A characterization of graphs with regular distance-$2$ graphs

For non-negative integers~$k$, we consider graphs in which every vertex has exactly $k$ vertices at distance~$2$, i.e., graphs whose distance-$2$ graphs are $k$-regular. We call such graphs $k$-metamour-regular motivated by the terminology in polyamory. While constructing $k$-metamour-regular graphs is relatively easy -- we provide a generic construction for arbitrary~$k$ -- finding all such graphs is much more challenging. We show that only $k$-metamour-regular graphs with a certain property cannot be built with this construction. Moreover, we derive a complete characterization of $k$-metamour-regular graphs for each $k=0$, $k=1$ and $k=2$. In particular, a connected graph with~$n$ vertices is $2$-metamour-regular if and only if $n\ge5$ and the graph is a join of complements of cycles (equivalently every vertex has degree~$n-3$), a cycle, or one of $17$ exceptional graphs with $n\le8$. Moreover, a characterization of graphs in which every vertex has at most one metamour is acquired. Each characterization is accompanied by an investigation of the corresponding counting sequence of unlabeled graphs.


Introduction
For a given graph, let us construct its distance-2 graph as follows: It has the same vertices as the original graph, and there is an edge between two vertices if these vertices are at distance 2 in the original graph. Here, distance 2 means that such two vertices are different, not adjacent, and have a common neighbor. An example is shown in Figure 1.1. Distance-2 graphs and properties of vertices at distance 2 have been heavily studied in the literature; see the survey in Section 1.2. We pursue the theme of characterizing all graphs whose distance-2 graphs are in a given graph class. We specifically set our focus on the graph class of regular graphs.
The research is motivated by the relationship concept polyamory, 1 where every person might be in a relationship with any number of other persons. Naturally, this can be modeled as a graph, where each vertex represents a person, and two vertices are adjacent if the corresponding persons are in a relationship. In polyamory, two persons that are both in a relationship with the same third person, but are not in a relationship with each other, are called metamours. 2 We adopt this terminology and use the term metamour of a vertex for a vertex at distance 2. Correspondingly, we call the distance-2 graph of a graph G, the metamour graph of G.
We have discussed the context and set-up the necessary vocabulary, so we are now ready to talk about the content and results of this article. We investigate graphs where each vertex has the same number of metamours. If this number is k, we say that the graph is k-metamour-regular. Reformulated, a k-metamour-regular graph is a graph whose distance-2 graph is k-regular. The leftmost connected component of the graph in Figure 1.1 shows an example of a 2-metamour-regular (sub)graph with six vertices. We ask:

Question. Can we find all k-metamour-regular graphs and give a precise description of how they look like?
Certainly not every graph satisfies this property, but some do. For example, for k = 2 it is not hard to check that in (connected) graphs with at least five vertices, where • every vertex has two neighbors, i.e., cyclic graphs, or • every vertex is adjacent to every other vertex but two, i.e., complements of cyclic graphs, each vertex indeed has exactly two metamours. Hence, these graphs are 2-metamour-regular.
The second construction above can be generalized, and in this article we provide a generic construction that allows to create k-metamour-regular graphs for any number k. One of our key results is that the vast majority of these graphs can indeed be built by this generic construction. To be more precise, only k-metamour-regular graphs whose metamour graph consists of at most two connected components cannot necessarily be constructed this way.
This key result lays the foundations for another main result of this article, namely the identification of all 2-metamour-regular graphs, so we answer the question above for k = 2. Our findings are as follows: Every 2-metamour-regular graph of any size falls either into one of the two groups (cyclic or complements of cyclic graphs) above or into the third group of • 17 exceptional graphs with at least six and at most eight vertices.
We provide a systematic and explicit description of the graphs in the first two groups. All 2-metamourregular graphs with at most nine vertices-this includes the 17 exceptional graphs of the third group-are shown in Figures 3.2 to 3.6. Summarized, we present a complete characterization of all 2-metamour-regular graphs. Note that as a consequence, exceptional graphs exist only for up to eight vertices.
In addition to the above result we derive several structural properties of k-metamour-regular graphs for any number k. We also characterize all graphs in which every vertex has no metamour (k = 0), exactly one metamour (k = 1), and at most one metamour. As a byproduct of every characterization including the one for k = 2, we are able to count the number of graphs with these properties. Moreover-and this might be the one sentence take-away message of this article-our findings imply that besides the graphs that are simple to discover (i.e., can be built by the generic construction), only a few (if any) small exceptional graphs are 0-metamour-regular, 1-metamour-regular, graphs where every vertex has most one metamour and 2-metamour-regular.

Outline
We now provide a short overview on the structure of this paper. The terms discussed so far are formally defined in Section 2. Moreover, there we introduce joins of graphs which are used in the systematic and explicit description of the graphs in the characterizations of metamour-regular graphs. The section also includes some basic properties related to those concepts. Section 3 is a collection of all results derived in this article. This is accompanied by plenty of consequences of these results and discussions. The proofs of all results are given in Sections 4 to 9. We conclude in Section 10 and provide many questions, challenges and open problems for future work.
So far not mentioned is the next section. There, literature related to this article is discussed.

Related literature
In this section we discuss concepts that have already been examined in literature and that are related to metamours (i.e., vertices having distance 2, or "neighbors of neighbors") and the metamour graphs induced by their relations. Metamour graphs are called distance-2 graphs in Iqbal, Koolen, Park and Rehman [18], and some authors also call them 2-distance graphs (e.g., Azimi and Farrokhi [3]) or 2nd distance graphs (e.g., Simić [25]). This notion also appears in the book by Brouwer, Cohen and Neumaier [9, p. 437]. The overall question is to characterize all graphs whose distance-n graph equals some graph of a given graph class. Simić [25] answers the question when the distance-n graph of a graph equals the line graph of this graph. Characterizing when the distance-2 graph is a path or a cycle is done by Azimi and Farrokhi [3], and when it is a union of short paths or and a union of two complete graphs by Ching and Garces [12]. Azimi and Farrokhi [4] also tackled the question when the distance-2 graph of a graph equals the graph itself. This question is also topic of the online discussion [28]. Bringing the context to our article, we investigate the above question for the graph class of regular graphs.
Moreover, vertices in a graph that have distance two, i.e., metamours, or more generally vertices that have a given specific distance, are discussed in the existing literature in many different contexts. The persons participating in the exchange [11] discuss algorithms for efficiently finding vertices having specific distance on trees. Moreover, the notion of dominating sets is extended to vertices at specific distances in Zelinka [31] and in particular to distance two in Kiser and Haynes [20].
Also various kinds of colorings of graphs with respect to vertices of given distance are studied. Typically, the corresponding chromatic number is analyzed, for instance Bonamy, Lévêque and Pinlou [5], Borodin, Ivanova and Neustroeva [7], and Bu and Wang [10] provide such results for vertices at distance two. Algorithms for finding such colorings are also investigated. We mention here Bozdağ, Çatalyürek, Gebremedhin, Manne, Boman and Özgüner [8] as an example. Kamga, Wang, Wang and Chen [19] study variants of so-called vertex distinguishing colorings, i.e., edge colorings where additionally vertices at distance two have distinct sets of colors. Their motivation comes from

Graph-theoretic definitions, notation & conventions
In this graph-theoretic article we use standard graph-theoretic definitions and notation; see for example Diestel [13]. We use the convention that all graphs in this article contain at least one vertex, i.e., we do not talk about the empty graph. Moreover, we use the following convention for the sake of convenience.
Convention 2.1. If two graphs are isomorphic, we will call them equal and use the equality-sign.
In many places it is convenient to extend adjacency to subsets of vertices and subgraphs. We give the following definition that is used heavily in Section 3.5 and Section 5.

Definition 2.2.
Let G be a graph, and let W 1 and W 2 be disjoint subsets of the vertices of G.
• We say that W 1 is adjacent in G to W 2 if there is a vertex v 1 ∈ W 1 adjacent in G to some vertex v 2 ∈ W 2 .
• We say that W 1 is completely adjacent in G to W 2 if every vertex v 1 ∈ W 1 is adjacent in G to every vertex v 2 ∈ W 2 .
By identifying a vertex v ∈ V (G) with the subset {v} ⊆ V (G), we may also use (complete) adjacency between v and a subset of V (G). Moreover, for simplicity, whenever we say that subgraphs of G are (completely) adjacent, we mean that the underlying vertex sets are (completely) adjacent.
We explicitly state the negation of adjacent: We say that W 1 is not adjacent in G to W 2 if no vertex v 1 ∈ W 1 is adjacent in G to any vertex v 2 ∈ W 2 . We will not need the negation of completely adjacent.
We recall the following standard concepts and terminology to fix their notation.
• For a set W ⊆ V (G) of vertices of a graph G, the induced subgraph G[W ] is the subgraph of G with vertices W and all edges of G that are subsets of W , i.e., edges incident only to vertices of W .
• A set µ ⊆ E(G) of edges of a graph G is called matching if no vertex of G is incident to more than one edge in µ. In particular, the empty set is a matching. The set µ is called perfect matching if every vertex of G is incident to exactly one edge in µ.
• For a set ν ⊆ E(G) of edges of a graph G, we denote by G−ν the graph with vertices V (G−ν) = V (G) and edges E( • The union of graphs G 1 and G 2 , written as • A path π of length n in a graph G is written as π = (v 1 , . . . , v n ) for vertices v i ∈ V (G) that are pairwise distinct.
• A cycle γ of length n or n-cycle γ in a graph G is written as • For a graph G, we write C(G) for the set of connected components of G. Note that each element of C(G) is a subgraph of G.
• We write G for the complement of a graph G, i.e., the graph with the same vertices as G but with an edge between vertices exactly where G has no edge.
• We use the complete graph K n for n ≥ 1, the complete t-partite graph K n 1 ,...,nt for n i ≥ 1, i ∈ {1, . . . , t}, the path graph P n for n ≥ 1, and the cycle graph C n for n ≥ 3.
Remark 2.3. We will frequently use the complement C n of the cycle graph C n for n ≥ 3. Note that • C 3 is the graph with 3 isolated vertices, • C 4 is the graph with 2 disjoint single edges, and • C 5 equals C 5 (see also Figure 3.2).
We close this section and continue with definitions and concepts that are specific for this article.

Metamours
We now formally define the most fundamental concept of this article, namely metamours.

Definition 2.4.
Let G be a graph.
• A vertex v of the graph G is a metamour of a vertex w of G if the distance of v and w on the graph G equals 2.
• The metamour graph M of G is the graph with the same vertex set as G and an edge between the vertices v and w of M whenever v is a metamour of w in G.
We can slightly reformulate the definition of metamours: A vertex v having a different vertex w as metamour, i.e., having distance 2 on a graph, is equivalent to saying that v and w are not adjacent and there is a vertex u such that both v and w have an edge incident to this vertex u, i.e., u is a common neighbor of v and w.
Clearly, there is no edge in a graph between two vertices that are metamours of each other. This is reflected in the relation between the metamour graph and the complement of a graph, and put into writing as the following observation.
Observation 2.5. Let G be a graph. Then the metamour graph of G is a subgraph of the complement of G.
The question whether the metamour graph equals the complement will appear in many statements of this article. The following simply equivalence is useful. Proposition 2.6. Let G be a connected graph with n vertices. Then the following statements are equivalent: (a) The metamour graph of G equals G.
(b) The graph G has diameter 2 or G = K n .

