javascript - 给定 X 个骰子输入，生成特定数字

Question

我正在尝试想出一个解决方案，我需要掷一些骰子(大小相同)并达到指定的数字。假设我已经进行了所有验证以确保数字有效并且理论上可以达到期望的结果，那么有人有一个好的算法来解决这个问题吗？请注意，它应该是随机出现的，而不仅仅是直线划分。

一些例子

roll 3 d6 and get 14 -> so it could output 5,3,6 or 6,6,2

roll 4 d20 and get 66 -> so it could output 16,14,19,17

我需要一个通用函数，它可以接受任意大小的骰子、任意数量的骰子以及所需的结果。

我的初步尝试如下，尽管这不会产生所需的输出(您现在可以忽略mod，这也允许修饰符)。此示例还缺少对可实现所需输出的验证，但这不是问题的一部分。

let desired = 19
let mod = 0
let dm = desired - mod
let n = 5;// number of dice
let d = 6 // dice sides
let nums = []

for(i =0; i< n; i++) {
    nums.push(Math.round(Math.random() * Math.round(d)) + 1)
}

let sum = nums.reduce((acc,val) => acc + val)


nums = nums.map(a => Math.round((a/sum) * dm))

let diff = dm - (nums.reduce((acc,val) => acc + val))
function recursive(diff) {
    let ran = nums[Math.random() * Math.round(nums.length -1)]
    if(nums[ran] + diff > d || nums[ran] + diff < 1) {
        recursive(diff)
    } else {
        nums[ran] += diff
    }
}
while(diff != 0) {
    recursive(diff)
    diff += diff < 0 ? 1 : -1;
}

alert(nums)

Answer 1

递归:

function foo(desired, rolls, sides, current) {
  if (rolls === 0) {
    return current.reduce((s, c) => s + c) === desired ? current : null;
  }
  
  const random = [];
  for (let i = 1; i <= sides; i++) {
    const randomIndex = Math.floor(Math.random() * (random.length + 1))
    random.splice(randomIndex, 0, i);
  }
  
  for (const n of random) {
    const result = foo(desired, rolls - 1, sides, [...current, n]);
    if (result) {
      return result;
    }
  }
}

console.log(foo(14, 3, 6, []))

非递归:

function foo(desired, rolls, sides) {
  const stack = [[]];
  while (stack.length) {
    const current = stack.pop();    
    const random = [];
    for (let i = 1; i <= sides; i++) {
      const randomIndex = Math.floor(Math.random() * (random.length + 1));
      random.splice(randomIndex, 0, i);
    }
  
    for (const n of random) {
      if (current.length === rolls - 1) {
        if (current.reduce((s, c) => s + c + n) === desired) {
          return [...current, n];
        }
      } else {
        stack.push([...current, n]);
      }
    }
  }   
}

console.log(foo(14, 3, 6));

非递归且内存消耗最小:

function foo(desired, rolls, sides) {
  const currentIndexes = Array(rolls).fill(0);
  const randoms = Array.from({ length: rolls }, () => {
    const random = [];
    for (let i = 1; i <= sides; i++) {
      const randomIndex = Math.floor(Math.random() * (random.length + 1));
      random.splice(randomIndex, 0, i);
    }
    return random;
  })
  while (true) {
    if (currentIndexes.reduce((s, idx, i) => s + randoms[i][idx], 0) === desired) {
      return currentIndexes.map((idx, i) => randoms[i][idx]);
    }
    for (let i = currentIndexes.length - 1; i >= 0; i--) {
      if (currentIndexes[i] < sides - 1) {
        currentIndexes[i] += 1;
        break;
      }
      currentIndexes[i] = 0;
    }
  }
}

console.log(foo(14, 3, 6));

非递归解决方案，通过根据前一卷计算最后一卷，实现最小内存消耗并提高性能。

function foo(desired, rolls, sides) {
  const currentIndexes = Array(rolls - 1).fill(0);
  const randoms = Array.from({ length: rolls - 1 }, () => {
    const random = [];
    for (let i = 1; i <= sides; i++) {
      const randomIndex = Math.floor(Math.random() * (random.length + 1));
      random.splice(randomIndex, 0, i);
    }
    return random;
  })
  while (true) {
    const diff = desired - currentIndexes.reduce((s, idx, i) => s + randoms[i][idx], 0);
    if (diff > 0 && diff <= sides) {
      return [...currentIndexes.map((idx, i) => randoms[i][idx]), diff];
    }
    for (let i = currentIndexes.length - 1; i >= 0; i--) {
      if (currentIndexes[i] < sides - 1) {
        currentIndexes[i] += 1;
        break;
      }
      currentIndexes[i] = 0;
    }
  }
}

console.log(foo(66, 4, 20));