TITLE: Exact coverability of $\mathbb{Z}_n$ by cyclic shifts of a given set -- easy? NP-complete?
QUESTION [14 upvotes]: Recently Ernest Davis asked me about the following computational problem: we're given as input a composite integer $n$, a divisor $k$ of $n$, and a subset $S \subset \mathbb{Z}_n$ of size k.  The problem is to decide whether $\mathbb{Z}_n$ can be covered with $n/k$ cyclic translations of $S$, i.e. sets of the form $S+a_i$ for various $a_i \in \mathbb{Z}_n$.  This is simply a special case of the NP-complete EXACT COVER problem---namely, where the available sets are all cyclic shifts of each other, and where all cyclic shifts are available.  My suspicion is that the special case is already NP-complete, while Ernest suspects that an $n^{O(1)}$ time algorithm exists.  I searched Google (and Garey&Johnson) and couldn't find leads -- would appreciate any thoughts or references!

REPLY [4 votes]: E.D., I think the conjecture is not true. If one tries it on the smallest non-periodic tiling (see below) it fails. The example is from N.G. de Bruijn, On the factorization of cyclic groups, Indag. Math. 15, 4, 1953.
The following Python program
# A + B is a tiling of Z mod 72
A = [0, 8, 16, 18, 26, 34]
B = [18, 54, 24, 60, 48, 12, 17, 41, 65, 45, 69, 21]
m = len(A)*len(B)


# Cheking tiling
S = m*[0]
for x in A:
    for y in B:
        S[(x+y) % m] += 1
if S == m*[1]: print("Yes, they tile")

L = [x % len(B) for x in B] # The elements of B mod len(B)

H = len(L)*[0] # Compute the histogram of L
for x in L: H[x] += 1

print(f"Frequencies of B mod {len(B)}: {H}")

prints
Yes, they tile
Frequencies of B mod 12: [4, 0, 0, 0, 0, 3, 2, 0, 0, 3, 0, 0]

I also believe the problem is polynomial. I do not even know if it is in co-NP (are there certificates for non-tiling?).