Metamour-degree & metamour-regularity
Having the concept of metamours, it is natural to investigate the number of metamours of a vertex. We formally define this "degree" and related concepts below. Definition 2.7. Let G be a graph.
• The metamour-degree of a vertex of G is the number of metamours of this vertex.
• The maximum metamour-degree of the graph G is the maximum over the metamour-degrees of its vertices.
• For k ≥ 0 the graph G is called k-metamour-regular if every vertex of G has metamour-degree k, i.e., has exactly k metamours.
We finally have k-metamour-regularity at hand and can now start to relate it to other existing terms. We begin with the following two observations. Observation 2.8. Let k ≥ 0, and let G be a graph and M its metamour graph. Then G is kmetamour-regular if and only if the metamour graph M is k-regular.
The number of vertices with odd degree is even by the handshaking lemma. Therefore, we get the following observation. (b) For every vertex of G, the sum of its degree and its metamour-degree equals n − 1.
Note that if k ≥ 0 and G is a connected k-metamour-regular graph with n vertices, then (b) states that the graph G is (n − 1 − k)-regular. We use this in Proposition 2.13.

Joins of graphs
Given two graphs, we already have defined the union of these graphs in Section 2.1. A join of graphs is a variant of that. We will introduce this concept now, see also Harary [14, p. 21], and then discuss a couple of simple properties of joins, also in conjunction with metamour graphs. Definition 2.11. Let G 1 and G 2 be graphs with disjoint vertex sets V (G 1 ) and V (G 2 ). The join of G 1 and G 2 is the graph denoted by G 1 ∇ G 2 with vertices V (G 1 ) ∪ V (G 2 ) and edges Some graphs in Figures 3.3 to 3.6 are joins of complements of cycle graphs. All of the joins of graphs in this paper are "disjoint joins". We use the convention that if the vertex sets V (G 1 ) and V (G 2 ) are not disjoint, then we make them disjoint before the join. We point out that the operator ∇ is associative and commutative.
Let us get to know joins of graphs in form of a supplement to Remark 2.3. We have for the complete multipartite graph K 3,3,...,3 .
There are connections between joins of graphs and metamour graphs that will appear frequently in the statements and results of this article. We now present first such relations.
(c) There are graphs G 1 and Proposition 2.13. Let k ≥ 0 and G be a connected k-metamour-regular graph with n vertices. Let M be the metamour graph of G. Then the following statements are equivalent: (a) The metamour graph M equals G.
Note that we have G = K n in (b) if and only if k = 0.

Characterizations & properties of metamour-regular graphs
It is now time to present the main results of this article and their implications. In this section, we will do this in a formal manner using the terminology introduced in Section 2. This section also includes brief sketches of the proofs of the main results. The actual and complete proofs of the results follow later, from Section 5 on to Section 9. Proof-wise the results on k-metamour-regular graphs for k ∈ {0, 1, 2} build upon the result for arbitrary k ≥ 0; this determines the order of the sections containing the proofs. We will in this section, however, start with k = 0, followed by k = 1 and k = 2 and only deal with general k later on.

0-metamour-regular graphs
As a warm-up, we start with graphs in which no vertex has a metamour. The following theorem is not very surprising; the only graphs satisfying this property are complete graphs. An alternative point of view is that of the metamour graph. The theorem simply implies that in the case of 0-metamour-regularity, the metamour graph is empty and also equals the complement of the graph itself. The latter property will occur frequently later on which also motivates its formulation in the following corollary. The characterization provided by Theorem 3.1 makes it also easy to count how many different 0-metamour-regular graphs there are and leads to the following corollary. The Euler transform, see Sloane and Plouffe [26], of this sequence gives the numbers m =0 (n) of unlabeled but not necessarily connected 0-metamour-regular graphs with n vertices. The number m =0 (n) equals the partition function p(n), i.e., the number of integer partitions 3 of n. The corresponding sequence starts with n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 m =0 (n) 1 2 3 5 7 11 15 22 30 42 56 77 101 135 176 and is A000041 in The On-Line Encyclopedia of Integer Sequences [23].
This completes the properties of 0-metamour-regular graphs that we bring here. We will, however, see in the following sections how these properties behave in context of other graph classes.

1-metamour-regular graphs
The next easiest case is that of graphs in which every vertex has exactly one other vertex as metamour. As the metamour relation is symmetric, these vertices always come in pairs. We write this fact down in the following proposition. Proposition 3.4. Let G be a graph with n vertices. Then the following statements hold: (a) If every vertex of G has at most one metamour, then the edges of the metamour graph of G form a matching, i.e., the vertices of G having exactly one metamour come in pairs such that the two vertices of a pair are metamours of each other.
(b) If G is 1-metamour-regular, then n is even and the edges of the metamour graph of G form a perfect matching.
By this connection of 1-metamour-regular graphs to perfect matchings, we can divine the underlying behavior. This leads to our main result of this section, a characterization of 1-metamour-regular graphs; see the theorem below. It turns out that one exceptional case, namely the graph P 4 (Figure 3.1), occurs. When excluding G = P 4 , then the graphs in the theorem are exactly the cocktail party graphs [30]. Let us again view this from the angle of metamour graphs. As soon as we exclude the exceptional case P 4 , the metamour graph and the complement of a 1-metamour-regular graph coincide; see the following corollary. Corollary 3.6. A connected graph with n ≥ 5 vertices is 1-metamour-regular if and only if its complement equals its metamour graph and this graph is 1-regular.
Note that a 1-regular graph with n vertices is a graph induced by a perfect matching of K n . In view of Proposition 2.13, we can extend the two equivalent statements.
As we have a characterization of 1-metamour-regular graphs (provided by Theorem 3.5) available, we can determine the number of different graphs in this class. Clearly, this is strongly related to the existence of a perfect matching; details are provided below and also in Section 7, where proofs are given. The Euler transform, see [26], of the sequence of numbers m =1 (2n) gives the numbers m =1 (2n) of unlabeled but not necessarily connected 1-metamour-regular graphs with 2n vertices. The sequence of these numbers starts with This sequence also counts how often a part 2 appears in all integer partitions 4 of n + 2 with parts at least 2. The underlying bijection is formulated as the following corollary.
Corollary 3.8. Let n ≥ 0. Then the set of unlabeled 1-metamour-regular graphs with 2n vertices is in bijective correspondence to the set of partitions of n + 2 with smallest part equal to 2 and one part 2 of each partition marked.

Graphs with maximum metamour-degree 1
Let us now slightly relax the metamour-regularity condition and consider graphs in which every vertex of G has at most one metamour. In view of Proposition 3.4, matchings play an important role again. Formally, the following theorem holds. As in the sections above, the obtained characterization leads to the following equivalent statements with respect to metamour graph and complement. Corollary 3.10. A connected graph with n ≥ 5 vertices has the property that every vertex has at most one metamour if and only if its complement equals its metamour graph and this graph has maximum degree 1.
Note that graphs with maximum degree 1 and n vertices are graphs induced by a (possibly empty) matching of K n . In view of Proposition 2.6, we can extend the two equivalent statements by a third saying that G has diameter 2 or G = K n .
Counting the graphs with maximum metamour-degree 1 relies on the number of matchings; see the relevant proofs in Section 8 for details. We obtain the following corollary. and we have m ≤1 (n) = n 2 + 1 for n ≥ 5.
The Euler transform, see [26], gives the sequence of numbers m ≤1 (n) of unlabeled but not necessarily connected graphs with maximum metamour-degree 1 and n vertices which starts with

2-metamour-regular graphs
We now come to the most interesting graphs in this article, namely graphs in which every vertex has exactly two other vertices as metamours. Also in this case a characterization of the class of graphs is possible. We first consider Observation 2.8 in view of 2-metamour-regularity. As a graph is 2-regular if and only if it is a union of cycles, the following observation is easy to verify. We are ready to fully state the mentioned characterization formally as the theorem below. We comment this result and discuss implications afterwards. Note that Theorem 3.13 generalizes the main result of Azimi and Farrokhi [3, Theorem 2.3] which only deals with metamour graphs being connected.
Theorem 3.13. Let G be a connected graph with n vertices. Then G is 2-metamour-regular if and only if n ≥ 5 and one of (a) G = C n 1 ∇ · · · ∇ C nt with n = n 1 + · · · + n t for some t ≥ 1 and n i ≥ 3 for all i ∈ {1, . . . , t}, The only graph of order 5 ( + one differently drawn copy ) where each vertex has exactly 2 metamours  A representation of every 2-metamour-regular graph with at most 9 vertices can be found in Figures 3.2, 3.3, 3.4, 3.5 and 3.6. For 10 vertices, all 2-metamour-regular graphs-there are 6 of them-are C 10 , C 10 , For rounding out Theorem 3.13, we have collected a couple of remarks and bring them now.
1. The smallest possible 2-metamour-regular graph has 5 vertices, and there is exactly one with five vertices, namely C 5 ; see Figure 3.2. This graph is covered by Theorem 3.13(a) as well as (b) because C 5 = C 5 .
2. Theorem 3.13(a) can be replaced by any other equivalent statement of Proposition 2.13. Figure 3.4: All 9 graphs of order 7 where each vertex has exactly 2 metamours  4. The graphs C n 1 , . . . , C nt of Theorem 3.13(a) satisfy . This means that the decomposition of the graph G = C n 1 ∇· · ·∇C nt reveals the metamour graph of G and vice versa.
5. For Theorem 3.13(a) as well as for (b) with n = 5, every graph satisfies that its complement equals its metamour graph. For all other cases, this is not the case. A full formulation of this fact is stated as Corollary 3.17.
The proof of Theorem 3.13 is quite extensive and we refer to Section 9 for the complete proof; at this point, we only sketch it.
Sketch of Proof of Theorem 3.13. Let G be 2-metamour-regular. We apply Theorem 3.20 (to be presented in Section 3.5) with k = 2, which leads us to one of three cases. The generic case is already covered by Theorem 3.20 combined with Observation 3.12.
If the metamour graph of G is connected, then we first rule out that G is a tree. If not, then the graph contains a cycle, and depending on whether G contains a cycle of length n or not, we get , H a 7 }, respectively. (This parts of the proof are formulated as Proposition 9.6, Lemma 9.2 and Lemma 9.1.) We prove these parts by studying the longest cycle in the graph G and step-by-step obtaining information between which vertices edges, non-edges and metamour relations need to be. Knowing enough, 2-metamour-regularity forces the graph to be bounded in the number of vertices. As this number is quite small, further case distinctions lead to the desired graphs. Furthermore, a separate investigation is needed when every vertex has degree 2 or n − 3 or a mixture of these to degrees; here n is the number of vertices of G. This leads to G = C n and n odd, G = C n and G ∈ {C 5 , H c 6 }, respectively. (These parts of the proof are formulated as Lemma 9.4, Lemma 9.3 and Lemma 9.5.) If the metamour graph of G is not connected, it consists of exactly two connected components (by Theorem 3.20), and the graph is split respecting the connected components of the metamour graph of G. Due to 2-metamour-regularity, Theorem 3.20 guarantees that the number of vertices of each piece is a most 2. Studying each configuration separately lead to the graphs G = C n and n even, The characterization provided by Theorem 3.13 has many implications. We start with the following easy corollaries. Corollary 3.15. Let G be a connected graph with n ≥ 9 vertices. Then G is 2-metamour-regular if and only if G is either C n or C n 1 ∇ · · · ∇ C nt with n = n 1 + · · · + n t for some t ≥ 1 and n i ≥ 3 for all i ∈ {1, . . . , t}. Corollary 3.16. Let G be a connected graph with n ≥ 9 vertices. Then G is 2-metamour-regular if and only if G is either 2-regular or (n − 3)-regular.
As before, we consider the relation of metamour graphs and complement more closely; see the following corollary. Again, we feel the spirit of Proposition 2.13. (c) The graph G has diameter larger than 2.
Theorem 3.13 makes it also possible to count how many different 2-metamour-regular graphs with n vertices there are. We provide this in the following corollary. The sequence of numbers m =2 (n) is A334275 in The On-Line Encyclopedia of Integer Sequences [23]. We apply the Euler transform, see Sloane and Plouffe [26], on this sequence and obtain the numbers m =2 (n) of unlabeled but not necessarily connected 2-metamour-regular graphs with n vertices. The sequence of these numbers starts with

