Randomly Generate Letters According to their Frequency of Use?

Question

How can I randomly generate letters according to their frequency of use in common speech?

Any pseudo-code appreciated, but an implementation in Java would be fantastic. Otherwise just a poke in the right direction would be helpful.

Note: I don't need to generate the frequencies of usage - I'm sure I can look that up easily enough.

Answer 1

One quick way to do it would be to generate a list of letters, where each letter appeared in the list in accordance with its frequency. Say, if "e" was used 25.6% of the time, and your list had length 1000, it would have 256 "e"s.

Then you could just randomly pick spots from the list by using (int) (Math.random() * 1000) to generate random numbers between 0 and 999.

Answer 2

I am assuming that you store the frequencies as floating point numbers between 0 and 1 that total to make 1.

First you should prepare a table of cumulative frequencies, i.e. the sum of the frequency of that letter and all letters before it.

To simplify, if you start with this frequency distribution:

A  0.1
B  0.3
C  0.4
D  0.2

Your cumulative frequency table would be:

A  0.1
B  0.4 (= 0.1 + 0.3)
C  0.8 (= 0.1 + 0.3 + 0.4)
D  1.0 (= 0.1 + 0.3 + 0.4 + 0.2)

Now generate a random number between 0 and 1 and see where in this list that number lies. Choose the letter that has the smallest cumulative frequency larger than your random number. Some examples:

Say you randomly pick 0.612. This lies between 0.4 and 0.8, i.e. between B and C, so you'd choose C.

If your random number was 0.039, that comes before 0.1, i.e. before A, so choose A.

I hope that makes sense, otherwise feel free to ask for clarifications!

Answer 3

Not even a pseudo-code, but a possible approach is as follows:

Let p1, p2, ..., pk be the frequencies that you want to match.

1. Calculate the cumulative frequencies: p1, p1+p2, p1+p2+p3, ... , 1
2. Generate a random uniform (0,1) number x
3. Check which interval of the cumulative frequencies x belongs to: if it is between, say, p1+..+pi and p1+...+pi+p(i+1), then output the (i+1)st letter

Depending on how you implement the interval-finding, the procedure is usually more efficient if the p1,p2,... are sorted in decreasing order, because you will usually find the interval containing x sooner.

Answer 4

What I would do is scale the relative frequencies as floating point numbers such that their sum is 1.0. Then I would create an array of the cumulative totals per letter, i.e. the number that must be topped to get that letter and all those "below" it. Say the frequency of A is 10%, b is 2% and z is 1%; then your table would look something like this:

0.000 A ; from 0% to 10% gets you an A
0.100 B ; above 10% is at least a B
0.120 C ; 12% for C...
...
0.990 Z ; if your number is >= 99% then you get a Z

Then you generate yourself a random number between 0.0 and 1.0 and do a binary search in the array for the first number smaller than your random number. Then pick the letter at that position. Done.

Answer 5

Using a binary tree gives you a nice, clean way to find the right entry. Here, you start with a frequency map, where the keys are the symbols (English letters), and the values are the frequency of their occurrence. This gets inverted, and a NavigableMap is created where the keys are cumulative probability, and the values are symbols. That makes the lookup easy.

  private final Random generator = new Random();

  private final NavigableMap<Float, Integer> table = 
    new TreeMap<Float, Integer>();

  private final float max;

  public Frequency(Map<Integer, Float> frequency)
  {
    float total = 0;
    for (Map.Entry<Integer, Float> e : frequency.entrySet()) {
      total += e.getValue();
      table.put(total, e.getKey());
    }
    max = total;
  }

  /** 
   * Choose a random symbol. The choices are weighted by frequency.
   */ 
  public int roll()
  {
    Float key = generator.nextFloat() * max;
    return table.higherEntry(key).getValue();
  }