You forgot to mention a crucial rule: you can't 'swipe' in a direction that doesn't move any tiles (effectively a 'pass'). This can force you to swipe in a direction that messes everything up.

I think you can get 2^17 since a 4 could appear in the last spot instead of a 2

Omg if this video was 20:48 long it would have killed people.

As mentioned, it is theoretically possible to get the 2^17 tile, not just the 2^16. The AI version I'm working on (will post soon to github) has a high score at the moment of 268,788, with the 16,384 tile. It often makes 8,192, but has only managed 16,384 once so far.

I got the 8192 tile once =). But then I failed miserably to get a higher score and stopped with less than 100k points =(

so that is actually the highest tile you can get to because assuming you get super lucky you can get another 4 tile instead of the 2 tile and so it is actually 2^17 is the absolute highest tile you can get i think

@SomeoneEvilTV We did originally say something like that, but it got waffley and we cut it. Here I say you need k free cells to achieve 2^k and the maximum achievable tile is therefore 2^16. This was assuming we only generate 2-tiles. I also show a formula to calculate the total score of making the 2^k tile which is (k-1)2^k. That means the maximum score is when you fill the board from 2^16 to 2, which is sum_i=1^16 (i-1)2^i = 1,835,012. If we include generating 4-tiles as well, you can actually go one step further and achieve the 2^17 tile. In that case the maximum score would be when we fill the board from 2^17 to 2^2. If I did that using 4-tiles only then that would double the maximum score from 1,835,012 to 3,670,024. If we use 2-tiles only, with a few exceptions, then I reckon the maximum score is sum_i=2^17 (i-1)2^i - 16*4 = 3,932,100. (You need to subtract 4 sixteen times because I need to generate sixteen 4-tile for free to fill the board from 2^17 to 4).

You are having trouble with 8192 because your strategy is slightly off. Keep your highest number in the bottom - right corner with numbers descending to the left, yes. But, your second row should have its highest number on the left with numbers descending to the right. On your third row keep the highest number on the right with numbers descending to the left. This strategy will allow you to snake doubles around the board and if followed with a friendly computer opponent will look just like your layout of the highest score possible.

What I find fascinating is how complicated the game sounds when explained, versus how easy it actually is to just start playing; you pick up the rules intuitively in no time.

The doge version is really hard on my computer, because it's just pink boxes; Doge doesn't appear, and the boxes don't change color when they come together. It's pretty epic lol

gets 4096 on a game about 2048 mathimaticians.

I just discovered this game a couple of days ago and I'm hooked. I thought I had watched every numberphile vid but had no idea that you guys had made a video about it. I don't want to spoil the fun I'm having so I will come back and watch the vid when I hit 2048

These two are amazing together! Please guys, I'd love to see some more collaborations in the future :)

Surely the best possible tile would be 2^17 right? Because if it was giving 10% 4 tiles and it have you the best possible outcomes then you would end up with a similar grid but that last 2 could be a 4 allowing you to just go one by one combining until you get to the 2^17 tile...I agree if it's only 2 tiles but giving best outcome with 10% 4 tiles could go higher.

i was hoping Brady would make a video about 2048 and then found this... thank you Steve and Dr Grimme for making this.

Super fantastic look back :D I had forgotten about this gem already!

I never thought about this game as difficult, but as I load up the AI program and it loses 5 times in a row, I start to feel like this victory is never going to be within my grasp

