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.
Cypress SnowPros Well.. that wasn't exactly on the video, it was a TH-cam's notation... Some people deactivate them on their settings and they could've never notice it...
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.
@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).
Few more weird derivatives that you might find interesting louhuang.com/2048-numberwang/ based off of some skit that I can't recall at the moment and www.csie.ntu.edu.tw/~b01902112/9007199254740992/ Instead of 2^11, it becomes 2^53 in a 8x8 board.
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
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
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.
Matthew Swallow I absolutely agree. I get to 8192 about 90 percent of the time with that strategy. The furthest I've gotten is 8192 with a 2048 as well.
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.
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)
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)
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.
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
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 found a way to save-scum the game and tried this out - it works. I have a 131072 tile who's number doesn't quite fit, and i'm about 5/8ths of the way to finally maxing out all the tiles. I must have died thousands of times along the way.
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
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).
Brady did, but it was just watching his professors play it and react. It wasn't all that interesting. I was hoping to see a video like this from Brady. This was just as good.
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
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
the biggest tile I have ever gotten is 4096, kept going to a score of approx 67 500 points. gotta get to 8192 :D btw i didn't use quite the same strategy as you... I'm always keeping the largest tiles in a corner but then a make more like a chain, like the one you showed all the way up to 2^16, so my basic "rule" when i play is: when i have a 128 tile (because that's pretty much when strategy strats to matter) i pick either up, down, left or right and I never go that way no matter what, so I store all the high numbers in a row, for example at the top, and then I make kind of like a chain going down with lower and lower numbers, if you're doing it good and you're row at top has the lowest number 2^k. 2^k is also the highest number you should have on the row below.
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.
James, (at 19:10 ) you CAN'T end up with only one tile. since at 512 + 512, you get a new tile. assuming you get no new tile at 2048, you get at best you get 3 tiles (step: tiles for 2028 'snake' / new tiles) 1: 4+4=8 / 2 2: 8+8=16 / 2,2 3: 16+16=32 / {2+2=4},2 4: 32+32=64 / 4,2,2 6: 64+64=128 / 4,{2+2=4},2 7: 128+128=256 / {4+4=8},2,2 8: 256+256=512 / 8,{2+2=4},4 9: 512+512=1024 / 8,{4+4=8},2 10: 1024+1024=2048 / {8+8=16},2 (no spawn) paradoxically, to get the lowest score, you'd need to be counterproductive during these last 10 steps ...
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.
If fours are assumed to be included in the maximum game score and the last tile was a four, then the largest tile could be 2^17, right? Then you could fill the board again with decreasing powers of two.
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
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 :)
the statement that 2^16 is te maximum tile is incorrect, the actual biggest tile that can be achieved is 2^17. you're thinking oh this guy is crazy, but asentienod there's a 10% of a 4 spawning so instead ogf having a 2 in the corner of the maximum score board if you're very lucky a 4 can spawn there this allowing you to combine the tiles to get 2^17
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...
shouldn't the highest tile be 2^17 if the last tile is 4 instead of 2 at 20:32 because then you could combine all the following tiles after that to get 2^17
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.
Played a bit of hard version - got to 256+128 tiles. The problem with it is that it's too predictable, making all it's "hardness" into readable patterns =)
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
Striker977 you can www.dropbox.com/s/l01blskzkccsu3i/Screenshot%202014-05-07%2019.03.56.png ( with save game/undo , and too k me over one month of playing in breaks )
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 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?
and...., 65536 is not the largest number you can go,, you can up to the maximum of 131,072 . this is because it might pop out 2^2 , which is 4, at the last box , from there you can combine everything and get 131,072,, like i just did :D
The biggest tile isn't 65536, it would be 131072, because there is a 10% chance that the 2 tile that spawns in the final gap is a 4, allowing you to complete that board at 21:00
I always enjoyed besting the Asians in class at school and college. It's nice to remind them that it is possible for someone else to be smarter than them
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
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.
My pronunciation, "dohj", was the most popular in this Slate poll www.slate.com/blogs/browbeat/2013/11/15/doge_pronunciation_how_do_you_pronounce_the_name_of_the_shibe_doge_meme.html . Fascinating stuff.