k-metamour-regular graphs
In this section, we present results that are valid for graphs with maximum metamour-degree k and k-metamour-regular graphs for any non-negative number k. We start with Proposition 3.19 stating that the join of complements of k-regular graphs is a k-metamour-regular graph.
Then G has metamour-graph M . In particular, if M is k-regular for some k ≥ 0, then G is kmetamour-regular.
We call this construction of a graph with given metamour graph generic construction. In particular this generic construction allows us to build k-metamour-regular graphs. We will not investigate further options to construct k-metamour-regular graphs (as, for example, with circulant graphs), as the above construction suffices for the ultimate goal of this paper to characterize all k-metamour-regular graphs for k ≤ 2. Last in our discussion of Proposition 3.19, we note that, as we have G = M 1 ∇ · · · ∇ M t , we can also extend Proposition 3.19 by the equivalent statements of Propositions 2.12 and 2.13.
Next we state the main structural result about graphs with maximum metamour-degree k as Theorem 3.20. A consequence of this main statement is that every k-metamour-regular graph is the join of complements of k-regular graphs as in Proposition 3.19, or has in its metamour graph only one or two connected components; see Corollary 3.21 for a full formulation. Theorem 3.20. Let k ≥ 0. Let G be a connected graph with maximum metamour-degree k and M its metamour graph. Then exactly one of the following statements is true:

In this case we have
and t ≥ 2, and any other equivalent statement of Proposition 2.12.

(c) The metamour graph M is not connected and no induced subgraph
In this case the metamour graph M has exactly two connected components and the following holds.
, G M is the graph G after deleting every edge between two vertices that are from different connected components of the metamour graph M . Then we have: (iv) If two different connected components of G M are adjacent in G, then these connected components are completely adjacent in G.

(v) If two vertices of different connected components
The complete and extensive proof of Theorem 3.20 can be found in Section 5; we again only sketch it at this point. Theorem 3.20. Suppose the metamour graph of G is not connected. We split the graph respecting the connected components of the metamour graph of G.

Sketch of Proof of
First, we lift adjacency of some vertices of different components of this split to all vertices and likewise, metamour relations between vertices of different components. By this structural property, we show on the one hand that the maximum metamour-degree k implies that each of the components has at most k vertices. On the other hand, by investigating an alternating behavior on shortest paths between different components, we deduce that the metamour graph of G has exactly two connected components.
Before gluing everything together, two more properties need to be derived: We show that within each of the components, the non-edges correspond exactly to metamour relations, and we bound the number of neighboring components of a component. This provides enough structure completing the proof.
Let us discuss the three outcomes of Theorem 3.20 in view of the characterizations provided in Sections 3.1 to 3.4. Toward this end note that in case (b), the graph G is obtained by the generic construction.
For 0-metamour-regular graphs due to Theorem 3.1 there is no graph that is not obtained by the generic construction. So only (b) happens, except if the graph consists of only one vertex in which case a degenerated (a) happens. For 1-metamour-regular graphs we know that there is only one graph not obtained by the generic construction by Theorem 3.5. This exceptional case is associated to (c), otherwise we are in (b). Finally Theorem 3.13 states that beside the generic case associated to (b), there is only the class with graphs C n and 17 exceptional cases of 2-metamour-regular graphs associated to (a) and (c).
At last in this section, we bring and discuss the full formulation of a statement mentioned earlier.
By this corollary every k-metamour-regular graph that has at least three connected components in its metamour graph is a join of complements of k-regular graphs and therefore can be built by the generic construction. As a consequence, it is only possible that a 2-metamour-regular graph is not obtained by the generic construction if its metamour graph has at most two connected components.

Proofs regarding foundations
We start by proving Proposition 2.6 which relates the metamour graph, the complement and the diameter of a graph.
Proof of Proposition 2.6. Suppose (a) holds. Let v and w be vertices in G. If v = w, then their distance is 0. If v and w are adjacent in G, then their distance is 1. Otherwise, there is no edge {v, w} in G. Therefore, this edge is in G = M , where M is the metamour graph of G. This implies that v and w are metamours. Thus, the distance between v and w is 2. Consequently, no distance in G is larger than 2, which implies that the diameter of G is at most 2. As a result, either the diameter of G is exactly 2, or it is at most 1. If the diameter is equal to 1, then all vertices of G are pairwise adjacent and therefore G = K n . Furthermore, there are at least two vertices at distance 1 and hence n ≥ 2. If the diameter is 0, then G = K 1 . Now suppose (b) holds. If G = K n , then its diameter is either equal to 0 if n = 1 or equal to 1 if n ≥ 2. Therefore, in this case the diameter of G is at most 2. Now let {v, w} be an edge in G. Then v = w and these vertices are not adjacent in G, so their distance is at least 2. As the diameter is at most 2, the distance of v and w is at most 2. Consequently their distance is exactly 2 implying that they are metamours. So the edge {v, w} is in M , and hence the complement of G is a subgraph of the metamour graph of G. Due to Observation 2.5, the metamour graph of G is a subgraph of the complement of G, hence the metamour graph of G and G coincide, so we have shown (a).
Next we prove Proposition 2.10 which relates the metamour graph, the complement, and the degree and metamour-degree of the vertices of a graph.
Proof of Proposition 2.10. We use that for a graph G with n vertices, the sum of the degrees of a vertex in G and in the complement G always equals n − 1.
If (a) holds, then the degree of a vertex in the metamour graph equals the degree in the complement G. This yields (b) by using the statement at the beginning of this proof.
If (b) holds, then the sum of the degree and the metamour-degree of a vertex equals the sum of the degree in G and the degree in G of this vertex by the statement at the beginning of this proof. Therefore, the metamour-degree of a vertex is equal to the degree in G of this vertex. Due to Observation 2.5, the metamour graph of G is a subgraph of G, and therefore the metamour graph of G equals G.
Finally we prove Propositions 2.12 and 2.13 that relate the metamour graph and joins.
Proof of Proposition 2.12. We start by showing that (a) implies (b). In M there are no edges between its different connected components. Therefore, in the complement M = G, there are all possible edges between the vertices of different components. This is equivalent to the definition of the join of graphs; the individual graphs in C(M ) are complemented, and consequently (b) follows.
For proving that (b) implies (c), we simply set G 1 = M 1 and G 2 = M 2 ∇ · · · ∇ M t . By the definition of the operator ∇, (c) follows.
We now show that (c) implies (a). Every pair of vertices of G 1 has a common neighbor in G 2 . This implies that the vertices of this pair are metamours if and only if they are not adjacent in G 1 . The same holds for any pair of vertices of G 2 by symmetry or due to commutativity of the operator ∇. As a consequence of this and because G = G 1 ∇ G 2 and the definition of the operator ∇, the metamour graph M and the complement of G coincide. As every possible edge from a vertex of G 1 to a vertex of G 2 exists in G, this complement G has at least two connected components, hence t ≥ 2. Finally, the equivalence of (a) and (d) follows from Proposition 2.10.
At this point we have shown all results form Section 2 and therefore have laid the foundations of the subsequent results.

Proofs regarding k-metamour-regular graphs
Next we give the proofs of results from Section 3.5 concerning graphs with maximum metamourdegree k and k-metamour-regular graphs that are valid for arbitrary k ≥ 0. Summarized, we have that u is a metamour of v if and only if u is in M i and adjacent to v in M i . This yields that the metamour graph of G is M = M 1 ∪ · · · ∪ M t which was to show.
The k-metamour-regularity follows directly from Observation 2.8.
Proof of Theorem 3.20. If the metamour graph M is connected, then we are in case (a) and nothing is to show. So suppose that the metamour graph is not connected. We partition the vertices of M (and therefore the vertices of G) into two parts V V * such that the vertex set of each connected component of M is a subset of either V or V * and such that neither V nor V * is empty, i.e., we partition by the connected components C(M ) of the metamour graph M . As M is not connected, it consists of at least two connected components, and therefore such a set-partition of the vertices of M is always possible. We now split up the graph G into the two Rephrased, we obtain G and G * from G by cutting it (by deleting edges) in two, but respecting and not cutting the connected components of its metamour graph M . Note that the formulation is symmetric with respect to G and G * , therefore, we might switch the two without loss of generality during the proof. This also implies that in the statements of the following claims we may switch the roles of G and G * .
There is a path π u ,u = ( We use induction to prove that , then u 1 and v +1 are metamours. But u 1 has all its metamours in G , a contradiction. Hence, {u 1 , v +1 } ∈ E(G), which finishes the induction. In particular, we have proven that {u 1 , v s } ∈ E(G). Now we prove that {u , v s } ∈ E(G) holds for all ∈ {1, . . . , r} by induction. This holds for = 1 by the above. Now assume {u , v s } ∈ E(G). We have {u , u +1 } ∈ E(G) because this edge is a part of the path π u ,u . If {u +1 , v s } ∈ E(G), then u +1 and v s are metamours, a contradiction since every metamour of u +1 is in G . Therefore, {u +1 , v s } ∈ E(G) holds and the induction is completed. As a result, we have Proof of B. The graph G is connected, so every connected component of G * is adjacent in G to G . By A this implies that G is completely adjacent in G to G * , i.e., all possible edges between G and G * exist. Therefore, G = G ∇ G * by the definition of the join of graphs.
By Proposition 2.12, the full decomposition into the components C(M ) follows.

