Generalized Graph Packing Problems Parameterized by Treewidth
Abstract
-Packing is the problem of finding a maximum number of vertex-disjoint copies of in a given graph . -Partition is the special case of finding a set of vertex-disjoint copies that cover each vertex of exactly once. Our goal is to study these problems and some generalizations on bounded-treewidth graphs. The case of being a triangle is well understood: given a tree decomposition of having treewidth tw, the -Packing problem can be solved in time , while Lokshtanov et al. [ACM Transactions on Algorithms 2018] showed, under the Strong Exponential-Time Hypothesis (SETH), that there is no algorithm for any even for -Partition. Similar results can be obtained for any other clique for . We provide generalizations in two directions:
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We consider a generalization of the problem where every vertex can be used at most times for some . When is any clique with , then we give upper and lower bounds showing that the optimal running time increases to . We consider two variants depending on whether a copy of can be used multiple times in the packing.
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If is not a clique, then the dependence of the running time on treewidth may not be even single exponential. Specifically, we show that if is any fixed graph where not every 2-connected component is a clique, then there is no algorithm for -Partition, assuming the Exponential-Time Hypothesis (ETH).
Keywords and phrases:
Graph Packing, Graph Partitioning, Parameterized Complexity, Treewidth, Pathwidth, pw-SETH, Single-Exponential Lower Bound, Slightly Superexponential Lower BoundCopyright and License:
2012 ACM Subject Classification:
Theory of computation Fixed parameter tractabilityEditors:
Anne Benoit, Haim Kaplan, Sebastian Wild, and Grzegorz HermanSeries and Publisher:
Leibniz International Proceedings in Informatics, Schloss Dagstuhl – Leibniz-Zentrum für Informatik
1 Introduction
Parameterized complexity theory has proven instrumental in systematically understanding the computational complexity of various combinatorial problems under different parameterizations. Parameterization by treewidth implies tractability for a large number of fundamental algorithmic problems. A prominent line of research has emerged around classifying the complexity of classical NP-hard graph problems under this parameterization framework [8, 7].
In their influential work, Lokshtanov et al. [8] studied six classical combinatorial problems for which parameterized algorithms are known where the running-time dependence on treewidth is single exponential. They showed that, under Strong Exponential Time Hypothesis (SETH), the running times are optimal in the base of the exponent. Following this work, significant efforts have been devoted to generalizing and extending these results. In particular, five out of six problems in [8] have been put into a wider context and generalized to an infinite family of problems: -Coloring was generalized into -homomorphism problems [1, 10, 9], Independent Set (equivalent to Vertex Cover), MaxCut and Odd Cycle Transversal [2] into -homomorphism deletion problem, and Dominating Set into general -dominating set problems [5]. Curticapean and Marx [3] showed tight lower bounds for the problem of counting perfect matchings, which was later extended into general factor problems. However, from the initial six problems studied by Lokshtanov et al. [8], the triangle packing problem has remained unaddressed by prior generalization efforts.
The running time for triangle packing stated in [8] is actually valid for any clique packing problem, instead of triangle with just three vertices. In this paper, we investigate further generalizations of the triangle packing problem. This line of research naturally gives rise to two conceptually distinct directions:
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1.
Allowing vertices to be covered multiple times: The motivation for this generalization stems naturally from closely related problems such as fractional packing, where vertices can inherently contribute fractionally or multiple times to different packings. To illustrate, Figure 1.1 shows a graph that does not admit a triangle partition, yet allows a collection of triangles that cover each vertex exactly twice. Motivated by this observation, we formally define and explore a generalized packing problem, where each vertex of can be covered at most times, for any fixed integer . Unlike case, this generalization demands careful consideration of the definition, specifically on whether a copy of may appear multiple times in the solution.
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2.
