fbs7 said:
Wow, many exquisite insights!
Let me ask you mathematicians this question: a pseudo-random sequence will repeat, even if after an absurd amount of time. But the digits of π do not repeat.
We know that neither of them are truly random, given that number n-th can be exactly calculated, although both cases "seem" to be random up to some measure.
But, given the fact that the digits of π do not repeat, does that make that sequence a bit more "random-ny" than a pseudo-random generator, if there's such thing? I fear I'm butchering proper mathematic-inglish again!
A pseudo-random sequence of numbers is just a sequence that has no readily apparent pattern, but is generated by a finite algorithm that takes no external inputs after the generation of the first M values (M is a fixed positive integer). Since the digits of pi are generated by an algorithm and its calculation takes no external inputs, they are a pseudo-random sequence.
So far as I am aware, it is not the case that a pseudo-random sequence must eventually repeat. It is just that non-repeating sequences become too slow to compute, the longer they continue, so the ones used by computers tend to be repeating for efficiency reasons. One way that I think will make a sequence non-repeating is as follows:
First, a definition: we say that a sequence of numbers, which may be finite or infinite, is '
repeating with lag L' if, for any j that is an index of the sequence such that j+L is also an index, the j-th and the (j+L)th element are equal.
The algorithm will use a simple, repeating, pseudo-random generator as its base. It generates positive integers.
Say we have already generated the first n numbers. Then for every k from 2 to (n+1), let m(n,k) be the quotient from dividing (n+1) by k and let r(n,k) be the quotient on dividing k by 2. For each k, there will be zero or one numbers such that, if that number is chosen, the last r(n,k) . m(n,k) elements out of the first (n+1) sequence elements, is repeating with lag m(n,k). Such a number is a 'forbidden number', as it creates a temporary repeat for a significant tail of the sequence thus far. Let F(n) be the set of all forbidden numbers at the step for generating the (n+1)th number.
Then generate a new number h(n+1) from the simple generator. Now chose as the (n+1)th number the closest positive integer to h(n+1) that is not in F(n).
I believe such a sequence will never repeat. But it will become progressively slower, because the number of checks that need to be made for each number increases as the sequence progresses.
But the sequence is pseudo-random because it satisfies the definition.