio being <=1.

Also, as TDM mentioned, you can have multiple merges per turn, losing multiple tiles in one turn. I suspect that in order to lose multiple tiles (say, 2 tiles) in one turn, there must have been one previous turn where you didn't merge anything, but I can't think of why. Either that, or I'm not fully understanding Ricardo's post.




Stefan said:


> Sa967St said:
> 
> 
> > Question: Does anyone know the probability of a getting a 4 instead of a 2?
> ...


Thanks!


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## Stefan (Mar 23, 2014)

Sa967St said:


> I was referring to proving the formula itself



That's what I thought, and Ricardo's first sentence does prove it.



Sa967St said:


> Also, as TDM mentioned, you have have multiple merges per turn, losing multiple tiles in one turn. I suspect that in order to lose multiple tiles (say, 2 tiles) in one turn, there must have been one previous turn where you didn't merge anything, but I can't think of why.



Really doesn't matter *when* the merges occurred, what matters is their overall number (that's what you have in the formula, after all).

Let's say you start with 4 tiles and during play, you gain 5 and lose 7. You end up with 4+5-7=*2*, no matter when when the gains and losses happened. In general:


```
End = Start + Gains - Losses
    = Start + Turns - Merges
```

The first equation is known from everyday life, and the second is what Ricardo pointed out. Just rearrange the result to get your version of the formula.


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## Stefan (Mar 24, 2014)

Sa967St said:


> I suspect that in order to lose multiple tiles (say, 2 tiles) in one turn, *there must have been one previous turn where you didn't merge anything*, but I can't think of why.



Actually there must even have been *two* such merge-less previous turns. If you never have such a turn then you always have two tiles on the board (the merged and the new) and thus can only do one merge. Only merge-less turns increase the number of tiles, and each only increases it by 1. Since two merges in one turn need four tiles on the board and you start with only two tiles, this requires two previous merge-less moves.

Little riddle: How many times during one game can you have exactly two tiles on the board?

Harder riddle (and I don't know the exact answer): What's the smallest possible sum of all tiles on the board when you reach the 2048 tile?


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## RicardoRix (Mar 24, 2014)

Stefan said:


> Little riddle: How many times during one game can you have exactly two tiles on the board?



Assuming you always get the '2' tile on a move, it's the new 2 tile with every other number combination.
so 2+2, 2+4, 2+8, 2+16, 2+32, etc.

Actually scrub that, I can't see how you get anything other than 2+2, and then 2+4. You've got too many 2's cluttering up the place.

For the hard riddle I'm going with 5 tiles: 2,2,8,8,2048.


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## TDM (Mar 24, 2014)

Can anyone beat that?


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## Jaycee (Mar 24, 2014)

^Just got a 248.


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## Stefan (Mar 24, 2014)

That was one of the early things I tried, back when I failed getting anywhere near the normal goal. Here I had a 164:
http://www.speedcubers.de/showthread.php?tid=11623&pid=183517#pid183517

Edit: 96 now:


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## TDM (Mar 24, 2014)

Stefan said:


> Edit: 96 now:


That's impressively bad  Do you have a tactic for this? I find it hard to not pair numbers...


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## Stefan (Mar 24, 2014)

TDM said:


> That's impressively bad  Do you have a tactic for this? I find it hard to not pair numbers...



I just got another 96. My tactic? Spamming up-down-up-down-up-down-etc and getting "lucky".


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## TDM (Mar 24, 2014)

Stefan said:


> I just got another 96. My tactic? Spamming up-down-up-down-up-down-etc and getting "lucky".


That's what I do, except I can't get lucky. When you can't move, do you start spamming left-right, or do one L/R move and then keep doing up-down?


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## Stefan (Mar 24, 2014)

TDM said:


> That's what I do, except I can't get lucky. When you can't move, do you start spamming left-right, or do one L/R move and then keep doing up-down?



One game only takes a few seconds that way, so just do it more often . In my second 96, the game was already over right after the U/D spamming.

Got an 88 now, but after the U/D spamming I could still do L/R a bit and I did stop and think about how to do it that (that's also how I got my first 96).


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## TDM (Mar 24, 2014)

Stefan said:


> One game only takes a few seconds that way, so just do it more often . In my second 96, the game was already over right after the U/D spamming.
> 
> Got an 88 now, but after the U/D spamming I could still do L/R a bit and I did stop and think about how to do it that (that's also how I got my first 96).


After getting 1000+ several times in a row...




It's very unusual to get the game to stop without having to do L/R, but you were right: keep going and you'll get lucky eventually


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## RicardoRix (Mar 24, 2014)

Stefan said:


> Actually there must even have been *two* such merge-less previous turns. If you never have such a turn then you always have two tiles on the board (the merged and the new) and thus can only do one merge. Only merge-less turns increase the number of tiles, and each only increases it by 1. Since two merges in one turn need four tiles on the board and you start with only two tiles, this requires two previous merge-less moves.
> 
> Little riddle: How many times during one game can you have exactly two tiles on the board?
> 
> Harder riddle (and I don't know the exact answer): What's the smallest possible sum of all tiles on the board when you reach the 2048 tile?



