Induced odd cycle packing number, independent sets, and chromatic number
aa r X i v : . [ c s . D M ] J a n Induced odd cycle packing number,independent sets, and chromatic number
Zdenˇek Dvoˇr´ak ∗ Jakub Pek´arek † January 9, 2020
Abstract
The induced odd cycle packing number iocp( G ) of a graph G is the max-imum integer k such that G contains an induced subgraph consisting of k pairwise vertex-disjoint odd cycles. Motivated by applications to geometricgraphs, Bonamy et al. [1] proved that graphs of bounded induced odd cyclepacking number, bounded VC dimension, and linear independence numberadmit a randomized EPTAS for the independence number. We show thatthe assumption of bounded VC dimension is not necessary, exhibiting a ran-domized algorithm that for any integers k ≥ t ≥ n -vertexgraph G of induced odd cycle packing number returns in time O k , t ( n k + ) anindependent set of G whose size is at least α ( G ) − n / t with high probability.In addition, we present χ -boundedness results for graphs with bounded oddcycle packing number, and use them to design a QPTAS for the indepen-dence number only assuming bounded induced odd cycle packing number. The graph classes defined by forbidden cycles or induced cycles of certain lengthsfigure prominently in the structural graph theory, motivated in particular by the ∗ Charles University, Prague, Czech Republic. E-mail: [email protected] . Sup-ported by project 17-04611S (Ramsey-like aspects of graph coloring) of Czech Science Founda-tion. † Charles University, Prague, Czech Republic. E-mail: [email protected] . Sup-ported by GAUK grant 118119 (Algorithms for graphs with restrictions on cycles). χ -boundedness. A class of graphsis χ -bounded if the chromatic number of the graphs from the class can be boundedby a function of the clique number (of course, there is no such function in general,due to numerous known constructions of triangle-free graphs of arbitrarily largechromatic number). The notion of χ -boundedness was introduced by Gyarf´as [6],who also proposed a number of influential questions on this topic. As an example,he conjectured that graphs without odd holes are χ -bounded; this conjecture wasonly recently confirmed by Scott and Seymour [8]. In a similar vein, Bonamy,Charbit, and Thomass´e [2] showed that graph classes that forbid induced cyclesof length that is a multiple of 3 have bounded chromatic number.The aforementioned ideas commonly appear in the context of geometricallydefined graph classes. The main motivation for our work comes from recent al-gorithmic results exploiting the property of not having two disjoint odd cyclesin certain geometric graph classes. Specifically, Bonamy et al. [1] (improvingupon an earlier work of Bonnet et al. [3]) show that the clique number can beapproximated arbitrarily well in the intersection graphs of disks, and in the in-tersection graphs of unit balls (more precisely, there exists a randomized e ffi cientpolynomial-time approximation scheme for this problem). The key insight theyuse is that the complements of such graphs do not contain the disjoint union of twoodd cycles as an induced subgraph, combined with additional properties derivedfrom the geometry of the problems (bounded VC dimension, linear lower boundon the independence number).More generally, the induced odd cycle packing number , iocp( G ) of a graph Gis the maximum integer k such that G contains an induced subgraph consisting of k pairwise vertex-disjoint odd cycles. Note that whenever two cycles are connectedby an edge, their vertices do not induce such a subgraph; furthermore, unlike theperfect 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 iocp( G ) < k are k-OC-free . Bonamy et al. [1] proved the following. Theorem 1 (Bonamy et al. [1]) . There exists a randomized algorithm taking asan input integers k , b , c , a ≥ and a k-OC-free graph G of VC dimension at most such that α ( G ) ≥ | V ( G ) | / b, and in time O k , b , c , a ( | V ( G ) | ) returns an independentset of G of size at least (1 − / a ) α ( G ) . Later (private communication), they proved that one can remove the assump-tion that the VC dimension is bounded, at the expense of making the exponent inthe time complexity depend on the desired precision, i.e., obtaining a PTAS withtime complexity O ( | V ( G ) | f ( k , b , a ) ) for some function f rather than an EPTAS. Asour first result, we show that it is not necessary to make this sacrifice, obtainingan EPTAS without the assumption of bounded VC dimension. We present theimproved argument in section 2.In section 3 we focus our attention on coloring. We show that k -OC-freegraphs are χ -bounded by a function polynomial in the clique number (with thedegree of the polynomial depending linearly on k ). Furthermore, our proof ofthis fact can be turned into an algorithm running in polynomial time for fixedmaximum clique size. In section 4 we follow by a lower bound on the χ -boundingfunction for the case when k =
1. Finally, in section 5, we show how these resultscan be combined to obtain a QPTAS for the maximum independent set in graphswith bounded induced odd cycle packing number.
