On Fine-Grained Exact Computation in Regular Graphs
aa r X i v : . [ c s . CC ] A ug On Fine-Grained Exact Computation in Regular Graphs
Saeed Akhoondian Amiri ∗ August 21, 2020
We show that there is no subexponential time algorithm for computing theexact solution of the maximum independent set problem in d -regular graphs,for any constant d >
2, unless ETH fails. We also discuss the extensionsof our construction to other problems and other classes of graphs, including5-regular planar graphs.
The independent set problem is a fundamental graph covering problem that asks for aset of pairwise nonadjacent vertices; we are interested to maximize the size of such a setand in particular find a maximum size such set. This is known as maximum independentset (MIS) problem.MIS is hard to approximate within any constant factor in general graphs[13], however,on the bounded degree or regular graphs, a simple greedy algorithm provides a constant-factor approximation of the optimum solution. Given its hardness, researchers havestudied the time complexity of the exact problem, i.e. they seek for fastest possibleexponential exact algorithm for the problem.Many NP-hard problems are solvable exactly in time ˜ O (2 n ) . In particular for theindependent set problem the trivial algorithm of testing all possible solutions yields sucha running time. There are several improvements over the trivial upper bound both ingeneral graphs and graphs of bounded degree [1, 4, 11, 12]; however all of them havea running time of the form ˜ O ( c n ) for a certain constant c <
2. Hence, all of them areexponential to n . On the other hand, there are NP-hard problems that are subexponen-tial time solvable; e.g. exploiting bidimensionality theory, it is possible to solve severalNP-hard problems in time 2 O ( √ n ) in excluded minor graphs [2]. Such differences raisethe question of what NP-hard problems have subexponential time algorithms?One of the main tools designed to better understand the exact complexity of hardcomputational problems is the Exponential Time Hypothesis (ETH) [7]. Assuming ETH, ∗ Department of Computer Science, University of Cologne, Germany, [email protected]. We hide a Poly( n ) factor in ˜ O notation. o ( n ) to solve the 3-SAT problem, where n standsfor the number of variables in the formula. It has been proven that under the sameassumption several other known problems, such as k -SAT [10] MIS in bounded degreegraphs [8] are ETH-hard.Both of the above results, for k -SAT and MIS on bounded degree graphs, are showingthe hardness of problem in a sparse instances of the input. The main challenge to provesuch a hardness is to transfer an arbitrary instance to an instance that is sparse or it hascertain structural property. If the transformation takes subexponential time then we canlink them to the existing known ETH-hard problems to show they are also ETH-hard.These reductions are best known as sparsification lemmas .In this work, we continue a similar spirit by providing a transformation from genericbounded degree graphs to d -regular graphs, a quite restricted well-structured graphs. Agraph G is d -regular if degree of every vertex of G is exactly d . We show that for everyinteger d > . Their reduction is a bit more involved than the two othersas they had to support the Hamiltonicity of the underlying graph. All of the above con-structions are specialized for their specific purpose and we do not see a direct extensionof mentioned papers to general graphs for every constant d > d -regular planar graph ( d > d = 1 is a matching and d = 2 is a disjoint union ofcycles. In both of them a linear time greedy algorithm gives an optimal solution: takea vertex of least degree into MIS and then together with its neighbor(s) remove themfrom the graph, repeat this process until reaching an empty graph.The closely related problems of finding and counting cliques and finding a minimumvertex cover have also been discussed in this paper. The simple gadget constructionfacilitates further customization to obtain lowerbounds on such problems.Before delving into the technical details, let us introduce the notations that are usedthroughout the paper. N denotes the set of natural numbers and for a set of integers { , . . . , k } we write [ k ]. Degree of a vertex v in a graph G = ( V, E ) is written as d v . Wewrite ∆ for the maximum degree of a graph. For a vertex v , the closed neighborhood of v is denoted by N [ v ] (it contains v and all of its neighbors). They used a claim in the book [5] that states MIS is NP-complete in cubic planar graphs. In the book,the authors cite the paper of Garey et al. [6]. To our understanding, this is an incorrect referencing.However later, Mohar [9] showed the hardness of the problem in the claimed class. Therefore theresult of Fleischner et al. [3] is valid. MIS has no Subexponential Algorithm on d -regular Graphs In this section, we show the ETH-hardness of finding MIS in d -regular graphs. Let G be a graph of maximum degree ∆. We construct a gadget G ∆ where all of itsvertices except one of them ( v ∆ ) has degree ∆. Then for every vertex v ∈ V ( G ), if∆ − d v = d > v to d copies G v , . . . , G dv of the G ∆ gadgets byadding edges from v to all of these gadgets. This way the resulted graph G ′ is a ∆-regulargraph. The construction of gadgets is such that from a maximum independent set in G ′ we can derive a maximum independent set of G .We may further assume that the maximum degree ∆ is an odd integer. Otherwise, weadd a complete graph on ∆ + 2 vertices to the original graph. The resulting graph hasan odd maximum degree ∆ + 1 and it is clear that in polynomial time we can constructMIS of the original graph from MIS of the new graph and vice verse. Hence, we havethe following assumption in the rest of paper. Assumption: ∆ is an odd number bigger than 1.