I think I can explain this smallest sum by working backwards. So where normally you always get a 2 tile added in, going backwards you must have a 2 tile to take away. So you need to divide down the 2048 tile as quick as possible so it starts to give you 2's to take away. So you need a few mid-range tiles to get you to this stage. The minimum for this would be: 2,2,8,8,2048,

The sequence going forward:

2,4,8,16,32,64,128,256,512,1024,
2,2,4,8,16,32,64,128,256,512,1024,
2,4,4,8,16,32,64,128,256,512,1024,
2,2,8,8,16,32,64,128,256,512,1024,
2,4,16,16,32,64,128,256,512,1024,
2,2,4,32,32,64,128,256,512,1024,
2,4,4,64,64,128,256,512,1024,
2,2,8,128,128,256,512,1024,
2,4,8,256,256,512,1024,
2,2,4,8,512,512,1024,
2,4,4,8,1024,1024,
2,2,8,8,2048,


----------



## uyneb2000 (Mar 25, 2014)




----------



## Rocky0701 (Mar 25, 2014)

Somebody made a cubing version of 2048? Sweet!


----------



## Stefan (Mar 25, 2014)

RicardoRix said:


> 2,4,8,16,32,64,128,256,512,1024



But is this state actually achievable? I'd instantly believe it if the board were long so you could really think about doing this simply in a straight line, but on the 4x4 board it's not that obvious.

And is 4,4,8,16,32,64,128,256,512,1024 maybe also achievable? It would lead to a smaller end result.

Also, what about finishes where you don't go straight up through the powers of two, one per turn? Where you don't just do one merge-towards-2048 per turn but several? For example, right before the last turn you do of course need two 1024-tiles, but you don't need to already have a 1024-tile before the next-to-last turn if that next-to-last turn builds both 1024-tiles at the same time. Maybe such other progressions could lead to a smaller end result?

(I btw really do have these questions and don't know the answers)

Your answer to my simpler riddle is not quite correct yet, btw, as you're still assuming only 2-tiles popping up. There's also a somewhat neat little insight behind that riddle, which explains why you can't get any boards with just two tiles except very few small ones.


----------



## Sa967St (Mar 25, 2014)

Done 3D and 4D. 



Spoiler: 3D













Spoiler: 4D











Edit: and 5D!



Spoiler: 5D


----------



## cmhardw (Mar 27, 2014)

Stefan said:


> Little riddle: How many times during one game can you have exactly two tiles on the board?



A quick lemma to show that (Number of turns taken) \( \geq \) (Number of Merges made) must hold true from the start of the game to your current board state:


Spoiler



2=B-T+M

where
B=number of tiles on the board
T=number of turns taken
M=number of merges

Assume T<M

then
M=T+k where k is a natural number
2=B-T+(T+k)
2=B+k
2-k=B

This would say that that board has fewer than 2 tiles. I am going by Sarah's statement earlier that this is impossible.

Therefore \( T \geq M \) must hold true.



Since \( T \geq M \) must hold true, then we can write:
T=M+j where j is a non-negative integer

2=B-(M+j)+M
2=B-j
2+j=B

And you have 2 tiles on your board only when j=0, which means when T=M.

That doesn't answer your question, but that is the extent of what I have noticed so far.

--edit--
Just trying different board states in my head I can't find any games where I have 2 tiles more than three times.


----------



## TDM (Mar 27, 2014)

Not doing it randomly helps.


----------



## Jaycee (Apr 1, 2014)

Second time getting the 4096, and I felt like I could have gotten the 8192, but I messed up while paying half of my attention to this and half of my attention to my friend who was talking to me in study hall.


----------



## LNZ (Apr 2, 2014)

I have downloaded this game from iTunes for my iPad and my PB is 7440.


----------



## Phillip1847 (Apr 2, 2014)

http://www.csie.ntu.edu.tw/~b01902112/9007199254740992/


----------



## TDM (Apr 2, 2014)

Phillip1847 said:


> http://www.csie.ntu.edu.tw/~b01902112/9007199254740992/


That would take a _very_ long time to complete... I tried to lose, got to two 1024s, and accidentally hit space. Even that was taking a while to do. To actually solve it would take... anyone want to calculate?


----------



## brian724080 (Apr 2, 2014)

Phillip1847 said:


> http://www.csie.ntu.edu.tw/~b01902112/9007199254740992/



I think the challenge is to lose ASAP


----------



## Stefan (Apr 2, 2014)

TDM said:


> I tried to lose, got to two 1024s, and accidentally hit space. Even that was taking a while to do.



Took me about 1:45. Game over at 9040, max tile 128 (seven of them).


----------



## angham (Apr 2, 2014)

my friend failed with highest tile as 8


----------



## Sa967St (Apr 2, 2014)

I think I'm finally done playing this.



Spoiler: 16384 Hex














Spoiler: Tetris 2048













Spoiler: Doge 2048













Spoiler: Numberwang 2048











All from http://2048.directory.


----------



## Jaycee (Apr 2, 2014)

That hex looks fun.


----------



## Sa967St (Apr 2, 2014)

Jaycee said:


> That hex looks fun.


It's actually quite frustrating, since all it takes is one unlucky entering block to ruin a game. If you have all the rows filled out and have to use a forbidden move, you just have to hope that your highest block can be moved back the next turn. It took me more than 10 tries, and each one usually took 1-2 hours.


----------



## Laradoodle4 (Apr 2, 2014)

Got this game yesterday, totally addicted, pb is 11672 and have not yet got 2048. It's incredibly frustrating as well.