B
As a consequence of B, we are finished with the proof in the case that So from now on we consider the case that every G[V (M i )] with M i ∈ C(M ) has at least two connected components. This is the set-up of statement (c).

C. Suppose we are in the set-up of (c). Let
Proof of C. As both G * 1 and G * 2 are adjacent in G to G 1 , they are both completely adjacent in G to G 1 due to A. Furthermore, G * 1 is not adjacent in G to G * 2 , i.e., no vertex of G * 1 is adjacent in G to any vertex of G * 2 , because they are in different connected components of G * . Hence, every vertex of G * 1 is a metamour of every vertex of G * 2 .

C D. Suppose we are in the set-up of (c). Then every connected component of G has at most k vertices and this connected component's vertex set is a subset of the vertex set of one connected component of the metamour graph M .
Proof of D. Let G 1 ∈ C(G ). As G is not connected but the graph G is connected, there is a path π from a vertex of G 1 to some vertex in some connected component of C(G ) other than G 1 . By construction of G and G * , the path π splits from start to end into vertices of G 1 , followed by vertices of some G * 1 ∈ C(G * ), followed by some vertices of G 2 ∈ C(G ), and remaining vertices. Therefore, we have connected components G 2 and G * 1 such that at least one vertex of G 1 is connected to some vertex of G * 1 and at least one vertex of G 2 is connected to some vertex of G * 1 . Then, due to C every vertex of G 1 is a metamour of every vertex of G 2 . From this, we now deduce two statements.
First, if we assume that G 1 contains at least k + 1 vertices, then every vertex of G 2 has at least k + 1 metamours, a contradiction to k being the maximum metamour-degree of G. Therefore, G 1 contains at most k vertices.
Second, every vertex of G 1 is adjacent in the metamour graph M to every vertex of G 2 , so G 1 is completely adjacent in M to G 2 . Therefore, all these vertices are in the same connected component of the metamour graph M . In particular, this is true for the set of vertices of G 1 as claimed, and so the proof is complete. D E. Suppose we are in the set-up of (c). Let v 1 and v 2 be two vertices of different connected components of G . Then every shortest path from v 1 to v 2 in G consists of vertices alternating between G and G * .
Note that u 1 and u r are both from G but from different connected components. Hence, π consists of at least two vertices from G and at least one vertex from G * .
Assume that the vertices of π are not alternating between G and G * . Then, without loss of generality (by reversing the enumeration of the vertices in the path π), there exist indices i < j and graphs { G, G} = {G , G * } with the following properties: Every vertex of the subpath π i,j = (u i , u i+1 , . . . , u j ) of π is of G, and the vertex u j+1 exists and is in G.
As π i,j is a path, all of its vertices are of the same connected component of G. As u j is adjacent to u j+1 and due to A, the vertex u j+1 is adjacent to every vertex of π i,j , in particular, adjacent to u i . But then u 1 , . . . , u i , u j+1 , . . . , u r is a shorter path between v 1 and v 2 , a contradiction. Hence, our initial assumption was wrong, and the vertices of π are alternating between G and G * . Proof of F. The metamour graph M is not connected, therefore it has at least two connected components. Assume it has at least three components. Then, without loss of generality (by switching G and G * ), the graph G contains vertices of at least two different connected components of M . Let v 1 and v 2 be two vertices of G that are in different connected components of M . If follows from D that v 1 and v 2 are in different connected components of G . Let {G 1 , G 2 } ⊆ C(G ) such that v 1 ∈ G 1 and v 2 ∈ G 2 . Let π = (u 1 , . . . , u r ) be a shortest path from u 1 = v 1 to u r = v 2 in G. Now consider two vertices u 2i−1 and u 2i+1 for i ≥ 1 of π. Both u 2i−1 and u 2i+1 are from G because π consists of alternating vertices from G and G * due to E and u 1 is from G . If both u 2i−1 and u 2i+1 are from the same connected component of G , then they are in the same connected component of the metamour graph M by D. If u 2i−1 and u 2i+1 are in different connected components of G , then they are metamours because they have the common neighbor u 2i and they are not adjacent. Therefore, they are in the same connected component of M as well. Hence, in any case u 2i−1 and u 2i+1 are in the same connected component of M . By induction this implies that u 1 = v 1 is in the same connected component of M as u r = v 2 , a contradiction to v 1 and v 2 being from different connected components of M . Hence, our assumption was wrong and the metamour graph consists of exactly two connected components. F G. Suppose we are in the set-up of (c). Then every connected component G is k-metamour-regular, then the connected component is a regular graph. Proof of G. Let G 1 be a connected component of G . As G is connected, there is an edge from a vertex v 1 of G 1 to G * and this is extended to every vertex of G 1 by A. Therefore, two different vertices of G 1 are metamours if and only if they are not adjacent in G . Restricting this to the subgraph G 1 yields the first statement.
Furthermore, by construction of G and G * , every metamour of v 1 is in G .
Let v be such a metamour, and suppose that v is not in G 1 . The vertices v 1 and v have a common neighbor u that has to be in G * as v 1 and v are in different connected components of G . The vertex u is completely adjacent in G to G 1 because of A. Hence, v is a metamour of every vertex of G 1 . As a consequence, every vertex of G 1 has the same number of metamours outside of G 1 , i.e., in G but not in G 1 . If no such pair of vertices v 1 of G 1 and v not of G 1 that are metamours exists, then every vertex of G 1 still has the same number of metamours outside of G 1 , namely zero. Now let us assume that G is k-metamour-regular. As every vertex of G 1 has the same number of metamours outside of G 1 , this implies that every vertex of G 1 must also have the same number of metamours inside G 1 . We combine this with the results of first paragraph and conclude that every vertex of G 1 is adjacent to the same number of vertices of G 1 , and hence G 1 is a regular graph. Proof of H. Let G * ⊆ C(G * ) be such that a connected components of G * is in G * if and only if it is adjacent in G to G 1 . G * 1 consists of k − d vertices and there is some vertex v of G 1 that is adjacent to some vertex of G * 1 . We have to prove that |G * | ≤ d + 2 in order to finish the proof. So assume |G * | > d + 2. Let v * be a vertex of some connected component G * 2 ∈ G * other than G * 1 . Then v * is adjacent to v due to A. Because of C, every vertex in any connected component in G * except G * 2 is a metamour of v * . The component G * 1 contains k − d vertices and there are at least d + 1 other components each containing at least one vertex. In total, v * has at least (k − d) + (d + 1) = k + 1 metamours, a contradiction to k being the maximum metamour-degree of G. Therefore, our assumption was wrong and |G * | ≤ d + 2 holds.

H
Now we are able to collect everything we have proven so far and finish the proof of statement (c). Due to F, the metamour graph M of G consists of exactly two connected components, and consequently the connected components of G M coincide with the union of the connected components of G and G * . Then D implies (i), G implies (ii), G implies (iii), A implies (iv), C implies (v) and H implies (vi). This completes the proof.
Proof of Corollary 3.21. As the metamour graph consists of at least 3 connected components, we cannot land in the cases (a) and (c) of Theorem 3.20. But then the statement of the corollary follows from (b). Now we have proven everything we need to know about graphs with maximum metamour-degree k and k-regular-metamour graphs for general k and can use this knowledge to derive the results we need in order to characterize all k-regular-metamour graphs for k ∈ {0, 1, 2}.

Proofs regarding 0-metamour-regular graphs
We are now ready to prove all results concerning 0-metamour-regular graphs from Section 3.1. In order to do so we first need the following lemma. Proof. Let n be the number of vertices of G. Clearly the statement is true for n = 1 as no other vertices are present and for n = 2 as the two vertices are adjacent due to connectedness. So let n ≥ 3, and let v be a vertex of G that has no metamour.
Assume there is a vertex w = v ∈ V (G) such that {v, w} ∈ E(G). G has a spanning tree T because G is connected. Let v = u 1 , u 2 , . . . , u r = w be the vertices on the unique path from v to w in T . Then {u i , u i+1 } ∈ E(G) for all i ∈ {1, . . . , r − 1}, so due to our assumption r ≥ 3 holds. In particular, {u 1 , u 2 } ∈ E(G). If {u 1 , u 3 } ∈ E(G), then both u 1 and u 3 are adjacent to u 2 , but not adjacent to each other and therefore would be metamours. But Now we are able to prove Theorem 3.1 that provides a characterization of 0-metamour-regular graphs.
Proof of Theorem 3.1. If G is 0-metamour-regular then every vertex of G has no metamour and hence G = K n due to Lemma 6.1. Furthermore, K n is 0-metamour-regular as every vertex is adjacent to all other vertices.
Next we prove the corollaries which yield an alternative characterization of 0-metamour-regular graphs and allow to count 0-metamour-regular graphs.
Proof of Corollary 3.2. Due to Theorem 3.1, a connected graph with n vertices is 0-metamour-regular if and only if is equal to K n . This is the case if and only if its complement has no edges. In this case the complement also equals the metamour graph.
Proof of Corollary 3.3. This is an immediate and easy consequence of the characterization provided by Theorem 3.1.

