Let G(V, E) denote a graph. A cut is a partition of the vertices V into two subsets V1 and V2. Any edge (u, v) ∈ E with u ∈ V1 and v ∈ V2 is said to be crossing the
cut and is a cut edge. The size of a cut is the total number of edges crossing the cut. In edge-weighted graphs, the size of the cut is defined to be sum of weights of
the edges crossing the cut. The max-cut problem is to find a cut of maximum size. The variation, in which each partition set is required to have the same cardinality, is called the max-bisection problem.
In a recent paper, Diaz et al. [32] prove that both the max-cut and the max- bisection problems are NP-hard on unit disk graphs. They use the fact that max-cut problem is NP-hard on graphs with bounded degree for ∆ ≥ 3. They show that max- cut on a graph with bounded degree 4 can be reduced to the max-cut problem on a unit disk graph in polynomial time. The complexity proof of max-bisection easily follows by a reduction from max-cut in unit disk graphs. More specifically, finding the max-cut in G is equivalent to finding a max-bisection in G0, which is formed by
two copies of G. Hunt el al. [54] investigate PTAS’s for various problems in unit disk graphs. They state that there is a PTAS for max-cut problem in λ-precision unit disk graphs (λ-precision implies that centers of unit disks are at least distance λ away from each other).
Jansen et al. [55] present a PTAS for max-bisection in unit disk graphs. By imposing grids on the plane, the graph is divided into subgraphs. If the subgraph is dense, then PTAS for dense instances of max-bisection is used as proposed by Arora et al. [9]. Otherwise, the solution is obtained by enumeration. Authors also use shifting to obtain different subgraphs. The solution for each subgraph S, all maximum (ns− p, p) partitions, where ns is the number of vertices in S, are used to
obtain an overall solution for the input graph via dynamic programming. We observe that this technique can also be used for unit ball graphs with some adjustments. Theorem 8. Max-bisection and max-cut problems both admit PTAS in unit ball graphs.
adjustments.
Obtaining subgraphs: The first step is to impose a 3-dimensional grid. Each cell of this grid is a cube with side length 2. The h-th yz-plane is at x = 2h, −∞ < h < ∞. Furthermore, the h-th yz-strip is the strip between the h-th and (h + 1)-st yz planes. The xz-planes and xz-strips, xy-planes and xy-strips are indexed similarly. Each strip is open on one side and closed on the other side. Thus, each unit ball is centered in exactly one strip.
For a fixed integer k, the subgraph Hi,j,l of the input graph G, −∞ < i, j, l < ∞,
is the subgraph induced by the centers of unit balls that lie in the intersection of the yz-strips i, i + 1, . . . , i + k, the xz-strips j, j + 1, . . . , j + k and the xy- strips l, l + 1, . . . , l + k. The number of vertices of Hi,j,l is denoted by ni,j,l. We
observe that the size of a maximum independent set on Hi,j,l is bounded by
6(k + 2)3/π. The unit balls in H
i,j,l are contained in a cube with side length
2(k + 1) + 2 = 2(k + 2). Dividing the volume of this cube by the volume of a unit ball, we can fit at most 6(k + 2)3/π non-adjacent unit balls in this region.
Lemma 4 of Jansen et al. [55] also holds for subgraph Hi,j,l. The lemma states
that there is a positive constant c such that if ni,j,l > c log(n), then the subgraph
Hi,j,l is dense. This is proved by showing that there exist at least ni,j,lπ/6(k +
2)3 maximal independent sets in H
i,j,l. If each maximal independent set is
considered as a vertex, since they are all maximal, we have a complete graph. Thus, the number of edges in Hi,j,l is Ω((ni,j,l)2). Thus, Corollary 4 of the same
paper [55] also holds. This corollary states that if ni,j,l > c log(n), then the size
of a max-bisection of Hi,j,l is Ω((ni,j,l)2). This directly shows that the max-cut
size of Hi,j,l is also Ω((ni,j,l)2).
partitions of Hi,j,l. If Hi,j,l is dense, a solution with an additive error of 2²(ni,j,l)2
is obtained by solving a polynomial integer program, which is presented by Jansen et al. [55] and Arora et al. [9]. Otherwise, an optimal solution is com- puted by enumeration.
Combining solutions to get an overall solution: The graph Gr,s,t for each r, s
and t, 0 ≤ r, s, t ≤ k, is defined as the union of all subgraphs Hi,j,l, where
i(modk + 1) = r, j(modk + 1) = s and l(modk + 1) = t. All maximum (nr,s,t−
p, p)-partitions of Gr,s,tare obtained by merging solutions of each subgraph Hi,j,l
of Gr,s,t. First, the subgraphs are ordered in increasing order of the sum i+j +l.
All partitions of consecutive pairs are computed first by using solutions of each subgraph, then the same is done for quadruples by using the solutions of pairs. In this fashion, all maximum partitions are computed for Gr,s,t.
Output: For max-bisection, we pick the largest bisection of Gr,s,t, 0 ≤ r, s, t ≤ k,
and for max-cut we pick the largest partition.
Lemma 11. The performance ratio of the above algorithm is (1 − 1
k+1)3(1 − 2²) for
a fixed integer k, and a given ² < 1.
Proof. The proof is inline with the two-dimensional case [55]. Suppose we deal with max-cut problem. Let C∗ be a max-cut of G. Observe that any graph G
r,s,t misses
some edges that are present in G. Particularly, those are the edges that have endpoints in different subgraphs Hi,j,l of Gr,s,t. For any edge e of C∗, there is at most one r,
0 ≤ r ≤ k, such that e cuts a yz-plane whose index modulo k + 1 is r for fixed s and t. The same holds for any fixed pair of (r, s, t). Thus, there exist a graph Gr,s,t, 0 ≤ r, s, t ≤ k, such that a max-cut of Gr,s,t has at least (1 − k+11 )3|C∗| edges.
and each such subgraph has a max-cut of Ω((ni,j,l)2), the solution obtained for each
Gr,s,t is a (1 − 2²)-approximation of the optimal solution. Therefore, for the input
graph G, we obtain a performance ratio of (1 − 1
k+1)3(1 − 2²).