----------



## TheNextFeliks (Apr 4, 2014)

Like 7th time. 939 moves and 933 merges. I usually take 15ish minutes. That might have been a little less.

At school yesterday: I saw someone playing this on their phone. Someone says "How does it feel to play a cheap knockoff?" Ugh. Felt like listening to MMAP about V-Cubes. Anyway the kid knew what he was doing. He had the good strategy. He only got to 1024 but he was doing great and probably won.


----------



## bran (Apr 4, 2014)

I think I should stop now


----------



## TDM (Apr 4, 2014)

bran said:


> I think I should stop nowView attachment 3837


wat
16384 pls


----------



## ajayd (Apr 5, 2014)

A friend at my school got to 8192 also. I want to see what the 16384 tile is like


----------



## brian724080 (Apr 5, 2014)

ajayd said:


> A friend at my school got to 8192 also. I want to see what the 16384 tile is like



Hack it in or lose your sanity


----------



## Jaycee (Apr 5, 2014)

My goal is to get to the 8192. I've gotten the 4096 five times, and I "win" (get 2048) in roughly half of games I play now. I will get it. Soon. I wonder how many people on SS have gotten it? :S


----------



## TDM (Apr 5, 2014)

brian724080 said:


> Hack it in or lose your sanity


It would probably be easy to reach on this, given a few hours...

E: 34976. 2048, 1024, 512, 128 as top row, also had another 128, a 64 and a couple of 32s. Finally best in school (although the other guy can cheat because of the version he has), but still no 4096 yet


----------



## LNZ (Apr 5, 2014)

Again, using the iPad app, my PB is now 7660. iTunes do say that their version is not original and give the original source site.


----------



## Jaycee (Apr 5, 2014)




----------



## TDM (Apr 5, 2014)

Jaycee said:


> Image


Told you it could be done on that  How long did that take?

And how far have people got on the hard version?


----------



## rj (Apr 6, 2014)

Some more variants.

http://phenomist.wordpress.com/2048-variants/

My personal favorite: chenglou.github.io/flappy-2048-side-by-side/


----------



## Jaycee (Apr 7, 2014)

A weekend's work. I intend on getting to 2^20 by the end of the week and then I'll be done.


----------



## mande (Apr 7, 2014)

2:51.23 timed single. Has anyone here gotten a better time than this?


----------



## TDM (Apr 7, 2014)

mande said:


> 2:51.23 timed single. Has anyone here gotten a better time than this?


Really? Assuming 4s appear 10% of the time, on average the total on the board increases by 2.2 each turn. You therefore did 903 turns in 171 seconds? That's 5.44 TPS. If you can really do that, that's crazy. I use my phone a lot now, and that'd be impossible to do on my phone... but even on a keyboard, that'd require you to think so fast.


----------



## Jaycee (Apr 7, 2014)

Update on my 8x8 grid: I showed a friend in class today and when I was moving my laptop to show him my hand mashed the arrow keys. So... I might try again in a couple weeks but I don't feel like starting that over lol.


----------



## TheNextFeliks (Apr 10, 2014)

8:53.06. Wanted to try timing. I feel like I did strategic spamming. It worked well.

http://sbeyer.github.io/2048/

Got 2048 with score 5776. Really easy. Took a minute, maybe.

Also got 7605 on Numberwang.

Hmm. Won't let me upload image.


----------



## TDM (Apr 10, 2014)

TheNextFeliks said:


> http://sbeyer.github.io/2048/
> 
> Got 2048 with score 5776. Really easy. Took a minute, maybe.


Yay, an easy version where 256 tiles can appear right where you need them!

Oops


----------



## brian724080 (Apr 10, 2014)

TheNextFeliks said:


> http://sbeyer.github.io/2048/
> 
> Got 2048 with score 5776. Really easy. Took a minute, maybe.
> 
> ...



I just don't get Numberwang


----------



## TDM (Apr 10, 2014)

4096 and still going.
E: 54824


40961024256164816824844242


----------



## Jaycee (Apr 10, 2014)

brian724080 said:


> I just don't get Numberwang



Same... the one time I tried it I just mashed random keys.


----------



## DAoliHVAR (Apr 10, 2014)

Jaycee said:


> Same... the one time I tried it I just mashed random keys.



i think you have to be brittish to get it or smth 
btw just god my first win after like ~30 tries all together
feels good mayne


----------



## Sa967St (Apr 10, 2014)

brian724080 said:


> I just don't get Numberwang


Just pay attention to the colours of the tiles. Everything else is nonsense. 

https://www.youtube.com/watch?v=qjOZtWZ56lc


----------



## Rocky0701 (Apr 11, 2014)

Anybody ever tried this version? I like it, but it takes forever. http://annimon.github.io/16384/


----------



## brian724080 (Apr 11, 2014)

Sa967St said:


> Just pay attention to the colours of the tiles. Everything else is nonsense.
> 
> https://www.youtube.com/watch?v=qjOZtWZ56lc



That's numberwang!


----------



## ThomasJE (Apr 11, 2014)

Rocky0701 said:


> Anybody ever tried this version? I like it, but it takes forever. http://annimon.github.io/16384/



I tried it for a few minutes, and then randomly pressed loads of buttons and got the 4096 tile. 47276 so far.

EDIT:

4823216128163841284242242

199084 points and still going...


