Professor Armstrong suggests the following procedure for generating a uniform random permutation:
PERMUTE-BY-CYCLIC (A)
1 n ← length [A]
2 offset ← RANDOM (1, n)
3 for i ← 1 to n
4 do dest ← i + offset
5 if dest > n
6 then dest ← dest -n
7 B[dest] ← A[i]
8 return B
Show that each element A[i] has a 1/n probability of winding up in any particular position in B. Then show that Professor Armstrong is mistaken by showing that the resulting permutation is not uniformly random.