Considering packings of arbitrary graphs: The second natural direction involves generalizing triangle packing by replacing triangles with an arbitrary fixed graph . This problem has already been studied in the literature under the name of -partition in [6]. Specifically, in [6] the authors prove the NP-Hardness of the -partition problem where has a connected component with at least three vertices. In this paper we let be a connected graph with at least three vertices and study the complexity of the -partition problem parameterized by treewidth.
1.1 Our Results
Let be a fixed graph and be an integer. Given a graph , a subgraph of is called a copy of if is isomorphic to . Moreover, if is a copy of in , we say that covers if .
Definition 1.
We say that a multiset (respectively, set) of subgraphs of is a (respectively, ) in if
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1.
each is isomorphic to for
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2.
each vertex of is covered at most times by the subgraphs in .
The collection is called a (respectively, ) of if each vertex is covered exactly times.
Observe that when is equal to , two copies of are not allowed to have any common vertices. Therefore, in this case, the notions of and coincide, and we write to simpify the notation. We define in a similar way. The first problem we introduce, , asks the maximum size of an in .
Input: A graph Output: The size of a largest in .
For a fixed , we show that can be solved in time for graphs of treewidth .
Theorem 2.
Let be an arbitrary graph such that it contains at least 3 vertices. Then, can be solved in time for all -vertex graphs where is the treewidth of .
Then, we define partitioning problems in which each vertex of must be covered the same number of times. For an arbitrary graph , we let which results in vertex disjoint copies of .
Input: A graph Question: Is there an of such that each vertex is covered exactly once?
Observe that is a special case of the problem , which implies an algorithm with running time for the problem on graphs of treewidth . We show that this running time cannot be improved for many choices of .
Theorem 3.
Let be an arbitrary graph with at least 3 vertices such that is not a block graph. Then, there is no algorithm for problem, that solves all instances in time where is the treewidth of , unless ETH fails.
When is a clique, we consider the more general problem in which a vertex can be covered more than once, at most times for some . Therefore, we distinguish between two variants of the problem: one where each clique can be selected at most once, and another without restriction. The problem asks the maximum size of a in .
Input: A graph Output: The size of a largest in .
Similar to the problem, the problem asks the maximum size of a in .
Input: A graph Output: The size of a largest in .
In the full version of the paper we show that admits a single-exponential time algorithm where the same algorithmic result also applies to with slight modifications.
Theorem 4.
Let and be integers. Then, can be solved in time for all -vertex graphs given together with a tree decomposition of width at most .
Moreover, we define partitioning problems where is a clique.
Input: A graph Question: Is there a of ?
Input: A graph Question: Is there a of ?
Similarly, we prove that this running time is optimal up to polynomial factors in the size of the input graph.
Theorem 5.
Let and be integers. If there exists an such that can be solved in time for all -vertex graphs given together with a path decomposition of width at most , then the -SETH fails.
Finally, we show that the lower bound result in Theorem 5 can also be transferred similarly.
Theorem 6.
Let and be integers. If there exists an such that can be solved in time for all -vertex graphs given together with a path decomposition of width at most , then the -SETH fails.
2 Technical Overview
In this section, we will give an overview of the techniques and ideas presented in the paper.
2.1 Preliminaries
A graph is -connected if for each such that it holds that is connected. A graph is also called biconnected if it is -connected, and a block is a maximal -connected component of . Similarly, a graph is called a block graph if every block of is a clique.
A vertex of a connected graph is a cutvertex if is disconnected. In this paper, we refer to both the blocks of a graph and the nodes of the corresponding block tree interchangeably, with a slight abuse of notation. While formally distinct, we find it convenient to treat them as equivalent entities for the sake of clarity and brevity. For a vertex and a block , we write if .
Definition 7.
For a function and a set , we let denote the restriction of to . Similarly, we let denote the restriction of to . Finally, for and a value , we let denote a function which is defined as
We use the Iverson bracket , which is defined to be if the proposition is true and otherwise.
