Question

Question spawned from this one. The problem can be formulated as follows:

Given two positive integers n and m, with m <= n, is there a way to find a suite of numbers, which cycles and covers all possible values from 0 to n?

As a basic example, if we take 3 as a number, for whatever number current between 0 and 3, we can compute the next value as:

next = (current+3) % 4

This will cycle. For instance: 1 -> 0 -> 3 -> 2 -> 1 etc. I found this solution by “chance” and it is even general ((i + n) % (n + 1) for any n), I cannot prove it mathematically. And it is a little too obvious.

Are there better ways to generate such a permutation?

Answer 1

I’m not sure what you intend m in the question to refer to, or how you’re defining “a suite of numbers”). However, one way of getting a cycle of number is to use a recursion (or iteration) of the form:

next = f(current)

for some function f. For example, linear congruential RNGs use the iteration:

x = ( a · x + c ) mod m   where 0 < a, c < m

They don’t always produce all values from 0 to m-1, but under certain circumstances they do:

c and m are relatively prime

a - 1 is divisible by every prime factor of m (not including m)

if m is divisible by 4, a - 1 is divisible by 4.

(This is the Hull-Dobell theorem.)

Note that a, c == 1 satisfies the above criteria for any m. Futhermore, if m is prime, any values of a and c satisify the criteria, and if m is a power of 2, then the criteria are satisfied by any a, c such that a == 1 mod 4 and c == 1 mod 2. However, for certain values of m (eg. 6), the only value of a which will work is 1.

This might not qualify as “stateless”, but I don’t think that there is any strictly stateless solution; for example, you might look for some function f such that:

f(0), f(1),... f(m-1)

is a permutation of

0, 1, ..., m-1

so that you could generate the cycle by calling f(i) for successive values of i. But that’s still a state, since you have to remember the last value of i you used,