Steve Mould Well I guess it must depend on how people treat the mood of the meme. I would have thought the name would be even more grammatically incorrect, so D-OH-g seemed right for me. "dohj" just doesn't seem to float my boaj- I mean boat
Thats because its MEANT to be pronounced 'dohj'. The meme originated from homestarrunner, where homestar calls strong bad his 'd-o-g-e' and strong bad pronounces it 'dohj'. To me d-oh-g sounds laughably wrong anyway.
21:00 I'm sorry to say, but you could get the 2^2 tile generated at the last square. this makes it possible to end up merging everything, and you can then reach the 2^17 tile, snaking it's way down to the 2^2 square at the end. and please don't kill me :(
the ripe off was still awesome and most games have similar concepts...they made the video on 2048 cus its more popular that is all. its not like they said 3s sucked just play 2048 its better
but 3s took at least a year of work to develop, and is being sold at a price. It's not free, but 2048 is - so 2048 is pulling customers that would have gone for threes. In other words, it's ripping off the developers of threes
oh im sorry your right every type of game only has 1 version and if that version isnt as popular as the one that cost money sucks to suck....really welcome to supply and demand free beats costs 10 times out of 10....i would understand your point if this was how all things where but seriously EVERY game ever has more than one version...angry birds is a rip off of crush the castle....flappy birds was a ripe of of the standard cave explorer type games....3s was good 2048 beat them thats all
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.
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.
Came here to say this. It takes luck, but you can do it. I saw a screenshot of a guy who undid moves by using save states set up the bored and then get a 4 generated in the last tile to get the 2^17th. So with 4's filling up your last cell and letting you reach the absolute highest tiers, you should be able to reach 3.5 million. 16*2^17 was about 2 million, so we can tack that on to the theoretical 1.5 million from the video. I think.
20:00 You're forgetting the 10% of the times the last tile will be a 4, meaning you can get a 131 072 tile and then fill up your board, ending with a 4.
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.
At 20:30 you place 2^1 in the top right. But if the computer was giving you the best possible tile each time it would have put another 4 there and you'd have been able to collapse down the chain and make a 2^17.
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.
My current score is 351168 with a 16384, a 8192 and 4096 tile lined up. I've make sure I don't become 'zoned out' whilst playing. That's where things go wrong, so I stop long before my finger has worn down to the bone. Playing it methodically by rigorously ensuring that the tiles form a 'snake' from one corner and zig zagging across and up and back again. Also making sure you don't fill up the top row so the only option is to swipe up. It's mostly a matter of endurance until the grid fills up and there's a few squares to operate in, whereupon it becomes more like chess. It's very satisfactory going from being on the brink of clogging up and having all the numbers collapse down to the bottom row - I get a little rush of dopamine before I re-engage with my peripheral vision and realise my tortoise has recently died from boredom.
When you did the 2048 video on numberphile, I tried it and found out that you can just keep the highest number in a corner and then in the decreasing order along the side. You can go on forever.
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
At 18:26 where James says he can remove 20% of the 2's, that's not right. Suppose you generate 100 tiles, and 10 of them are 4's, as expected. That gives you 110 2's and so you can remove just 20/110 of the 2's, or 18.1818% of them.
As for a theoretical max, let's suppose a game that can give you 4s, but let's say you get astronomically (un)lucky and that only happens on the last tile that would otherwise cause you to lose the game with the 2¹⁶ tile that you start showing at 20:00. Then you can do one more round and get a 2¹⁷ tile, can you not?
I definitely get what you said about becoming addicted to the game - it gets exponentially harder and longer all the time, until all of a sudden, where did my afternoon go...
You really should've cut 3:40 from this video Also, having just a 2048 tile would be impossible, because say you had a field with two 4's and an 8, 16, 32, 64, 128, 256, 512, and 1024. Then, you would need to merge those tiles, meaning each merge would spawn a new 2/4 tile. Even in some case where you had only two 1024 tiles (which would be impossible for the same reason), merging them would spawn another tile.The real question is how few tiles you could have when the 2048 tile is created, but I doubt you could get a better scenario than the one I just outlined, because 4 is the highest lowest tile you could have on the board without merging to make it.
Watched this video about a week ago. Have had the 'Impossible' version open on my computer since. Finally got to 256 (and then of course promptly got 'Game Over'). Not to be competitive, but if James Grimes never got past 128 on the 'Impossible' setting I'll be extremely pleased with myself...
