I'm trying to figure out the time-complexity of the problem I describe below, which I call the semi-ordered Eulerian tour problem or the SOET problem. Either finding an efficient algorithm for this problem or proving that this is actually in NP-hard would settle this question for me. Below I describe the problem, give two examples of instances to the problem and explain what I have realized so far. To be precise, my questions are:


* What is the time-complexity of this problem?
* Is this problem NP-hard?
* If it exists, what is an algorithm that solves this problem in polynomial time in terms of both $n$ and $k$? (see below)


**The SOET problem**


Given a 4-regular graph $G=(V,E)$, i.e. a graph where each vertex have degree 4, the problem is to find a semi-ordered Eulerian tour (defined below). Note that the graph is not assumed to be simple, so multi-edges and self-loops are allowed.


*Definition (semi-ordered Eulerian tour):*
Let $S\subseteq V$ be a subset of the vertices in $G$ and $s=s\_1s\_2\dots s\_k$ be a permutation of $S$. The Eulerian tour $U$ is a semi-ordered Eulerian tour with respect to $S$ if, $U=As\_1Bs\_2\dots s\_kXs\_1Ys\_2\dots s\_kZ$ for some $s$ and where $A,B\dots,X,Y,\dots,Z$ are words with letters in $V\setminus S$. That is, $U$ traverses $s$ in some order once and then again in the same order. To be clear, which specific $s$ that satisfies this is unimportant.


Note that since $G$ is 4-regular an Eulerian tour always exists and furthermore it will pass each vertex exactly twice.


**Examples**


Here I give two examples of 4-regular graphs. The first one, [G\_1](https://i.stack.imgur.com/te1yc.jpg), there is a SOET with respect to $\{a,b,c,d\}$, since there is a Eulerian tour $U=abcdaebced$. In the second example, [G\_2](https://i.stack.imgur.com/y1SYR.jpg), there is none.


**Current status**


I have realized that this problem is at least fixed-parameter tractable in terms of $k$, i.e. there is an algorithm with time-complexity $\mathcal{O}(f(k)p(n))$, where $k=|S|$, $n=|V|$ and $p$ is some polynomial. This can be seen by mapping the problem to $k!k^3$ edge-disjoint paths problems (DPP) where you try to find paths between pairs of vertices in $S$. This has to be done for all permutations of the vertices in $S$, the reason for the factor $k!$ in the number of DPP:s. Furthermore each vertex has to be split into two vertices and connected by edges in one out of three ways (vertically, horizontally or diagonally), to actually give a Eulerian tour. This are in total another $3^k$ DPP:s. Note that this does not necessarily exclude the possibility of an algorithm which is also polynomial in $k$.


*Note regarding duplicate:*


*I have now also asked [the same question](https://cstheory.stackexchange.com/questions/39307/complexity-of-finding-semi-ordered-eulerian-tours-in-a-4-regular-graph) on Theoretical Computer Science since after two weeks of no response here I thought that this might be the better place for this question.*



Complexity of finding semi-ordered Eulerian tours in a 4-regular graph
The SOET problem is NP-Complete as we show in the following paper:


Axel Dahlberg, Jonas Helsen, Stephanie Wehner, **How to transform graph states using single-qubit operations: computational complexity and algorithms**, 
 [CoRR abs/1805.05306 (2018)](https://arxiv.org/abs/1805.05306).


See Corollary 3.4.1


