So I’m not sure if this question belongs here or maybe Math overflow. In any case, my question is about information theory.
Let’s say I have a 16 bit word. There are 65,536 unique configurations of 1’s and 0’s in that number. What each one of those configurations represents is unimportant as depending on your notation (2’s complement vs signed magnitude etc.) the same configuration can mean different things.
What I’m wondering is are there any techniques to store more information than that in a 16 bit word?
My original ideas were like odd/even parity or something but then I realized that’s already determined by the configuration… i.e. there is no extra information encoded in that. I’m beginning to wonder if no such thing exists.
EDIT For example, let’s say some magical computer (thinking quantum or something here) could understand 0,1,a. Then obviously we have 3^16 configurations and can now store more than the numbers [0 – 65,536]. Are there any other properties of a 16 bit word that you can mess with in order to encode extra information in your bit stream?
EDIT2 I am really struggling to put this into words. Right now when I look at a 16 bit word in the computer, the property which conveys information to me the relative ordering of individual 1’s and 0’s. Is there another property or way of looking at a 16 bit word which would allow more than 2^16 unique “configurations”? (Note it would no longer be a configuration, but 2^16 xxxx’s where xxxx is a noun describing an instance of that property). The only thing I can really think of is something like if we looked at the number of 1 to 0 transitions or something rather than whether each bit was actually a 1 or 0? Now transitions does not yield more than 2^16 combinations because it is ultimately solely dependent on the configuration of 1’s and 0’s. I’m looking for properties that would derive from the configuration of 1’s and 0’s AND something else thus resulting in MORE than 2^16. Does anyone even know what this would be called if it did exist?
EDIT3 Ok I got it. My question boils down to this: How do we prove that the configuration of 1’s and 0’s in a word completely defines it? I.E. How do we prove that you need no other information besides the bitmap to show equality between two 16 bit words?
FINAL EDIT
I have an example… If instead of looking at the presence of 1’s and 0’s we look at transition between bits we can store 2^16 alphabet characters. If the bit to left is the same, treat it as a 1, if it transitions, treat it as a 0. Using the 16 bit word as a circularly linked list type structure where each link represent 0/1 we basically for a 16 bit word out of the transition between bits. That is an exact example of what I was looking for but that results in 2^16, nothing better. I am convinced that you cannot do better and am marking the correct answer =(
The amount of information in a particular configuration of 16 0/1s is determined by the probability of this configuration (this is called self-information). This can be bigger than 16 bits if the configuration is less likely than 1/(2^16), but that means that some other configurations are more likely than 1/(2^16) and so will contain less information than 16 bits.
To take into account all the possible configurations, you have to use the expected value of self-information (called entropy) of individual configurations. This value will reach its maximum when the probabilities of all configurations are equal (that is 1/(2^16)) and then it will be exactly 16 bits.
So the answer is no, you cannot store more than 16 bits of information in 16 0/1s.
See
EDIT It is important to realize that bit does not stand for 0 or 1, but it is a unit of information, that is -log_2 P(w) where P(w) is the probability of a particular configuration.