So you've addressed all the 'easy' questions. What I'm wondering is: What's the highest tile you can theoretically create against a perfect --tile always in the worst place-- adversary? A more general version of that question is: What are the characteristics of the grid after a game where both players use a perfect strategy (infinite-recursion minimax algorithm). That's probably an intractable problem. Like, if you've solved that, you've simply solved the game. Still, I'd be interested to hear your intuitions. What are your highest lower-bound and lowest upper-bound for the highest tile-value at the end of a perfect game? (Show your work.)
When I programmed a computer to play the game, I did a brute force search 5 levels deep, choosing the swipe that resulted in the highest expected value of empty spaces. It got to 4096 a lot.
By my calculations, with 100% optimal conditions, the maximum number is, yes 131,072, and the MAXIMUM POSSIBLE SCORE is 3,932,100 with only fifteen 4's ever being given and only and always when it is the last possible tile. The Point total equation would be: (sum i=3 to 17 of [(2^i)(i-1)]) - (4*15), since there would be a total of 15 "free" 4's. It's begins at i=3 because the lowest non-free tile would be an 8 (2^3) created from 2+2=4,+4=8, giving 12 points. The next tile would be 16 giving 44 points, 48 minus the "free" 4. And it would continue from there until the last tile of 131,072.
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
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.
Um, actually, if you got the best game, you'd be looking at a 2^17 tile. The best score would have up to the point James showed going by 2s alone, with the last tile being a 2^2, allowing them to join all of the tiles to 2^17, and then get all the way upto a 2^16 with the final tile needed being a 2^2, and then a 2^15 with the final tile needed being a 2^2, and so on. If I'm not mistaken, the best possible score should be something like 3932100. I thought it should be 4 less, but then I considered that I hadn't added the final 4 to the total which means it shouldn't be subtracted from the total either. Lowest possible score to get a 2048 would supposedly be 17792, which would only be reachable by only getting "4" tiles and never combining tiles when unnecessary.
If a 4 didn't occasionally pop up instead of a 2, the game would be really easy, you could just swipe left up left up left up until you beat the game (or left down, right down, or right up). So the 4 was added for a reason
May I say that in your "maximum" score board, you count on the game spawning a 2 in the sixteenth slot on the board. If however, the game took the approximately 10% spawn rate of a 4 tile and spawned THAT, then you could reach one higher tile. This tile would be 2 to the 18th power or 131072.
"this is one of the games you have dreams about it" ... or nightmares ... Actually I used to play this game for quite a while, beat it on 2048, then on 4096, then my personal best I think was somewhere around 6k ... and you know the higher you try to go the more you have to spend time on the early parts. Actually I quickly got that I don't need to think for the first 128 or 256 (and often to 512 with just few conscious corrections) ... Few months later I got major head surgery (fortunately didn't involve my brain directly, but was removing a benign tumor from my right acoustic nerve /Vestibular Schwannoma/) ... anyway, after 14 hours of surgery and another few hours afterwards, just when my consciousness started to return, I remember I was dreaming of playing this game, but instead of 4x4 square I was playing on 20x20 square, and with every more I felt I got in a more difficult situation, which most often meant that I'll lose the game in few turns (actually in my dream most of the field was still empty, so in-fact I was very far from any difficult situation, but I guess dreaming of a concussed brain is allowed to drop the ball from time to time :)) and will have to start over again ... It was incredibly unpleasant feeling ... for which reason I haven't played the game ever since ... actually watching this video is the first thing related to this game I've done yet :)
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.
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
I like your thinking
Cypress SnowPros Yeah, I posted that comment before I saw that
Cypress SnowPros Well.. that wasn't exactly on the video, it was a TH-cam's notation... Some people deactivate them on their settings and they could've never notice it...
My best score is e^(iτ)
@Caffine Molecule no. e^iπ is 1 (in radians) e^i(2π) =2.
I got the 8192 tile once =). But then I failed miserably to get a higher score and stopped with less than 100k points =(
That exact same thing happened to me!
I managed to continue a bit more 131 k is my best.Hurts so much when you lose :D
I got to the 65536 tile (2^16) and a high score of 1,116,856
please quit playing practice mode that doesn't count
Arda Sev Yeah I got that on practice, but I've gotten 200k plus on classic. It's just fun seeing how far you can go
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.
How did this project go?
Please inform us of the progress and the github link, thanks
@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).