Let us start with a mild variation on a crucial part of the argument of Bonamyet al. [1], which we state in more generality than we actually need (a weightedsetting), Lemma 3 below. The odd girth of a graph G is the length of shortest oddcycle appearing in G . We need an observation on neighborhoods of such shortestodd cycles. Lemma 2.
Let g be an odd integer, let G be a graph of odd girth at least g, andlet C be a shortest odd cycle in G. Let t ≤ ( g − / be a non-negative integer, andsuppose that every vertex of G is at distance at most t from C. Let z be a vertex ofC, let A denote the set of vertices of C at distance at most t from z, and let R bethe set of vertices of G at distance at most t from A. Then G − R is bipartite.Proof.
Note that C is geodesic in G , i.e., the distance between any two verticesof C is the same in C as in G (otherwise, G would contain a path Q between twovertices x and y of C shorter than the distance between x and y in C , and C ∪ Q would contain an odd cycle shorter than C ), and in particular C is an inducedcycle. 3et F be a forest of shortest paths from vertices in G to V ( C ), and for each v ∈ V ( G ), let f ( v ) denote the vertex in which the component of F containing v intersects C . Note that ( F ∪ C ) − z is also a forest, and thus it has a proper 2-coloring ψ . We claim that the restriction of ψ is a proper 2-coloring of G − R .Suppose for a contradiction there exists an edge uv ∈ E ( G − R ) with ψ ( u ) = ψ ( v ).Since u , v < R , we have f ( u ) , z , f ( v ), and thus there exists a unique path P between u and v in ( F ∪ C ) − z . Since ψ ( u ) = ψ ( v ), the cycle P + uv has odd length,and since C is a shortest odd cycle in G , we have | E ( P ) | + ≥ | C | ≥ g ≥ t + P contains both f ( u ) and f ( v ). Let P ′ be the subpath of P between f ( u ) and f ( v ); we have | E ( P ′ ) | ≥ | E ( P ) | − t ≥ | C | − t −
1. Since P ′ is a subpathof the path C − z and | E ( C − z ) | = | C | −
2, we can by symmetry assume that thedistance between f ( u ) and z is at most t , and thus f ( u ) ∈ A . But then u ∈ R , whichis a contradiction. (cid:3) Given an assignment w : V ( G ) → Z + of weights to vertices of G , let for eachset X ⊆ V ( G ) define w ( X ) = P v ∈ X w ( v ) and let α w ( G ) be the maximum of w ( X )over all independent sets X in G . Lemma 3.
There exists an algorithm that, for input integers k ≥ and b ≥ , ann-vertex graph G of induced odd cycle packing number at most k and odd girth atleast b (8 b − , and an assignment w : V ( G ) → Z + of weights to vertices, returnsin time O ((2 b ) k n ) an independent set X ⊆ V ( G ) such that w ( X ) ≥ (1 − k / b ) α w ( G ) .Furthermore, every such graph satisfies α w ( G ) ≥ (1 − k / b ) w ( V ( G )) / .Proof. If G is bipartite, then we can find an independent set in G of largest weightvia a maximum flow algorithm in O ( n ); and considering the heavier of the twocolor classes of G , we have α w ( G ) ≥ w ( V ( G )) / k >
0. We find (in O ( n ) by BFS from each vertex)a shortest odd cycle C in G , necessarily of length at least 2 b (8 b − Z = { z , . . . , z b } be a set of vertices of C at distance at least 8 b − i ∈ { , . . . , b } let R i denote the set of vertices of G at distance at most4 b − z i . Let L i denote the set of vertices of G at distance exactly i from C .We claim that iocp( G − L i − R i ) ≤ k − ≤ i ≤ b . Indeed, let G bethe subgraph of G induced by vertices at distance less than i from C and G thesubgraph induced by vertices at distance greater than i from C , so that G − L i isdisjoint union of G and G . Since G contains the odd cycle C whose verticeshave no neighbors in G , we have iocp( G ) ≤ iocp( G − L i ) − ≤ k −
