Determining if an input undirected graph is Hamiltonian, i.e., if it
has a cycle that visits every vertex exactly once, is one of the most
famous NP-complete problems. We consider the following generalization of
Hamiltonian cycles: for a fixed set \(S\) of natural numbers, we want to visit
each vertex of a graph \(G\) exactly
once and ensure that any two consecutive vertices can be joined in \(k\) hops for some choice of \(k \in S\). Formally, an \(S\)-Hamiltonian cycle is a permutation
\((v_0,\ldots,v_{n-1})\) of the
vertices of \(G\) such that, for \(0 \leq i \leq n-1\), there exists a walk
between \(v_i\) and \(v_{i+1 \bmod n}\) whose length is in \(S\). (We do not impose any constraints on
how many times vertices can be visited as intermediate vertices of
walks.) Of course Hamiltonian cycles in the standard sense correspond to
\(S=\{1\}\). We study the \(S\)-Hamiltonian cycle problem of deciding
whether an input graph \(G\) has an
\(S\)-Hamiltonian cycle. Our goal is to
determine the complexity of this problem depending on the fixed set
\(S\). It is already known that the
problem remains NP-complete for \(S=\{1,2\}\), whereas it is trivial for
\(S=\{1,2,3\}\) because any connected
graph contains a \(\{1,2,3\}\)-Hamiltonian cycle. Our work
classifies the complexity of this problem for most kinds of sets \(S\), with the key new results being the
following: we have NP-completeness for \(S =
\{2\}\) and for \(S = \{2,
4\}\), but tractability for \(S = \{1,
2, 4\}\), for \(S = \{2, 4,
6\}\), for any superset of these two tractable cases, and for
\(S\) the infinite set of all odd
integers. The remaining open cases are the non-singleton finite sets of
odd integers, in particular \(S = \{1,
3\}\). Beyond cycles, we also discuss the complexity of finding
\(S\)-Hamiltonian paths, and show that
our problems are all tractable on graphs of bounded cliquewidth.