Few more weird derivatives that you might find interesting
louhuang.com/2048-numberwang/
based off of some skit that I can't recall at the moment
and
www.csie.ntu.edu.tw/~b01902112/9007199254740992/
Instead of 2^11, it becomes 2^53 in a 8x8 board.
Are you still playing 2048, if so what version do you use?
Omg if this video was 20:48 long it would have killed people.
Doxal Dise
*24:29
@@awawpogi3036 They said IF it was. The only way their point makes sense is if they knew how long it was.
If it was 34:08, it would have been subtly more crazy.
Thank you.
@@TheArmyofWin Yes! 2048 seconds
I was looking for this comment lol
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
Was going to post this. Definitely 2^17 is possible, but only 10% chance of that happening and if it didn't the person would utterly loose it!
Xander Michels and they were assuming nice ai so I totally agree
Xander Michels they have an AI that got to that number
you can undo and get it
The proper version only has 3 undo's though that makes that non really an option as you're gonna need that on your way there :p
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.
You can get a 2^17 (131072) if a four spawns at the end rather than a two.
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
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.
Matthew Swallow I absolutely agree. I get to 8192 about 90 percent of the time with that strategy. The furthest I've gotten is 8192 with a 2048 as well.
In our house, we call it the conveyor belt
It's The Snake over here
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.
gets 4096 on a game about 2048
mathimaticians.
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!
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
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.
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)
+Austin McMinn thanks for the detailed comment Austin. Good to see a strategy written out like that. I think that's basically my strategy.
:D Glad you appreciated it and sorry for being two and a half years late.
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)
Nice dude, what a legend score nonetheless. I've only gotten to 2^15.
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.
I got the 131072 in practice mode. My score is actually only 3867164 becasue it's hard the get every possible point :D
Why would you care lul
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
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.
That's correct and it is what I thought when I saw that.
I found a way to save-scum the game and tried this out - it works. I have a 131072 tile who's number doesn't quite fit, and i'm about 5/8ths of the way to finally maxing out all the tiles. I must have died thousands of times along the way.
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
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).
Thank you for understanding that. I share that same opinion.
i was hoping Brady would make a video about 2048 and then found this...
thank you Steve and Dr Grimme for making this.
Brady did, but it was just watching his professors play it and react. It wasn't all that interesting. I was hoping to see a video like this from Brady. This was just as good.
The hard version requires a change in strategy: keep the high numbered tile in the center four squares, because the AI screws you less....
Ok, 4:35 4098 should really be 4096 surely. If Grime got 4098 then math obviously broke and the universe would have melted.
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.
www.doctorwho.tv/games/thirteen/
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
I got 16384 actually.. kinda sad...
Criangulien Amazing!
Really? I've only gotten 8192 lol.
!too Me
I was playing 5 by 5 table and once i combine two 16384 my galaxy s7 edge crashed :(
Don't feel too bad. I've got a 32,768 tile.
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
These two are amazing together! Please guys, I'd love to see some more collaborations in the future :)
the biggest tile I have ever gotten is 4096, kept going to a score of approx 67 500 points. gotta get to 8192 :D
btw i didn't use quite the same strategy as you... I'm always keeping the largest tiles in a corner but then a make more like a chain, like the one you showed all the way up to 2^16, so my basic "rule" when i play is: when i have a 128 tile (because that's pretty much when strategy strats to matter) i pick either up, down, left or right and I never go that way no matter what, so I store all the high numbers in a row, for example at the top, and then I make kind of like a chain going down with lower and lower numbers, if you're doing it good and you're row at top has the lowest number 2^k. 2^k is also the highest number you should have on the row below.
"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.
TH-cam:
2014: nah
2015: nah
2016: nah
2017: nah
2018: nah
2019: YEET
2020: even more YEET
@@kstergiou3 2021: EVEN MORE YEET
James, (at 19:10 ) you CAN'T end up with only one tile. since at 512 + 512, you get a new tile. assuming you get no new tile at 2048, you get at best you get 3 tiles
(step: tiles for 2028 'snake' / new tiles)
1: 4+4=8 / 2
2: 8+8=16 / 2,2
3: 16+16=32 / {2+2=4},2
4: 32+32=64 / 4,2,2
6: 64+64=128 / 4,{2+2=4},2
7: 128+128=256 / {4+4=8},2,2
8: 256+256=512 / 8,{2+2=4},4
9: 512+512=1024 / 8,{4+4=8},2
10: 1024+1024=2048 / {8+8=16},2 (no spawn)
paradoxically, to get the lowest score, you'd need to be counterproductive during these last 10 steps ...
