有什么更好的方法来设计随机生成器的内置统计信息?

上下文 :随机句子生成器

1)功能 generateSentence() 生成随机语句 作为字符串返回( 工程罚款 )

2)功能 calculateStats() 输出唯一的 上述函数理论上可以生成字符串(也可以很好地工作 在里面 这个模型 ,所以请务必阅读免责声明,我不想浪费您的时间)

3)功能 generateStructure() 单词列表 Dictionnary.lists 随着时间的推移不断增长

主发电机功能的快速模型:

function generateSentence() {
  var words = [];
  var structure = generateStructure();

  structure.forEach(function(element) {
    words.push(Dictionnary.getElement(element));
  });

  var fullText = words.join(" ");
  fullText = fullText.substring(0, 1).toUpperCase() + fullText.substring(1);
  fullText += ".";
  return fullText;
}

var Dictionnary = {
  getElement: function(listCode) {
    return randomPick(Dictionnary.lists[listCode]);
  },
  lists: {
    _location: ["here,", "at my cousin's,", "in Antarctica,"],
    _subject: ["some guy", "the teacher", "Godzilla"],
    _vTransitive: ["is eating", "is scolding", "is seeing"],
    _vIntransitive: ["is working", "is sitting", "is yawning"],
    _adverb: ["slowly", "very carefully", "with a passion"],
    _object: ["this chair", "an egg", "the statue of Liberty"],
  }
}

// returns an array of strings symbolizing types of sentence elements
// example : ["_location", "_subject", "_vIntransitive"]
function generateStructure() {
  var str = [];

  if (dice(6) > 5) {// the structure can begin with a location or not
    str.push("_location");
  }

  str.push("_subject");// the subject is mandatory

  // verb can be of either types
  var verbType = randomPick(["_vTransitive", "_vIntransitive"]);
  str.push(verbType);

  if (dice(6) > 5) {// adverb is optional
    str.push("_adverb");
  }

  // the structure needs an object if the verb is transitive
  if (verbType == "_vTransitive") {
    str.push("_object");
  }

  return str;
}

// off-topic warning! don't mind the implementation here,
// just know it's a random pick in the array
function randomPick(sourceArray) {
  return sourceArray[dice(sourceArray.length) - 1];
}

// Same as above, not the point, just know it's a die roll (random integer from 1 to max)
function dice(max) {
  if (max < 1) { return 0; }
  return Math.round((Math.random() * max) + .5);
}

在某种程度上,我想知道它能输出多少不同的唯一字符串,我写了一些类似的东西(同样,非常简单):

function calculateStats() {// the "broken leg" function I'm trying to improve/replace
  var total = 0;
  // lines above : +1 to account for 'no location' or 'no adverb'
  var nbOfLocations = Dictionnary.lists._location.length + 1;
  var nbOfAdverbs = Dictionnary.lists._adverb.length + 1;

  var nbOfTransitiveSentences = 
    nbOfLocations *
    Dictionnary.lists._vTransitive.length *
    nbOfAdverbs *
    Dictionnary.lists._object.length;
  var nbOfIntransitiveSentences =
    nbOfLocations *
    Dictionnary.lists._vIntransitive.length *
    nbOfAdverbs;

  total = nbOfTransitiveSentences + nbOfIntransitiveSentences;
  return total;
}

(附带说明:不要担心名称空间污染、输入参数的类型检查或诸如此类的事情,为了示例的清晰性,我们假设这是在一个气泡中。)

重要免责声明 这不是关于修复我发布的代码。这是一个模型,它的工作,因为它是。真正的问题是 “随着未来可能的结构的复杂性,以及列表的大小和多样性,对于这些类型的随机结构,什么是更好的计算策略,而不是我的笨拙。 计算器() 函数,很难维护,很可能处理天文大数*,而且容易出错?”

*在真正的工具中,目前有351120个独特的结构,对于句子…总数已经超过(10次方80)一段时间了。