2.2 Gadgets
In this paper, we derive our lower bounds via reductions from well-studied base problems whose intractability is established under standard complexity hypotheses. Each reduction makes use of compact, purpose-built gadgets – small graphs whose behavior can be precisely engineered. Embedding these gadgets into our constructions yields intuitive, transparent proofs that highlight the underlying ideas without excessive technical overhead.
More formally, we define a gadget as a graph with designated portal vertices , where the vertices are called internal vertices. We say that a graph is an extension of a gadget if contains as an induced subgraph, where each internal vertex of satisfies
In other words, in an extension , only the portal vertices of can have neighbors outside of .
We also require gadgets to behave in a structured way. Whenever a copy of intersects the gadget, it must lie entirely within the gadget when connected to a larger graph. We formalize this in the following definition.
Definition 8.
Let be a gadget and be the set of its portal vertices. For a graph and , we say that is if, for any extension of and any / of , the following holds: If there exists that contains an internal vertex of , then all vertices in must belong to . Formally,
Next, we define the relation realized by a gadget.
Definition 9.
Let be a graph, and be a gadget with portal vertices . We say dist--realizes (respectively, arb--realizes) a relation if the following holds:
We say that strict--realizes if it both dist- and arb--realizes .
Remark 10.
When , the notions of dist--realization and arb--realization coincide. In this case, we simply say that the relation is -realized by , instead of using the term “strict-”.
We say a relation is for and if for each , the weight of is equivalent to , i.e. . Moreover, a relation has weight if . Recall that a gadget is a small, engineered graph used to enforce specific behaviors in a reduction. We streamline the lower bound constructions by building general-purpose gadgets that can realize any relation. The descriptions of the gadgets are deferred to the full version of the paper. This way, we avoid repetitive constructions and keep the focus on the core ideas.
Lemma 11.
Let be an arbitrary graph, be an integer and be a relation that is for some . Then, there exists a gadget that -realizes the relation and the size of is bounded by some function of . Moreover, for relations with constant weight, it holds that .
When , we require the relation to have more structure, i.e., the relation should be . However, this restriction can be easily handled in our lower bound constructions.
Lemma 12.
Let , be a clique, be a constant and be a relation. Then, there exists a gadget that strict--realizes the relation . Moreover, the size of is bounded by some function of .
2.3 Single-Exponential Lower Bound
In this section, we give an overview of how the above-described gadgets are used to prove Theorem 6. Classically, when proving conditional lower bounds based on SETH, one needs to find a reduction from SAT. However, this usually involves carrying out repetitive, unnecessary work that is not specific to the problem one is working on. One of the strengths of the framework introduced by Lampis [7] is removing the need for such repetitive constructions.
The primal graph of a CSP instance is a graph that has a vertex for each variable of , and there is an edge between if and appear together in a constraint. The pathwidth of the CSP instance is defined to be the pathwidth of its primal graph. Similarly, by path decomposition of , we mean a path decomposition of the primal graph of . The following lemma from [7] implies that we can assume the path decomposition to be nice.
Lemma 13 (Lemma 2.1 in [7], restated for CSPs).
There is a linear-time algorithm that takes as input a CSP formula with variables and constraints and a path decomposition of its primal graph of width and outputs a nice path decomposition of containing at most bags, as well as an injective function from the set of constraints of to such that for each constraint , contains all the variables of .
The following conjecture from [7], called -SETH, will form the basis of our hardness results.
Conjecture 14 (Conjecture 1.1 from [7].).
For all we have the following: there exists no algorithm which takes as input a instance on variables and a path decomposition of its primal graph of width and correctly decides if is satisfiable in time .
Moreover, we have the following result from [7], which proves the equivalence of falsifying -SETH and finding a faster algorithm for 2-CSP.
Theorem 15 (Theorem 3.2 from [7], shortened).
For each the following statements are equivalent:
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1.
The -SETH is false.
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2.
There exist and an algorithm that takes as input a -CSP instance on alphabet , together with a path decomposition of , and decides if is satisfiable in time .