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.
If fours are assumed to be included in the maximum game score and the last tile was a four, then the largest tile could be 2^17, right? Then you could fill the board again with decreasing powers of two.
What if the last number would havve been a 4? THen you would get 2^17?
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
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 :)
the statement that 2^16 is te maximum tile is incorrect, the actual biggest tile that can be achieved is 2^17. you're thinking oh this guy is crazy, but asentienod there's a 10% of a 4 spawning so instead ogf having a 2 in the corner of the maximum score board if you're very lucky a 4 can spawn there this allowing you to combine the tiles to get 2^17
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...
Nevermind... i just realized the "hard" version features an autoplay mode now.. and he destroyed it within a single 128 tile...
shouldn't the highest tile be 2^17 if the last tile is 4 instead of 2 at 20:32
because then you could combine all the following tiles after that to get 2^17
I subscribed as I was scared about my life.
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.
Swiiiiiiiiipe!
Played a bit of hard version - got to 256+128 tiles. The problem with it is that it's too predictable, making all it's "hardness" into readable patterns =)
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
I got a 4 at the end ;)
***** Problem is that you can't get pass 65k :P
Striker977
you can www.dropbox.com/s/l01blskzkccsu3i/Screenshot%202014-05-07%2019.03.56.png ( with save game/undo , and too k me over one month of playing in breaks )
KillingPae that is pretty fucking awesome
***** He got a 4 at the end not a 2.....
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 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?
+flemingnexus He already stated the assumption that only "2" tiles are generated
and...., 65536 is not the largest number you can go,, you can up to the maximum of 131,072 . this is because it might pop out 2^2 , which is 4, at the last box , from there you can combine everything and get 131,072,, like i just did :D
32768 was the biggest tile I've ever reached (on the android version where you can undo your latest step tho)
The biggest tile isn't 65536, it would be 131072, because there is a 10% chance that the 2 tile that spawns in the final gap is a 4, allowing you to complete that board at 21:00
I'm asian and I can barely get 1024:(
Patient Zero Finally, we have corrupted your gene pool. :)
Now I can get 4096 but I rarely play anymore
Plot twist: He's a 1 year old asian
I always enjoyed besting the Asians in class at school and college. It's nice to remind them that it is possible for someone else to be smarter than them
Your mum has been cheating!
Weird watching this 8 years on and remembering playing this obsessively for a while.
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
I love James' "We don't have time to get into ALL that (Doctor Who lore)" face at 11:30
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.
My pronunciation, "dohj", was the most popular in this Slate poll www.slate.com/blogs/browbeat/2013/11/15/doge_pronunciation_how_do_you_pronounce_the_name_of_the_shibe_doge_meme.html . Fascinating stuff.
Steve Mould Well I guess it must depend on how people treat the mood of the meme. I would have thought the name would be even more grammatically incorrect, so D-OH-g seemed right for me. "dohj" just doesn't seem to float my boaj- I mean boat
Boaj! That made me laugh. Yeah d-oh-g has a certainly logic to it. I just said it the way I heard other people say it. I'm a sheep.
Tehkillerguy Yeah, most people tend to say something like dohj, which was really dumb to me, I pronounce it with the soft G as well.
Thats because its MEANT to be pronounced 'dohj'. The meme originated from homestarrunner, where homestar calls strong bad his 'd-o-g-e' and strong bad pronounces it 'dohj'. To me d-oh-g sounds laughably wrong anyway.
21:00 I'm sorry to say, but you could get the 2^2 tile generated at the last square. this makes it possible to end up merging everything, and you can then reach the 2^17 tile, snaking it's way down to the 2^2 square at the end. and please don't kill me :(
Does anybody realize 2048 was a ripoff of "threes" which is the same as this but with combining threes
if you watch the video they do mention 3s
ljlightfire1 but the video is still on 2048, not threes, and condoning the ripoff... that's not okay
the ripe off was still awesome and most games have similar concepts...they made the video on 2048 cus its more popular that is all. its not like they said 3s sucked just play 2048 its better
but 3s took at least a year of work to develop, and is being sold at a price. It's not free, but 2048 is - so 2048 is pulling customers that would have gone for threes. In other words, it's ripping off the developers of threes
oh im sorry your right every type of game only has 1 version and if that version isnt as popular as the one that cost money sucks to suck....really welcome to supply and demand free beats costs 10 times out of 10....i would understand your point if this was how all things where but seriously EVERY game ever has more than one version...angry birds is a rip off of crush the castle....flappy birds was a ripe of of the standard cave explorer type games....3s was good 2048 beat them thats all
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.