1. Furthermore,by Lemma 2, the graph G − R i is bipartite, and thus iocp( G − L i − R i ) = iocp( G − R i ) + iocp( G − R i ) ≤ k −
1. 4ence, we can call the algorithm recursively for the graphs G i = G − L i − R i for all i ∈ { , . . . , b } with k replaced by k −
1. As we branch into 2 b branchesand recurse to the depth k , this gives total time complexity O ((2 b ) k n ). Out ofthe independent sets obtained from the recursive calls, we return the heaviest one.Consider a heaviest independent set I in G . Since the sets L , . . . , L b are pairwisedisjoint, and the sets R , . . . , R b are pairwise disjoint as well, we have b X i = w ( I ∩ ( L i ∪ R i )) ≤ w ( I ) , and thus there exists i ∈ { , . . . , b } such that w ( I ∩ ( L i ∪ R i )) ≤ w ( I ) / b . Hence, α w ( G i ) ≥ (1 − / b ) α w ( G ). Since the independent set returned from recursion hasweight at least (1 − ( k − / b ) α w ( G i ), we end up returning a set of weight at least(1 − ( k − / b )(1 − / b ) α w ( G ) ≥ (1 − k / b ) α w ( G ) as required.An analogous argument shows there exists j ∈ { , . . . , b } such that w ( V ( G j )) ≥ (1 − / b ) w ( V ( G )). Since α w ( G j ) ≥ (1 − ( k − / b ) w ( V ( G j )) /
2, we have α w ( G ) ≥ (1 − ( k − / b )(1 − / b )) w ( V ( G )) / ≥ (1 − k / b ) w ( V ( G )) / (cid:3) For a set S of graphs, an S -packing in a graph G is a set X of pairwise-vertexdisjoint induced subgraphs of G , each isomorphic to a graph belonging to S (notethat we allow edges between members of X , unlike in the definition of iocp). Let V ( X ) = S X ∈X V ( X ). For an integer g ≥
3, let S g denote the set of all odd cycles oflength less than g . A maximal S g -packing X in an n -vertex graph G can be foundin time O ( n ) by repeatedly finding a shortest induced odd cycle and deleting itfrom G ; observe that G − V ( X ) has odd girth at least g . Lemma 3 is used to dealwith graphs without odd cycles of length less than g ; so, it remains to handle thegraphs containing an S g -packing covering a large fraction of the vertices.For a set Y ⊆ V ( G ), let N ( Y ) denote the set of vertices of V ( G ) \ Y with aneighbor in Y . For positive integers n ≥ r and a positive real number s ≤ n , let q ( n , r , s ) = (cid:16) n −⌈ s ⌉ r (cid:17)(cid:16) nr (cid:17) be the upper bound on the probability that a uniformly random r -element subsetof { , . . . , n } is disjoint from a fixed subset of size at least s . Lemma 4.
There exists a randomized algorithm that, for input integers k , r , p ≥ (with p ≥ k) and an n-vertex graph G (with n ≥ r) of induced odd cycle packingnumber at most k, returns in time O ( n + (4 p ) k n + rn ) an independent set I, nduced subgraphs H , . . . , H m (for some m ≤ n) of G, each containing an oddcycle as a connected component, and induced subgraphs G , . . . , G r of G, suchthat at least one of the following claims holds:(a) | I | ≥ (1 − k / p ) α ( G ) , or(b) there exists i ∈ { , . . . , m } such that α ( H i ) ≥ (1 − / p ) α ( G ) , or(c) with probability at least − q (cid:0) n , r , k p α ( G ) (cid:1) , there exists j ∈ { , . . . , r } such that α ( G j ) = α ( G ) and | V ( G j ) | < (cid:0) − k p (cid:1) n.Proof. Let g = p (16 p − X be a maximal S g -packing in G . Let I bean independent set in G − V ( X ) found using the algorithm from Lemma 3 with b = p . Let X = { C , . . . , C m } and for i ∈ { , . . . , m } , let H i = G − N ( C i ). Let U = { u , . . . , u r } be a subset of r vertices of G chosen uniformly at random, andfor j ∈ { , . . . , r } , let G j = G − N ( { u j } ).Suppose first that | V ( X ) | ≤ k p n . Since k p ≤ .