Proofs regarding 1-metamour-regular graphs
In this section we present the proofs of the results from Section 3.2. They lead to a characterization of 1-metamour-regular graphs.
Proof of Proposition 3.4. Whenever a vertex v ∈ V (G) is a metamour of a vertex w = v ∈ V (G) then also w is a metamour of v. Therefore, supposing that v has exactly one metamour, so has w and the vertices v and w form a pair such that the two vertices of the pair are metamours of each other. This also leads to an edge from v to w in the metamour graph of G. As every vertex has at most one metamour, the edge from v to w is isolated in the metamour graph of G, so v and w have no other adjacent vertices in the metamour graph of G. Therefore, the edges of the metamour graph form a matching, which yields (a).
Suppose now additionally that G is 1-metamour-regular. Then we can partition the vertices of G into pairs of metamours. Hence, n is even and the edges of the metamour graph form a perfect matching, so (b) holds.
For proving Theorem 3.5, we need some auxiliary results. We start by showing that the graphs mentioned in the theorem are indeed 1-metamour-regular. Proof. This is checked easily.
The following proposition is slightly more general than needed in the proof of Theorem 3.5 and will be used later on. Proof. As the matching µ is perfect, every vertex v of G is incident to one edge in µ. Thus, by the proof of Proposition 7.2, every v has exactly one metamour.
We are now ready for proving Theorem 3.5.
Proof of Theorem 3.5. The one direction of the equivalence follows directly from Proposition 7.1 and Proposition 7.3, so only the other direction is left to prove.
Suppose we have a graph G with n vertices that is 1-metamour-regular. Due to Proposition 3.4(b), n is even, and the set of edges of the metamour graph of G forms a perfect matching. In particular, each connected component of the metamour graph consists of two adjacent vertices.
If the metamour graph is connected, then it consists of only two adjacent vertices and n = 2. This can be ruled out easily, so we have n ≥ 4 and the metamour graph is not connected. Now we can use Theorem 3.20 and see that one of the two cases (b) and (c) applies.
In the second case (c) the metamour graph of G consists of two connected components, so the metamour graph consists of n = 4 vertices with two edges that form a perfect matching. It is easy to see that G = P 4 or G = C 4 = K 4 − µ for some perfect matching µ of K 4 are the only two possibilities in this case.
As a result, we obtain in any case G = P 4 or G = K n − µ for a perfect matching µ of K n , which is the desired result.
To finish this section we prove the three corollaries of Theorem 3.5.
Proof of Corollary 3.6. Due to the characterization of 1-metamour-regular graphs of Theorem 3.5, we know that a connected graph with n ≥ 5 vertices is 1-metamour-regular if and only if it is equal to K n − µ for some perfect matching µ of K n . This is the case if and only if the complement is the graph induced by µ. Furthermore, a graph is induced by a perfect matching if and only if it is 1-regular. To summarize, a connected graph with n ≥ 5 vertices is 1-metamour-regular if and only if its complement is a 1-regular graph. In this case the complement also equals the metamour graph, which implies the desired result.
Proof of Corollary 3.7. We use the characterization provided by Theorem 3.5. So let us consider 1-metamour-regular graphs. Such a graph has at least n ≥ 4 vertices, and n is even. Every perfect matching µ of K n results in the same unlabeled graph K n − µ; this brings to account 1. For n = 4, there is additionally the graph P 4 . In total, this gives the claimed numbers.
Proof of Corollary 3.8. Let G be an unlabeled graph with n pairs of vertices that each are metamours. We first construct a pair (λ 1 + · · · + λ t , s), where λ 1 + · · · + λ t is a partition of n with λ i ≥ 2 for all i ∈ {1, . . . , t} and s is a non-negative integer bounded by r λ which is defined to be the number of i ∈ {1, . . . , t} with λ i = 2. Let . . , t}, and let us assume that λ 1 ≥ · · · ≥ λ t . Then n = λ 1 + · · · + λ t , so this is a partition of n. As there is no graph G i with only 1 metamour-pair, λ i ≥ 2 for all i ∈ {1, . . . , t}. We define s to be the number of i ∈ {1, . . . , t} with G i = P 4 . We clearly have s ≤ r λ .
Conversely, let a pair (λ 1 + · · · + λ t , s) as above be given. For every i ∈ {1, . . . , t} with λ i ≥ 3 there is exactly one choice for a 1-metamour-regular graph G i with 2λ i vertices by Theorem 3.5. Now consider parts 2 of λ 1 + · · · + λ t . We choose any (the graphs are unlabeled) s indices and set G i = P 4 . Then we set G i = C 4 for the remaining r λ − s indices. The graph G = G 1 ∪ · · · ∪ G t is then fully determined. Thus, we have a found a bijective correspondence.
We still need to related our partition of n to the partition of n+2 of Corollary 3.8. A partition of n+2 is either n+2 = (n+2), n ≥ 1, in which case no additional part 2 appears, or n+2 = λ 1 +· · ·+λ t +2 for a partition n = λ 1 + · · · + λ t . Here one additional part 2 appears. Therefore, every pair (λ 1 + · · · + λ t , s) from above maps bijectively to a partition λ 1 + · · · + λ t + 2 of n + 2 together with a marker of one of the r λ + 1 parts 2 in this partition that is uniquely determined by s (by some fixed rule that is not needed to be specified explicitly). This completes the proof of Corollary 3.8.

Proofs regarding graphs with maximum metamour-degree 1
Next we prove the results of Section 3.3 on graphs with maximum metamour-degree 1. We start with the proof of the characterization of these graphs.
Proof of Theorem 3.9. It is easy to see that in the graphs K 1 and K 2 no vertex has any metamour, so the condition that each vertex has at most one metamour is satisfied. Furthermore, by Propositions 7.1 and 7.2, every vertex has indeed at most one metamour in the remaining specified graphs. Therefore, one direction of the equivalence is proven, and we can focus on the other direction. So, let G be a graph in which every vertex has at most one metamour. Due to Theorems 3.1 and 3.5, it is enough to restrict ourselves to graphs G, where at least one vertex of G has no metamour and at least one vertex of G has exactly one metamour. We will show that n ≥ 3 and that G = K n − µ for some matching µ that is not perfect and contains at least one edge.
Let V 0 ⊆ V (G) and V 1 ⊆ V (G) be the set of vertices of G that have no and exactly one metamour, respectively, and let v ∈ V 0 . Due to Lemma 6.1, every vertex in V 0 , in particular v, is adjacent to all other vertices. Furthermore, by Proposition 3.4, the vertices in V 1 induce a matching µ in both the metamour graph and the complement of G. This matching µ contains at least one edge because V 1 is not empty, and µ is not perfect because V 0 is not empty. Furthermore, this implies that V 1 contains at least two vertices and in total that n ≥ 3.
Let w and w be two vertices in V 1 that are not metamours. Since v is a common neighbor of both w and w , this implies that {w, w } ∈ E(G). Hence, all possible edges except those in µ are present in G and therefore G = K n − µ.
Next we prove the two corollaries of Theorem 3.9.
Proof of Corollary 3.10. Due to Theorem 3.9, in a connected graph G with n ≥ 5 vertices every vertex has at most one metamour if and only if G = K n − µ for some matching µ of K n . This is the case if and only if the complement is the graph induced by µ. Furthermore, a graph is induced by a matching if and only if it has maximum degree 1. To summarize, a connected graph with n ≥ 5 vertices has maximum metamour-degree 1 if and only if its complement is a graph with maximum degree 1. In this case the complement also equals the metamour graph, which implies the desired result.
Proof of Corollary 3.11. We use the characterization provided by Theorem 3.9. So let us consider graphs with maximum metamour-degree 1. For n ∈ {1, 2}, we only have K 1 and K 2 , so m ≤1 (n) = 1 in both cases.
So let n ≥ 3. Every perfect matching µ of K n having the same number of edges results in the same graph K n − µ. A matching can contain at most n/2 edges and each choice in {0, . . . , n/2 } for the number of edges is possible. This brings to account n/2 + 1. For n = 4, there is additionally the graph P 4 . In total, this gives the claimed numbers.

Proofs regarding 2-metamour-regular graphs
This section is devoted to the proofs concerning 2-metamour-regular graphs from Section 3.4. It is a long way to obtain the final characterization of 2-metamour-regular graphs of Theorem 3.13, so we have outsourced the key parts of the proof into several lemmas and propositions.
For the proofs of Lemma 9.1, Lemma 9.2, Lemma 9.5 and Proposition 9.7 we provide many figures. Every proof consists of a series of steps, and in each of the steps vertices and edges of a graph are analyzed: It is determined whether edges are present or not and which vertices are metamours of each other. The figures of the actual situations show subgraphs of the graph (and additional assumptions) in the following way: Between two vertices there is either an edge or a non-edge or nothing drawn. If nothing is drawn, then it is not (yet) clear whether the edge is present or not. A metamour relation might be indicated at a non-edge. Note that we frequently use the particular graphs defined by Figures 3.3, 3.4 and 3.5.

Graphs with connected metamour graph
The proof of the characterization of 2-metamour-regular graphs in Theorem 3.13 is split into two main parts, which represent whether the metamour graph of G is connected or not in order to apply the corresponding case of Theorem 3.20. Figure 9.1: Subgraphs of the situations in the proof of A If the metamour graph of a graph with n vertices is connected, then according to Observation 3.12 the metamour graph equals C n . Here we make a further distinction between graphs that do and that do not contain a cycle of length n as a subgraph. First, we characterize all 2-metamour-regular graphs whose metamour graph is connected and that do not contain a cycle of length n. Lemma 9.1. Let G be a connected 2-metamour-regular graph with n vertices • whose metamour graph equals the C n , • that is not a tree, and • that does not contain a cycle of length n.

r}, be a longest cycle in
G. In all the figures accompanying the proof, the longest cycle is marked by . By assumption, we have r < n. For proving the lemma, we have to show that G ∈ {H a 6 , H b 6 , H a 7 }. As a cycle has length at least 3, we have r ≥ 3 for the length of the cycle γ. As we also have n > r, we may assume n ≥ 4.
We start by showing the following claims.

A. A vertex u in G that is not in the cycle γ is adjacent to at most one vertex in γ.
If u is adjacent to a vertex v in γ, then u is a metamour of each neighbor of v in γ.

Proof of A.
Let u ∈ V (G) be a vertex not in γ. We assume that u is adjacent to v 1 (without loss of generality by renumbering) and some v j in the cycle γ. We first show that v 1 and v j are not two consecutive vertices in γ. So let us assume that they are, i.e., j = 2 (see Figure 9.1(a)) or j = r which works analogously). Then (v 1 , u, v 2 , . . . , v r , v 1 ) would be a longer cycle which is a contradiction to γ being a longest cycle. Hence, v 1 and v j are not consecutive vertices in γ. Then r ≥ 4 as there need to be at least one vertex between v 1 and v j on the cycle on each side. If r = 4, then v j = v 3 and we are in the situation shown in Figure 9.1(b). There, (u, v 1 , v 2 , v 4 , v 3 , u) is a 5-cycle which contradicts that the longest cycle is of length 4. Therefore, r = 4 cannot hold.
If r > 4, then u is a metamour of v 2 , v r , v j−1 and v j+1 , because it has a common neighbor (v 1 or v j ) with these vertices and is not adjacent to them. At least one of v j−1 and v j+1 is different from v 2 and v r , so |{v 2 , v r , v j−1 , v j+1 }| ≥ 3. This contradicts the 2-metamour-regularity of G.
Therefore, we have shown that u is adjacent to at most one vertex in γ. Now suppose u is adjacent to a vertex v in γ. Then u is not adjacent to any neighbor of v in γ and therefore a metamour of every such neighbor. There exists a vertex w in G but not in γ that is adjacent to (without loss of generality) v 1 , but not to any other v j , j = 1, in γ.

Proof of B.
As r < n, there exists a vertex w not in the cycle γ. The graph G is connected, so there is a path from a vertex of γ to w . Therefore, there is also a vertex w not in γ which is adjacent to a vertex v i . By renumbering, we can assume without loss of generality that i = 1.
As the vertex w is adjacent to v 1 , w is not adjacent to any other v j by A.

B
At this point, we assume to have a vertex w as in B; the situation is shown in Figure 9.2.
C. The graph G contains the edge {v 2 , v r }.

Proof of C.
Assume that there is no edge between v 2 and v r ; see Figure 9.3. Then v 2 and v r are metamours of each other, and consequently (w, v 2 , v r , w) forms a 3-cycle in the metamour graph of G. This contradicts that the metamour graph of G is C n and n ≥ 4.

