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from random import randint

def run(goal, runs):
  total = 0
  for x in xrange(runs):
    sequence = []
    for i in xrange(3):
      sequence.append(randint(0,1))
    while sequence[-3:] != goal:
      sequence.append(randint(0,1))

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var quote = String.fromCharCode(39);
var source = [
  'var quote = String.fromCharCode(39);',
  'var source = [',
  '  ',
  '];',
  'console.log(source[0]);',
  'console.log(source[1]);',
  'source.forEach(function(line) {',
  '  console.log(source[2] + quote + line + quote + ",");',

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Proposition: If $f : A \to \mathbb{R}$ is continuous at $c \in A$, $kf$ is continuous at $c$ for $k \in \mathbb{R}$ Proof: Let $\epsilon > 0$ be given. Since $f$ is continuous at $c$, $\exists \delta > 0$ such that if $|x - c| < \delta$ then $|f(x) - f(c)| < \epsilon / (|k| + 1)$. Thus if $|x - c| < \delta$:

$$ |kf(x) - kf(c)| = |k||f(x) - f(c)| \leq \frac{|k|\epsilon}{|k| + 1} = \frac{|k|}{|k|+1}\epsilon < \epsilon $$

If we want to show that the product of two continuous functions is continuous, we need the following lemma:

Lemma: If $f : A \to \mathbb{R}$ is continuous at $c \in A$, $f$ is bounded in a neighborhood of $c$. Proof: Let $\epsilon = 1$. As $f$ is continuous at $c$ we know there exists some $\delta &gt; 0$ such that if $|x - c| &lt; \delta$, then $|f(x) _ f(c)| &lt; \epsilon = 1$. By the triangle inequality, we can go on to say that $|f(x)| - |f(c)| \leq |f(x) - f(c)| &lt; 1$, therefore $|f(x)| &lt; 1 + |f(c)| = M$. Thus if $|x - c| &lt; \delta$, then $|f(x)| &lt; M$, i.e. it is bounded.

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var _ = require('underscore');

var freqVector = function(word, letters) {
  var freq = _.groupBy(word.split(''), function(l) {return l;});
  return _.map(letters, function(l) {
    return freq[l] ? freq[l].length : 0;
  }); 
};

var dot = function(v1, v2) {

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function counter(times, callback) {
  var ct = 0;
  return function() {
    ct++;
    if (ct === times)
      callback();
  };
};

// register a counter function