mark and recapture for generative text

mark_and_recapture.js

// This is using the "Bayesian" formula taken from
// http://en.wikipedia.org/wiki/Mark_and_recapture
// to provide a crude estimate of "how many" outputs the grammar might generate.
//
// As I understand it, it is quite easy to technically have infinite outputs,
// and this technique "wrongly" will always give a finite answer.
// However, informally, we want a lot of visible variety,
// and technically-infinite doesn't provide guidance towards
// more visible variety, but abundance will.
//
// The idea is that as you work on a generative-text thing,
// you try to keep driving the abundance number up and up,
// (along with other good qualities, of course).

  var seenFirst = [];
  var effort = 1000;
  for (var i = 0; i < effort; i += 1) {
    seenFirst.push(myGrammar.flatten("#origin#"));
  }
  var firstSample = [];
  seenFirst.forEach(function(item) {
    if (firstSample.indexOf(item) < 0) {
      firstSample.push(item);
    }
  });
  var K = firstSample.length;
  var seenSecond = [];
  for (var i = 0; i < effort; i += 1) {
    seenSecond.push(myGrammar.flatten("#origin#"));
  }
  var secondSample = [];
  var k = 0;
  seenSecond.forEach(function(item) {
    if (secondSample.indexOf(item) < 0) {
      secondSample.push(item);
      if (firstSample.indexOf(item) >= 0) {
        // we saw this one before
        k += 1;
      }
    }
  });
  var n = secondSample.length;
  var li = document.createElement("li");
  li.appendChild(document.createTextNode("abundance is " +
                      Math.floor((K-1)*(n-1)/(k-2)) +
                      " plus or minus " +
                      Math.sqrt((K-1)*(n-1)*(K-k-1)*(n-k+1)/((k-2)*(k-2)*(k-3))).toP\
recision(4)
                      ));
  document.getElementById("results").appendChild(li);