----------



## Sa967St (Apr 11, 2014)

Phillip1847 said:


> http://www.csie.ntu.edu.tw/~b01902112/9007199254740992/



I shouldn't have started this. :/



Spoiler: 524288











I needed just another 2 or 3 hours to get to 2^20, but my laptop had to force restart. I'm not playing this again.


----------



## TDM (Apr 11, 2014)

Second 4096; first was on my phone and idk how to upload from there.


----------



## Jont828 (Apr 11, 2014)

Oh. My. God. I LOVE THIS GAME!!!  My PB is about 26k. You have to keep your highest block in the bottom left corner and keep the bottom row filled (try to use only left and down moves) so that your higher numbers will gravitate towards it.


----------



## TDM (Apr 11, 2014)

Jont828 said:


> Oh. My. God. I LOVE THIS GAME!!!  My PB is about 26k. You have to keep your highest block in the bottom left corner and keep the bottom row filled (try to use only left and down moves) so that your higher numbers will gravitate towards it.


I've seen most people use the bottom row. Am I the only one who uses the top (except Sarah)?
Also, Jont828, try to keep the numbers going descending from left to right in that first row, and in the second make them descending from right to left (although you probably won't be able to fill it until after 2048).


----------



## Jaycee (Apr 12, 2014)

I use top. I prefer keeping my biggest tile in (0,4) aka the top left corner, but I can also do (4,4) aka the top right if I must.


----------



## uberCuber (Apr 12, 2014)

TDM said:


> I've seen most people use the bottom row. Am I the only one who uses the top (except Sarah)?



I use the top as well.


----------



## DAoliHVAR (Apr 15, 2014)

uberCuber said:


> I use the top as well.



whenever i have big merges i use whatever side has the best arangment of numbers
also i have beat the game 6 times now
it seems that after the first win it gets exponentially easier


----------



## Sa967St (Apr 16, 2014)

2048 in 3:41.58 using the version with the timer that Dan Cohen made. http://venim.info/2048/






The last 512 was pretty slow since I had a 2 trapped at the top. I think I should be able to sub3 soon.


----------



## TDM (Apr 16, 2014)

Sa967St said:


> video


How do you think so fast? You look like your spamming moves as fast as possible and actually getting somewhere...


----------



## brian724080 (Apr 16, 2014)

TDM said:


> How do you think so fast? You look like your spamming moves as fast as possible and actually getting somewhere...



Strategic spamming, to some degree


----------



## Robert-Y (Apr 16, 2014)

@Sarah: Nice 

Think less and just spam more! Maybe you're accuracy might decrease but you should get faster times...


----------



## DAoliHVAR (Apr 16, 2014)

damn sarah noice
update 
wins:8
and this happened also


----------



## soup (Apr 16, 2014)

5:36.75. A bit on the slow side.


----------



## porkynator (Apr 16, 2014)

I tried speedsolving for the first time today and I got a sub4 after a couple of tries (3:59.64) 
But I like going for high scores, although a good game takes too long (PB: 71k)


----------



## Jaycee (Apr 16, 2014)

7:36.33, (DNF), 7:32.67, 7:29.84, 7:41.29 = 7:35.03 (mean - DNF)


----------



## brian724080 (Apr 17, 2014)

Totally gonna try speedsolving this (no, just no)


----------



## Chree (Apr 18, 2014)

Had a 5 hour flight yesterday. I don't think this actually counts because I used The Undo Button once. It was under 100k anyway.


----------



## TDM (Apr 18, 2014)

Chree said:


> Had a 5 hour flight yesterday. I don't think this actually counts because I used The Undo Button once. It was under 100k anyway.


wat
Also, I use both my undos right at the start so I'm not tempted to use them later.


----------



## GamerEliza (Apr 18, 2014)

I didn't even notice there was a undo button, fml.


----------



## Rocky0701 (Apr 18, 2014)

GamerEliza said:


> I didn't even notice there was a undo button, fml.


Haha same for me, that should help.


----------



## PeelingStickers (Apr 18, 2014)

This is slightly harder to speedsolve on a phone :/


----------



## kunparekh18 (Apr 21, 2014)

Won for the first time yesterday. Had been trying it for a couple of days. It was on the original site, so no undos used.


----------



## Rpotts (Apr 21, 2014)

so much spamming


----------



## 10461394944000 (Apr 21, 2014)

was considering solving a 2048 sliding puzzle and posting it in this thread because lel

but solving a 683x3 doesnt sound fun


----------



## mark49152 (Apr 21, 2014)

Man this is addictive. 36140 on second try. Somehow got my 2048 and 1024 diagonally separated and it was downhill from there.


----------



## cmhardw (Apr 21, 2014)

I finally got the 4096! Woo Hoo!


----------



## Meep (Apr 22, 2014)




----------



## Bhargav777 (Apr 22, 2014)

My friend just got this. Anyone with a better score? Uwr?


----------



## Meep (Apr 22, 2014)

Bhargav777 said:


> My friend just got this. Anyone with a better score? Uwr?



That's in practice mode, where you can keep undoing your last move. The RNG spawns your next number in a different spot each time that you can effectively get a perfect board with enough patience.


----------



## kunparekh18 (Apr 22, 2014)

Meep said:


> That's in practice mode, where you can keep undoing your last move. The RNG spawns your next number in a different spot each time that you can effectively get a perfect board with enough patience.



I thought there were just 2 Undos per game?