2048 Strategy: 1.) Pick a direction that will be your highest priority. This direction will NEVER CHANGE. For me, I choose the down direction. 2.) Pick a second and third highest priority for directions. These will alternate. Your second highest priority direction will determine your corner. These directions should be opposite eachother. For myself, I choose left as my second highest and right as my third highest priorities. This means my bottom left corner is my, "corner" 3.) Your fourth highest priority direction, THE DIRECTION YOU NEVER WANT TO GO is always opposite of your highest priority position. For instance, I always choose down for my highest priority, so up will be my lowest priority and it is a move I almost never want to make. Strategy: Get your largest number into your "corner" and have the second largest and third largest etc. numbers in descending order away from your largest number in the direction opposite to your second direction priority. This is a little confusing so that just means that if your priorities are 1. down, 2. left, 3. right, and 4. up that your largest number should be bottom left and then the numbers should descend so that your fourth largest number is on the bottom right. You should do this until the row or column does not move and combine AT ALL by moving in the directions of your first three priority moves. So for me, the bottom row would not move if I made a left, right, or down move. Now, your second and third priorities are changed. You get a new "corner" that is the one bow away in the 4th priority direction of your fourth largest number. For me, it would be the square just above the bottom right. When this new row or column would not move if you were to make any of your top three priority moves, you will make a new corner just above the eighth largest number and this strategy repeats. TL;DR Form a chain of numbers from largest to smallest. Keep your largest number in a corner and then have the remaining numbers descend down a row or column, reach the end, move one unit out, and then move down the row or column until the end. (if you moved in rows initially, keep going in rows, if you moved in columns, keep moving in columns)

21:52 Assuming the very last tile given was a 4, then the highest tile createable would be higher than 2^16. perhaps 2^17.

On the highest possible score: On the last two tiles you could combine 2 + 2 into 4 and then the game could spawn you another 4, and by combining the series upwards the highest number would be 2^17 and the highest score a lot higher

So, by my math, I get a max score of 3,932,100. The 131,072 tile is possible with a final 4 (which would be worthless, since it didn't come from a pair of 2's), and assuming that's the only 4, it's worth a maximum of (2^k)(k-1)-4 = (2^17)(16)-4 = 2,097,152. Then a subsequent 65,536 would ultimately need a 4 as well; (2^16)(15)-4=983,040. The final tile you'd make in the series would be an 8- a created 4 plus a generated 4- worth 12 points (2^3)(2)-4. The tile after that wouldn't score, and the game would be over. The full sum of 12+44+124+...+983,036+2,097,148=3,932,100.

KZfaq: 2014: nah 2015: nah 2016: nah 2017: nah 2018: nah 2019: YEET

In one of my early games, the first day I was playing, I got two 1024 tiles on the board with their corners touching, but I wasn't able to slide either row over so I could combine them

Really enjoyed this video and I love that the sound effect for swiping is 'swwiiiipe!'

The Dr. Who version should be changed so John Hurt is 9, Chris is 10, David is 11, Matt is 12 and Peter is 13.

"I speak maths fluently"

My favorite has to be the rainbow-colored one ("prism"), because you can stare in one place and take it all in at once, really quickly. It doesn't let you play past 2048, unfortunately, but the colors cycle (so 4096 would be the same color as 2) and it lends itself more to speed play than distance.

In the version that I played you could also get 1's, in which case you will also get points for combining into 2's. And if you then end up with only a 2048 tile, then you would have got 2048 point for each of the layers from 2^1 to 2^11, which equals 11*2^11 = 22528 for creating the the 2048 tile

The hard version requires a change in strategy: keep the high numbered tile in the center four squares, because the AI screws you less....

I got 16384 actually.. kinda sad...

That first strategy is what i also came up with while playing on a PC. Because then it is just press and hold arrow down until no more numbers spawn in, then press left (i like having them sorted to the left) to collabs all the 2, 4, 8 and so on. Then resuming with arrow down. It really gets you through the boring part really fast. Things change a bit later on because high number are quite static. To resist the urge to combine like two 512 and give up the order you were trying to keep up is key to winning in my opinion ( "uh... whatever, i just make another 512" ) This also gets you to tile higher then 2^11 or 2^12. Same works with the fibonacci variant although you have to alter your strategy earlier in the game. Worst thing is a big bumber amidst the ones you are currently manipulating.

I'd be great at this. I used to play something similar in my stash chest in Diablo 2 when gem collecting. I'm glad I stayed a while and listened.

What if the last number would havve been a 4? THen you would get 2^17?

This is NOT a rip-off of Threes, any more than Sonic is a rip-off of Mario. Threes' creators may have created a new sub-genre (debatable) for these games, but 2048 has entirely different rules and mechanics. As such, it cannot sensibly be called a rip-off or clone. Personally I prefer the simplicity of 2048, and being a programmer powers of two sit better in my brain. James' strategy matches mine, but I'm yet to reach 2048 (annoyingly close with 1024 + 512 + 256 though).