For a vertex v of degree d v < ∆ we construct δ v = ∆ − d v distinct gadgets H v , . . . , H vδ v asfollows (all of them have the same structure). In the following we explain the constructionof a single gadget, let say H .First create (∆ − / K , . . . , K (∆ − / with partitions ofsize ∆ −
1. We name the partitions of the i ’th bipartite graph A i , B i , for i ∈ [(∆ − / − a , . . . , a (∆ − / , b , . . . , b (∆ − / to the gadget H . Connect all verticesof partition A i (resp. B i ) to a i (resp. b i ). Then connect all a i , b i ’s ( i ∈ [(∆ − / h . The construction of H is completed. By construction, every vertex except h , has degree ∆. The degree of h is 2(∆ − / − h is the vertex that connectsour gadget to the graph G .Whenever it is necessary, if a gadget H is the j ’th gadget of a vertex v , to distinguishdifferent gadgets, we add indices v, j to H and all of the aforementioned vertices andpartitions. E.g. instead of a vertex h we may write h vj .The construction of the auxiliary graph G ′ is pretty simple: take G as a base, thenfor every v ∈ G connect all of its gadgets, i.e. H vj ’s, to v by adding edges { h vj , v } for j ∈ [ δ v ]. Let us make some observation on G ′ . First observe that every vertex of G ′ hasdegree exactly ∆.We formalize the second observation for bounding order of G ′ in the following. Observation 1.
Order of an attached gadget to any vertex is O (∆ ) . Since there are atmost ∆ such gadgets attached to a vertex v , G ′ has O (∆ | V ( G ) | ) vertices. As the numberof edges of each gadget is at most ∆ times more than its vertices, G ′ has O (∆ | V ( G ) | + | E ( G ) | ) edges. .3 From an MIS in G ′ to an MIS in G The main observation on each individual gadget is the following (we ignore the indices ofthe gadget for simplicity). In any MIS of G ′ , for a gadget H , from each bipartite graph K i in H we have to take one of its partitions: A i or B i , entirely into the MIS. The designof H is such that, after the previous selection we can take either of the sets a i ’s or b i ’sin the solution. But then we are not able to take the vertex h in the MIS. Consequently,vertex v (a vertex of G that is connected to the gadget H in G ′ ) is freely available tojoin MIS later. Hence, the existence of v in MIS merely depends on the structure of G ,not its connected gadgets. We prove these claims formally in the following.First we explain how to construct an MIS in a single gadget H . Lemma 2.
Let H be a gadget constructed as above, then it has an MIS I , such that thevertex h I and the size of I is (∆ − / − .Proof. We first constructively show that an independent set of the claim size and struc-ture exists; then we prove it is a maximum independent set. To construct I , take allvertices in partitions A i ( i ∈ [(∆ − / I , then add all vertices with labels b i to I . The size of I is as claimed, it does not contain a vertex h , and it is an independentset of H . It lefts to show there is no independent set I ′ of larger size in H .Since K i , the i ’th complete bipartite graph of the gadget, is a complete bipartitegraph, we can take at most ∆ − − K i ’s, let call it K , an MIS I ′ of H has at most t ≤ ∆ − K . If t >
0, then w.l.o.g. suppose the selectedvertices of K are in its B part . Since t > b in H − K it means b is not in the independent set, hence if we takethe entire B part of K into the independent set, the resulting set is still an independentset and larger than I ′ , a contradiction.It remains to show the claim holds for the case of t = 0. t = 0 means no vertex of K is in I ′ , then we should have both a i , b i ∈ I (otherwise we add one side of K to I ′ and make a larger independent set). If this is the case, we remove a i from I and add allvertices of the A partition of K to the independent set to make it larger, a contradiction.Therefore, in any maximum independent set I ′ , for every bipartite graph K i , entiretyof one of its partitions is in I ′ . For the remaining undecided vertices, observe that wemay take at most (∆ − / G and G ′ by the followinglemma. Lemma 3.