Motivated by Theorem 15, Section 3 presents a reduction from the -CSP problem to the Single-Clique Partitioning(c,d) problem. The core idea is to encode each variable’s possible assignments using a set of vertices whose coverage count in a represents the chosen value. We then introduce gadgets each enforcing a single constraint of the -CSP instance via carefully specified relations so that only coverage patterns corresponding to satisfying assignments are possible. By designing these gadgets to interact locally, we ensure the pathwidth of the resulting instance increases by at most a constant, completing the reduction.
2.4 Slightly Superexponential Lower Bound
We now give a high-level description of the proof of Theorem 3. Let be a graph with at least vertices that is not a block graph. Next, we will describe the lower bound for the problem which is stated in Theorem 3. Recall that is solvable in slightly super-exponential time for any fixed graph , i.e., there is an algorithm with running time .111The notation suppresses polynomial factors in the input size ; that is, for some constant . We show that this running time is optimal under ETH, i.e. there exists no algorithm for with running time . To that end, we use an auxiliary problem for which such a lower bound was presented in [4]. Specifically, we define the problem where given a graph on a vertex set , we ask whether there exists an independent set in that contains exactly one vertex from each row and each column.
Input: A graph on the vertex set Question: Is there an independent set of such that induces a permutation on ?
The following hardness result was presented for the analogous problem in [4] which translates easily to . Below we state it for .
Theorem 16 (Theorem 14.14 in [4]).
Unless ETH fails, cannot be solved in time .
To prove Theorem 3, we reduce the problem to . The construction relies on structural properties of . Let be a block in with a minimum separator such that has at least two connected components. Since contains a non-clique block, such and exist. We partition into two subsets, and , and create copies of each. The construction in the proof of Theorem 3 ensures that any includes copies of , each covering exactly one copy of and one of . This yields configurations, corresponding to all permutations of a set of size . Additionally, for each edge in the instance, we introduce a gadget to prevent from being included in the set of vertices induced by this permutation. Moreover, these gadgets are designed to preserve pathwidth by interacting only locally. The detailed proof of Theorem 3 is deferred to the full version of the paper.
2.5 Algorithmic Results
Recall that a tree decomposition of a graph is a pair , where is a tree whose every node is assigned a bag , satisfying the following properties:
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For every vertex , there exists at least one bag such that .
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For every edge , there exists a bag such that .
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For every vertex , the set of nodes induces a connected subtree of .
The width of a tree decomposition is the size of its largest bag minus one, and the treewidth of is the minimum width over all possible tree decompositions of . A nice tree decomposition is a rooted tree decomposition in which each node is one of the four types: leaf, introduce, forget or join. We refer the reader to [4] for more details.
The algorithmic results in Theorems 2 and 4 follow a fairly standard dynamic programming framework over tree decompositions. We define a suitable set of states for each bag in the decomposition, capturing the essential information needed to extend partial solutions. The main challenge of the approach lies in carefully designing these states and proving the correctness of the update rules that determine how states transition from one bag to the next. Usually what determines the running time is the state representation and transitions to the specific structural constraints of our problem.
In particular, the single-exponential algorithm for the problem defines the state of a bag based on how many times each vertex in the bag is covered by a partial solution. This yields states for a bag of size , naturally leading to a running time of .
In contrast, the slightly superexponential algorithm for the problem requires keeping track of a partition of the bag, in which each part corresponds to a partial copy of . This results in possible states for a bag of size , leading to an overall running time of . The algorithms are presented in detail in the full version of the paper.
3 Lower Bounds for Clique Partitioning Problems
In this section we prove Theorems 5 and 6. Theorem 15 says that in order to prove a conditional lower bound based on -SETH, one can start the reduction from the -CSP problem. In the following, we will present a reduction from the -CSP problem to the . Subsequently, we will describe another reduction from to problem, and prove Theorems 5 and 6.
