Javascript 帮助 - Nim 游戏

Question

我是编码新手，我正在为一个 Java 脚本类(class)项目工作，我需要一些帮助。我必须制作一个名为 Nim 的小游戏，它有简单的规则，而且我已经有了一个可以运行的游戏，但我需要在不改变堆或石头数量的情况下让计算机更具挑战性。赢得太容易了。我大麦弄清楚了如何让它发挥作用，但我不确定我能做些什么来让它更具挑战性。任何指针将不胜感激。

快速解释:3 堆，每堆 5 block 石头。显示为:“A 堆 oooooo，B 堆 oooooo，C 堆 oooooo”玩家和计算机通过输入堆字母和石子数量轮流移除石子。(例如:B4，A2，C3)最后一个pulse石子输了。

function pile(name, stones) {
  this.name = name;
  this.stones = stones;
  display = function(name, stones) {
    console.log(this.name + " " + this.stones);
  }
}
var pileA = new pile("A", "ooooo");
var pileB = new pile("B", "ooooo");
var pileC = new pile("C", "ooooo");

function displayBoard() {
  console.log("Game OF NIM!");
  console.log(pileA, pileB, pileC);
}

function yourMove() {
  displayBoard();
  var moveData = prompt("Enter the letter of the pile and the number of stones to take.\n " 
    + "Ex.B5 or A2 ");
    if (moveData[0] == 'A') {
      var x = pileA.stones.length - moveData[1];
      pileA.stones = pileA.stones.slice(0, x);
      return pileA.stones;
    } else if (moveData[0] == 'B') {
      var x = pileB.stones.length - moveData[1];
      pileB.stones = pileB.stones.slice(0, x);
      return pileB.stones;
    } else if (moveData[0] == 'C') {
      var x = pileC.stones.length - (moveData[1] + 1);
      pileC.stones = pileC.stones.slice(0, x);
      return pileC.stones;
    }
  }

  function computersMove() {
    displayBoard();
    alert("Now it is the computer's turn!");
    if (pileA.stones != "") {
      pileA.stones = pileA.stones.slice(0, 0);
      return pileA.stones;
    } else if (pileB.stones != "") {
      pileB.stones = pileB.stones.slice(0, 0);
      return pileB.stones;
    } else if (pileC.stones != "") {
      pileC.stones = pileC.stones.slice(0, 0);
      return pileC.stones;
    }
    if (pileA.stones == "" && pileB.stones == "" && pileC.stones == "") {
      console.log("Computer Wins!!!");
    }
  }
  while (pileA.stones != "" || pileB.stones != "" || pileC.stones != "") {
    yourMove();
    if (pileA.stones == "" && pileB.stones == "" && pileC.stones == "") {
      console.log("You win!!!");
    }
    if (pileA.stones != "" || pileB.stones != "" || pileC.stones != "") {
      computersMove();
    }
  }
  alert('Game Over!');

Answer 1

'Nim 游戏'有 a simple deterministic solution .以下算法计算给定堆栈/堆/堆数组的最佳移动:

// Returns the optimal move or undefined if no winning move is possible:
function optimal_move(stacks) {
  var stacks_xor = stacks.reduce((r, e) => r ^ e, 0);
  var is_endgame = stacks.reduce((r, e) => r + (e > 1), 0) < 2;
  var move = stacks.reduce((move, stack, i) => {
    var take = stack - (is_endgame ^ stack ^ stacks_xor);
    return take > move[1] ? [i, take] : move;
  }, [0, 0]);
  return move[1] > 0 ? move : undefined;
}

console.log(optimal_move([3, 4, 5])); // [0, 2] 'remove 2 from stack 0'

如果没有获胜的可能，我建议您执行随机 Action ，直到人类玩家出错。

为了使游戏更具娱乐性，将随机选择一定百分比(取决于难度)的移动而不是最优移动。