I hate Doctor Who
Davros, is that you?
RFC3514 Nah, he's clearly a Dalek.
:(
James is too nice to correct Steve and it creates an awkward tension that Steve isn't aware of in this 2048 math video!
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.
Well, what is when a 4 appears at the end, would your maximum not be 2^17?
Came here to say this. It takes luck, but you can do it. I saw a screenshot of a guy who undid moves by using save states set up the bored and then get a 4 generated in the last tile to get the 2^17th. So with 4's filling up your last cell and letting you reach the absolute highest tiers, you should be able to reach 3.5 million. 16*2^17 was about 2 million, so we can tack that on to the theoretical 1.5 million from the video. I think.
Yeah
20:00 You're forgetting the 10% of the times the last tile will be a 4, meaning you can get a 131 072 tile and then fill up your board, ending with a 4.
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.
I've gotten the 524288 tile several times. I play on a 8x8 grid though.
At 20:30 you place 2^1 in the top right. But if the computer was giving you the best possible tile each time it would have put another 4 there and you'd have been able to collapse down the chain and make a 2^17.
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.
Rather than the minimax algorithm, this would be best solved with stochastic optimization, e.g. a Monte Carlo method.
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.
My current score is 351168 with a 16384, a 8192 and 4096 tile lined up. I've make sure I don't become 'zoned out' whilst playing. That's where things go wrong, so I stop long before my finger has worn down to the bone.
Playing it methodically by rigorously ensuring that the tiles form a 'snake' from one corner and zig zagging across and up and back again. Also making sure you don't fill up the top row so the only option is to swipe up.
It's mostly a matter of endurance until the grid fills up and there's a few squares to operate in, whereupon it becomes more like chess. It's very satisfactory going from being on the brink of clogging up and having all the numbers collapse down to the bottom row - I get a little rush of dopamine before I re-engage with my peripheral vision and realise my tortoise has recently died from boredom.
Really enjoyed this video and I love that the sound effect for swiping is 'swwiiiipe!'
When you did the 2048 video on numberphile, I tried it and found out that you can just keep the highest number in a corner and then in the decreasing order along the side. You can go on forever.
Also you can do it super quickly
40650 score in 30 minutes but I messed up by swiping when my row wasn’t stuck
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
At 18:26 where James says he can remove 20% of the 2's, that's not right. Suppose you generate 100 tiles, and 10 of them are 4's, as expected. That gives you 110 2's and so you can remove just 20/110 of the 2's, or 18.1818% of them.
As for a theoretical max, let's suppose a game that can give you 4s, but let's say you get astronomically (un)lucky and that only happens on the last tile that would otherwise cause you to lose the game with the 2¹⁶ tile that you start showing at 20:00. Then you can do one more round and get a 2¹⁷ tile, can you not?
19:12--it says James's best score is around 75000, but he said an achievable high score is around 66000. he beat his own recommendations!
I definitely get what you said about becoming addicted to the game - it gets exponentially harder and longer all the time, until all of a sudden, where did my afternoon go...
Paused at 5:17. Searched for app "20" and first suggestion was "2048". Came back to video after 2hrs to CURSE YOU.
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 really should've cut 3:40 from this video
Also, having just a 2048 tile would be impossible, because say you had a field with two 4's and an 8, 16, 32, 64, 128, 256, 512, and 1024. Then, you would need to merge those tiles, meaning each merge would spawn a new 2/4 tile. Even in some case where you had only two 1024 tiles (which would be impossible for the same reason), merging them would spawn another tile.The real question is how few tiles you could have when the 2048 tile is created, but I doubt you could get a better scenario than the one I just outlined, because 4 is the highest lowest tile you could have on the board without merging to make it.
Watched this video about a week ago. Have had the 'Impossible' version open on my computer since. Finally got to 256 (and then of course promptly got 'Game Over'). Not to be competitive, but if James Grimes never got past 128 on the 'Impossible' setting I'll be extremely pleased with myself...