3.1 Lower Bounds
We first prove Theorem 5. To that end, intuitively, we show that a fast algorithm for the problem implies a fast algorithm for the -CSP problem.
Lemma 17.
Let and be integers. Suppose there exists an such that can be solved in time for all -vertex graphs given together with a path decomposition of width at most . Then, there exist , an integer and an algorithm that takes as input a -CSP instance on alphabet , together with a path decomposition of , and decides if is satisfiable in time .
Proof.
Let , be as in the lemma statement and let be the hypothetical algorithm for . Moreover, we let denote the clique and let be the smallest integer that is a multiple of such that
| (3.1) |
Observe that is a constant that only depends on and .
Let and . We will now present a reduction from 2-CSP with alphabet size to . The idea is as follows: in order to make use of regular relations, we will represent each variable of the 2-CSP instance by vertices. Note that each of the vertices can be covered between and times, which can be thought of as the state of a vertex. In total, vertices combined give rise to states. These states can also be visualized as vectors . Next, we consider the following subset of vectors
Observe that we have , because we can append to each vector in at most many ’s so that the new vector constructed this way satisfies . Hence there exists an injective map where we define . We use the set to simulate many assignments to a variable of the 2-CSP instance. Observe that , as a relation, is .
Construction of the instance.
Let be a 2-CSP instance with variables , contraints and alphabet . Let be a nice path decomposition of width . Finally, let be a function that maps each constraint of to a bag such that contains the variables occurring in . We will construct an instance of the problem as follows:
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1.
For each , define to be the smallest integer such that . Similarly, let be the largest integer such that . For each and , introduce vertices
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2.
Define the relation
Observe that is by construction. The same also holds for , because for each such that for , we have
where the last equivalence holds because and are both equivalent to . Introduce two gadgets and that arb--realize the relation and , respectively, which exist by Lemma 12. Then, identify the portal vertices of with , and similarly, identify the portal vertices of with the vertices .
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3.
Let . We say that represents for if . In that case, we define to be the relation, and and to be the indices of the variables associated with the constraint . We define the new relation
Observe that is because for each we have
since is a multiple of . Hence, we can introduce a gadget by Lemma 12 that arb--realizes . Then, we identify the portal vertices of with the vertices
(3.2) Next, we define the relation where
(3.3) and let
Observe that is because for each we have
By Lemma 12, there exists a gadget that arb--realizes . Next, for each such that , we introduce a gadget that arb--realizes the relation and identifies the portal vertices of with
(3.4) Finally, for each and such that , we define the gadget that covers at step as
This is the whole construction of the instance which we call .
Equivalence of the instances.
We now prove that is satisfiable if and only if admits a , by establishing each direction separately.
Suppose that there exists an assignment to such that is satisfied. In the following, we will describe a of . For each , define . Now let . Observe that since for each , it holds that there exists a of that covers according to . Moreover, there exists a of that covers according to . In both cases, the internal vertices of the gadgets are covered exactly times.
Now let . For all , there exists a of that covers
| (3.5) |
according to , because . Moreover, if represents for some , and are the variables corresponding to , then there exists a that covers
| (3.6) |
according to . Again, in both cases, the internal vertices of the gadgets are covered exactly times. Next, we prove that the remaining vertices are covered times as well.
Claim 18.
It holds that for each , such that and , is covered exactly times by .
Proof.
We prove the claim by induction on . Let , and suppose that . Since , we have that . Consider the vertices , which are covered according to by . Moreover, is covered according to by which follows from ˜3.5 or ˜3.6, depending on whether or not, respectively. All in all, is covered exactly times for .
Now suppose that the claim holds for , and we will prove the claim for . Let such that . Consider the vertices , which are covered according to by . This follows from ˜3.5 and 3.6. Now, observe that we have either or . In both cases, by using the arguments in the case, one can conclude that is covered according to by . Therefore, all in all, it holds that each vertex is covered exactly times for . Now for the reverse implication, suppose that has a . Since each gadget used in the construction of is , this implies that for each gadget there is a that covers its interval vertices exactly times, and its portal vertices according to the relation associated with it. In particular, for each , let denote the vector such that covers the vertices according to .
