Complexity of digraph homomorphism to an oriented cycle

Question

Given a fixed directed graph (digraph) $D$, the $D$-COLORING decision problem asks whether an input digraph $G$ has a homomorphism to $D$. (A homomorphism of $G$ to $D$ is a mapping $f$ of $V(G)$ to $V(D)$ that preserves the arcs, that is, if $uv$ is an arc of $G$, then $f(u)f(v)$ is an arc of $D$.)

The class of $D$-COLORING problems is strongly connected to the Dichotomy Conjecture for CSPs stated by Feder and Vardi (accessible on citeseer).

In this 2001 paper (accessible on the author's page, here), Feder proves a dichotomy theorem when $D$ is an oriented cycle (by oriented cycle I mean an undirected cycle where each edge is replaced by a single arc, that can be oriented arbitrarily), in other words, he shows that for any oriented cycle $D$, $D$-COLORING is either polynomial-time solvable or NP-complete.

Unfortunately, Feder's classification is highly nontrivial and not explicit, as the complexity of many cases is related to the complexity of certain restricted variants of SAT that depend on the orientation. By looking at the paper, I have not been able to determine the answer to my question:

Question: What is the smallest size of an oriented cycle $D$ such that $D$-COLORING is NP-complete?

The answer might be stated somewhere in the literature, but I could not find it.

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Edit: Let me give more details on Feder's classification. Feder shows that any NP-complete oriented cycle must be balanced, that is, have the same number of arcs in both directions (hence it has even order). Then, consider the "levels" induced by the orientation (start to go around the cycle at an arbitrary vertex; if an arc goes right, you go up by 1, if an arc goes left, you go down by 1). Then, if there is at most one "top-bottom run", it is polynomial. If there are at least 3 such "runs" and the cycle is a core, it is NP-complete. (In András' example from the comments, there are three such "runs", but the cycle is not a core.) The most tricky cases are those with two "top-bottom runs". Some are hard, some polynomial, and Feder relates them to special SAT problems to obtain a dichotomy. I guess one could make a case-by-case analysis and see what kind of SAT problem it corresponds to...

As an intermediate question: What is the smallest oriented cycle that has three "top-bottom" runs and is a core? Such an example would be NP-complete by the above discussion.

Answer 1

There recently has been some activity in this area. We wrote a program to determine the smallest NP-complete oriented trees, the smallest ones have 20 vertices. More details can be found here https://arxiv.org/abs/2205.07528.

I used the same program to find the smallest NP-complete oriented cycle. It has 26 vertices and it is also the smallest oriented cycle without a majority polymorphism:

There is only one other NP-hard oriented cycle with 26 vertices and it is just the above one with all edges reversed.

Could you verify with the top-bottom run (not sure what the definition is) criterion that this cycle is NP-hard?

Answer 2

For the intermediate question (a core with three top-bottom runs), how about this?

Some notation: I will be describing runs by words in $\{l,r\}^*$, with e.g. $llrl$ corresponding to a subgraph $\cdot\leftarrow\cdot\leftarrow\cdot\to\cdot\leftarrow\cdot$. The level increases on $r$ arcs and decreases on $l$ arcs, and I assume that its minimum is $0$. Some straightforward constraints are:

• There cannot be a run consisting only of $l$s or only of $r$s, because otherwise there is an obvious homomorphism from $D$ to this run (mapping each node of $D$ to the one with the same level). This also implies that the maximum level must be at least $3$.
• If the maximum level were $3$, then all top-bottom (resp bottom-top) runs would be of the form $llr(lr)^ill$ (resp. $rrl(rl)^irr)$; again it is not very hard to find a homomorphism from $D$ to the run which minimizes $i$.

However, for maximum level $4$ there is a solution, of length $36$: consider $D$ given by $(rrrlrrlllrll)^3$. This has the required top-bottom runs and is a core (see below). By the above constraints, it is necessarily minimal, since each run only has a single "backwards" edge.

To convince ourselves that this is a core, let's first name the vertices ($v_1,\ldots,v_{36}$). The bottom (i.e. level $0$) vertices are $v_1,v_{13},v_{25}$. Any homomorphism $\varphi$ from $D$ to a subgraph must preserve levels, and in particular $\varphi(v_1)\in\{v_1,v_{13},v_{25}\}$; modulo the obvious automorphism $v_i\mapsto v_{i+12}$, it is enough to consider the case $\varphi(v_1)=v_1$. Consider the neighbourhood of $v_1$ in $D$ (annotated with levels):

$v_{34}(1)\to v_{35}(2)\leftarrow v_{36}(1)\leftarrow v_1(0)\to v_2(1)\to v_3(2)\to v_4(3)\leftarrow v_5(2)\to v_6(3)\to v_7(4)$

Starting with $\varphi(v_1)=v_1$, we have $\varphi(v_2)\in\{v_{36},v_2\}$. But if $\varphi(v_2)=v_{36}$, then $\varphi(v_3)=v_{35}$, and we have no possible value for $\varphi(v_4)$. We get $\varphi(v_2)=v_2,\varphi(v_3)=v_3,\varphi(v_4)=v_4$. Next $\varphi(v_5)\in\{v_3,v_5\}$, but for $\varphi(v_5)=v_3$ we get $\varphi(v_6)=v_4$, with no possible value for $\varphi(v_7)$. So $\varphi$ must be the identity on the entire run $v_1\to\ldots\to v_7$, and by repeating the same argument for the remaining runs, the same is true on all of $D$. In particular, $\varphi$ does not map $D$ onto a proper subgraph.