1, Lemma 3 implies α ( G ) ≥ α ( G − V ( X )) ≥ . − k / b ) n ≥ . n , and thus | V ( X ) | ≤ k p α ( G ). Consequently, α ( G − V ( X )) ≥ (cid:0) − k p (cid:1) α ( G ), and by Lemma 3, we have | I | ≥ (cid:0) − k p (cid:1) α ( G − V ( X )) ≥ (1 − k / p ) α ( G ), implying (a) holds.Hence, we can assume | V ( X ) | > k p n . Let J be a largest independent set in G .If there exists i ∈ { , . . . , m } such that | N ( C i ) ∩ J | ≤ | J | / p , then (b) holds. Hence,assume that | N ( C i ) ∩ J | > | J | / p for every i ∈ { , . . . , m } ; and consequently, foreach such i , there exists b i ∈ V ( C i ) such that | N ( b i ) ∩ J | > | J | pg . Let B = { b , . . . , b m } and note that m ≥ | V ( X ) | / g > k pg n . By double-counting the number of edges of G between B and J , we have X v ∈ J deg v ≥ X v ∈ J | N ( v ) ∩ B | = X b ∈ B | N ( b ) ∩ J | > m | J | pg > k p g | J | n . Let J ′ consist of vertices of J of degree greater than k p g n ≥ k p n . Since J is a largest independent set in G , for every v ∈ J we have α ( G − N ( v )) = α ( G ).Hence, if U ∩ J ′ , ∅ , then (c) holds. We have X v ∈ J ′ deg v > k p g | J | n − X v ∈ J \ J ′ deg v ≥ k p g | J | n , n , we have | J ′ | > k p g | J | = k p g α ( G ) ≥ k p α ( G ). Hence, the probability that U ∩ J ′ = ∅ is at most q (cid:0) n , r , k p α ( G ) (cid:1) . (cid:3) By recursing in the branches given by (b) and (c), we obtain a randomizedEPTAS under assumption that α ( G ) = Ω ( | V ( G ) | ). Theorem 5.
There exists a randomized algorithm that, for input integers k ≥ and t ≥ and an n-vertex graph G of induced odd cycle packing number at most k,returns in time O k , t ( n k + ) an independent set of G whose size is at least α ( G ) − n / twith probability at least / .Proof. Let p = kt . Let d ≥ (cid:0) − p (cid:1) d < / p .Let r be the smallest integer such that q (cid:0) n , r , p n (cid:1) ≤ d for every n ≥
0; notethat such an integer exists, since q (cid:0) n , r , p n (cid:1) ≤ (cid:0) − p (cid:1) r . Let us nowdescribe a recursive procedure that, applied to an induced subgraph G ′ of G and anon-negative integer k ′ ≤ k such that iocp( G ′ ) ≤ k ′ , returns (with high probability)an independent set of G ′ of size at least α ( G ′ ) − nk ′ / p .If | V ( G ′ ) | ≤ nk ′ / p , then we return an empty set (or any other independent setin G ′ ). If k ′ =
0, then G ′ is bipartite and we can find a largest independent set in G ′ via a maximum flow algorithm in O ( n ). Hence, suppose that k ′ ≥
1. Applythe algorithm from Lemma 4 to G ′ (using k ′ as k ) to obtain an independent set I and induced subgraphs H , . . . , H m , G , . . . , G r . For i ∈ { , . . . , m } , let C i be acomponent of H i isomorphic to an odd cycle; clearly iocp( H i − V ( C i )) ≤ k ′ −
1; let I i be an independent set in H i obtained by a recursive call for H i − V ( C i ) with k ′ replaced by k ′ −
1, together with an independent set of size α ( C i ) = ⌈| C i | / ⌉ in C i .For j ∈ { , . . . , r } , if | V ( G j ) | < (cid:0) − p (cid:1) | V ( G ′ ) | , then let J j be the independentset in G j obtained by a recursive call for G j , otherwise let J j = ∅ . We return thelargest of the sets I , I , . . . , I m , J , . . . , J r .Let us analyze this procedure when applied to G with k ′ = k . As the recursionstops whenever the graph has at most n / p vertices and in the recursive calls onthe subgraphs G j , we decrease