C
At this point, we have the situation shown in Figure 9.4. In the next steps we will rule out possible values of r.
Proof of D. Our initial situation is shown in Figure 9.5(a). Suppose there is an additional vertex v 1 of G adjacent to v 1 ; see Figure 9.5(b). Then by A, v 1 has metamours v 2 and v 3 . Therefore, (w, v 2 , v 1 , v 3 , w) is a 4-cycle in the metamour graph of G. As this  Figure 9.5: Subgraph of the situation in the proof of D cycle does not cover v 1 , we have a contradiction to the metamour graph being the single cycle C n for n > r = 3. Therefore, there is no additional vertex adjacent to v 1 .
At this point, we know that w is a metamour of both v 2 and v 3 ; see again Figure 9.5(a). We now look for the second metamour of v 2 and v 3 , respectively. As we ruled out an additional vertex adjacent to v 1 , there need to be an additional vertex adjacent to v 2 or to v 3 .
Without loss of generality (by symmetry), suppose there is an additional vertex v 3 of G adjacent to v 3 ; see Figure 9.5(c). Then by A, v 3 has metamours v 1 and v 2 . Therefore, these two vertices are the two metamours of v 3 . There cannot be an additional vertex v 3 of G adjacent to v 3 , because due to the same arguments as for v 3 this vertex would be a metamour of v 2 , hence v 2 would have three metamours, and this contradicts the 2-metamour-regularity of G.
Suppose there is no additional vertex adjacent to v 2 . Then, in order to close the metamour cycle containing (v 1 , v 3 , v 2 , w, v 3 ), there needs to be a path from v 3 to w. This implies the existence of a cycle longer than r = 3, therefore cannot be. So there is an additional vertex adjacent to v 2 ; the situation is shown in Figure 9.5(d).
By the same argument as above, v 1 and v 3 are the two metamours of v 2 . Therefore, (w, v 2 , v 3 , v 1 , v 2 , v 3 , w) is a 6-cycle in the metamour graph of G and n = 6. This is the graph G = H a 6 . We can only add additional edges between the vertices w, v 2 and v 3 , but this would lead to a cycle of length larger than 3. So there are no other edges present. There cannot be any additional vertex because this vertex would need to be in a different cycle in the metamour graph, contradicting that the metamour graph is the C n . D As a consequence of D, the proof is finished for r = 3, because then G = H a 6 . What is left to consider is the case r ≥ 4 and consequently n ≥ 5. The situation is again as in Figure 9  As v 1 has distance 2 on the cycle γ to both v 3 and v r−1 , and is not adjacent to these vertices, the vertices v 3 and v r−1 are metamours of v 1 .

E
At this point, we have the situation shown in Figure 9.6. Note, that it is still possible that v 3 = v r−1 .

F. We cannot have r = 4.
Proof of F. As r = 4, we have v 3 = v r−1 . This situation is shown in Figure 9.7.
Suppose there is an additional vertex v 3 of G adjacent to v 3 . Then by A, v 3 has metamours v 2 and v 4 . Therefore, (w, v 2 , v 3 , v 4 , w) is a 4-cycle in the metamour graph of G. This is a contradiction to the metamour graph being C n and n ≥ 6, so there is no additional vertex adjacent to v 3 . This implies that we cannot have a vertex at distance 1 from v 3 other than v 2 and v 4 . Now suppose there is an additional vertex v 2 of G adjacent to v 2 . Again by A, v 2 has metamours v 1 and v 3 . Therefore, (v 2 , v 1 , v 3 , v 2 ) is a 3-cycle in the metamour graph of G. This is again a contradiction to the metamour graph being C n and n ≥ 6, so there is no additional vertex adjacent to v 2 either. Likewise, by symmetry, there is also no additional vertex adjacent to v 4 .
As v 2 and v 4 are the only neighbors of v 3 , we cannot have a vertex at distance 2 from v 3 other than v 1 . This means that there is no second metamour of v 3 which contradicts the 2-metamour-regularity of G. F At this point, we can assume that r ≥ 5 as the case r = 4 was excluded by F, and consequently also n ≥ 6. The situation is still as in Figure 9.6. G. We have r ≤ 6. Specifically, either r = 5, or r = 6 and there is an edge {v 2 , v 5 } in G. In the second case, the two metamours of the vertex v 2 are w and v 4 .
Proof of G. As r ≥ 5, the two metamours of v 1 are on the cycle γ, namely the distinct vertices v 3 and v r−1 ; see Figure 9.8(a). Figure 9.8: Subgraphs of the situations in the proof of G Suppose {v 2 , v r−1 } ∈ E(G). We again distinguish between two cases. If r ≥ 6, then w, v 4 and v r−2 are metamours of v 2 . In this case, the 2-metamour-regularity of G implies that v r−2 = v 4 and therefore r = 6; see Figure 9.8(c). If r < 6, then by the findings so far, we must have r = 5, and therefore we are also done in this case. G By G we are left with the two cases r = 5 and r = 6. One possible situation for r = 5 and the situation for r = 6 are shown in Figure 9.8(b) and (c), respectively, and we will deal with these two situations now.
Proof of H. The full situation for r = 5 is shown in Figure 9.9(a). Clearly the situation is symmetric in the potential edges {v 2 , v 4 } and {v 3 , v 5 }, so we have to consider the three cases that both, one and none of these two edges are present.
First let us assume that neither {v 2 , v 4 } nor {v 3 , v 5 } is an edge; see Figure 9.9(b). Then v 2 and v 4 as well as v 3 and v 5 are metamours, so we have the 6-cycle (w, v 2 , v 4 , v 1 , v 3 , v 5 , w) in the metamour graph of G. This is the graph G = H b 6 . There cannot be any additional vertex because this vertex Figure 9.10: Subgraphs of the situations in the proof of I would need to be in a different cycle in the metamour graph contradicting that the metamour graph is the C n . There also cannot be any additional edges because all edges and non-edges are already determined.
Next let us assume that there is exactly one of the edges {v 2 , v 4 } and {v 3 , v 5 } present in G, without loss of generality let {v 3 , v 5 } ∈ E(G); see Figure 9.9(c). At this point we know that v 3 and v 1 as well as v 5 and w are metamours, and we are looking for the second metamours of v 3 and v 5 . As the vertices w, v 1 and v 4 already have two metamours each, there need to be additional vertices for these metamours.
Statement A implies that there is no additional vertex of G adjacent to v 5 as v 1 has already the two metamours v 3 and v 4 . Likewise, by symmetry, there is no additional vertex adjacent to v 2 . Moreover, by the same argument, there is also no additional vertex adjacent to v 3 as v 4 has the two metamours v 1 and v 2 .
Therefore, there need to be an additional vertex v 4 adjacent to v 4 . By A, v 4 has metamours v 3 and v 5 . This gives the 7-cycle (w, v 2 , v 4 , v 1 , v 3 , v 4 , v 5 , w) in the metamour graph of G and the graph G = H a 7 . There cannot be any additional vertex because this vertex would need to be in a different cycle in the metamour graph contradicting that the metamour graph is the C n . There also cannot be any additional edges because all edges and non-edges are already determined.
At last, let us consider the case that both of the edges {v 2 , v 4 } and {v 3 , v 5 } are present in G. We already know that w is a metamour of v 2 and are now searching for the second metamour of v 2 . There does not exist a vertex v 1 adjacent to v 1 in G but not in γ, because this would induce a C 4 in the metamour graph by the same arguments as in the proof of D. Furthermore, there cannot be a vertex v 2 in G but not in γ that is adjacent to v 2 , due to the fact that this vertex would be a third metamour of v 1 , a contradiction. By symmetry, there is no vertex of G without γ adjacent to v 5 . If there would be a vertex v 3 in G but not in γ which is adjacent to v 3 , then due to A, this vertex would have v 2 , v 4 and v 5 as a metamour, a contradiction to the 2-metamour-regularity of G. Again by symmetry, there is no vertex of G without γ adjacent to v 4 . Therefore, v 2 cannot have a second metamour in G and this case cannot happen.
H Statement H finalizes the proof for r = 5. Hence, r = 6 is the only remaining value for r we have to consider. I. We cannot have r = 6.
Proof of I. As r = 6, there is an edge {v 2 , v 5 } in G by G. The initial situation is shown in Figure 9.10(a).
Suppose v 3 and v 5 are not adjacent. Then (v 1 , v 3 , v 5 , v 1 ) is a 3-cycle in the metamour graph of G. This contradicts that the metamour graph is C n and n > r = 6, so we can assume {v 3 , v 5 } ∈ E(G). Likewise, suppose that v 4 and v 6 are not adjacent. Then (w, v 2 , v 4 , v 6 , w) is a 4-cycle in the metamour graph of G. This contradicts that the metamour graph is C n and n > r = 6, so we can assume {v 4 , v 6 } ∈ E(G). The current situation is shown in Figure 9.10(b).
Statement A implies that there is no additional vertex of G adjacent to v 2 as v 1 has already the two metamours v 3 and v 5 . By symmetry, there is also no additional vertex adjacent to v 6 . By the same argumentation as above, there is no additional vertex adjacent to v 1 as well as to v 3 because of vertex v 2 and its metamours. Moreover, we slightly vary the argumentation to show that there cannot be an additional vertex adjacent to v 5 . Suppose there is an additional vertex v 5 of G adjacent to v 5 . Then, v 5 is not adjacent to v 2 as we have shown above, so v 5 is as well a metamour of v 2 . This contradicts the 2-metamour-regularity of G again.
The vertex v 4 has v 2 as metamour. We are now searching for its second metamour. It cannot be w or v 1 as these vertices have already two other metamours each. It cannot be any of v 3 , v 5 or v 6 either as all of them are adjacent to v 4 . Moreover, the second metamour of v 4 cannot be adjacent to v 3 , v 5 or v 6 , as we above ruled additional neighbors to these vertices out. Therefore, there has to be an additional vertex v 4 adjacent to v 4 . By A, this vertex v 4 has metamours v 3 and v 5 . This results in the 4-cycle (v 1 , v 3 , v 4 , v 5 , v 1 ) in the metamour graph of G and contradicts our assumption that this graph is C n and n > r = 6.

I
We have now completed the proof of Lemma 9.1 as in all cases we were able to show that G ∈ {H a 6 , H b 6 , H a 7 } holds.
After characterizing all 2-metamour-regular graphs whose metamour graph is connected and that do not contain a cycle of length n, we can now focus on 2-metamour-regular graphs whose metamour graph is connected and that contain a cycle of length n. Here, we make a further distinction depending on the degree of the vertices and begin with the following lemma.

Lemma 9.2. Let G be a connected 2-metamour-regular graph with n vertices
• whose metamour graph equals the C n , • that contains a cycle of length n, and • that has a vertex of degree larger than 2 and smaller than n − 3.
Proof. Let γ be a cycle of length n in G. First, we introduce some notation. Let v be a vertex of G, and let u and u be the two metamours of v. We explore the vertices on the cycle γ starting with v: The set of vertices on both sides of v strictly before u and u are called the fellows of v. The remaining set of vertices strictly between u and u on γ is called the opponents of v; see Figure 9.11. In other words for each vertex v of G the set of vertices of G can be partitioned into v, its fellows, its metamours and its opponents. We start with the following claims.

