1 Introduction
The graph classes defined by forbidden cycles or induced cycles of certain lengths figure prominently in the structural graph theory, motivated in particular by the Strong Perfect Graph Theorem [4] which shows that perfect graphs are characterized by forbidden odd holes and their complements. The most interesting graph parameters in the context of these graph classes are the chromatic number, the independence number, and the clique number: while they are NPhard to approximate within any fixed precision [9] in general graphs, using semidefinite programming they can be determined in polynomial time for perfect graphs [5].
Perfect graphs also motivate the concept of boundedness. A class of graphs is bounded if the chromatic number of the graphs from the class can be bounded by a function of the clique number (of course, there is no such function in general, due to numerous known constructions of trianglefree graphs of arbitrarily large chromatic number). The notion of boundedness was introduced by Gyarfás [6], who also proposed a number of influential questions on this topic. As an example, he conjectured that graphs without odd holes are bounded; this conjecture was only recently confirmed by Scott and Seymour [8]. In a similar vein, Bonamy, Charbit, and Thomassé [2] showed that graph classes that forbid induced cycles of length that is a multiple of 3 have bounded chromatic number.
The aforementioned ideas commonly appear in the context of geometrically defined graph classes. The main motivation for our work comes from recent algorithmic results exploiting the property of not having two disjoint odd cycles in certain geometric graph classes. Specifically, Bonamy et al. [1] (improving upon an earlier work of Bonnet et al. [3]) show that the clique number can be approximated arbitrarily well in the intersection graphs of disks, and in the intersection graphs of unit balls (more precisely, there exists a randomized efficient polynomialtime approximation scheme for this problem). The key insight they use is that the complements of such graphs do not contain the disjoint union of two odd cycles as an induced subgraph, combined with additional properties derived from the geometry of the problems (bounded VC dimension, linear lower bound on the independence number).
More generally, the induced odd cycle packing number, of a graph G is the maximum integer such that contains an induced subgraph consisting of pairwise vertexdisjoint odd cycles. Note that whenever two cycles are connected by an edge, their vertices do not induce such a subgraph; furthermore, unlike the perfect graph case, we work with all odd cycles, not just with odd holes (i.e., iocp takes into account also triangles). We say that graphs with are OCfree. Bonamy et al. [1] proved the following.
Theorem 1 (Bonamy et al. [1]).
There exists a randomized algorithm taking as an input integers and a OCfree graph of VC dimension at most such that , and in time returns an independent set of of size at least .
Later (private communication), they proved that one can remove the assumption that the VC dimension is bounded, at the expense of making the exponent in the time complexity depend on the desired precision, i.e., obtaining a PTAS with time complexity for some function rather than an EPTAS. As our first result, we show that it is not necessary to make this sacrifice, obtaining an EPTAS without the assumption of bounded VC dimension. We present the improved argument in section 2.
In section 3 we focus our attention on coloring. We show that OCfree graphs are bounded by a function polynomial in the clique number (with the degree of the polynomial depending linearly on ). Furthermore, our proof of this fact can be turned into an algorithm running in polynomial time for fixed maximum clique size. In section 4 we follow by a lower bound on the bounding function for the case when . Finally, in section 5, we show how these results can be combined to obtain a QPTAS for the maximum independent set in graphs with bounded induced odd cycle packing number.
2 EPTAS assuming linear independence number
Let us start with a mild variation on a crucial part of the argument of Bonamy et al. [1], which we state in more generality than we actually need (a weighted setting), Lemma 3 below. The odd girth of a graph is the length of shortest odd cycle appearing in . We need an observation on neighborhoods of such shortest odd cycles.
Lemma 2.
Let be an odd integer, let be a graph of odd girth at least , and let be a shortest odd cycle in . Let be a nonnegative integer, and suppose that every vertex of is at distance at most from . Let be a vertex of , let denote the set of vertices of at distance at most from , and let be the set of vertices of at distance at most from . Then is bipartite.
Proof.
Note that is geodesic in , i.e., the distance between any two vertices of is the same in as in (otherwise, would contain a path between two vertices and of shorter than the distance between and in , and would contain an odd cycle shorter than ), and in particular is an induced cycle.
Let be a forest of shortest paths from vertices in to , and for each , let denote the vertex in which the component of containing intersects . Note that is also a forest, and thus it has a proper coloring . We claim that the restriction of is a proper coloring of . Suppose for a contradiction there exists an edge with . Since , we have , and thus there exists a unique path between and in . Since , the cycle has odd length, and since is a shortest odd cycle in , we have . In particular, contains both and . Let be the subpath of between and ; we have . Since is a subpath of the path and , we can by symmetry assume that the distance between and is at most , and thus . But then , which is a contradiction. ∎
Given an assignment of weights to vertices of , let for each set define and let be the maximum of over all independent sets in .
Lemma 3.
There exists an algorithm that, for input integers and , an vertex graph of induced odd cycle packing number at most and odd girth at least , and an assignment of weights to vertices, returns in time an independent set such that . Furthermore, every such graph satisfies .
Proof.
If is bipartite, then we can find an independent set in of largest weight via a maximum flow algorithm in ; and considering the heavier of the two color classes of , we have .
Hence, we can assume . We find (in by BFS from each vertex) a shortest odd cycle in , necessarily of length at least . Let be a set of vertices of at distance at least from one another, and for let denote the set of vertices of at distance at most from . Let denote the set of vertices of at distance exactly from .
We claim that for . Indeed, let be the subgraph of induced by vertices at distance less than from and the subgraph induced by vertices at distance greater than from , so that is disjoint union of and . Since contains the odd cycle whose vertices have no neighbors in , we have . Furthermore, by Lemma 2, the graph is bipartite, and thus .
Hence, we can call the algorithm recursively for the graphs for all with replaced by . As we branch into branches and recurse to the depth , this gives total time complexity . Out of the independent sets obtained from the recursive calls, we return the heaviest one. Consider a heaviest independent set in . Since the sets , …, are pairwise disjoint, and the sets , …, are pairwise disjoint as well, we have
and thus there exists such that . Hence, . Since the independent set returned from recursion has weight at least , we end up returning a set of weight at least as required.
An analogous argument shows there exists such that . Since , we have as required. ∎
For a set of graphs, an packing in a graph is a set of pairwisevertex disjoint induced subgraphs of , each isomorphic to a graph belonging to (note that we allow edges between members of , unlike in the definition of iocp). Let . For an integer , let denote the set of all odd cycles of length less than . A maximal packing in an vertex graph can be found in time by repeatedly finding a shortest induced odd cycle and deleting it from ; observe that has odd girth at least . Lemma 3 is used to deal with graphs without odd cycles of length less than ; so, it remains to handle the graphs containing an packing covering a large fraction of the vertices.
For a set , let denote the set of vertices of with a neighbor in . For positive integers and a positive real number , let
be the upper bound on the probability that a uniformly random element subset of is disjoint from a fixed subset of size at least .
Lemma 4.
There exists a randomized algorithm that, for input integers (with ) and an vertex graph (with ) of induced odd cycle packing number at most , returns in time an independent set , induced subgraphs , …, (for some ) of , each containing an odd cycle as a connected component, and induced subgraphs of , such that at least one of the following claims holds:

