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.

  • chemicalsx 12 years ago

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

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.

deletes 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.

  • chemicalsx 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?

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.

  • deletes 12 years ago

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

TrainedMonkey 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_massa 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.

    • TrainedMonkey 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.

      • gus_massa 12 years ago

        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);
            }
          };
vishaldpatel 12 years ago

Alternative challenge: least number of moves to game over.

  • chemicalsx 12 years ago

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

hleszek 12 years ago

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