----------



## Rocky0701 (Apr 22, 2014)

Meep said:


> That's in practice mode, where you can keep undoing your last move. The RNG spawns your next number in a different spot each time that you can effectively get a perfect board with enough patience.


Yes, you can get to 32,000 so i highly doubt that he actually got 16000 because that would only leave 2 squares lefts to combine numbers.


----------



## Meep (Apr 22, 2014)

kunparekh18 said:


> I thought there were just 2 Undos per game?



It depends which one you download. A friend of mine has the same one in the screenshot he posted, and it says 'Practice mode' in it.


----------



## Rpotts (Apr 22, 2014)

My fingers are more sore than they've ever been from cubing.


----------



## TDM (Apr 22, 2014)

Meep said:


> image


How...?


Meep said:


> That's in practice mode, where you can keep undoing your last move. The RNG spawns your next number in a different spot each time that you can effectively get a perfect board with enough patience.


How can you do practice mode?


----------



## Bhargav777 (Apr 22, 2014)

Patience^15. Is 16384 possible without undos?


----------



## gogozerg (Apr 22, 2014)

2048 in 3 minutes 10 seconds.

4096 not too difficult.

8192 easy with undos, of course.


----------



## Meep (Apr 22, 2014)

Bhargav777 said:


> Patience^15. Is 16384 possible without undos?



Definitely possible; I managed to get 8192 and 4096, and lost trying to make the next 4096. ):

Using undos, I believe it's possible to get 131072 (Assuming you get a 4 at a very specific moment).


----------



## Jaycee (Apr 22, 2014)

A couple weeks ago, on the 8x8 grid, I did manage to get to a 131072 and a 65536 on the same board, and I had the goal of getting to 2^20, but I gave up because I got bored of it. Maybe I'll try again!

Oh also I got my 2048 speedsolve time down to 6:30. And another thing. I'm sitting in English class right now, and I can see 5 people playing 2048 from where I am right now. And to think I was one of the people who helped make it popular in my school xD


----------



## Bhargav777 (Apr 22, 2014)

Meep said:


> Definitely possible; I managed to get 8192 and 4096, and lost trying to make the next 4096. ):
> 
> Using undos, I believe it's possible to get 131072 (Assuming you get a 4 at a very specific moment).



Cool! Are there any apk files with a timer that you know?


----------



## Sa967St (Apr 22, 2014)

@Meep: O_O



Jaycee said:


> A couple weeks ago, on the 8x8 grid, I did manage to get to a 131072 and a 65536 on the same board, and I had the goal of getting to 2^20, but I gave up because I got bored of it. Maybe I'll try again!


I wouldn't recommend it unless you want to kill *a lot* of time. I spent a full week playing it on and off, where I would spent the majority of my free time on it, and I was just a few hours away from getting 2^20, but my laptop had to force restart (post). I spent over 5 hours each day playing it, often at full speed. It's hard to lose unless you scramble the board, so you'll want to just keep going.


----------



## Rpotts (Apr 23, 2014)

I'm done forever. 

I look up and it appears Sarah got almost 4 times as far. How...


----------



## Meep (Apr 23, 2014)

Bhargav777 said:


> Cool! Are there any apk files with a timer that you know?



Unfortunately not.  Mostly because I don't have an Android device.



Sa967St said:


> I wouldn't recommend it unless you want to kill *a lot* of time. I spent a full week playing it on and off, where I would spent the majority of my free time on it, and I was just a few hours away from getting 2^20, but my laptop had to force restart (post). I spent over 5 hours each day playing it, often at full speed. It's hard to lose unless you scramble the board, so you'll want to just keep going.



That's why you write something to play for you 

I ran it for about a week and it got this far:


Spoiler


----------



## Sa967St (Apr 23, 2014)

Bhargav777 said:


> My friend just got this. Anyone with a better score? Uwr?



With unlimited undos on http://quaxio.com/2048/;






Here's what it looked like about 15 moves before:

http://i.imgur.com/7OoCqdA.png

Still playing it.


----------



## Bhargav777 (Apr 23, 2014)

Sa967St said:


> With unlimited undos on http://quaxio.com/2048/;
> 
> 
> 
> ...



That "let me continue from there, pweaseeeee" moment!


----------



## Jaycee (Apr 23, 2014)

Sa967St said:


> Here's what it looked like about 15 moves before:
> 
> http://i.imgur.com/7OoCqdA.png



So perfect ;__;


----------



## Mikel (Apr 25, 2014)

I got the 4096 tile a few days back. I guess you could say I dosed the gj or something like that.


----------



## gogozerg (Apr 25, 2014)

2:58.35


----------



## DAoliHVAR (Apr 25, 2014)

3rd 4096 and i haven't lost yet
gonna try and get the 8000


----------



## Iggy (Apr 26, 2014)

I finally got to 2048 a few days ago. This game is seriously addictive


----------



## Sa967St (Apr 27, 2014)

Sa967St said:


> With unlimited undos on http://quaxio.com/2048/;
> 
> http://i.imgur.com/ofFBNbN.png
> 
> ...



I think this is as far as I can get.

http://i.imgur.com/3FmHCdX.png

I've undone the last 4096 block three times to get unstuck to get down to the final 64 block. In this version, undoing and redoing doesn't change what the given block (2 or 4) is, or where it appears, so I can't get any further without undoing a lot and hoping for the best.

Edit: Spoke too soon. 







The next block I got was a 2, unfortunately.