the number of vertices by the factor of at most1 − p , this amounts to at most r d calls for the initial value of k . Inside each ofthe calls, we also recurse with k replaced by k − n induced subgraphs H − V ( C ), . . . , H m − V ( C m ). Hence, the total number of calls to the procedurefor various induced subgraphs of G and values of k ′ is at most (cid:0) r d n (cid:1) k = O k , t ( n k ).The time complexity of each of the calls is dominated by the time complexity ofthe algorithm from Lemma 4, and thus the total time complexity is O k , t ( n k + ) asrequired. 7et us now argue about correctness. If α ( G ′ ) ≤ nk ′ / p , then any independentset in G ′ is a valid answer. In case k ′ =
0, we return an exact largest independentset. In these cases, we say that the current call is a final one. Assuming k ′ ≥ α ( G ′ ) > nk ′ / p ≥ n / p , we now have the following possibilities depending onwhich outcome of Lemma 4 holds.(a) | I | ≥ (1 − k ′ / p ) α ( G ′ ). In this case the returned independent set has size atleast | I | ≥ α ( G ′ ) − k ′ p α ( G ′ ) ≥ α ( G ′ ) − nk ′ / p . We also say that the current callis a final one.(b) Or, there exists i ∈ { , . . . , m } such that α ( H i ) ≥ (1 − / p ) α ( G ′ ); in this case,we say that the recursive call for H i − V ( C i ) is an unconditionally correctbranch . Note that if the recursive call to H i − V ( C i ) returns an independentset of size at least α ( H i − V ( C i )) − n ( k ′ − / p , then the returned independentset has size at least α ( C i ) + α ( H i − V ( C i )) − n ( k ′ − / p = α ( H i ) − n ( k ′ − / p ≥ α ( G ′ ) − n / p − n ( k ′ − / p = α ( G ′ ) − nk ′ / p .(c) Or, since α ( G ′ ) > n / p , with probability at least 1 − q (cid:0) n , r , p n (cid:1) ≥ − d ,there exists j ∈ { , . . . , r } such that α ( G j ) = α ( G ′ ) and | V ( G j ) | < (cid:0) − p (cid:1) | V ( G ′ ) | ; in this case we say that the recursive call for G j is a con-ditionally correct branch . Note that if the recursive call to G j returns anindependent set J j of size at least α ( G j ) − nk ′ / p , then the returned indepen-dent set has size at least | J j | ≥ α ( G j ) − nk ′ / p = α ( G ′ ) − nk ′ / p .Let us trace the run of the procedure when applied to G with k ′ = k , till we reacheither a final node or a node with no correct branch, taking an unconditionallycorrect branch when available. Note that if we reach a final node, then by theabove analysis, the returned independent set will have size at least α ( G ) − nk / p = α ( G ) − n / t . Furthermore, whenever we take a conditionally correct branch, thenumber of vertices decreases by a factor of at most 1 − p , and thus on theconsidered path, there are at most d nodes without an unconditionally correctbranch. Since the probability that a non-final node has neither an unconditionallynor a conditionally correct branch is at most d , the probability that we stop in anon-final node is at most 1 / (cid:3) Note we can iterate the algorithm a times and return the largest of the returnedsets, reducing the probability of a result worse than α ( G ) − n / t to 2 − a .8 χ -boundedness The classes of graphs with bounded induced odd packing number are χ -bounded,that is, their chromatic number is bounded by a function of their maximum cliquesize. Let us start with the triangle-free case, where we need to show an absolutebound on the chromatic number. Lemma 6.