, or

there exists such that , or

with probability at least , there exists such that and .
Proof.
Let . Let be a maximal packing in . Let be an independent set in found using the algorithm from Lemma 3 with . Let and for , let . Let be a subset of vertices of chosen uniformly at random, and for , let .
Suppose first that . Since , Lemma 3 implies , and thus . Consequently, , and by Lemma 3, we have , implying (a) holds.
Hence, we can assume . Let be a largest independent set in . If there exists such that , then (b) holds. Hence, assume that for every ; and consequently, for each such , there exists such that . Let and note that . By doublecounting the number of edges of between and , we have
Let consist of vertices of of degree greater than . Since is a largest independent set in , for every we have . Hence, if , then (c) holds. We have
and since each vertex has degree less than , we have . Hence, the probability that is at most . ∎
By recursing in the branches given by (b) and (c), we obtain a randomized EPTAS under assumption that .
Theorem 5.
There exists a randomized algorithm that, for input integers and and an vertex graph of induced odd cycle packing number at most , returns in time an independent set of whose size is at least with probability at least .
Proof.
Let . Let be the smallest integer such that . Let be the smallest integer such that for every ; note that such an integer exists, since . Let us now describe a recursive procedure that, applied to an induced subgraph of and a nonnegative integer such that , returns (with high probability) an independent set of of size at least .
If , then we return an empty set (or any other independent set in ). If , then is bipartite and we can find a largest independent set in via a maximum flow algorithm in . Hence, suppose that . Apply the algorithm from Lemma 4 to (using as ) to obtain an independent set and induced subgraphs , …, , , …, . For , let be a component of isomorphic to an odd cycle; clearly ; let be an independent set in obtained by a recursive call for with replaced by , together with an independent set of size in . For , if , then let be the independent set in obtained by a recursive call for , otherwise let . We return the largest of the sets , , …, , , …, .
Let us analyze this procedure when applied to with . As the recursion stops whenever the graph has at most vertices and in the recursive calls on the subgraphs , we decrease the number of vertices by the factor of at most , this amounts to at most calls for the initial value of . Inside each of the calls, we also recurse with replaced by to at most induced subgraphs , …, . Hence, the total number of calls to the procedure for various induced subgraphs of and values of is at most . The time complexity of each of the calls is dominated by the time complexity of the algorithm from Lemma 4, and thus the total time complexity is as required.
Let us now argue about correctness. If , then any independent set in is a valid answer. In case , we return an exact largest independent set. In these cases, we say that the current call is a final one. Assuming and , we now have the following possibilities depending on which outcome of Lemma 4 holds.