----------



## Robert-Y (Apr 28, 2014)

Does anyone think that "blind spamming" is ok for a timed run to the 2048 tile? I can get sub 2 quite easily these days with this technique at the start, with enough tries. I was thinking about filming a sub 2 run, but personally I don't really like the idea, so I don't want to encourage people to do it, in order to get faster times. If someone get bashed and bashed their keys to get to the 2048 tile in like sub 1 without any thinking, that would just be stupid...


----------



## brian724080 (Apr 28, 2014)

Sa967St said:


>



I don't understand how you can possible get to this without messing up (forced to do a down move).


----------



## TDM (Apr 28, 2014)

brian724080 said:


> I don't understand how you can possible get to this without messing up (forced to do a down move).


If you're in that situation, undo, add in a move somewhere, and continue.


----------



## Bh13 (Apr 28, 2014)

Almost got to 8192 without undos. had a 4096, 2048, 1024, and 256 tile. Final score of 76,000


----------



## rj (Apr 28, 2014)

Wow. My high is 70k.


----------



## TheNextFeliks (Apr 29, 2014)

I lost in 8.35 with a 256. I feel accomplished.

Edit: 5.99 with 128. Wow.

Edit2: 2.90 highest was 64 lol. I'd love to see someone beat this.


----------



## DAoliHVAR (Apr 30, 2014)

hey peepz
anyone here know of another method of solving apart from the "keep the things at the top row" one XD


----------



## TDM (Apr 30, 2014)

TheNextFeliks said:


> I lost in 8.35 with a 256. I feel accomplished.
> 
> Edit: 5.99 with 128. Wow.
> 
> Edit2: 2.90 highest was 64 lol. I'd love to see someone beat this.


6.38 with 128, close. Only done a few tries though. It took me a while to realise how you got that TPS.
E: 256 in something like 3 seconds (not sure exactly what), then it took until 6.90 to lose.
E2: 6.89. Improvement.
E3: 6.04 with 128, getting close to your 128 time.
E5: 5.00 w/ 128.
E6: Exactly 10.00 with a 256 and 128 next to each other.


----------



## TheNextFeliks (Apr 30, 2014)

TDM said:


> 6.38 with 128, close. Only done a few tries though. It took me a while to realise how you got that TPS.
> E: 256 in something like 3 seconds (not sure exactly what), then it took until 6.90 to lose.
> E2: 6.89. Improvement.
> E3: 6.04 with 128, getting close to your 128 time.
> ...



2.73. Score was 586. One 64, one 32. I just spam hard.

Edit: Lolwut. Dat tps.


----------



## TDM (May 1, 2014)

TheNextFeliks said:


> 2.73. Score was 586. One 64, one 32. I just spam hard.
> 
> Edit: Lolwut. Dat tps.
> View attachment 4016


How...!? I try to lose as fast as possible and haven't had a time under 5 seconds yet.


----------



## Jaycee (May 1, 2014)

You're crazy, next Feliks O.O That turnspeed must be insane.

Original 2048: I'm getting closer every day to the 8192. Yesterday I got a 4096, 2048, 1024, and 512 on the same board. I'm way to late for this to actually be impressive but I set this as a goal for myself a while ago so I will make it happen xD


----------



## Ninja Storm (May 1, 2014)

I can DNF in sub2, but only with a mechanical keyboard. I've gotten a 1.69 before. 

I can't do it on rubber dome/scissor switch, though.


----------



## Rocky0701 (May 1, 2014)

Lol i love how you guys are timing yourselfs to the extent that Ninja Syorm has begun to see how fast he can DNF.


----------



## TDM (May 1, 2014)

DNF in 3.39, highest tile was 32. Score was 484.


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## Sa967St (May 2, 2014)

Just got my first 8192 on the regular version. Has anyone gotten 16384?


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## Ninja Storm (May 2, 2014)

Best DNF on camera: 2.41

video here


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## Rocky0701 (May 2, 2014)

Ninja Storm said:


> Best DNF on camera: 2.41
> 
> video here


I still think it's funny how you're trying to time yourself to the extent of seeing how fast you can DNF haha


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## Jaycee (May 2, 2014)

Right. There. Ended up like this.


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## Chree (May 2, 2014)

Sa967St said:


> Just got my first 8192 on the regular version. Has anyone gotten 16384?



Haven't hit 16384 yet... but did get my first (legit, no undos used) 8192 today.







Last piece was that unlucky 4 in the top right. A 2 would have changed my life.


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## TDM (May 2, 2014)

Chree said:


> Last piece was that unlucky 4 in the top right. A 2 would have changed my life.


Am I the only one who thinks this game would be so much easier if only 2s appeared? 4s are nice sometimes, but a lot of the time they just get in the way...


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## Robert-Y (May 2, 2014)

My best and only average of 5:

2:38.92, DNF, 2:37.92, 2:08.18, 2:02.96 => 2:28.34

(Obviously this is for the time taken to create a 2048 tile, not to DNF lol)

This average is over 30 seconds above my average success >_<

Anyone else tried for an average of 5? It was really tough for me..


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## Jaycee (May 2, 2014)

Literally getting closer every day.

This is even more painful because I only lost because my finger twitched and I hit an incorrect key near the end.


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## gogozerg (May 5, 2014)

Robert-Y said:


> Anyone else tried for an average of 5? It was really tough for me..



No, to many DNFs, and my best is only 2:49.77.


