What is the best way to find the period of a (repeating) list in Mathematica?

Question

What is the best way to find the period in a repeating list?

For example:

a = {4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2}

has repeat

Answer 1

I don't know how to solve it in mathematica, but the following algorithm (written in python) should work. It's O(n) so speed should be no concern.

def period(array):
    if len(array) == 0:
        return False
    else:
        s = array[0]
        match = False
        end = 0
        i = 0

        for k in range(1,len(array)):
            c = array[k]

            if not match:
                if c == s:
                    i = 1
                    match = True
                    end = k
            else:
                if not c == array[i]:
                    match = False
                i += 1

        if match:
            return array[:end]
        else:
            return False

# False         
print(period([4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2,1]))
# [4, 5, 1, 2, 3]            
print(period([4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2]))
# False
print(period([4]))
# [4, 2]
print(period([4,2,4]))
# False
print(period([4,2,1]))
# False
print(period([]))