The maximum k-clique problem in a graph G is equivalent to the maximum clique problem in Gk. By Lemma 7 of Chapter III, Gk is not necessarily a unit disk graph.
There are some strong indications that this problem is NP-hard in unit disk graphs. We claim this as a conjecture and give complexity results for some geometric graphs which may be helpful in verifying our conjecture.
For ² ∈ [0, 1], a unit ²-quasi-disk is a connected compact set Q of the plane such that there exists a point P such that D(P, 1 − ²) ⊆ Q ⊆ D(P, 1), where D(C, r) denotes the disk of radius r centered at C [24]. Figure 9 illustrates this concept.
ε − 1
1
Fig. 9 A unit ²-quasi-disk.
Ceroi [24] proved that the maximum clique problem on the class of intersection graphs of unit ²-quasi-disks is NP-hard for any ² > 0. The proof is done by a reduction from the maximum independent set problem on cubic graphs.
Lemma 13. Given a unit disk graph G, Gk is an intersection graph of unit ²-quasi-
disks with ² = k−1 k .
Proof. Recall that the intersection model, the containment model and the proximity model are all equivalent with some scaling modifications on the unit distance. It is easier to prove this lemma by considering the containment model. Given a unit disk graph G with unit distance 1 for the containment model, it is easy to see that for a given vertex v any neighbor of v in Gk is at most distance k from v in G,
there may exist vertices that are not adjacent to v but are at distance less than k, as in Figure 10. Thus, the actual neighborhood region is a subset of vertices covered by the disk of radius k. Let O denote the point that corresponds to vertex v. Since G is a unit disk graph, all vertices that are at most distance 1 from v are contained in the neighborhood region. Let Q denote the set of points, such that each point, if added as a new vertex to G, would be a neighbor of v in Gk. Then,
we have D(O, 1) ⊆ Q ⊆ D(O, k). When we scale all the coordinates of the points by multiplying by 1
k, we get D(O, 1 − ²) ⊆ Q ⊆ D(O, 1) with ² = k−1k , where for
convenience we use the same notations as before for transformed O and Q. Now, we need to show that Q is a connected compact set. It is compact since it is closed and is a subset of a compact set D(O, k). Now, assume that Q is not connected. Let u be a vertex in a compact set Q2 and let the vertex v belong to the compact set Q1 such
that Q1
S
Q2 = Q and Q1
T
Q2 = ®. Let T denote the shortest path from v to u in
G (see Figure 11). Then there exist vertices t1 ∈ Q1 and t2 ∈ Q2 in T such that they
are adjacent in G. As shown in Figure 12, consider the disk D(t1, dist(t1, t2)). Since
t2 is contained in Q, any vertex that lies on this disk also belongs to Q as it has a
shortest path to v of length less than k, since dist(v, t1) < dist(v, t2) ≤ k. Therefore,
we have Q1
S Q2
S
D(t1, dist(t1, t2)) ⊆ Q. Observe that not having any vertices of G
in this disk does not violate any property. Thus, Q is connected.
(0 ,0 ) (0 , 1 ) (0 ,2 ) (0 ,3 .1 )
a b c d
Pa t h T v w u Q 2 Q 1 Fig. 11 Q1 S
Q2 = Q contains the k-distance neighborhood of vertex v in G for
k > 0. Q 2 t1 t2 D (t 1,d ist (t1, t2 )) Q 1
By Lemma 13, the power graphs of unit disk graphs form a subclass of intersection graphs of unit ²-quasi-disks, on which the maximum clique problem is NP-hard.
String graphs are intersection graphs of curves in the plane. Each vertex corre- sponds to a curve and two vertices are joined by an edge if the corresponding curves intersect. Intersection graphs of ellipses are defined in a similar fashion. An inter- val graph is the intersection graph of a multiset of intervals on the real line. It has one vertex for each interval in the set, and an edge between every pair of vertices corresponding to intervals that intersect. Multiple interval graphs are a natural gen- eralization of interval graphs, where each vertex may have more than one interval associated with it. A circle graph is a graph whose vertices can be associated with chords of a circle such that two vertices are adjacent if and only if the corresponding chords in the circle intersect. A graph is chordal if each of its cycles of four or more nodes has a chord, which is an edge joining two nodes that are not adjacent in the cycle. An equivalent definition is that any chordless cycles have at most three nodes. The maximum clique problem is polynomial-time solvable in interval graphs, chordal graphs and circle graphs. It is proved in [22] that the maximum clique problem is NP-hard in t-interval graphs for t ≥ 3 by a reduction from the maximum 2-DNF satisfiability problem. Let ρ denote the ratio of the larger over the smaller radius of an ellipse. Ambuhl and Wagner [8] prove that the maximum clique problem is APX-hard in intersection graphs of ellipses for any 1 < ρ < ∞. This means that there is a constant c such that there is no approximation algorithm with ratio better than c. Thus, there is no PTAS. The proof is done by a reduction from the MAX5OCC2SAT problem (given a boolean formula in conjunctive normal form with at most two literals per clause and at most for five occurrences of every variable, find an assignment of truth values to the variables that satisfies the maximum number of clauses). Observe that for ρ = 1, we have unit disk graphs for which the maximum
clique problem is polynomial-time solvable. Thus, even a small change in ρ makes a huge complexity difference. The same is true for ρ = ∞, in which case the graph becomes an interval graph.
Kratochvil and Kubena [60] show that the maximum clique problem on the intersection graph of convex sets in the plane is NP-hard. The reduction for this proof is done from the maximum clique problem on co-planar graphs by showing that for every planar graph, one can assign convex sets in the plane to its vertices in such a way that two of the sets are disjoint if and only if the corresponding vertices are adjacent. Although it is easy to see that the kth power of a unit disk graph is not
necessarily an intersection graph of convex sets, this result is important in terms of providing a general idea on more general intersection graphs.