Every triangle-free graph G satisfies χ ( G ) ≤ + iocp ( G ) . Further-more, if G has odd girth at least , then χ ( G ) ≤ + iocp ( G ) , and if G has girthat least , then χ ( G ) ≤ + iocp ( G ) .Proof. We prove the claim by induction on the induced odd cycle packing number.If iocp( G ) =
0, then G is bipartite and χ ( G ) ≤
2, hence suppose that iocp( G ) > C be a shortest odd cycle in G , which is necessarily induced. Since iocp( G − N [ V ( C )]) ≤ iocp( G ) −
1, we can color G − N [ V ( C )] by the induction hypothesis,and it su ffi ces to show how to color G [ N [ V ( C )]] using at most 5 extra colors, or atmost 4 extra colors if G has odd girth at least 7, or using at most one extra color if G has girth at least 7.Let A = { z , z , z } be a set consisting of three consecutive vertices of C and let R = N [ A ]. By Lemma 2, G [ N [ V ( C )] \ R ] is bipartite. Since G is triangle-free, theneighborhood of any vertex is an independent set, and N ( { z , z } ) is disjoint from N ( z ). If G has odd girth at least 7, then N ( { z , z } ) is an independent set as well;hence, we can use two new colors to color G [ N [ V ( C )] \ R ], one color for N ( z )(which includes z and z ) and one color for N ( { z , z } ) (which includes z ), usingfour extra colors in total. Otherwise, we use two new colors to color G [ N [ V ( C )] \ R ], one color for N ( z ), one color for N ( z ), and one color for N ( z ) \ N ( z ), usingfive extra colors in total.In the case that G has girth at least 7, we claim that N ( V ( C )) is an independentset. Indeed, suppose for a contradiction vertices w , z ∈ N ( C ) are adjacent, and let w ′ and z ′ be neighbors of w and z in C , respectively. Since G has girth at least 7,the distance between w ′ and z ′ in C is greater than three. However, then C + w ′ wzz ′ contains an odd cycle shorter than C , which is a contradiction. Hence, we can usethree of the colors used on G − N [ V ( C )] to color C and one extra color to color N ( V ( C )). (cid:3) For the case G contains triangles, let f (0 , ω ) = ω ,and for k ≥
1, let us inductively define f ( k , ω ) = ω + (2 + k ) (cid:16) ω (cid:17) + f ( k − , ω ) (cid:16) ω (cid:17) . Theorem 7.
Every graph G satisfies χ ( G ) ≤ f ( iocp ( G ) , ω ( G )) . roof. We prove the claim by induction on the induced odd cycle packing number.If iocp( G ) =
0, then G is bipartite and χ ( G ) ≤
2, hence suppose that iocp( G ) > K be a largest clique in G , and for each v ∈ V ( G ) \ K , let A ( v ) denote the setof vertices of K to which v is not adjacent; the maximality of K implies A ( v ) , ∅ .Let A ′ ( v ) be an arbitrary subset of A ( v ) of size min(3 , | A ( v ) | ). For a set X ⊆ K with | X | ∈ { , , } , let B ( X ) = { v ∈ V ( G ) \ K : A ′ ( v ) = X } . If | X | =
1, then themaximality of K implies B ( X ) is an independent set; for each 1-element set X , weuse one color for all vertices of X ∪ B ( X ). If | X | =
2, then the maximality of K implies G [ B ( X )] is triangle-free, and by Lemma 6, we can use 2 + G ) colorsto color G [ B ( X )]. Finally, if | X | =
3, then iocp( G [ B ( X )]) ≤ iocp( G [ B ( X )]) − B ( X ) are non-adjacent to the triangle induced by X ; hence, wecan use f (iocp( G ) − , ω ( G )) colors to color G [ B ( X )]. Summing the numbers ofcolors over all choices of X , we conclude that at most f (iocp( G ) , ω ( G )) colors areused to color G . (cid:3) Let us remark that we can obtain the coloring as in Theorem 7 in polynomialtime: instead of choosing K as a largest clique, the inspection of the proof showsthat it su ffi ces to choose one which cannot be enlarged by adding at most threeand removing at most two vertices, and such a clique can be found in polynomialtime. χ -bounding function In the previous section we showed that every graph G has chromatic number atmost f (iocp( G ) , ω ( G )), where f is of order roughly ω ( G ) G ) . We do not knowwhether this upper bound is tight. In this section we show that there exist graphswith induced odd cycle packing number one whose chromatic number is almostquadratic in ω ( G ). We give a probabilistic construction. In order to carry out theprobabilistic calculations, we will use the following bounds. Lemma 8 (See [7], Cherno ff Bound) . Suppose X is a sum of independent { , } -variables. For any δ > ,Prob [ X ≤ (1 − δ ) E [ X ]] ≤ e − δ E [ X ]2 . Lemma 9 (See [7], Talagrand’s Inequality II) . Let X be a non-negative randomvariable, not identically , which is determined by n independent trials T , . . . ,T n , and satisfying the following for some integers c , r > : Changing the outcome of any one trial can change X by at most c. • For any non-negative integer d, if X ≥ d, then there is a set of at most rdtrials whose outcomes certify that X ≥ d.Then for every non-negative t ≤ E [ X ] ,Prob h | X − E [ X ] | > t + c p r E [ X ] i ≤ e − t c r E [ X ] . We are now ready to give the construction (a variation on a standard construc-tion of triangle-free graphs with no large independent set).
