# Properties and Recognition of Atom Graphs
## Atom Graph Subclasses
### Atom Graphs of \(\mathcal{G}_{nine}\)
It follows that each minimal separator of \(AG(G)\) is the intersection of two maximal cliques of \(AG(G)\), and therefore is of size \(1\) since otherwise, there would be a pair \(\{S,S^{\prime}\}\) of clique minimal separators of \(G\) and an edge of \(AG(G)\) in \(\mathcal{A}_{S}\cap\mathcal{A}_{S^{\prime}}\) which would be associated with both \(S\) and \(S^{\prime}\). It follows that \(AG(G)\) is a block graph. **Lemma 6**. : _For each connected block graph \(G\), \(AG(G)=CG(G)\)._  
Proof. : As \(G\) is chordal, \(AG(G)\) is a subgraph of \(CG(G)\) with the same node set. Let us show that each edge of \(CG(G)\) is an edge of \(AG(G)\). Let \(XY\) be an edge of \(CG(G)\), let \(T\) be an atom tree of \(G\) and let \(X^{\prime}Y^{\prime}\) be an edge of \(P_{T}(X,Y)\). As \(1\leq|X\cap Y|\leq|X^{\prime}\cap Y^{\prime}|=1\) with \(X\cap Y\subseteq X^{\prime}\cap Y^{\prime}\), it follows that \(X\cap Y=X^{\prime}\cap Y^{\prime}\), and therefore \(XY\) is an edge of \(AG(G)\) by Characterization 5. To prove that a block graph is an atom graph of a block graph, or equivalently by Lemma 6, a clique graph of a block graph, we will use a well-known technique allowing to compute a chordal graph whose clique graph is isomorphic to a given graph. For each graph \(G=(V,E)\), RetroCG\((G)\) denotes the intesection graph of the set \(\mathcal{K}(G)\cup\{\{x\},x\in V\}\). Let \(G=(V,E)\) be a connected graph, and let \(G^{\prime}=\text{RetroCG}(G)\). If \(\text{CG}(G)\) is chordal and for each maximal clique \(\mathcal{K}\) of \(\text{CG}(G)\), the elements of \(\mathcal{K}\) have a non-empty intersection then \(G^{\prime}\) is chordal, \(G\) is isomorphic to \(\text{CG}(G^{\prime})\) and the maximal cliques of \(G^{\prime}\) are the sets \(N_{G^{\prime}}[\{x\}]\) for each element \(x\) of \(V\). We recall the following properties of \(\text{RetroCG}(G)\), which are easy to check:  
(1) For each \(x\in V\), \(N_{G^{\prime}}[\{x\}]\) is a maximal clique of \(G^{\prime}\),  
(2) For each \(\{x,y\}\subseteq V\)\(xy\) is an edge of \(G\) if and only if \(N_{G^{\prime}}[\{x\}]\cap N_{G^{\prime}}[\{y\}]\
eq\emptyset\),  
(3) If for each maximal clique \(\mathcal{K}\) of \(\text{CG}(G)\), the elements of \(\mathcal{K}\) have a non-empty intersection then the maximal cliques of \(G^{\prime}\) are the sets \(N_{G^{\prime}}[\{x\}]\) for each element \(x\) of \(V\),  
(4) \(\text{CG}(G)\) is chordal if and only if \(G^{\prime}\) is chordal. It follows from these properties that under the hypothesis of the lemma, \(G^{\prime}\) is chordal, and the mapping associating each vertex \(x\) of \(G\) with \(N_{G^{\prime}}[\{x\}]\) is an isomorphism from \(G\) to \(\text{CG}(G^{\prime})\). Let \(G=(V,E)\) be a connected block graph, and let \(G^{\prime}=\text{RetroCG}(G)\). Then \(G^{\prime}\) is a block graph and \(G\) is isomorphic to \(\text{CG}(G^{\prime})\). Proof. : By Lemma 6\(\text{CG}(G)=\text{AG}(G)\), so by Lemma 5\(\text{CG}(G)\) is chordal. Let \(\mathcal{K}\) be a maximal clique of \(\text{CG}(G)\), let us show that the elements of \(\mathcal{K}\) have a non-empty intersection. If \(G\) has a unique atom then it is trivial, otherwise by Lemma 5 again it is the case since all the elements of \(\mathcal{K}\) contain a given clique minimal separator \(S\) of \(G\). Hence, by Lemma 7\(G^{\prime}\) is chordal, \(G\) is isomorphic to \(\text{CG}(G^{\prime})\) and the maximal cliques of \(G^{\prime}\) are the sets \(N_{G^{\prime}}[\{x\}]\) for each element \(x\) of \(V\). Let us show that each minimal separator of \(G^{\prime}\) is of size \(1\). We assume for contradiction that it is not the case. Then, there is a pair \(\{x,y\}\) of \(V\) and a pair \(\{K,L\}\) of \(\mathcal{K}(G)\) in \(N_{G^{\prime}}[\{x\}]\cap N_{G^{\prime}}[\{y\}]\), i.e., such that \(\{x,y\}\subseteq K\cap L\). It follows that \(KL\) is an edge of \(\text{CG}(G)\), i.e., of \(AG(G)\), and therefore is associated with a minimal separator of \(G\) of size at least \(2\), which contradicts the fact that \(G\) is a block graph. Hence, \(G^{\prime}\) is a block graph. Proof.