Given an integer k , there is an independent set I ′ of G ′ of size at least k + Σ v ∈ V ( G ) (∆ − d v ) · ((∆ − / − if and only if there is an independent set I ofsize at least k in G . Moreover, we can construct I from I ′ and vice verse in linear time. Clearly if a vertex from the B part of K is in any independent set of H then no vertex from its A partcan contribute to that independent set, as K is a complete bipartite graph. roof. The only if direction is straightforward: initialize I ′ = I then add all maxi-mum independent sets of all gadgets, computed by the approach explained in the proofof Lemma 2, to I ′ . The size of I ′ is as claimed. On the other hand, none of the verticesof gadgets that are connected to the vertices of G are in I ′ . It means there is no con-flict between choices in gadgets and vertices in I , hence I ′ is an independent set of theclaimed size.For the if part, by Lemma 2 there are at most Σ v ∈ V ( G ) (∆ − d v ) · (∆ − / − I ′ that are in G ′ − G . Hence, at least k vertices I = { u , . . . , u k } of I ′ arebelonging to both G and G ′ , thus I is an independent set of size k in G .The main theorem is the consequence of the previous lemmas and the sparsificationlemma for the independent set problem. Theorem 4.
There is no algorithm with running time o ( | E | ) to solve the maximumindependent set problem in ∆ -regular graphs unless ETH fails.Proof. Johnson and Szegedy [8] showed that there is no subexponential time algorithmto compute an MIS in graphs of degree at most 3, unless ETH fails. As explained earlier,in the description for Assumption 2.1, w.l.o.g. we can assume the input graph has anodd maximum degree. Thus, in Lemma 3 we provided a reduction from an independentset problem in graphs of maximum degree ∆ to ∆-regular graphs. By Observation 1 thesize of each gadget is Poly(∆) (independent of the order of G ), thus our reductions arefine grain, therefore the theorem follows. The gadget construction simply extends to vertex cover and clique problems. On theother hand, another extension is to set up a similar lower bound in planar graphs. Ourgadgets are not planar but it is easy to modify the most interior part of the gadgets(the bipartite graphs) to obtain planar gadgets. We explain the case of 5-regular planargraphs then we talk about the extension to the maximum clique problem.
Regular Planar Graphs
As discussed earlier, it is well known that the MIS problem is hard in 3 , a i , b i as we had, however, instead of bipartite graphs in the gadget, weinsert a modified Icosahedron as drawn in the Fig 1), we call this graph X . Lemma 5. X has a maximum independent set of size and both vertices a, b will be inany MIS.Proof. One can observe that vertices a, b, f, k all together form an independent set ofthe claimed attributes, We prove that this is the only MIS of X by showing that in5 b cd efghi jk l Figure 1: Graph X is obtained by deleting an edge { a, b } from an Icosahedron. The vertices a, b, k, f (in red) form an independent set of size 4 and this is the only independent setof size 4 in X . Every other independent set has size less than 4. any MIS, both vertices a, b are present. Except a, b , every other vertex has degree 5and every two non-adjacent vertices share at most 2 neighbors. Hence, if there are twovertices x, y ∈ V ( X ) − { a, b } in an MIS I , then | N [ x ] ∪ N [ y ] | ≥ − X except at most two of them, let call them u, v , are in the closedneighborhood of x, y . Clearly both u, v are in I otherwise size of I is less than 4. If { u, v } = { a, b } we are done; otherwise, w.l.o.g. let suppose u / ∈ { a, b } . Since | N [ u ] | = 6and u is not neighbor of x, y , u covers at least 6 − − N [ x ] ∪ N [ y ].It means x, y, u together are neighbor of all vertices of X , hence, v cannot be in I , acontradiction.The rest of the proof is straightforward from above lemma and our general construc-tion. Construct a gadget H by taking 2 copies X , X of X and adding a vertex h . Thenconnect a i , b i ∈ X i to h (we added indices to vertices of X to distinguish the disjointcopies of them). Now we can attach copies of a gadget H to every vertex that has adegree less than 5 in a given planar graph. Theorem 6.
MIS problem is NP-hard in -regular planar graphs.Proof. By result of Mohar [9] we know that MIS problem is NP-hard in cubic planargraphs. Given a cubic planar graph G construct a 5-regular planar graph G ′ as explainedabove. G has an independent set of size k if and only if G ′ has an independent set ofsize k + 4Σ v ∈ V ( G ) (5 − d v ). Since G ′ is a 5-regular graph the claim follows in a same lineas for general graphs. Triangles and Cliques
The gadgets do not have a triangle as a subgraph, on the other hand, the originalconnections in the graph G are untouched, hence there is a clique on at least k ≥ G ′ if and only if there is a clique of order k in G . Since the transformation from6 to G ′ happens in linear time on graphs of bounded degree, essentially every hardnessresult in graphs of bounded degree, for finding triangles or small cliques extends to theregular graphs. In this work, we showed the maximum independent set problem has no subexponentialalgorithm in d -regular graphs. Our construction with simple modifications extends toother covering problems and also to other classes of graphs. We believe this work couldease the way to obtain fine-grain reductions for other problems.We considered the independent set problem, one of the most basic problems were itssparsification lemma is known. Another interesting direction is to consider the k -SATproblem when the corresponding graph has the same degree for all variables and clauses. Acknowledgment:
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