(This problem is an investigation on a computer.)

• Create an identity matrix (for example, in Octave, I=eye(5)) and a permutation vector (example p=[3 4 1 2 5])
• Create a permutation matrix by permuting the columns (example P=I(:,p))
• Compute matrix powers (P,P^2,P^3,P^4,...)
• Which is the smallest positive power that returns P to the identity ?
• Find five different p's, each with the property that P^k = I, for k = 1,2,3,4,5 and k is the smallest such value
• Hint: when k = 1, the answer is p=[1 2 3 4 5]. When k=5, the answer is p=[2 3 4 5 1]. Now you should find the answers when k=2,3, and 4

== Solution ==

I am sure there exists better solutions.  Anyhow, here is one solved via Python and Octave:

First I generated the 120 permutations via Python, and save the result to a file, permute_5.txt
import itertools
for i in itertools.permutations([1,2,3,4,5]):
    print ' '.join(map(str, i))
Next is the Octave code to find the answers based on the permutation file.
I know the max value of k is 6 via experimentation.
[a b c d e] = textread('permute_5.txt','%d %d %d %d %d');
I = eye(5);
distinct_ks=[0 0 0 0 0 0];
for i = 1:size(a)(1)
    p=[a(i) b(i) c(i) d(i) e(i)];
    P=I(:,p);
    k=1;
    while (!all(all(I == P^k)))
        k = k + 1;
    endwhile
    if distinct_ks(k)
        k
        p
    endif
    distinct_ks(k) = distinct_ks(k) + 1;
endfor
distinct_ks
Here is a summary of the interesting results:
P^1 == I when p = [1 2 3 4 5]
P^2 == I when p = [1 2 3 5 4]
P^3 == I when p = [1 2 4 5 3]
P^4 == I when p = [1 3 4 5 2]
P^5 == I when p = [2 3 4 5 1]
P^6 == I when p = [2 1 4 5 3]
Distributions:
k       : 1    2    3    4    5    6
P^k == I: 1   25   20   30   24   20

# posted by rot13(Unafba Pune) @ 9:35 PM