. In this case the returned independent set has size at least . We also say that the current call is a final one.

Or, there exists such that ; in this case, we say that the recursive call for is an unconditionally correct branch. Note that if the recursive call to returns an independent set of size at least , then the returned independent set has size at least .

Or, since , with probability at least , there exists such that and ; in this case we say that the recursive call for is a conditionally correct branch. Note that if the recursive call to returns an independent set of size at least , then the returned independent set has size at least .
Let us trace the run of the procedure when applied to with , till we reach either a final node or a node with no correct branch, taking an unconditionally correct branch when available. Note that if we reach a final node, then by the above analysis, the returned independent set will have size at least . Furthermore, whenever we take a conditionally correct branch, the number of vertices decreases by a factor of at most , and thus on the considered path, there are at most nodes without an unconditionally correct branch. Since the probability that a nonfinal node has neither an unconditionally nor a conditionally correct branch is at most , the probability that we stop in a nonfinal node is at most . ∎
Note we can iterate the algorithm times and return the largest of the returned sets, reducing the probability of a result worse than to .
3 boundedness
The classes of graphs with bounded induced odd packing number are bounded, that is, their chromatic number is bounded by a function of their maximum clique size. Let us start with the trianglefree case, where we need to show an absolute bound on the chromatic number.
Lemma 6.
Every trianglefree graph satisfies . Furthermore, if has odd girth at least , then , and if has girth at least , then .
Proof.
We prove the claim by induction on the induced odd cycle packing number. If , then is bipartite and , hence suppose that and the claim holds for all graphs with smaller induced odd cycle packing number. Let be a shortest odd cycle in , which is necessarily induced. Since , we can color by the induction hypothesis, and it suffices to show how to color using at most extra colors, or at most extra colors if has odd girth at least , or using at most one extra color if has girth at least .
Let be a set consisting of three consecutive vertices of and let . By Lemma 2, is bipartite. Since is trianglefree, the neighborhood of any vertex is an independent set, and is disjoint from . If has odd girth at least , then is an independent set as well; hence, we can use two new colors to color , one color for (which includes and ) and one color for (which includes ), using four extra colors in total. Otherwise, we use two new colors to color , one color for , one color for , and one color for , using five extra colors in total.
In the case that has girth at least , we claim that is an independent set. Indeed, suppose for a contradiction vertices are adjacent, and let and be neighbors of and in , respectively. Since has girth at least , the distance between and in is greater than three. However, then contains an odd cycle shorter than , which is a contradiction. Hence, we can use three of the colors used on to color and one extra color to color . ∎
For the case contains triangles, let for every positive integer , and for , let us inductively define .
Theorem 7.
Every graph satisfies .
Proof.
We prove the claim by induction on the induced odd cycle packing number. If , then is bipartite and , hence suppose that and the claim holds for all graphs with smaller induced odd cycle packing number. Let be a largest clique in , and for each , let denote the set of vertices of to which is not adjacent; the maximality of implies . Let be an arbitrary subset of of size . For a set with , let . If , then the maximality of implies is an independent set; for each element set , we use one color for all vertices of . If , then the maximality of implies is trianglefree, and by Lemma 6, we can use colors to color . Finally, if , then , since all vertices in are nonadjacent to the triangle induced by ; hence, we can use colors to color . Summing the numbers of colors over all choices of , we conclude that at most colors are used to color . ∎
Let us remark that we can obtain the coloring as in Theorem 7 in polynomial time: instead of choosing as a largest clique, the inspection of the proof shows that it suffices to choose one which cannot be enlarged by adding at most three and removing at most two vertices, and such a clique can be found in polynomial time.
4 Lowerbound for bounding function
In the previous section we showed that every graph has chromatic number at most , where is of order roughly . We do not know whether this upper bound is tight. In this section we show that there exist graphs with induced odd cycle packing number one whose chromatic number is almost quadratic in . We give a probabilistic construction. In order to carry out the probabilistic calculations, we will use the following bounds.
Lemma 8 (See [7], Chernoff Bound).
Suppose is a sum of independent variables. For any ,
Lemma 9 (See [7], Talagrand’s Inequality II).
Let
be a nonnegative random variable, not identically
, which is determined by independent trials , …, , and satisfying the following for some integers :
Changing the outcome of any one trial can change by at most .