A. Every vertex of G is adjacent to each of its fellows.
Proof of A. Let v 1 be a vertex of G and γ = (v 1 , . . . , v n , v 1 ). Suppose v p is the vertex with smallest index p that is not adjacent to v 1 . We have to show that v p is a metamour of v 1 . The index p exists because v 1 is not adjacent to its metamours. Moreover, this index satisfies p > 2 as v 2 is adjacent to v 1 because they are consecutive vertices on γ. Thus, v 1 and v p have v p−1 as common neighbor and are therefore metamours. By symmetry, the vertex v q with largest index q that is not adjacent to v 1 , is also a metamour of v 1 . Note that as v 1 has exactly two metamours, v p and v q are these metamours, so v 1 is adjacent to each of its fellows.  Figure 9.12: Subgraph of the situation between C and D B. Every vertex of G is either adjacent to each of its opponents, or not adjacent to any of its opponents.
Proof of B. It is enough to show that if a vertex v 1 of G is adjacent to at least one opponent of v 1 , then it is adjacent to every opponent of v 1 . Let γ = (v 1 , . . . , v n , v 1 ), and let W be a subset of the opponents of v 1 that consists of consecutive vertices of γ, say from v i to v j for some i ≤ j, such that each of these vertices is adjacent to v 1 , and W is maximal (with respect to inclusion) with this property. Note that the set W is not empty because of our assumption. Clearly none of the vertices in W is a metamour of v 1 . However v i−1 and v j+1 are metamours of v 1 because of their common neighbors v i and v j , and the maximality of W . Therefore, as v 1 has exactly two metamours, W equals the set of opponents of v 1 which was to show.

B
Now we are ready to start with the heart of the proof of Lemma 9.2. Suppose v 1 is a vertex of G with 2 < deg(v 1 ) < n − 3. In order to complete the proof we have to show that G = H b 7 . Let γ = (v 1 , . . . , v n , v 1 ) be a cycle of length n, and let v p and v q be the metamours of v 1 with p < q. In the following claims we will derive several properties of G. v 1 is adjacent to its fellows v 2 , . . . , v p−1 , v q+1 , . . . , v n and not adjacent to any metamour or opponent v p , . . . , v q . Furthermore, p + 1 < q holds, i.e., there exists at least one opponent of v 1 .

Proof of C.
Clearly v 1 is not adjacent to its metamours v p and v q . Furthermore, v 1 is adjacent to all its fellows v 2 , . . . , v p−1 , v q+1 , . . . , v n by A. This together with deg(v 1 ) < n − 3 implies that v 1 has an opponent to which it is not adjacent, so p + 1 < q. Then by B, v 1 is not adjacent to any of its opponents.  Figure 9.14: Subgraph of the situation in the proof of F Now deg(v 1 ) > 2 together with C imply that v 1 has at least one fellow different from v 2 and v n . Without loss of generality (by renumbering the vertices in the opposite direction of rotation along γ) assume that v q+1 is a fellow of v 1 different from v n , so in other words we assume q + 1 < n. The situation is shown in Figure 9.12.
We will now prove several claims about edges, non-edges and metamours of G.  In particular, v 1 is adjacent to v q−1 which contradicts C. Therefore, v q and v q+2 are not adjacent and because of their common neighbor v q+1 , metamours.
Proof of G. The two metamours of v q are v 1 and v q+2 because of F. This implies that v 2 , . . . , v q−1 are fellows of v q and therefore adjacent to v q because of A.
G Figure 9.15 shows the current situation. Proof of H. The vertex v q−1 is not adjacent to any of v 2 , . . . , v p−1 due to D. Furthermore, v q−1 has the common neighbor v q with each of these vertices because of G. So every vertex v 2 , . . . , v p−1 is a metamour of v q−1 . This implies |{v 2 , . . . , v p−1 }| ≤ 1 because v q−1 also has v q+1 as metamour and has in total exactly two metamours. Moreover, as v 2 is adjacent to v 1 , v 1 and v 2 are not metamours, thus v 2 and v p cannot coincide. This implies p > 2 has to hold. In consequence, we obtain p = 3 implying v p−1 = v 2 has to hold.  If v p is adjacent to v i for i ∈ {q + 2, . . . , n}, then v p+1 and v i are metamours because they have v p as a common neighbor, and they are not adjacent due to D. This is a contradiction as v p+1 already has the two metamours v p−1 and v q+1 because of E and an implication of I. As a result, v p is not adjacent to any of v q+2 , . . . , v n .

(a) between H and
If v p would be adjacent to v q+1 , then v p and v q+2 are metamours because of the common neighbor v q+1 and because they are not adjacent by the above. But then, due to F, (v p , v 1 , v q , v q+2 , v p ) is a cycle in the metamour graph which does not contain all vertices, a contradiction to our assumption. Therefore, v p is not adjacent to v q+1 . The vertex v p is adjacent to v q due to G, therefore v q is a common neighbor of v p and v q+1 , and hence these vertices are metamours. J Figure 9.17(a) shows the situation.
K. It holds that q + 2 = n, so there are exactly two fellows of v 1 on the cycle γ between v q and v 1 . Furthermore, the vertices v p−1 and v q+2 are metamours of each other, and v p−1 is adjacent to all vertices except its metamours.
Proof of K. The vertex v p−1 is a metamour of v q−1 due to H, and it is adjacent to v q because of G. This together with p + 1 = q − 1 by I implies that v p−1 is adjacent to one of its opponents, namely v q .
Then by B and A, this implies that v p−1 is adjacent to all vertices except its metamours.
If v p−1 is adjacent to a vertex v i for i ∈ {q + 2, . . . , n}, then v p and v i are metamours because they have v p−1 as common neighbor and are not adjacent due to J. But v p already has the two metamours v 1 and v q+1 due to J, a contradiction. As a result, v p−1 is not adjacent to any vertex of v q+2 , . . . , v n . Now assume q + 2 < n, so the vertex v q+3 exists. Due to the fact that v p−1 is adjacent to all vertices except its metamours and that it has v p+1 as metamour by E, it follows that it is adjacent to at least one of v q+2 and v q+3 . But we showed that v p−1 is not adjacent to any of these two vertices, a contradiction. Therefore, q + 2 = n holds. Furthermore, v p−1 is not adjacent to v q+2 , and therefore these two vertices are metamours of each other. Next we consider all cases of 2-metamour-regular graphs whose metamour graph is connected, that contain a cycle of length n and whose degrees are not as in the previous lemma.

Lemma 9.3. Let G be a connected 2-metamour-regular graph with n vertices
• whose metamour graph equals the C n , • that contains a cycle of length n, and • in which every vertex has degree n − 3. Then Proof. If a vertex v of G has degree n − 3, then v is adjacent to all but two vertices. These two vertices are exactly the metamours of v. This implies that G equals the complement of the metamour graph. Hence, G = C n as the metamour graph of G is the C n .
Lemma 9.4. Let G be a connected 2-metamour-regular graph with n vertices • whose metamour graph equals the C n , • that contains a cycle of length n, and • in which every vertex has degree 2.
Then G = C n and n is odd.
Proof. Let γ be a cycle of length n in G. If every vertex of G has degree 2, then every vertex in the induced subgraph G[γ] has degree 2 as γ contains every vertex by assumption. As G[γ] is connected, it equals C n . In total this implies G = G[γ] = C n . It is easy to see that if n is even, then the metamour graph consists of exactly two cycles of length n 2 which contradicts our assumption. Therefore, n is odd. Figure 9.18: Subgraph of the situation in the proof of A • whose metamour graph equals the C n , • that contains a cycle of length n, • in which every vertex has degree 2 or n − 3, and • that has a vertex of degree 2 and a vertex of degree n − 3.

A.
We have 5 ≤ n ≤ 7 and the metamours of v 1 are v 3 and v n−1 .

Proof of A.
Clearly v 1 is only adjacent to v 2 and v n . Hence, v 3 and v n−1 have to be the two metamours of v 1 and G contains at least 5 different vertices, so n ≥ 5.
If v 2 is adjacent to some v i for i ∈ {4, . . . , n − 2}, then v 1 is a metamour of v i due to the common neighbor v 2 ; see Figure 9.18. Hence, v 2 is not adjacent to any vertex v 4 , . . . , v n−2 . However, because deg(v 2 ) = n − 3, the vertex v 2 is adjacent to every vertex but its two metamours. This implies that |{v 4 , . . . , v n−2 }| ≤ 2, because v 2 has at most two metamours among v 4 , . . . , v n−2 . As a result, we have n ≤ 7.

A
This implies that n = 5, n = 6 and n = 7 are the only cases to consider. We do so in the following claims.
B. If n = 5, then G = C n .  Proof of C. If n = 6, then v 3 and v n−1 = v 5 are the two metamours of v 1 .
If v 3 is not adjacent to v 5 , then v 3 and v 5 are metamours because of their common neighbor v 4 ; see Figure 9.20(a). But then (v 1 , v 3 , v 5 , v 1 ) is a cycle in the metamour graph that does not contain all vertices, a contradiction to our assumption. Hence, v 3 and v 5 are adjacent; see Figure 9.20(b). Then v 6 is the only option left as the second metamour of v 3 , and v 2 is the only option left as the second metamour of v 5 .
If v 2 and v 6 are not adjacent, then they are metamours because of their common neighbor v 1 . But then (v 1 , v 3 , v 6 , v 2 , v 5 , v 1 ) is a cycle in the metamour graph that does not contain v 4 , a contradiction. So v 2 and v 6 are adjacent; see Figure 9.20(c).
But then v 4 has to have v 2 and v 6 as metamours, because they are the only options left. Hence, we obtain G = H c 6 . As a result, v 5 is only adjacent to v 4 and v 6 , and the vertices v 3 and v 7 have to be the two metamours of v 5 ; see Figure 9.21(b). In particular, v 5 is not adjacent to v 2 , and the vertices v 5 and v 2 are not metamours. This implies deg(v 2 ) < n − 3 = 4 which is a contradiction to deg(v 2 ) = 4. Hence, n = 7 is not possible.

D
To summarize, in the case that not all vertices of G have the same degree in {2, n − 3}, G = C 5 and G = H c 6 are the only possible graphs due to A, B, C and D. This finishes the proof of Lemma 9.5.
Eventually, we can collect all results on 2-metamour-regular graphs that have a connected metamour graph in the following proposition.
What is left to consider is the situation that there are two vertices with different degrees in G. This is done in Lemma 9.5, and we conclude G ∈ {C 5 , H c 6 } in this case. This completes the proof.

