4x3版2048被完全解出，最优策略平均得分超5万。
Strongly Solving 2048 4x3
- 按棋盘数字和分层状态空间，利用状态年龄不变性加速搜索
- 最优策略在常见初始状态下平均得分约50724.26
- 首次完整求解4x3版2048，适合博弈论与强化学习研究者
2048是一款基于4×4网格的随机单人游戏，玩家每次选择上下左右之一，合并相邻同数值方块以获得分数。本文针对一种新变体2048-4x3（4行3列，共12格）进行了强求解。该变体比原版少一整行。在最常见的初始状态（两个2号方块）下，最优策略的期望得分为约50724.26。研究识别出可到达状态数为1,152,817,492,752，后续状态数为739,648,886,170。核心方法是根据棋盘上所有方块数值之和定义‘状态年龄’，该值在有效操作后保持不变，在环境随机响应中增加2或4。据此，可按年龄分组状态空间，并仅依赖前一龄状态来枚举当前龄的所有状态及其后续状态。同时，通过逆序年龄遍历，可逐层确定状态价值。
原文摘要 · Abstract (English)
2048 is a stochastic single-player game involving 16 cells on a 4 by 4 grid, where a player chooses a direction among up, down, left, and right to obtain a score by merging two tiles with the same number located in neighboring cells along the chosen direction. This paper presents that a variant 2048-4x3 12 cells on a 4 by 3 board, one row smaller than the original, has been strongly solved. In this variant, the expected score achieved by an optimal strategy is about $50724.26$ for the most common initial states: ones with two tiles of number 2. The numbers of reachable states and afterstates are identified to be $1,152,817,492,752$ and $739,648,886,170$, respectively. The key technique is to partition state space by the sum of tile numbers on a board, which we call the age of a state. An age is invariant between a state and its successive afterstate after any valid action and is increased two or four by stochastic response from the environment. Therefore, we can partition state space by ages and enumerate all (after)states of an age depending only on states with the recent ages. Similarly, we can identify (after)state values by going along with ages in decreasing order.
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