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2048 x 2 = 4096 (martijnkorteweg.github.io)
22 points by chemicalsx on March 14, 2014 | hide | past | favorite | 24 comments


> 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.


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


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.


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


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.



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


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


you are the man.


It could just go to a 5x5 grid for 8192


Dogs?


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.


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


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.


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.


(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.


The new tiles are only 2 (90%) and 4 (10%). From the source https://github.com/gabrielecirulli/2048/blob/master/js/game_...

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



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


Is 6100[1] good?

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


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


Alternative challenge: least number of moves to game over.


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


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




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