Ok, 4:35 4098 should really be 4096 surely. If Grime got 4098 then math obviously broke and the universe would have melted.

20:34 only happens if a '2' tile spawns during that time as the last type, if you get a '4' tile then you can continue on the chain and reach 2^17th...if my thinking is right

Aw, come on. You guys should totally do a video like this about the game Threes. It isn't isomorphic at all, so I'd really like to see some strategies or at least some attempt at working some out. I know you mentioned it at the end, but please do more. I'd love that.

At first, I was like.. I'm not going to make it 2 minutes into this video, it's 24 minutes long.. Uh.. but it was really well done and interesting! Watched the whole thing. Great stuff. I'm happy to have wasted time watching this :)

Im still addicted to this game. I love playing other versions of this game, like the Hexagon version

The best way to go about it so to keep all the biggest numbers in sort of like an assembly line, starting in one corner. Say, you have the 2048 tile, make sure it's at the very corner. The your next job is to make another 2048 tile so it can combine with the original 2048 tile. In order to do that, the numbers leading up to 2048 should be closest to the original tile. The randomised placing of the tiles is the main challenge because you have to constantly devise a plan to have all the leading numbers in an assembly line to combine them easily. So when you do have the complete assembly of numbers leading up to 2048, it'll look something like snake. The sequence could look like 2,2,4,8 on the top row right below the 8 tile should be 16. So the second row would be 128, 64, 32, 16. Then under 128 should be 256. Third row would look like 256, 512, 1024, 2048. That's just the third row. So we can make it higher than 2048. Under the 2048 should be 4096. Fourth row should now look like 32768, 16384, 8192, 4096. Ideally, the best combination of tiles should look like this: 2,2,4,8 128, 64, 32, 16 256, 512, 1024, 2048 32768, 16384, 8192, 4096 You should have this so you can continue with having a bigger number.

I subscribed as I was scared about my life.

Swiiiiiiiiipe!

Well, what is when a 4 appears at the end, would your maximum not be 2^17?

Thanks for the strategy. I tried the game for the first time on watching this video and went past 2048 before the video was over :D

I noticed an error in your maths. Or rather logic. You said that the highest possible tile would be 2^ 16 and the reason is that you can't get higher is because you don't have something to merge the lowest tile with. But you could have a 4 generated in the slot where you had 2^1 and so that means that you could actually then merge it with the other 4, and merge that with all the other tiles and finally get to 2^17! Which I believe is about 131k?

32768 was the biggest tile I've ever reached (on the android version where you can undo your latest step tho)

Weird watching this 8 years on and remembering playing this obsessively for a while.

You can never have "just the 2048" tile on your screen, because when you combine that final chain, new "2" and "4" tiles appear for each match. There will be at most 10 "junk" tiles (some of them may combine in the process) left over at that point.

on the original version I had 2048, but on your phone there is a version with a 'undo' option, there I have a block of 65.536 and a score over 1.9 million. two other friends have a block of 32.768. so with a 'undo' option it isn't impossible, moreover easy to get 8.192

Soo... you have an AI that auto-plays awesomely the game... and also you have another AI that places the tile on the worst possible space.... What would happen if you merge those two?.. Who would "won"?? I've played the "hard" version and i can say it is a d*ck ... but also, the auto-play AI seems to make it without any strategy... that'd be interesting...

I got the 4096 tile one time. My score for that round was 68,968 (I also had a 2048 and a 512 but the rest got filled up with 2s and 4s in the worst possible places) My strategy is to build a chain that zigzags back and forth with tiles in order. Sometimes I can chain together two whole rows of this into one tile! That's the best feeling.

