我要实现的目标是结合使用N-dices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是骰子数,M是可能的总数组合-通过6 ^ N获得).到目前为止,我已经编写了以下代码：
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
>我错过了如何在保持m-n结构的同时调用函数内部的数组
谢谢你的帮助
解决方法:
定长组合
Recursion是一项功能性遗产,因此将其与功能性风格配合使用将产生最佳效果.递归就是将大问题分解为较小的子问题,直到达到基本情况为止.
下面,我们使用建议的Array.prototype.flatMap,但为尚未支持它的环境提供了一个polyfill.当n = 0时,我们已经达到基本情况,并且返回空结果.归纳的情况是n＞ 1. 0,其中选项将添加到较小的问题组合(选择,n-1)的结果中–我们说这里的问题较小,因为n-1更接近n = 0的基本情况
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 ] ]在程序中使用组合
编写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
如果您想知道flatMap和map的工作方式,我们可以自己实现它们.完全递归.
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 ] ]削弱
好的,因此组合使我们能够确定重复的固定选择集的可能组合.如果我们有2个独特的骰子并想获得所有可能的掷骰子怎么办？
const results =
rollDice (attackDice, defenceDice) ???
我们可以先调用rollDice(attackDice,1),再调用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唯一骰子的方法,即使每个骰子的侧面数量不同也是如此.在下面,我向您展示了我们编写的组合的两个版本以及为获得未开发潜力而进行的必要更改
// 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)
)使用此新版本的组合,我们可以滚动任意数量的任何大小的骰子-即使在此程序中可能是物理上不可能的三面骰子也可以^ _ ^
// 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,您可以掷相同骰子的集合
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' ]
// ]
利用程序的这种潜能,您可以将rollDice编写为
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' ]
// ]
或搭配各种骰子
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' ]
// , ...
// ]
进入高层次
很高兴看到我们如何仅用JavaScript中的几个纯函数就能完成很多工作.但是,上述实现是缓慢的,并且在其可以产生多少组合方面受到严重限制.
下面,我们尝试确定七个6面骰子的组合.我们预计6 ^ 7会产生279936个组合
const dice =
[ attackDice, attackDice, attackDice, attackDice, attackDice, attackDice, attackDice ]
rollDice (...dice)
// => ...
根据上面选择的组合的实现,如果它不会导致您的环境无限期挂起,则将导致堆栈溢出错误
为了提高性能,我们提供了Javascript提供的高级功能：generators.下面,我们重写组合,但这一次使用与生成器交互所需的命令式样式.
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', '♢' ]
// ]上面,我们使用Array.from将所有组合急切地收集到一个结果中.使用发电机时,这通常不是必需的.相反,我们可以在生成值时使用它们
在下面,我们使用for...of与每个组合从生成器中直接进行交互.在此示例中,我们显示了包含J或♡的任何组合
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 ♡
当然,这里还有更多的潜力.我们可以在for块中编写我们想要的任何内容.在下面,我们添加一个附加条件以使用继续跳过皇后区Q
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 ♡
也许最强大的功能是我们可以停止生成带有中断的组合.在下面,如果遇到国王K,我们立即停止发电机
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
您可以在这些条件下变得很有创意.在心脏中开始或结束的所有组合怎么样
for (const [ a, b, c, d, e ] of combinations (d2, d2, d2, d2, d2))
{
if (a === '♡' && e === '♡')
console.log (a, b, c, d, e)
}
// ♡ ♤ ♤ ♤ ♡
// ♡ ♤ ♤ ♡ ♡
// ♡ ♤ ♤ ♧ ♡
// ...
// ♡ ♢ ♢ ♡ ♡
// ♡ ♢ ♢ ♧ ♡
// ♡ ♢ ♢ ♢ ♡
并向您展示生成器适用于大型数据集
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
我们甚至可以编写高阶函数来与生成器一起使用,例如我们自己的过滤器函数.在下面,我们找到三个20边骰子的所有组合,这些骰子形成一个Pythagorean triple-3个整数,这些整数组成一个有效直角三角形的边长
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 ]
或将Array.from与映射功能一起使用,以将每个组合同时转换为新结果并将所有结果收集到数组中
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' }
// ]
什么功能？
函数式编程很深入.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', '♢' ]
// ]