Question

我正在尝试提出一种解决方案，其中我需要掷出多个骰子（大小均相同）并达到指定的数目。如果我已完成所有验证以确保数字有效，并且理论上可以得出所需结果，那么有人能解决这个问题吗？请注意，它应该随机出现，而不仅仅是平分。

一些例子

掷3 d6并得到14->这样它可以输出5,3,6或6,6,2

掷4 d20并获得66->这样它可以输出16,14,19,17

我需要一个通用函数，该函数可以接受任何大小，任何数量的掷骰子以及所需的结果。

下面是我的最初尝试，尽管没有产生期望的输出（您现在可以忽略mod，这也是允许使用修饰符）。此示例也缺少验证是否可以实现所需的输出，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));

具有最少内存消耗的非递归：

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));

一种非递归解决方案，它可以根据以前的滚动计算最后一个滚动，从而最大限度地减少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));

Answer 2

红宝石溶液：

def foo(count, dim, desired, results = [])
  return results if count == 0
  raise ArgumentError if count > desired
  raise ArgumentError if count * dim < desired

  max_roll = (dim <= desired - count) ? dim : desired - count + 1
  min_roll = [(desired - (count-1) * dim), 1].max
  roll = (rand(min_roll..max_roll))
  results << roll

  foo(count - 1, dim, desired - roll, results)

  results
end

puts foo(3, 6, 11).inspect
puts foo(2, 6, 11).inspect
puts foo(4, 4, 11).inspect

结果：

[3, 4, 4]
[5, 6]
[2, 3, 4, 2]

所以基本上它是递归函数。对于每个步骤：

掷骰子（允许的范围为min_roll..max_roll）
调用相同的函数，但通过骰子减少已消耗数量的计数，并通过roll值扩展结果数组

请注意一件事：使用此行为，结果开头可能会有更大的数字。为了避免这种混乱，只是在函数末尾混洗了结果

给定X个骰子输入，生成特定数量

2 个答案: