javascript递归骰子组合并将结果存储在矩阵中 [英] javascript recursive dice combination and store the result in a matrix

问题描述

我想要实现的是N-dices发布组合的经典结果之一，但是将结果保存在具有MN字段的矩阵中（其中N是骰子的数量，M是总数）可能组合的数量 - 通过6 ^ N获得。到目前为止，我已经编写了以下代码：

what I'm trying to achieve is one of the classical results of the combination of the launch of N-dices,but saving the results in a matrix with M-N fields (where N is the number of dices and M is the total number of possible combinations - obtained by 6^N). So far I've written the following code:

function Dice (commonFace, singleFace){
    this.diceFaces = ["critical", commonFace, commonFace, singleFace, "support1", "support2"]
    this.numCases = function(){
        return Math.pow(this.diceFaces.length, numberDices)
    }
}
//create the attack dice
var attackDice = new Dice("smash", "fury");

//create the defence dice
var defenceDice =  new Dice("block", "dodge");



//create a function that rolls the dice results and returns the number of results results
function rollDiceResults(diceTypeRolled, numberDicesRolled) {

    //total possible results of the rolls of that number of dices
    var totalPossibilites = diceTypeRolled.numCases(numberDicesRolled);

    //store the dice results
    var diceResults = new Array;


    function rollDice(diceType, iteration, array) {
        if (iteration == 1) {
            //return the base case
            for (i = 0; i < diceType.diceFaces.length; i++) {
                array[i] = (diceType.diceFaces[i]);
            }
        } else {
            //continue
            for (i = 0; i < diceType.diceFaces.length; i++) {

                array[i] = diceType.diceFaces[i];
                rollDice(diceType, iteration - 1, tempResult);
            }

        }
    }


    for (i = 0; i < numberDicesRolled; i++) {
        rollDice(diceTypeRolled, numberDicesRolled, diceResults);
    }

}

我得到的是

• 函数声明中的错误


• 我错过了如何调用数组在函数内部，同时保持mn结构

感谢您的帮助

推荐答案

固定长度组合

递归是一种功能性遗产，因此使用功能样式将产生最佳效果。递归就是将一个大问题分解为较小的子问题，直到达到基本情况。

Recursion is a functional heritage and so using it with functional style will yield the best results. Recursion is all about breaking a large problem down into smaller sub problems until a base case is reached.

下面，我们使用建议的 Array.prototype.flatMap 但是对于尚不支持它的环境包含polyfill。当 n = 0 我们的基本情况已经达到并且我们返回空结果。归纳案例是 n> 0 其中选项将被添加到较小问题组合的结果中（选项，n - 1）–我们说这个问题较小，因为 n - 1 更接近基本情况 n = 0

Below, we use the proposed Array.prototype.flatMap but include a polyfill for environments that don't support it yet. When n = 0 our base case has been reached and we return the empty result. The inductive case is n > 0 where choices will be added to the result of the smaller problem combination (choices, n - 1) – We say this problem is smaller here because n - 1 is closer to the base case of n = 0

Array.prototype.flatMap = function (f)
{
  return this.reduce ((acc, x) => acc.concat (f (x)), [])
}

const combinations = (choices, n = 1) =>
  n === 0
    ? [[]]
    : combinations (choices, n - 1) .flatMap (comb =>
        choices .map (c => [ c, ...comb ]))
        
const faces =
  [ 1, 2, 3 ]
  
// roll 2 dice
console.log (combinations (faces, 2))
// [ [ 1, 1 ], [ 2, 1 ], [ 3, 1 ], [ 1, 2 ], ..., [ 2, 3 ], [ 3, 3 ] ]

// roll 3 dice
console.log (combinations (faces, 3))
// [ [ 1, 1, 1 ], [ 2, 1, 1 ], [ 3, 1, 1 ], [ 1, 2, 1 ], ..., [ 2, 3, 3 ], [ 3, 3, 3 ] ]

在您的计划中使用组合

Using combinations with your program

写 rollDice 看起来像这样

const rollDice = (dice, numberOfDice) =>
  combinations (dice.diceFaces, numberOfDice)

console.log (rollDice (attackDice, 2))
// [ [ 'critical', 'critical' ]
// , [ 'smash', 'critical' ]
// , [ 'smash', 'critical' ]
// , [ 'fury', 'critical' ]
// , [ 'support1', 'critical' ]
// , [ 'support2', 'critical' ]
// , [ 'critical', 'smash' ]
// , [ 'smash', 'smash' ]
// , ...
// , [ 'critical', 'support2' ]
// , [ 'smash', 'support2' ]
// , [ 'smash', 'support2' ]
// , [ 'fury', 'support2' ]
// , [ 'support1', 'support2' ]
// , [ 'support2', 'support2' ]
// ]

没有依赖

如果你很好奇 flatMap 和 map 正在运行，我们可以自己实现它们。纯粹的递归，贯穿始终。

If you're curious how flatMap and map are working, we can implement them on our own. Pure recursion, through and through.

const None =
  Symbol ()

const map = (f, [ x = None, ...xs ]) =>
  x === None
    ? []
    : [ f (x), ...map (f, xs) ]
  
const flatMap = (f, [ x = None, ...xs ]) =>
  x === None
    ? []
    : [ ...f (x), ...flatMap (f, xs) ]

const combinations = (choices = [], n = 1) =>
  n === 0
    ? [[]]
    : flatMap ( comb => map (c => [ c, ...comb ], choices)
              , combinations (choices, n - 1)
              )
          
const faces =
  [ 1, 2, 3 ]
  
// roll 2 dice
console.log (combinations (faces, 2))
// [ [ 1, 1 ], [ 2, 1 ], [ 3, 1 ], [ 1, 2 ], ..., [ 2, 3 ], [ 3, 3 ] ]

// roll 3 dice
console.log (combinations (faces, 3))
// [ [ 1, 1, 1 ], [ 2, 1, 1 ], [ 3, 1, 1 ], [ 1, 2, 1 ], ..., [ 2, 3, 3 ], [ 3, 3, 3 ] ]

Nerfed

好的，所以组合允许我们确定重复的固定选项的可能组合。如果我们有2个唯一的骰子并希望获得所有可能的掷骰怎么办？

Ok, so combinations allows us to determine the possible combinations of a repeated, fixed set of choices. What if we had 2 unique dice and wanted to get the all the possible rolls?

const results = 
  rollDice (attackDice, defenceDice) ???

我们可以调用 rollDice（attackDice，1）然后 rollDice（defenceDice，1）然后以某种方式组合答案。但是有更好的方法;一种允许任意数量的独特骰子的方式，即使每个骰子上有不同数量的边。下面，我向您展示两个版本的组合我们写了一些必要的更改来访问未开发的潜力

We could call rollDice (attackDice, 1) and then rollDice (defenceDice, 1) and then somehow combine the answers. But there's a better way; a way that allows for any number of unique dice, even with a varying number of sides on each die. Below, I show you the two versions of combinations we wrote alongside the necessary changes to access untapped potential

 // version 1: using JS natives
const combinations = (choices, n = 1) =>
const combinations = (choices = None, ...rest) =>
  n === 0
  choices === None
    ? [[]]
    : combinations (choices, n - 1) .flatMap (comb =>
    : combinations (...rest) .flatMap (comb =>
        choices .map (c => [ c, ...comb ]))

// version 2: without dependencies
const combinations = (choices = [], n = 1) =>
const combinations = (choices = None, ...rest) =>
  n === 0
  choices === None
    ? [[]]
    : flatMap ( comb => map (c => [ c, ...comb ], choices)
              , combinations (choices, n - 1)
              , combinations (...rest)
              )

这个新版本的组合，我们可以滚动任意数量的任意大小的骰子 - 即使是在这个程序中物理上不可能的3面骰子也可以^ _ ^

With this new version of combinations, we can roll any number of dice of any size – even the physically impossible 3-sided die is possible in this program ^_^

// version 3: variadic dice
const combinations = (choices = None, ...rest) =>
  choices === None
    ? [[]]
    : flatMap ( comb => map (c => [ c, ...comb ], choices)
              , combinations (...rest)
              )

const d1 =
  [ 'J', 'Q', 'K' ]

const d2 =
  [ '♤', '♡', '♧', '♢' ]

console.log (combinations (d1, d2))
// [ [ 'J', '♤' ], [ 'Q', '♤' ], [ 'K', '♤' ]
// , [ 'J', '♡' ], [ 'Q', '♡' ], [ 'K', '♡' ]
// , [ 'J', '♧' ], [ 'Q', '♧' ], [ 'K', '♧' ]
// , [ 'J', '♢' ], [ 'Q', '♢' ], [ 'K', '♢' ]
// ]

当然你可以滚动一个同一个骰子的集合

And of course you can roll a collection of the same dice

console.log (combinations (d1, d1, d1))
// [ [ 'J', 'J', 'J' ]
// , [ 'Q', 'J', 'J' ]
// , [ 'K', 'J', 'J' ]
// , [ 'J', 'Q', 'J' ]
// , [ 'Q', 'Q', 'J' ]
// , [ 'K', 'Q', 'J' ]
// , [ 'J', 'K', 'J' ]
// , ...
// , [ 'K', 'Q', 'K' ]
// , [ 'J', 'K', 'K' ]
// , [ 'Q', 'K', 'K' ]
// , [ 'K', 'K', 'K' ]
// ]

利用您的程序挖掘这种潜力，您可以写 rollDice 作为

Tapping into this potential with your program, you can write rollDice as

const rollDice = (...dice) =>
  combinations (...dice.map (d => d.diceFaces))

console.log (rollDice (attackDice, defenceDice))
// [ [ 'critical', 'critical' ]
// , [ 'smash', 'critical' ]
// , [ 'smash', 'critical' ]
// , [ 'fury', 'critical' ]
// , [ 'support1', 'critical' ]
// , [ 'support2', 'critical' ]
// , [ 'critical', 'block' ]
// , [ 'smash', 'block' ]
// , ...
// , [ 'support2', 'support1' ]
// , [ 'critical', 'support2' ]
// , [ 'smash', 'support2' ]
// , [ 'smash', 'support2' ]
// , [ 'fury', 'support2' ]
// , [ 'support1', 'support2' ]
// , [ 'support2', 'support2' ]
// ]

或者各种骰子

const rollDice = (...dice) =>
  combinations (...dice.map (d => d.diceFaces))

console.log (rollDice (defenceDice, attackDice, attackDice, attackDice))
// [ [ 'critical', 'critical', 'critical', 'critical' ]
// , [ 'block', 'critical', 'critical', 'critical' ]
// , [ 'block', 'critical', 'critical', 'critical' ]
// , [ 'dodge', 'critical', 'critical', 'critical' ]
// , [ 'support1', 'critical', 'critical', 'critical' ]
// , [ 'support2', 'critical', 'critical', 'critical' ]
// , [ 'critical', 'smash', 'critical', 'critical' ]
// , [ 'block', 'smash', 'critical', 'critical' ]
// , [ 'block', 'smash', 'critical', 'critical' ]
// , [ 'dodge', 'smash', 'critical', 'critical' ]
// , [ 'support1', 'smash', 'critical', 'critical' ]
// , ...
// ]

走高端

很高兴看到我们如何通过一些纯粹的成就来实现这一目标JavaScript中的函数。然而，上述实施是缓慢的，并且在它可以产生多少组合方面受到严重限制。

It's cool to see how we can accomplish so much with just a few pure functions in JavaScript. However, the above implementation is slow af and severely limited in terms of how many combinations it could produce.

下面，我们尝试确定七个六面骰子的组合。我们预计6 ^ 7会导致279936种组合

Below, we try to determine the combinations for seven 6-sided dice. We expect 6^7 to result in 279936 combinations

const dice =
  [ attackDice, attackDice, attackDice, attackDice, attackDice, attackDice, attackDice ]

rollDice (...dice)
// => ...

取决于组合的实施你选择上面，如果它不会导致你的环境无限期挂起，它将导致堆栈溢出错误

Depending on the implementation of combinations you picked above, if it doesn't cause your environment to hang indefinitely, it will result in a stack overflow error

为了提高性能，我们达到了很高的水平Javascript提供的级别功能： generator 。下面，我们重写组合，但这次使用了与生成器交互所需的一些命令式样式。

To increase performance here, we reach for a high-level feature provided by Javascript: generators. Below, we rewrite combinations but this time using some imperative style required to interact with generators.

const None =
  Symbol ()

const combinations = function* (...all)
{
  const loop = function* (comb, [ choices = None, ...rest ])
  {
    if (choices === None)
      return
    else if (rest.length === 0)
      for (const c of choices)
        yield [ ...comb, c ]
    else
      for (const c of choices)
        yield* loop ([ ...comb, c], rest)
  }
  yield* loop ([], all)
}

const d1 =
  [ 'J', 'Q', 'K', 'A' ]
  
const d2 =
  [ '♤', '♡', '♧', '♢' ]

const result =
  Array.from (combinations (d1, d2))

console.log (result)
// [ [ 'J', '♤' ], [ 'J', '♡' ], [ 'J', '♧' ], [ 'J', '♢' ]
// , [ 'Q', '♤' ], [ 'Q', '♡' ], [ 'Q', '♧' ], [ 'Q', '♢' ]
// , [ 'K', '♤' ], [ 'K', '♡' ], [ 'K', '♧' ], [ 'K', '♢' ]
// , [ 'A', '♤' ], [ 'A', '♡' ], [ 'A', '♧' ], [ 'A', '♢' ]
// ]

以上，我们使用 Array.from 急切地将所有组合收集到一个结果中。使用发电机时通常不需要这样做。相反，我们可以使用值作为生成它们

Above, we use Array.from to eagerly collect all the combinations into a single result. This is often not necessary when working with generators. Instead, we can use the values as they're being generated

下面，我们使用 for ... of 在发生器出来时直接与每个组合进行交互。在此示例中，我们显示任何包含 J 或♡

const d1 =
  [ 'J', 'Q', 'K', 'A' ]

const d2 =
  [ '♤', '♡', '♧', '♢' ]

for (const [ rank, suit ] of combinations (d1, d2))
{
  if (rank === 'J' || suit === '♡' )
    console.log (rank, suit)
}
// J ♤ <-- all Jacks
// J ♡
// J ♧
// J ♢
// Q ♡ <-- or non-Jacks with Hearts
// K ♡
// A ♡

但当然这里有更多的潜力。我们可以在中为块写出我们想要的任何内容。下面，我们使用继续 c $ c>

But of course there's more potential here. We can write whatever we want in the for block. Below, we add an additional condition to skip Queens Q using continue

const d1 =
  [ 'J', 'Q', 'K', 'A' ]

const d2 =
  [ '♤', '♡', '♧', '♢' ]

for (const [ rank, suit ] of combinations (d1, d2))
{
  if (rank === 'Q')
    continue
  if (rank === 'J' || suit === '♡' )
    console.log (rank, suit)
}
// J ♤
// J ♡
// J ♧
// J ♢
// K ♡ <--- Queens dropped from the output
// A ♡

也许这里最强大的是我们可以停止与 break 生成组合。下面，如果遇到King K ，我们会立即停止发电机

And perhaps the most powerful thing here is we can stop generating combinations with break. Below, if a King K is encountered, we halt the generator immediately

const d1 =
  [ 'J', 'Q', 'K', 'A' ]

const d2 =
  [ '♤', '♡', '♧', '♢' ]

for (const [ rank, suit ] of combinations (d1, d2))
{
  if (rank === 'K')
    break
  if (rank === 'J' || suit === '♡' )
    console.log (rank, suit)
}
// J ♤
// J ♡
// J ♧
// J ♢
// Q ♡ <-- no Kings or Aces; generator stopped at K

你可以根据条件变得非常有创意。对于（const [a，b，c，d，e]，在心中开始或结束的所有组合怎么样？

You can get pretty creative with the conditions. How about all the combinations that begin or end in a Heart

for (const [ a, b, c, d, e ] of combinations (d2, d2, d2, d2, d2))
{
  if (a === '♡' && e === '♡')
    console.log (a, b, c, d, e)
}

// ♡ ♤ ♤ ♤ ♡
// ♡ ♤ ♤ ♡ ♡
// ♡ ♤ ♤ ♧ ♡
// ...
// ♡ ♢ ♢ ♡ ♡
// ♡ ♢ ♢ ♧ ♡
// ♡ ♢ ♢ ♢ ♡

并显示生成器适用于大型数据集

And to show you generators work for large data sets

const d1 =
  [ 1, 2, 3, 4, 5, 6 ]

Array.from (combinations (d1, d1, d1, d1, d1, d1, d1)) .length
// 6^7 = 279936

Array.from (combinations (d1, d1, d1, d1, d1, d1, d1, d1)) .length
// 6^8 = 1679616

我们甚至可以写得更高 - order函数用于生成器，例如我们自己的过滤器函数。下面，我们找到三个20面骰子的所有组合，形成一个 Pythagorean triple - 3个整体构成有效直角三角形边长的数字

We can even write higher-order functions to work with generators such as our own filter function. Below, we find all combinations of three 20-sided dice that form a Pythagorean triple - 3 whole numbers that make up the side lengths of a valid right triangle

const filter = function* (f, iterable)
{
  for (const x of iterable)
    if (f (x))
      yield x
}

const d20 =
  [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 ]

const combs =
  combinations (d20, d20, d20)

const pythagoreanTriple = ([ a, b, c ]) =>
  (a * a) + (b * b) === (c * c)

for (const c of filter (pythagoreanTriple, combs))
  console.log (c)

// [ 3, 4, 5 ]
// [ 4, 3, 5 ]
// [ 5, 12, 13 ]
// [ 6, 8, 10 ]
// [ 8, 6, 10 ]
// [ 8, 15, 17 ]
// [ 9, 12, 15 ]
// [ 12, 5, 13 ]
// [ 12, 9, 15 ]
// [ 12, 16, 20 ]
// [ 15, 8, 17 ]
// [ 16, 12, 20 ]

或使用 Array.from ，带有映射函数，可以将每个组合同时转换为新结果并收集数组中的所有结果

Or use Array.from with a mapping function to simultaneously transform each combination into a new result and collect all results in an array

const allResults =
  Array.from ( filter (pythagoreanTriple, combs)
             , ([ a, b, c ], index) => ({ result: index + 1, solution: `${a}² + ${b}² = ${c}²`})
             )

console.log (allResults)
// [ { result: 1, solution: '3² + 4² = 5²' }
// , { result: 2, solution: '4² + 3² = 5²' }
// , { result: 3, solution: '5² + 12² = 13²' }
// , ...
// , { result: 10, solution: '12² + 16² = 20²' }
// , { result: 11, solution: '15² + 8² = 17²' }
// , { result: 12, solution: '16² + 12² = 20²' }
// ]

功能是什么？

功能编程很深。潜入！

const None =
  Symbol ()

// Array Applicative
Array.prototype.ap = function (args)
  {
    const loop = (acc, [ x = None, ...xs ]) =>
      x === None
        ? this.map (f => f (acc))
        : x.chain (a => loop ([ ...acc, a ], xs))
    return loop ([], args)
  }
 
// Array Monad (this is the same as flatMap above)
Array.prototype.chain = function chain (f)
  {
    return this.reduce ((acc, x) => [ ...acc, ...f (x) ], [])
  }

// Identity function
const identity = x =>
  x

// math is programming is math is ...
const combinations = (...arrs) =>
  [ identity ] .ap (arrs)

console.log (combinations ([ 0, 1 ], [ 'A', 'B' ], [ '♡', '♢' ]))
// [ [ 0, 'A', '♡' ]
// , [ 0, 'A', '♢' ]
// , [ 0, 'B', '♡' ]
// , [ 0, 'B', '♢' ]
// , [ 1, 'A', '♡' ]
// , [ 1, 'A', '♢' ]
// , [ 1, 'B', '♡' ]
// , [ 1, 'B', '♢' ]
// ]

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