A question for you: If a player were to swipe at random directions each turn, 2 2s (only one 2 if the empty tile is less than or equal to 8) spawn on empty tiles every swipe, and there is a 10% chance that it the game spawn 4 instead of 2, what's the probability for that player to reach 2048? (If you understand what I mean, please answer)

I'm asian and I can barely get 1024:(

Great game. I think fans will like Blokkology too. It's a game for fans of lateral thinking puzzles - oh and maths too of course :) Play at www.blokkology.com

About that annotation in 20:39, there is a trick (or cheat, depending on point of view) which makes it's very easy. When you undo your move there is a chance that new number will appear in different cell, futhermore it can be 2 or 4. Without it, it might take a few months to create 2^17 :) I have actually made 2^17 in 3 different games (and then sooner or later I failed) and these 3 games took me about half a year (casual playing, 30-60 minutes, not every day)

HAHAHAHA "subscribe or I will kill you" that was hilarious! I just stumbled upon your channel because this video was linked to the Ricci Flow video by #phile and I'm so excited you guys have your own channels! This is awesome because you guys are always in #phile videos and I can't get enough of these math vids... and his science and space ones as well are great as well! KEEP UP THE GREAT WORK AND KEEP 'EM COMING GUYS!!!!!! P.S. please don't kill me, I've subscribed.

Hold... hold up a second did he just... did he say... Doozse? Instead of what everyone would normally assume would be pronounced D-OH-g? Good lord. It's like when people say may-may or me-me to what obviously must be "meem" Someone should study these "assumed pronunciations" and relate them to psychology just for the fun of it.

I hate Doctor Who

The problem with encoding the AI to combine as many tiles as possible is that eventually you will end up with a scenario in which you have to swipe up because it is your only possible move. In that case you run the risk of generating a two beneath your highest number. Therefore in the long run in which case this might occur many times it is best to generate new numbers when you know the highest numbered tiles won't move up so as to prevent that case from ever happening.

Didn't know you have a channel as well. Susbcribed!

Does anybody realize 2048 was a ripoff of "threes" which is the same as this but with combining threes

James Grime and Steve Mould in the same video?! Someone is reading my maths dream journal.

I managed to get to 1024 on the evil 2048, after many attempts. After a while you start to see the obvious traps it sets, like filling up all the rows so you have to go up, letting it take the corner from your big tile.

the highest you can get is 2^17 becouse if your lukky a 4 spawn in at the end making it posible to at them all together

You can get a 2^17 (131072) if a four spawns at the end rather than a two.

10:21 - yeah! Exactly, couple of times I was dreaming of those tiles..

You can just see James's delight when Steve mucks up :D

Well you can get to the 131072 if you are luckie enough to spawn a 4 instead of a two at the very last space cause in that case you could collide the entire row down to 2^17. there are versions of the game with different undo-options and this has been achieved in a version with undo (by me) but by the rules of the game its impossible to score any higher even with incredible luck or an undo-option. so if you want to check, the (my) highscore through this option it's 2065048

Hey!! I guess someone mentioned this but have you ever tried the 2048 Fibonacci?? is so much fun and it makes you to think more than the regular game, you play adding the fibonacci numbers like 1+1=2,+1=3,+2=5,+3,=8 and so on

Wouldn't it be possible to get 2^17 if you have the exact board shown in this video for the maximum score, and you get lucky and a 4 spawns in place of the 2 in the top right corner of the screen?

The game sometimes gives you a 4 block, so if at the end of your prefect board you got a 4 block (2^2) instead of a 2 (2^1) block, you would be able to add everything up and get a 2^17 block at the end and the prefect end board would look like your board but with everything to a power one higher than you have. SO if you are really lucky and get that 4 block you can finish with a high block of 131072, and if I have done my math correctly a high score of 4194306 or (2^1-1)*(2^17) + (2^2-1)*2^16 + (2^3-1)*2^15 + (2^4-1)*2^14 + (2^5-1)*2^13+(2^6-1)*2^12+(2^7-1)*2^11+(2^8-1)*2^10+(2^9-1)*2^9+(2^10-1)*2^8+(2^11-1)*2^7+(2^12-1)*2^6+(2^13-1)*2^5+(2^14-1)*2^4+(2^15-1)*2^3+(2^16-1)*2^2+(2^17-1)*2

Great vid. I got 20116 but had less on the board left? I need to look at this again.

