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

Does this work for you?

period[a_] := 
   Quiet[Check[
      First[Cases[
         Table[
            {k, Equal @@ Partition[a, k]}, 
            {k, Floor[Length[a]/2]}], 
         {k_, True} :> k
         ]], 
      $Failed]]

Strictly speaking, this will fail for things like

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

although this can be fixed by using something like:

(Equal @@ Partition[a, k]) && (Equal @@ Partition[Reverse[a], k])

(probably computing Reverse[a] just once ahead of time.)

Answer 2

I propose this. It borrows from both Verbeia and Brett's answers.

Do[
  If[MatchQ @@ Equal @@ Partition[#, i, i, 1, _], Return @@ i],
  {i, #[[ 2 ;; Floor[Length@#/2] ]] ~Position~ First@#}
] /. Null -> $Failed &

It is not quite as efficient as Vebeia's function on long periods, but it is faster on short ones, and it is simpler as well.

Answer 3

#include <iostream>
#include <vector>
using namespace std;


int period(vector<int> v)
{
    int p=0; // period 0

    for(int i=p+1; i<v.size(); i++)
    {

        if(v[i] == v[0])
        {
            p=i; // new potential period


            bool periodical=true;
            for(int i=0; i<v.size()-p; i++)
            {
                if(v[i]!=v[i+p])
                {
                    periodical=false;
                    break;
                }
            }
            if(periodical) return p;


            i=p; // try to find new period
        }
    }

    return 0; // no period

}


int main()
{

    vector<int> v3{1,2,3,1,2,3,1,2,3};
    cout<<"Period is :\t"<<period(v3)<<endl;


    vector<int> v0{1,2,3,1,2,3,1,9,6};
    cout<<"Period is :\t"<<period(v0)<<endl;

    vector<int> v1{1,2,1,1,7,1,2,1,1,7,1,2,1,1};
    cout<<"Period is :\t"<<period(v1)<<endl;
    return 0;
}