By induction, one can show that for each and such that , the tuple is covered by according to where . Hence, we let . Next, we will prove that is a satisfying assignment for . To that end, let and . To show that is satisfied by , let and be the variables associated with . By the above discussion, we know that and is covered according to and , respectively. By the definition of , and satisfy . Therefore, the assignment satisfies all constraints of , and is satisfied if and only if has a .
Pathwidth and size of the constructed instance.
To bound the pathwidth of , we will create a path decomposition by following . Specifically, for each , we first create a new path decomposition where each is a copy of and we replace each with the vertices . Note that the size of each is at most for .
In the following, we will add the remaining vertices of to the bags in such that is a valid path decomposition of . Let . For each such that , we replace the vertices with as follows. After the bag ,we first insert the bag , and then add another bag where we replace in with , and finally, another bag where we remove the vertices from . For a fixed , we keep adding the bags iteratively for each such that , until all vertices in all the gadgets are contained in the bag decomposition. Note that, by our construction, edges that are adjacent to a vertex in are also covered by either or .
Finally, for each and , we add the gadgets to the tree decomposition in a similar way. Since the size of each and is a function of , it is bounded by a constant. All in all, the pathwidth of increases at most by a constant. We have
Finally, since and are bounded by a polynomial of , it follows from the construction that
Running Time.
Constructing the graph from takes time polynomial in . By our assumption on , the whole reduction takes time
where the inequality follows from ˜3.1 and is a constant.
The proof of Theorem 5 follows from Theorems 15 and 17. We note here that the same construction can be used to prove Theorem 6, because the gadgets used in Lemma 17 strict--realize their relations (see Definitions 9 and 12). However, by presenting a simple reduction, we demonstrate that a fast algorithm for implies a fast algorithm for , which is used to prove Theorem 6 in a more formal way.
Lemma 19.
Let and be integers. Suppose there exists an such that can be solved in time for all -vertex graphs given together with a path decomposition of width at most . Then, there exist and such that can be solved in time for all -vertex graphs given together with a path decomposition of width at most .
Proof.
Let and be as in the statement of the lemma. Moreover, let be the hypothetical algorithm for .
Construction of the instance.
Let be an instance of . Moreover, we let denote the clique and construct a instance as follows. Let have the same vertex set as . We call these vertices the original vertices of . Moreover, for each clique in of size , we add a gadget that dist--realizes the relation . We identify the portal vertices of with . This is the whole construction.
Equivalence of Instances.
We will now prove that admits a if and only if admits a .
Suppose that has a which is denoted by . For each clique in , let denote the number of occurences of in . We construct a for as follows. For each clique in , consider a of such that the portal vertices of are covered times. Add this to . Since the gadgets are disjoint, the copies of are also disjoint. The internal vertices of the gadgets are covered exactly times. Moreover, the original vertices of are also covered times, since each equality gadget simulates a clique. Hence is a valid and therefore is a yes instance.
Now suppose that has a . Then, for each clique in , let denote the number of times covers its portal vertices. We construct a by including each exactly times in . Moreover, covers each vertex exactly times, hence it is a valid . Therefore, is a yes-instance.
Running Time.
First, we show that the pathwidth of is . Take the path decomposition of and for each clique , let denote the bag that contains the vertices of . Note that since is a clique, there exists such a bag. Moreover, we may assume each bag is unique for each clique , duplicating bags if necessary. Then, we simply add the vertices of to the bag , increasing its size at most by a constant. Therefore, we get a new path decomposition for where the size of a bag is at most . Hence, it holds that .
Constructing the graph takes time polynomial in , and we also have . Therefore, the whole reduction takes time
where is a constant.
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