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## TheNextFeliks (May 5, 2014)

Lol. That was fun. My record is 1.71 with a 64, two 16s and the rest were 2s and 4s.


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## LNZ (May 5, 2014)

Today, I got to the 2048 tile. The score was 20780.

I use the iPad app to play this game. Some claim that this is a KO of the original.


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## gogozerg (May 6, 2014)

4096 in 7:33.

Too bad the embedded timer stops when you hit 2048.


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## Chree (May 6, 2014)

My most efficient 2048 tile yet, and my first sub 20K


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## mehul (May 6, 2014)

I made a tile of 8192, in Practice Mode (I used a lot of undos). I used that method of of sticking all the numbers on the left and making the big number on top.


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## rj (May 8, 2014)

Good job. Question: Is there anyone who can do this blind?


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## TheNextFeliks (May 8, 2014)

rj said:


> Good job. Question: Is there anyone who can do this blind?



I did numberwang one BLD. You just have to spam lol.


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## ryanj92 (May 8, 2014)

rj said:


> Good job. Question: Is there anyone who can do this blind?



You can really do something that generates random stuff blind other than just spamming and hoping for the best 
If you fancy trying to visualise a full board,there's a deterministic one which places the 2 in the first available square...


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## mns112 (May 8, 2014)

get 2048 +

I'm 8192 and still going


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## TDM (May 12, 2014)

Chree said:


> My most efficient 2048 tile yet, and my first sub 20K


19988 


I think I'll stop here. (not that I have any choice...)


8192409620481024326425616281644242

171468 points.

(E: No undos)


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## LNZ (May 15, 2014)

The 2048 iPad and iPhone app now has 48 challenges in addition to the main game.

I have completed 4 of the 48 challenges.


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## Ninja Storm (May 15, 2014)

4:28.51 2048

Getting faster.


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## TDM (May 15, 2014)

First timed success: 4:44.94. Slowed down by carelessly moving the 1024 tile out the corner. Oh and eating an apple at the same time.


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## Lapinsavant (May 15, 2014)

80080 points with 4096 2048 1024 512 256 128 64 32 16 and another 32 but failed the 8192


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## cmhardw (May 19, 2014)

Katie got the 8192 tile today! Yay!


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## gogozerg (May 26, 2014)

2:11.67
Getting faster...


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## porkynator (May 26, 2014)

28 Points (no undos of course).
Has anyone done less?
(I'll post the screenshot if asked).


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## gogozerg (May 27, 2014)

A few times under 2 minutes now.
Best 1:48.67.


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## TDM (May 27, 2014)

gogozerg said:


> A few times under 2 minutes now.
> Best 1:48.67.


How do people do it so fast? I end up DNFing all the time. I've only had two fast successes so far out of 20+ attempts, most of them finishing within a minute or 30 seconds. I know it's spam TPS... but in which directions?


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## rj (May 31, 2014)

TDM said:


> How do people do it so fast? I end up DNFing all the time. I've only had two fast successes so far out of 20+ attempts, most of them finishing within a minute or 30 seconds. I know it's spam TPS... but in which directions?



I did a 1 tps solve in 4 mins. It's not hard. If it is, something's wrong.


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## TDM (May 31, 2014)

rj said:


> I did a 1 tps solve in 4 mins. It's not hard. If it is, something's wrong.


Is that even possible? I did a lot more than 1 TPS when I got my second success and if I remember correctly, there were no large tiles.


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## Sa967St (Jun 13, 2014)

Sa967St said:


> Just got my first 8192 on the regular version. Has anyone gotten 16384?


Me! 







Final score: http://i.imgur.com/b443U0c.png


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## cmhardw (Jun 18, 2014)

Sa967St said:


> Me!



Congrats Sarah!


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## rj (Jun 26, 2014)

TDM said:


> Is that even possible? I did a lot more than 1 TPS when I got my second success and if I remember correctly, there were no large tiles.



Oh yes. 440 moves is plenty.


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## uberCuber (Jun 27, 2014)

rj said:


> Oh yes. 440 moves is plenty.



1) 1 tps for 4 minutes is 240 moves, not 440 moves.
2) It is mathematically impossible to get a 2048 in even 440 moves. You can stop blatantly lying, now.


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## Sa967St (Jun 27, 2014)

Probably one of the most challenging versions: http://joezeng.github.io/144/


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## cmhardw (Jun 27, 2014)

Sa967St said:


> Probably one of the most challenging versions: http://joezeng.github.io/144/



Very fun! I can see getting hooked on this!

Also, congrats on getting the 144


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## brian724080 (Jun 27, 2014)

Sa967St said:


> Probably one of the most challenging versions: http://joezeng.github.io/144/



This version is awesome


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## rj (Jun 27, 2014)

uberCuber said:


> 1) 1 tps for 4 minutes is 240 moves, not 440 moves.
> 2) It is mathematically impossible to get a 2048 in even 440 moves. You can stop blatantly lying, now.



Dangit. I always misjudge tps and movecount.


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## LNZ (Jun 28, 2014)

On the Apple app store, there is a 2048 game where you can play on a 4x4, 5x5 or 6x6 grid.


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## gogozerg (Jun 30, 2014)

Did a few solves today, I got sub 1:40 at last.
Did anybody break the 1:30 wall?


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## TDM (Jun 30, 2014)

gogozerg said:


> Did a few solves today, I got sub 1:40 at last.
> Did anybody break the 1:30 wall?


