2048 x 2 = 4096

http://martijnkorteweg.github.io/4096/

by chemicalsx • 12 years ago

22 24 12 years ago

martijnkorteweg.github.io

deletes2• 12 years ago

I though I would see this sooner. This might be barely solvable, but the next power( after 4096 ) is almost impossible to reach and there should be some innovation in gameplay.

chemicalsx3• 12 years ago

main reason for me was that i kept on beating the 2048 one, so lets see if this one is beatable :)

wtpiu• 12 years ago

you are the man.

dwd• 12 years ago

The lack of challenge was a good cure for addiction...

croddin• 12 years ago

It could just go to a 5x5 grid for 8192

josu• 12 years ago

Dogs?

ponytech• 12 years ago

It had already been done : https://news.ycombinator.com/item?id=7386557

benjamincburns• 12 years ago

NOTE: This site is the official version of 2048. You can play it on your phone via http://git.io/2048. All other apps or sites are derivatives or fakes, and should be used with caution.

Might be good of you to remove that bit.

chemicalsx3• 12 years ago

good note, adjusted the page to give credit where credit is due

TrainedMonkey2• 12 years ago

4096 = 2^12. So you would need to utilize at least 12 out of 16 squares (if done perfectly) in order to beat it.

Since there are 16 squares total it is not hard to put upped limit on highest possible variation of the game that you can actually beat at 2^15 = 32768. Unless somebody expands the field of course.

gus_massa2• 12 years ago

Actually, the smaller number that appears is 2, and if you are lucky the last two numbers you get are 4. The smaller number must be repeated to start the massive collapse. So the greatest winnable version is 4+4+8+16+...+65536 = 131072 = 2^17.

TrainedMonkey2• 12 years ago

(4 + 4 + 8 + 16) + (32 + 64 + 128 + 256) + (512 + 1024 + 2048 + 4096) + (8,192 + 16,384 + 32,768 + 65,536)

You are correct good sir, in the analysis above I eliminated luck whatsoever and set all numbers appearing to 2 make an argument. My analysis still holds with that restriction.

I've seen numbers as large as 128 appearing, so if we go that route greatest winnable version would be even higher. I suspect that maximum number appearing is based on max number on the board now, so while we are speculating, it could be possible to make endless game by dynamically adjusting window of appearing numbers.

vishaldpatel• 12 years ago

Alternative challenge: least number of moves to game over.

chemicalsx3• 12 years ago

feels that that would be the true score of the game, least number of moves to win.

shurcooL• 12 years ago

I mean no negativity here, but isn't it neat how this is sorta like Call of Duty 5 or whatever sequel they're on now? Where people use terms like "milking the franchise."

Clearly, these forks/derivate works have some value (both to creators, and those playing), but I find the parallel to be interesting.

deletes2• 12 years ago

I don't see it. The 2^n games are free of money and adds.

ryanSrich• 12 years ago

Is 6100[1] good?

1.) http://i.imgur.com/c62H5We.png

hashberry• 12 years ago

Well, to beat the game you need a score of ~41000. So you are in the bottom 15%.

jthacker• 12 years ago

I actually did the same %s/2048/4096/g in the code the other day after getting 2048. And I finally beat it! http://i.imgur.com/pC51sNT.png

I suppose the next step is to make it endless and see how high you can go.

shmageggy• 12 years ago

2048-AI can solve this one too, and I think with a little refinement 8192 is definitely possible.

See the screenshot at the bottom: http://stackoverflow.com/a/22389702/1056032

eie• 12 years ago

I beat this with first try. I feel like I could almost always win with my strategy... http://imgur.com/HtC2mOh

hleszek• 12 years ago

I'll just wait for the 4096 AI version to play...

gibsonhu• 12 years ago

You can play 4096 on android

https://play.google.com/store/apps/details?id=com.runningwit...

2048 x 2 = 4096

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