Theorem 10.
There exists a family of graphs with induced odd cycle packing num-ber at most one and with arbitrarily large clique number such that every graph Hin this family satisfies χ ( H ) = Ω (cid:16) ω ( H )log ω ( H ) (cid:17) .Proof. Let G be a random G ( n , p ) graph for p = / k , where k is a su ffi cientlylarge even integer and n = k / A , A ⊂ V ( G ) are disjoint and have size three, and G contains allnine edges with one end in A and the other end in A ; in this case, we say that theset of these nine edges forms a K , . We say a set of edges of G is bad if it can bepartitioned into subsets of size 9, each of which forms a K , . Finally we say a setof edges is subbad if it is a subset of a bad set.We construct a graph G by deleting a maximal bad set B from G . By themaximality of B , no set of edges of G forms a K , . Let H be the complementof G . Consider any disjoint (odd) cycles in H . If there were no edge betweenthese cycles, then any pair of triples of vertices taken one from each cycle wouldinduce a complement of a supergraph of K , in G , which is a contradiction.Consequently iocp( H ) ≤
1. Furthermore, an analogous argument shows α ( H ) ≤
5, and thus χ ( H ) ≥ n / = Ω ( k ).Therefore, it su ffi ces to argue that ω ( H ) = α ( G ) = O ( k log k ) with non-zeroprobability. Let us consider a set S ⊆ V ( G ) of size at least k and define thefollowing random variables. X = | E ( G [ S ]) | X = max (cid:8) | Z | : Z ⊆ E ( G [ S ]), Z is subbad in G (cid:9) The probability that S is an independent set in G is at most Prob[ X ≤ X ].Indeed, if S is independent, then E ( G [ S ]) ⊆ B is subbad, and thus X = X . Let11 = (cid:16) | S | (cid:17) = Ω ( k ), so that E [ X ] = s / k . Using the Cherno ff bound (Lemma 8), weobtain Prob (cid:20) X ≤ s k (cid:21) = Prob[ X ≤ E [ X ] / ≤ e − E [ X = e − s k . Let Z = { e ∈ E ( G [ S ]) : some set containing e forms a K , in G } . Clearly, if Z ⊆ E ( G [ S ]) is subbad, then Z ⊆ Z , and thus X ≤ | Z | . For distinct vertices x , y ∈ V ( G ), the probability that xy is an edge and some set containing xy forms a K , in G is at most n k = k , and thus E [ X ] ≤ E [ | Z | ] ≤ s k . Let δ = s k − E [ X ] ≥ X ′ = X + δ , so that E [ X ′ ] = s k . Note that flipping the existence of asingle edge in G changes X ′ by at most 9. Furthermore, when X ′ ≥ d , there existat most 9 d edges in G whose presence certifies this is the case. Hence, we canapply the Talagrand’s inequality (Lemma 9) with c = r = t = E [ X ′ ] = s k .Note that 60 c p r E [ X ′ ] ≤ E [ X ′ ] for k large enough, since s = Ω ( k ). Hence, wehave Prob " X ′ > s k = Prob[ X ′ > E [ X ′ ]] ≤ e − E [ X ′ = e − s k . It follows thatProb[ X ≤ X ] ≤ Prob[ X ≤ X ′ ] ≤ Prob (cid:20) X ≤ s k (cid:21) + Prob " X ′ > s k ≤ e − s k + e − s k < e − | S | k for k large enough.Therefore, for any set S ⊆ V ( G ) of size q ≥ k , the probability that S is anindependent set in G is at most e − q k . Hence, the probability that G containsan independent set of size q is less than nq ! e − q k ≤ neq ! q e − q k ≤ ( ke ) q e − q k = exp (cid:16) q log( ke ) − q k (cid:17) , that is, smaller than 1 when q ≥ k log( ke ). (cid:3) As noted in the proof, the probabilistic construction used is unnecessarily re-strictive, excluding all disjoint cycles regardless of parity. Similarly all cycles orpaths of length ≥ QPTAS assuming only bounded iocp
The fact that triangle-free graphs with bounded iocp have bounded chromaticnumber has the following easy consequence.