So you've addressed all the 'easy' questions.
What I'm wondering is: What's the highest tile you can theoretically create against a perfect --tile always in the worst place-- adversary?
A more general version of that question is: What are the characteristics of the grid after a game where both players use a perfect strategy (infinite-recursion minimax algorithm). That's probably an intractable problem. Like, if you've solved that, you've simply solved the game.
Still, I'd be interested to hear your intuitions. What are your highest lower-bound and lowest upper-bound for the highest tile-value at the end of a perfect game? (Show your work.)
When I programmed a computer to play the game, I did a brute force search 5 levels deep, choosing the swipe that resulted in the highest expected value of empty spaces. It got to 4096 a lot.
+Tim-J.Swan I'm doing that just now. Expectimax, AI. But it's so hard coming up with a good heuristic...!
You mean it's hard to come up with something better? Supposedly there is a strategy which snakes the pieces, but I don't know how to do it.
+Tim-J.Swan Haha, I found out why it was so hard... Two major BUGS! But I killed and ate them, and now it works (Y)
By my calculations, with 100% optimal conditions, the maximum number is, yes 131,072, and the MAXIMUM POSSIBLE SCORE is 3,932,100 with only fifteen 4's ever being given and only and always when it is the last possible tile. The Point total equation would be: (sum i=3 to 17 of [(2^i)(i-1)]) - (4*15), since there would be a total of 15 "free" 4's. It's begins at i=3 because the lowest non-free tile would be an 8 (2^3) created from 2+2=4,+4=8, giving 12 points. The next tile would be 16 giving 44 points, 48 minus the "free" 4. And it would continue from there until the last tile of 131,072.
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
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.
Um, actually, if you got the best game, you'd be looking at a 2^17 tile. The best score would have up to the point James showed going by 2s alone, with the last tile being a 2^2, allowing them to join all of the tiles to 2^17, and then get all the way upto a 2^16 with the final tile needed being a 2^2, and then a 2^15 with the final tile needed being a 2^2, and so on. If I'm not mistaken, the best possible score should be something like 3932100. I thought it should be 4 less, but then I considered that I hadn't added the final 4 to the total which means it shouldn't be subtracted from the total either. Lowest possible score to get a 2048 would supposedly be 17792, which would only be reachable by only getting "4" tiles and never combining tiles when unnecessary.
James Grime and Steve Mould in the same video?! Someone is reading my maths dream journal.
If a 4 didn't occasionally pop up instead of a 2, the game would be really easy, you could just swipe left up left up left up until you beat the game (or left down, right down, or right up). So the 4 was added for a reason
The highest possible square is 2^17, in the case the 16th vacant square gets filled in with a 4 instead of a 2.
Im still addicted to this game.
I love playing other versions of this game, like the Hexagon version
Are you still going?
May I say that in your "maximum" score board, you count on the game spawning a 2 in the sixteenth slot on the board. If however, the game took the approximately 10% spawn rate of a 4 tile and spawned THAT, then you could reach one higher tile. This tile would be 2 to the 18th power or 131072.
"this is one of the games you have dreams about it" ... or nightmares ...
Actually I used to play this game for quite a while, beat it on 2048, then on 4096, then my personal best I think was somewhere around 6k ... and you know the higher you try to go the more you have to spend time on the early parts. Actually I quickly got that I don't need to think for the first 128 or 256 (and often to 512 with just few conscious corrections) ...
Few months later I got major head surgery (fortunately didn't involve my brain directly, but was removing a benign tumor from my right acoustic nerve /Vestibular Schwannoma/) ...
anyway, after 14 hours of surgery and another few hours afterwards, just when my consciousness started to return, I remember I was dreaming of playing this game, but instead of 4x4 square I was playing on 20x20 square, and with every more I felt I got in a more difficult situation, which most often meant that I'll lose the game in few turns (actually in my dream most of the field was still empty, so in-fact I was very far from any difficult situation, but I guess dreaming of a concussed brain is allowed to drop the ball from time to time :)) and will have to start over again ...
It was incredibly unpleasant feeling ... for which reason I haven't played the game ever since ... actually watching this video is the first thing related to this game I've done yet :)
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.
My best score is 20156 for making 2048
but my highest other tile was only a 32 (the rest being 2 & 4)
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