For any nonnegative integer , if , then there is a set of at most trials whose outcomes certify that .
Then for every nonnegative ,
We are now ready to give the construction (a variation on a standard construction of trianglefree graphs with no large independent set).
Theorem 10.
There exists a family of graphs with induced odd cycle packing number at most one and with arbitrarily large clique number such that every graph in this family satisfies .
Proof.
Let be a random graph for , where is a sufficiently large even integer and .
Suppose are disjoint and have size three, and contains all nine edges with one end in and the other end in ; in this case, we say that the set of these nine edges forms a . We say a set of edges of is bad if it can be partitioned into subsets of size , each of which forms a . Finally we say a set of edges is subbad if it is a subset of a bad set.
We construct a graph by deleting a maximal bad set from . By the maximality of , no set of edges of forms a . Let be the complement of . Consider any disjoint (odd) cycles in . If there were no edge between these cycles, then any pair of triples of vertices taken one from each cycle would induce a complement of a supergraph of in , which is a contradiction. Consequently . Furthermore, an analogous argument shows , and thus .
Therefore, it suffices to argue that with nonzero probability. Let us consider a set of size at least and define the following random variables.
The probability that is an independent set in is at most . Indeed, if is independent, then is subbad, and thus . Let , so that . Using the Chernoff bound (Lemma 8), we obtain
Let . Clearly, if is subbad, then , and thus . For distinct vertices , the probability that is an edge and some set containing forms a in is at most , and thus . Let and let , so that . Note that flipping the existence of a single edge in changes by at most . Furthermore, when , there exist at most edges in whose presence certifies this is the case. Hence, we can apply the Talagrand’s inequality (Lemma 9) with for . Note that for large enough, since . Hence, we have
It follows that
for large enough.
Therefore, for any set of size , the probability that is an independent set in is at most . Hence, the probability that contains an independent set of size is less than
that is, smaller than when . ∎
As noted in the proof, the probabilistic construction used is unnecessarily restrictive, excluding all disjoint cycles regardless of parity. Similarly all cycles or paths of length are excluded. From the point of view of vertex 6tuples, all of these structures exhibit very similar properties. It would seem that achieving a distinction between these patterns and the ones that are necessary to avoid requires more global conditions and thus a much more refined approach.
5 QPTAS assuming only bounded iocp
The fact that trianglefree graphs with bounded iocp have bounded chromatic number has the following easy consequence.
Lemma 11.
For all integers , every graph with vertices satisfies at least one of the following conditions:

has an independent set of size at least , or

every maximal packing of triangles in contains a triangle such that , or

contains a vertex of degree at least such that .
Proof.
Let be a maximal packing of triangles in . The graph is trianglefree, and by Lemma 6, . Consequently, . Suppose that does not have any independent set of size at least , and thus . Let be a largest independent set in . If for some , then the second outcome of the lemma holds.
Otherwise, for each , there exists such that . Consequently, there exists a vertex such that , and thus . Since , we have . ∎
Combining this lemma with Theorem 5, we obtain a QPTAS for the maximum independent set in graphs of bounded induced odd cycle packing number.
Theorem 12.
There exists a randomized algorithm that, for input integers and and an vertex graph of induced odd cycle packing number at most , returns in time an independent set of whose size is at least with probability at least .
Proof.
If (so is bipartite), we return the largest maximum independent set obtained via a maximum flow algorithm. Otherwise, we find a maximal packing of triangles in greedily, and return the largest of the independent sets obtained by

running the algorithm from Theorem 5 times with ,

for each , running the algorithm recursively for with replaced by and adding a vertex of to the returned independent set, and

for each of degree at least , running the algorithm recursively for .
Each recursive call either decreases or decreases the number of vertices by a factor of at most , implying the total number of the calls of the procedure is at most . Iterating the algorithm from Theorem 5 times for an induced subgraph of ensures we fail to find an independent set of size at least with probability at most . Hence, with probability at least (for large enough—for small , we can just find the largest independent set by brute force), we can assume that throughout the run of the algorithm, in part (a) for an induced subgraph of , at least one of the returned independent sets has size at least .
If , then in (a) we return an independent set of size at least . If contains a vertex of degree at least such that , then in (c) the corresponding recursive call gives an independent set of size at least .
If neither of these conditions holds, Lemma 11 implies there exists a triangle such that . The recursive call in (b) returns an independent set of of size at least , and since , the addition of a vertex of turns into an independent set of size at least
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