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A map graph is a contact graph of internally-disjoint regions of the plane, where the contact can be even a point. Namely, each vertex is represented by a simple connected region and two vertices are connected by an edge iff the corresponding regions touch. Map graphs are introduced in [2] to allow the representation of graphs containing large cliques in a readable way.

A clique planar graph is a graph \(G=(V,E)\) that admits a representation where each vertex \(u \in V\) is represented by an axis-parallel unit square R(u) and where, for some partition of V into vertex-disjoint cliques \(S=\{c_1, \dots , c_k\}\), each edge (u, v) is represented by the intersection between R(u) and R(v) if u and v belong to the same clique (intersection edges) or by a non-intersected curve connecting the boundaries of R(u) and R(v) otherwise (link edges); see Fig. 1(c). Clique planar graphs are introduced in [1], where it is mainly addressed the case where the clique partition S is given.

Figure 2 provides an example of a graph that is both a map graph and a clique planar graph. In [1] it is argued that there are graphs that admit a clique-planar representation while not admitting any representation as a map graph, and vice versa. In this poster we exhibit such counterexamples, establishing that neither of the classes of map graphs and of clique planar graphs is contained in the other.

Fig. 1.
figure 1

A graph (a) that is both a map graph (b) and a clique planar graph (c) (Color figure online).

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Fig. 2.
figure 2

A graph (a) that is clique planar (b) but not a map graph (c) (Color figure online).

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FormalPara Lemma 1

There exists a clique planar graph that is not a map graph.

Fig. 3.
figure 3

A map graph that is not clique planar.

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FormalPara Proof

Consider the graph G of Fig. 2(a). Observe that G is not planar since vertices 1, 3, 4, 5, and 6 (filled red in Fig. 2(a)) form a \(K_5\) subdvision. However, graph G is clique planar (see Fig. 2(b)). If G were also a map graph, some edges could be represented by regions sharing a point. Since only the two triangles 1, 2, 3 and 4, 5, 6 could be represented in such a way, this would imply the planarity of a graph \(G'\) obtained from G by possibly augmenting one or both of such triangles to wheels, which is not planar as \(G' \subseteq G\) (see Fig. 2(a) and (c)).    \(\square \)

FormalPara Lemma 2

There exists a map graph that is not a clique planar graph.

Proof. Consider a graph \(G_h=(V,E)\) composed by three sets \(V_1\), \(V_2\), and \(V_3\) of h vertices each, where the graph induced by \(V_1 \cup V_2\) is a clique and the graph induced by \(V_2 \cup V_3\) is a clique. Figure 3 shows that \(G_h\) is a map graph. Observe that in any partition S of V into vertex-disjoint cliques there are at least h / 2 vertices in \(V_2\) that do not fall into the same clique with the vertices of \(V_1\) or \(V_3\). The link edges among such vertices induce a \(K_{\frac{h}{2},h}\). Hence, for \(h=6\) the clique planarity of \(G_6\) would imply the planarity of \(K_{3,3}\).    \(\square \)

By Lemmas 1 and 2 we have the following.

FormalPara Theorem 1

Map Graphs \(\not \subseteq \) Clique Planar Graphs \(\wedge \) Clique Planar Graphs \(\not \subseteq \) Map Graphs.