I don't get how people can do it so fast... iirc, Rob Yau has a sub-1:30, but I don't know of anyone else who has got a success that fast. There could just be someone I don't know of though.


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## kcl (Jun 30, 2014)

TDM said:


> I don't get how people can do it so fast... iirc, Rob Yau has a sub-1:30, but I don't know of anyone else who has got a success that fast. There could just be someone I don't know of though.



Jay is getting near that speed iirc


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## gogozerg (Jul 22, 2014)

1:37, slightly faster.
But I was about 20 seconds late in the first half. Sub-1:20 looks possible.


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## G2013 (Jul 22, 2014)

I almost get to the 16384 tile!!!


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## siva.shanmukh (Oct 8, 2014)

Damn me! So close and yet so far.


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## Randomno (Oct 8, 2014)

I played this a lot a few months ago, got to 32768 using undos, don't really play much now. I got more into Piano Tiles eventally.


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## ~Adam~ (Oct 8, 2014)

Randomno said:


> I played this a lot a few months ago, got to 32768 using undos



Why would you waste so much time cheating to a good score?


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## Randomno (Oct 8, 2014)

cube-o-holic said:


> Why would you waste so much time cheating to a good score?



I didn't cheat. The app I was using had a version with an undo button, and I used it.


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## ~Adam~ (Oct 8, 2014)

Randomno said:


> I didn't cheat. The app I was using had a version with an undo button, and I used it.



I guess congratulations are in order then.

You really don't see using an undo button as cheating?


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## Randomno (Oct 8, 2014)

cube-o-holic said:


> I guess congratulations are in order then.
> 
> You really don't see using an undo button as cheating?



Not if it's in the game... It was called "Practice" mode anyway.


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## TDM (Oct 8, 2014)

Randomno said:


> Not if it's in the game... It was called "Practice" mode anyway.


It wasn't in the original game.


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## Randomno (Oct 8, 2014)

TDM said:


> It wasn't in the original game.



If you use the original, you can open a new tab and then go to that one later on if you mess up your game on the original. That _is_ cheating though.


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## ~Adam~ (Oct 8, 2014)

Randomno said:


> If you use the original, you can open a new tab and then go to that one later on if you mess up your game on the original. That _is_ cheating though.



I understand now. It's not cheating to undo moves but it is cheating to change to a game where you can undo moves.


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## Randomno (Oct 8, 2014)

cube-o-holic said:


> I understand now. It's not cheating to undo moves but it is cheating to change to a game where you can undo moves.



Whaa?

Can we just say your original post was "Why would you waste so much time undoing moves for a good score?" or something?


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## ~Adam~ (Oct 8, 2014)

Randomno said:


> Whaa?
> 
> Can we just say your original post was "Why would you waste so much time undoing moves for a good score?" or something?



Yes. Please answer that question. There is no point IMO getting a good score if you undo moves to accomplish it. It's similar to taking apart a scrambled cube, rebuild solved then start and stop a timer as quickly as possible and note down the time.


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## Randomno (Oct 8, 2014)

cube-o-holic said:


> Yes. Please answer that question. There is no point IMO getting a good score if you undo moves to accomplish it. It's similar to taking apart a scrambled cube, rebuild solved then start and stop a timer as quickly as possible and note down the time.



Cos I like to kill time, and also to see the 131k tile.


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## ~Adam~ (Oct 8, 2014)

Randomno said:


> Cos I like to kill time, and also to see the 131k tile.



So you took a screen shot of the momentous occasion?


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## Randomno (Oct 8, 2014)

cube-o-holic said:


> So you took a screen shot of the momentous occasion?



I'm not at it yet, and might not be forever.


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## ~Adam~ (Oct 8, 2014)

Randomno said:


> I played this a lot a few months ago, got to 32768 using undos, don't really play much now. I got more into Piano Tiles eventally.





Randomno said:


> I'm not at it yet, and might not be forever.



Ok then


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## Garf (Nov 25, 2021)

This is a theory session on what I call the 2048 parity problem.
If you don’t already know what 2048 is, it is an online game/puzzle that spawns in new blocks in a 4x4 grid, and you have to slide them together to make larger numbers. Each number is multiplied by 2, ex 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048. The only two blocks that get spawned in are 2 and 4, so it is a little unpredictable where the number will spawn and what number it is.
anyway, the way I play 2048 is that I some numbers in the beginning and slide them to the top left, creating even larger numbers. However, I can use only left and up, sometimes right, but rarely down. What happens is that the game will sometimes end up where I cannot move left and up, or even rarer, left, up, and right, forcing me to move left and spawning a number in the top left position, or moving down and the said thing happens. What is the best solution for this problem so that you can keep playing for as long as possible, yet save that top left corner? This can also happen with the 2nd, 3rd or even 4th rows, although it is rare.


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## xyzzy (Nov 25, 2021)

That doesn't exactly have anything to do with parity. (?)

The spawn rates are 90% for 2 and 10% for 4, iirc, so while it's random, it's still fairly predictable. Try to keep your board in a state where you can merge in large numbers when you need to, I guess. (It's a bit difficult to concretely describe strategies because it ultimately boils down to "think about what can happen, and try to avoid bad things happening", where determining how good/bad a board state is largely based on intuition you build up from playing lots of games and dying lots of times. There isn't really a concrete rule.)


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