Graphs with disconnected metamour graph
After characterizing all graphs that are 2-metamour-regular and that have a connected metamour graph, we now turn to 2-metamour-regular graphs that do not have a connected metamour graph. In this case either statement (b) or statement (c) of Theorem 3.20 is satisfied. In the case of (b) there is nothing left to do, because it provides a characterization. In the other case we determine all graphs and capture them in the following proposition. Proposition 9.7. Let G be a connected 2-metamour-regular graph with n vertices. Suppose statement (c) of Theorem 3.20 is satisfied. Then n ≥ 6 and one of (a) G = C n and n is even, or Proof. Theorem 3.20(c) implies that the metamour graph is not connected. First observe that by Observation 3.12, each connected component of the metamour graph of a 2-metamour-regular graph is a cycle.
The proof is split into several claims. As a first step, we consider the number of vertices of G.
A. We have n ≥ 6.

Proof of A.
As the metamour graph of G is not connected, the metamour graph contains at least two connected components, which are cycles. Each cycle has to contain at least three vertices, so n ≥ 6. We introduce the following notion: The signature σ of a graph is the tuple of the numbers of vertices of its connected components, sorted in descending order. If follows from (i) of Theorem 3.20(c) that all connected components of G and G * have at most 2 vertices. As a consequence, the signatures σ(G ) and σ(G * ) have entries in {1, 2}. Note that in case a connected component has two vertices, then these vertices are adjacent, i.e., this component equals P 2 .
We perform a case distinction by the signatures of the graphs G and G * ; this is stated as the following claims. We start with the case that at least one connected component of G or G * has two vertices.
B. If the first (i.e., largest) entry of σ(G ) is 2, then either σ(G ) = (2, 2) or σ(G ) = (2, 1, 1). In the latter case, the two vertices of the two connected components containing only one vertex do not have any common neighbor in G.
Proof of B. Suppose we have G 1 ∈ C(G ) with |V (G 1 )| = 2. Let v 1 be one of the two vertices of G 1 .
Then v 1 is adjacent to the other vertex of G 1 , therefore it must have its two metamours in another component of G .
Let us assume that a metamour of v 1 is in a connected component G 2 of G that consists of two vertices. Then every vertex of G 1 is a metamour of every vertex of G 2 due to (v) of Theorem 3.20(c).
(a) in proof of C Figure 9.22: Subgraphs of the situations in the proofs of C and D This implies that the four vertices of G 1 and G 2 form a C 4 in the metamour graph and consequently that M = C 4 . Therefore, G cannot contain other vertices and σ(G ) = (2, 2).
Let us now assume that the two metamours of v 1 are in different connected components G 2 and G 3 of G that consist of only one vertex each. Then also the other vertex of G 1 is a metamour of the two vertices in G 2 and G 3 due to (v) of Theorem 3.20(c). Hence, these four vertices form again a C 4 in the metamour graph and consequently M = C 4 . As a result, G cannot contain any more vertices, so σ(G ) = (2, 1, 1).
If the two vertices of G 2 and G 3 have a common neighbor, then they are metamours of each other because they are not adjacent. This is a contradiction to the fact that the vertex of G 2 already has two metamours; they are in G 1 . Hence, the vertices of G 2 and G 3 do not have any common neighbor.

B
By B we can deduce how the graphs G and G * look like, if one of their connected component contains 2 vertices. We will now continue by going through all possible combinations of signatures of G and G * implied by B. In every case, we have to determine which edges between vertices of G and G * exist and which do not exist in order to specify the graph G.
Due to (iv) of Theorem 3.20(c), we know that as soon as there is an edge in G between a connected component of G and a connected component of G * , then there are all possible edges between these two components in G. This implies, now rephrased in the language introduced in Section 2.1, adjacency of two connected components of G M is equivalent to complete adjacency of these components. Therefore, we equip G M = G ∪ G * with a graph structure: The set C(G M ) is the vertex set and the edge set-we simply write it as E(G M )-is determined by the adjacency relation above. Note that this graph is bipartite.
The components G 1 and G 2 need a common neighbor in G * with respect to G M because their vertices are metamours; see the proof of B. So, we assume without loss of generality (by renumbering the connected components of G * ) that needs to be adjacent to at least one connected component of G because G is connected, so assume without loss of generality (by renumbering the connected components of G ) that {G 2 , G * 2 } ∈ E(G M ). Now there is only the edge between G 1 and G * 2 left to consider. If The potential edges, which status is still undetermined, are {G 2 , G * 1 }, {G 2 , G * 3 } and {G 3 , G * 1 }. At this stage, if each of these pairs is a non-edge in G M , then it is easy to see that whenever two vertices should be metamours they are metamours, to be precise all components of G have the common neighbor G * 2 and all components of G * have the common neighbor G 1 . Therefore, we have enough edges in G M , so that additional edges between components of G and G * can be included without interfering with the metamours.

Assembling results & other proofs
With all results above, we are now able to prove the main theorem of this section which provides a characterization of 2-metamour-regular graphs.
Proof of Theorem 3.13. Let G be 2-metamour-regular and M its metamour graph. We apply Theorem 3.20 with k = 2. This leads us to one of three cases.
Case If we are in case (a) of Theorem 3.20, then the metamour graph M is connected and we apply Proposition 9.6. If we are in case (c) of Theorem 3.20, then the metamour graph consists of exactly two connected components and we can apply Proposition 9.7. Collecting all graphs coming from these two propositions yields the remaining graphs of (a), (b) and (c) of Theorem 3.13.
For the other direction, Proposition 3.19 implies that the graph G in (a) of Theorem 3.13 is a 2-metamour-regular graph for t ≥ 2. Furthermore, it is easy to check that all other mentioned graphs are 2-metamour-regular, which proves this side of the equivalence and completes the proof.
Finally we are able to prove the following corollaries of Theorem 3.13.
Proof of Corollary 3.15. The result is an immediate consequence of Theorem 3.13.
Proof of Corollary 3.16. Corollary 3.15 provides a characterization of all 2-metamour-regular graphs with n ≥ 9 vertices. It is easy to see that all of these graphs are either 2-regular (in the case of C n ) or (n − 3)-regular. This proves one direction of the equivalence.
For the other direction first consider a connected 2-regular graph on n vertices. Clearly, this graph equals C n , therefore this graph is 2-metamour-regular. If a connected graph is (n − 3)-regular, then its complement G is a 2-regular graph. As a result, each connected component of G is a cycle graph. Let C n 1 , . . . , C nt be the connected components of G. It is easy to see that then n i ≥ 3 and n = n 1 + · · · + n t hold. In consequence, G = C n 1 ∇ · · · ∇ C nt holds and therefore G is 2-metamour-regular. This completes the proof.
Proof of Corollary 3.17. The statement of the corollary is a direct consequence of Theorem 3.13.
Proof of Corollary 3.18. We use the characterization provided by Theorem 3.13. So let us consider 2-metamour-regular graphs. Such a graph has at least n ≥ 5 vertices.
In case (a), there is one graph per integer partition of n into a sum, where each summand is at least 3. Note that the graph operator ∇ is commutative which coincides with the irrelevance of the order of the summands of the sum. There are p 3 (n) many such partitions.
Case (b) gives exactly one graph for each n ≥ 5. The graph C 5 is counted in both (a) and (b); see first item of Remark 3.14. Case (c) brings in additionally 8 graphs for n = 6, 6 graphs for n = 7 and 3 graphs for n = 8.
In total, this gives the claimed numbers.
This completes all proofs of the present paper.

Conclusions & open problems
In this paper we have introduced the metamour graph M of a graph G: The set of vertices of M is the set of vertices of G and two vertices are adjacent in M if and only if they are at distance 2 in G, i.e., they are metamours. This definition is motivated by polyamorous relationships, where two persons are metamours if they have a relationship with a common partner, but are not in a relationship themselves. We focused on k-metamour-regular graphs, i.e., graphs in which every vertex has exactly k metamours. We presented a generic construction to obtain k-metamour-regular graphs from k-regular graphs for an arbitrary k ≥ 0. Furthermore, in our main results, we provided a full characterization of all k-metamour-regular graphs for each k ∈ {0, 1, 2}. These characterizations revealed that with a few exceptions, all graphs come from the generic construction. In particular, • for k = 0 every k-metamour-regular graph is obtained by the generic construction.
• For k = 1 there is only one exceptional graph that is k-metamour-regular and not obtained by the generic construction.
• In the case of k = 2 there are 17 exceptional graphs with at most 8 vertices and a family of graphs, one for each number of vertices at least 6, that are 2-metamour-regular and cannot be created by the generic construction.
Additionally, we were able to characterize all graphs where every vertex has at most one metamour and give properties of the structure of graphs where every vertex has at most k metamours for arbitrary k ≥ 0. Every characterization is accompanied by counting for each number of vertices how many unlabeled graphs there are. The obvious unanswered question is clearly the following.

Question 10.1. What is a characterization of k-metamour-regular graphs for each k ≥ 3?
This is of particular interest for k = 3. As our generic construction yields k-metamour-regular graphs for every k ≥ 0, we clearly already have determined a lot of 3-metamour-regular graphs. It would, however, be lovely to determine all remaining graphs. Another interesting question is about fixed maximum metamour-degree.

Question 10.2. What is a characterization of all graphs that have maximum metamour-degree k?
We have answered this question for k ∈ {0, 1} and would be delighted to know the answer in general, but as first steps specifically for k = 2 and k = 3.
It would also be interesting to find some structure in the graphs that are k-metamour-regular and cannot be obtained with our generic construction. In particular, we ask the following.
When dealing with metamour graphs, one question to ask is whether it is possible to characterize all graphs whose metamour graph has a certain property. In the present paper we have started to give an answer for the feature that the metamour graph is k-regular. But what about other graph classes? Of course it would be interesting to answer the following questions. Question 10.4. Is it possible to characterize all graphs whose metamour graph is in some graph class like planar, bipartite, Eulerian or Hamiltonian graphs or like graphs of a certain diameter, girth, stability number or chromatic number?
Another question of interest concerns constructing graphs, namely given a graph M , is there a graph G such that M is the metamour graph of G? If M is not connected, then the answer is easy and also provided in this paper, namely G = M is such a graph. However, if M is connected this question is still open and an answer more complicated. This give rise to the following question.
Question 10.5. What is a characterization of the class of graphs with the property that each graph in this class is the metamour graph of some graph?
Motivated by [28] we ask the following.

Question 10.6. What is a characterization of the class of graphs, where every graph is isomorphic to its metamour graph?
Going into another direction, one can also think about random graphs like the graphs from the Erdős-Rényi model G(n, p). Question 10.7. Given a random graph of G(n, p), which properties does its metamour graph have? Is there a critical value for p (depending on n) such that the metamour graph is connected?
In enumerative and probabilistic combinatorics the following question arise. Most of the results and open questions focus on the number of vertices of the graph and metamour graph respectively, as these two numbers match. But it would be interesting to know how the number of edges of the metamour graph of a graph relates to the number of edges in this graph. Specifically, we ask the following questions.