Lemma 11.
For all integers p ≥ , every graph G with n vertices satisfies at leastone of the following conditions: • G has an independent set of size at least n + iocp ( G ) , or • every maximal packing of triangles in G contains a triangle T such that α ( G − N ( T )) ≥ (1 − / p ) α ( G ) , or • G contains a vertex v of degree at least n p such that α ( G − N ( v )) = α ( G ) .Proof. Let X = { T , . . . , T m } be a maximal packing of triangles in G . The graph G − V ( X ) is triangle-free, and by Lemma 6, χ ( G − V ( X )) ≤ + G ). Con-sequently, α ( G ) ≥ α ( G − V ( X )) ≥ n − m + G ) . Suppose that G does not have anyindependent set of size at least n + G ) , and thus m ≥ n /
6. Let J be a largestindependent set in G . If | N ( T i ) ∩ J | ≤ | J | / p for some i ∈ { , . . . , m } , then the secondoutcome of the lemma holds.Otherwise, for each i ∈ { , . . . , m } , there exists v i ∈ V ( T i ) such that | N ( v i ) ∩ J | > | J | p . Consequently, there exists a vertex v ∈ J such that | N ( v ) ∩ { v , . . . , v m }| ≥ m p ,and thus deg v ≥ m p ≥ n p . Since v ∈ J , we have α ( G − N ( v )) = α ( G ). (cid:3) Combining this lemma with Theorem 5, we obtain a QPTAS for the maximumindependent set in graphs of bounded induced odd cycle packing number.
Theorem 12.
There exists a randomized algorithm that, for input integers k ≥ and p ≥ and an n-vertex graph G of induced odd cycle packing number atmost k, returns in time n O ( k + p log n ) an independent set of G whose size is at least (1 − k / p ) α ( G ) with probability at least / .Proof. If k = G is bipartite), we return the largest maximum independent setobtained via a maximum flow algorithm. Otherwise, we find a maximal packingof triangles X in G greedily, and return the largest of the independent sets obtainedby(a) running the algorithm from Theorem 5 n times with t = (4 + k ) p ,(b) for each T ∈ X , running the algorithm recursively for G − N [ T ] with k replaced by k − T to the returned independent set,and 13c) for each v ∈ V ( G ) of degree at least n p , running the algorithm recursivelyfor G − N ( v ).Each recursive call either decreases k or decreases the number of vertices by afactor of at most (cid:0) − p (cid:1) , implying the total number of the calls of the proce-dure is at most n O ( k + p log n ) . Iterating the algorithm from Theorem 5 n times foran induced subgraph G ′ of G ensures we fail to find an independent set of sizeat least α ( G ′ ) − n / t with probability at most 2 − n . Hence, with probability at least1 − n O ( k + p log n ) − n > / n large enough—for small n , we can just find thelargest independent set by brute force), we can assume that throughout the runof the algorithm, in part (a) for an induced subgraph G ′ of G , at least one of thereturned independent sets has size at least α ( G ′ ) − n / t .If α ( G ) ≥ n + k , then in (a) we return an independent set of size at least α ( G ) − n / t ≥ α ( G ) − (4 + k ) α ( G ) / t = (1 − / p ) α ( G ). If G contains a vertex v of degreeat least n p such that α ( G − N ( v )) = α ( G ), then in (c) the corresponding recursivecall gives an independent set of size at least (1 − k / p ) α ( G − N ( v )) = (1 − k / p ) α ( G ).If neither of these conditions holds, Lemma 11 implies there exists a triangle T ∈ X such that α ( G − N ( T )) ≥ (1 − / p ) α ( G ). The recursive call in (b) returnsan independent set I of G − N [ T ] of size at least (1 − ( k − / p ) α ( G − N [ T ]),and since α ( G − N ( T )) = α ( G − N [ T ]) +
1, the addition of a vertex of T turns I into an independent set of size at least (1 − ( k − / p )( α ( G − N ( T )) − + ≥ (1 − ( k − / p ) α ( G − N ( T )) ≥ (1 − ( k − / p )(1 − / p ) α ( G ) ≥ (1 − k / p ) α ( G ).Hence, the algorithm is correct. (cid:3) Considering Theorem 12, it is of course natural to ask whether the maximumindependent set problem admits a PTAS ob graphs with bounded induced oddcycle packing number.
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