3 ms·
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 har
by TrainedMonkey 13y 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 13y agoActually, 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 13y 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 13y agoThe new tiles are only 2 (90%) and 4 (10%). From the source https://github.com/gabrielecirulli/2048/blob/master/js/game_manager.js#L43 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); } };