Hi i d'nt know if someone suggested it but. What if in the maximum board u get a 4 instead of a 2 in the last block wont it Mean that the maximum block is 2^17??? And getting the maximum score To about twice the one u calculated???? 16x2^17

2^17 is the highest tile reachable. if the board showed the same sequence in the video (2^16, 2^15 etc) but with 2^1 actually being a 4, then they can form 2^17.

I once turned my game into a 12x12 grid because I was testing the options (FYI: 12x12: VERY hard to lose) and I managed to get the 2048, 4096, 8192, 16384, 32768 AND 65536 tiles. After the 512 tile, all the tiles just loop different shades of yellow.

is there a way to remove the you win screen?? i dont like it being there i wanna get further without the annoyance my score so far is 21312 haha

you deserve so many more subs

I laughed so much at the end when you sign off

But can't you have several 2^15 tiles on the board when you get the 2^16? and several 2^16-x?

It is actually possible to get the 2^17 tile. When you have te whole snake and there spawns a 4 tile in the upper right corner, you can get it!

If field is big enough, randomness in game will do all game for you. Cause you need only to repeat same sequence of moves every time. It will eventualy consentrates in some kind shaped figure.

I love James' "We don't have time to get into ALL that (Doctor Who lore)" face at 11:30

It is possible to make 131 072 if a the very last moment instead of a 2 appearing as you said a 4 appears. One of my friend made it in a version with infinite undos

The score for the nth tile is: (log(n, 2)-1)*n Where the second argument in the log funtion is the number base.

The highest possible square is 2^17, in the case the 16th vacant square gets filled in with a 4 instead of a 2.

Alright, no need to threaten me:) I subscribed as soon as I saw this channel existed!

Me and my friend actually both reached 16384 last year using the method of keeping biggest numbers in bottom right and the biggest after that to the left of it

Assuming that the value of new tiles generated is independent of the previous one, the probability of getting a 4 tile is 1/10. So we can in theory reach the 131072 tile. I find out that the probability to get the higher score possible in this game (3932100, all tiles between 2^17 and 2^2) is approximately 1.985e-6013, since we need only 2 tiles generated except 16 times, in a particular order.

Thanks for knowing me such game exists. I've tried and got 20228 for 2048 with random deals. It's great to challenge with higher numbers ;) Definitely, as someone said here, move in 3 directions (unless you know what you're doing) That means, know your further moves to the "release".

This is great, you should do more vids together!

A friend of mine spend the best part of a year playing in practice mode (Where you are allowed to undo movies) with the aim of getting to the point were you cannot add any more tiles.. He actually reached it! He also then kept undoing that last move until he got the last tile to be a 4 and joined up all the way down to 2^(17) .. I am not sure if he went on to get 2^(1), 2^(3), 2^(4) .... 2^(17) I don't think so though.

There exists a version of Tetris that always gives you the worst possible Tetromino (block), which some of you might find interesting :) It's called "Bastet" (after the Egyptian goddess :3) Cool video, guys! I didn't even know Steve had a channel; another one for me to subscribe to, haha.

u can get one higher than 65536 if the last tile that spawns is a 4. because then u can merge everything together again.

Time strategy...(top) --left,down,left,down....left,down (stop when no tiles move); left ,up,left,up....left,up,left,up (stop when no tiles move), goto :top....This stategy best fits the regular version, it presses the larger numbers to the left side (or right side which ever you choose)...and "proteccts" the opposing side by padding it with low numbers like 2's and 4's, makeing them more likely to combine.

After trying to think about it and only getting mabye 1000, I decided to try counter clockewise motions (I think I slipped up a time or two) but I got to 3032 on the second try with that method. Interesting.

Wouldn't it also be possible to have multiple 2^15's that cannot be combined because there are tiles in the way? That would higher the maximum score.

20:55 But what if you hit your 10% possibility of hiting four ( 2^2), then you would end up with 2^17 in end. thanks to answer

Just went to a 512 while rewatching this. Still awesome to play.

The position of the new tiles are not completely random, they tend to appear at the edge you swiped from.

Actually, if you’re lucky enough, surely you could get 2^17 as if the last one is a 4 then you could combine them all to get the 2^17 tile. This means that you could actually get an even higher score than you thought