For example, a single letter is one of 26 symbols (A, B, …, Z), a single digit is one of 10 (0, 1, …, 9), and two digits is one of 1026, and the decoding is equally straightforward.This particular scheme isn’t terribly efficient (because 27-100 remain unused), but it is “lossless” because you’ll always recover your original letter. Similarly, the set of every possible twenty-seven letter words has 26 different permutations in it (from aaaaaaaaaaaaaaaaaaaaaaaaaaa to zzzzzzzzzzzzzzzzzzzzzzzzzzz).

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So, when you try to go back to the old set, it’s impossible to tell which symbol is the right one.

As a general rule of thumb, count up the number of possible symbols (which can be numbers, letters, words, anything really) and make sure that the new set has more.

Here the log was done in base 2, but it doesn’t have to be; if you did the log in base 26, you’d know the average number of letters needed per symbol.

In base 2 the entropy is expressed in “bits”, in base e (the natural log) the entropy is expressed in “nats”, and in base the entropy is in “slices”.

If the nth symbol in your set shows up with probability P, then the entropy in bits (the average number of bits per symbol) is: .

The entropy tells you both the average information per character and the highest density that can be achieved.

The Shannon entropy gives us an absolute minimum to how much space is required for a block of data, regardless of how clever you are about encoding it.