 heads, the third is tails, and SO on. To illustrate the concept of minimal information measure further, consider the case of a die. As we know, a die is a cube with six spotted faces, corresponding to the event source X = {1, 2, 3, 4, 5, 6}. The faces have equal probabilities of showing up as the die rolls out and stops (assuming no other physical possibility). The equiprobability translates into the relation = 1) = p(x = 2) = ... = p(x = = 1/6. The information contained in any die-roll measurement is, therefore, I(X) = -log>(1/6)=log>(6): 2 2.584 bits. (3.9) The above result is puzzling because it shows that the bit information can be a noninteger, i.e., any positive real number. But how can one form a message string with 2.584 bits? The explanation is not to confuse between the information measure (the bit) and the number n of 1/0 symbols that is actually required to form the message. What the above result tells us is that the die-roll information can be coded by any string of n symbols, satisfying n 2.584. This means that the minimum string length is n = 3. It does, indeed, take three binary bits to represent all numbers from one to six, using the same symbol or block length. 8 Next, we are going to show that binary coding is not the only way to encapsulate this information. Indeed, the full information of a single die roll can also be coded through a block of three YES or NO symbols, which provide the answer to three independent questions Q1, Q2, and Q3. Here is an example of the three questions: Q1: Is the result x even? If YES, thenx E {2, 4, 6} = U1, if NO, thenx E {1, 3, 5} = U2. Q2: Is the result X strictly greater than 3? If YES, then. E {4, 5, 6} = V1; if NO, thenx E {1, 2, 3} = V2. Q3: Is the result x divided by 3? If YES, thenx E {3, 6} = W1; if NO, thenx E {1, 2, 4, 5} = W2. The combined answers to the three questions yield the value of X (in ensemble language, one writes x = with i, j, k = 1 or 2). For instance, we have 5 = U2n V1 n W2. It is straightforward to verify that the code correspondence 