On the Vertex -Path Cover

On the Vertex 𝑘-Path Cover Boˇstjan Breˇsar∗, Marko Jakovac∗ Faculty of Natural Science and Mathematics, University of Maribor, Slovenia Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia

J´an Katreniˇc†and Gabriel Semaniˇsin‡

ˇ arik University, Koˇsice, Slovakia Institute of Computer Science, P.J. Saf´

Andrej Taranenko

Faculty of Natural Science and Mathematics, University of Maribor, Slovenia Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia

December 5, 2013

Abstract A subset 𝑆 of vertices of a graph 𝐺 is called a 𝑘-path vertex cover if every path of order 𝑘 in 𝐺 contains at least one vertex from 𝑆. Denote by 𝜓𝑘 (𝐺) the minimum cardinality of a 𝑘-path vertex cover in 𝐺. In this paper present an upper bound for 𝜓3 of graphs with given average degree. We also give a lower bound for 𝜓𝑘 of regular graphs. For the Cartesian products of two paths we give an asymptotically tight bound for 𝜓𝑘 and the exact value for 𝜓3 .

1

Introduction

Let 𝐺 be a graph and 𝑘 be a positive integer. Then 𝑆 ⊆ 𝑉 (𝐺) is the vertex 𝑘-path cover of 𝐺 if every path on 𝑘 vertices in 𝐺 contains a vertex from 𝑆. We denote by 𝜓𝑘 (𝐺) the cardinality of a minimum vertex 𝑘-path cover in 𝐺. This graph invariant was recently introduced in [4], where the motivation for this problem arises in ensuring data integrity communication in wireless sensor networks using the 𝑘-generalized Canvas scheme [13]. ∗

Supported by the Ministry of Science and Higher Education of Slovenia under the grants J1-2043 and P1-0297. † The research of the author supported in part by Slovak APVV grant SK-SI-0010-08 and Slovak VEGA grant 1/0035/09. ‡ The research of the author supported in part by Slovak APVV grant SK-SI-0010-08, APVV-0007-07 and by the Agency of the Slovak Ministry of Education for the Structural Funds of the EU, under project ITMS:26220120007.

1

The vertex 𝑘-path cover can be regarded as the generalization of the vertex cover. Indeed, 𝜓2 (𝐺) equals to the size of minimum vertex cover, and 𝜓2 (𝐺) = ∣𝑉 (𝐺)∣ − 𝛼(𝐺) where 𝛼(𝐺) stands for the independence number. The value of 𝜓3 (𝐺) corresponds to the concept of dissociation number of a graph [17], defined as follows. A subset of vertices in a graph 𝐺 is called dissociation if it induces a subgraph with maximum degree 1. The number of vertices in a maximum cardinality dissociation set in 𝐺 is called the dissociation number of 𝐺 and is denoted 𝑑𝑖𝑠𝑠(𝐺). Clearly 𝜓3 (𝐺) = ∣𝑉 (𝐺)∣ − 𝑑𝑖𝑠𝑠(𝐺). The problem of computing 𝑑𝑖𝑠𝑠(𝐺) has been introduced by Yannakakis [17], who also proved it to be NP-hard in the class of bipartite graphs. The dissociation number problem was also studied in [1, 2, 6, 8, 14, 12], see [14] also for a survey. Recently, Tu and Zhou [15] presented a 2-approximation for 3-path vertex cover problem even for the weighted case of the problem. An exact algorithm for computing 𝜓3 (𝐺) in running time 𝑂(1.5171𝑛 ) for a graph of order 𝑛 was presented in [10]. The problem of computing 𝜓𝑘 (𝐺) is NP-hard for each 𝑘 ≥ 2, but polynomial for instance in trees, as shown in [4]. The authors investigate upper bounds on the value of 𝜓𝑘 (𝐺) and provide several estimations and exact values of 𝜓𝑘 (𝐺). It is also proven that 𝜓3 (𝐺) ≤ (2𝑛 + 𝑚)/6, for every graph 𝐺 with 𝑛 vertices and 𝑚 edges. In this paper we present a more general result showing that for an arbitrary graph 𝐺 with 𝑛 vertices and 𝑚 edges and an integer 𝑘 such that 𝑘𝑛 𝑚 𝑘≤𝑚 𝑛 ≤ 𝑘 + 1 we have 𝜓3 (𝐺) ≤ 𝑘+2 + (𝑘+1)(𝑘+2) . In Section 3 we consider bounds on 𝑑-regular graphs and show that for an arbitrary integer 𝑘 ≥ 2 and 𝑑−𝑘+2 𝑑-regular graph 𝐺, 𝑑 ≥ 𝑘 − 1 we have 𝜓𝑘 (𝐺) ≥ 2𝑑−𝑘+2 ∣𝑉 (𝐺)∣. Section 4 is devoted to the vertex 𝑘-path cover number of Cartesian products of graphs, in particular on grids. We conclude this section with the following straightforward values of 𝜓𝑘 . Proposition 1.1. Let 𝑘 ≥ 2 and 𝑛 ≥ 𝑘 be positive integers. Then ⌊ ⌋ ∙ 𝜓𝑘 (𝑃𝑛 ) = 𝑛𝑘 , ∙ 𝜓𝑘 (𝐶𝑛 ) = ⌈ 𝑛𝑘 ⌉, ∙ 𝜓𝑘 (𝐾𝑛 ) = 𝑛 − 𝑘 + 1.

2

Upper bounds on degree of vertices

First, recall the result of [4] which is a consequence of Lov´asz’s decomposition [11] of a graph with maximum degree Δ into subgraphs of maximum degree 1. 2

Lemma 2.1 ([4]). Let 𝐺 be a graph of maximum degree Δ. Then 𝜓3 (𝐺) ≤

⌈ Δ−1 2 ⌉ ⌈ Δ+1 2 ⌉

∣𝑉 (𝐺)∣.

In the following theorem we give a tight upper bound in terms of average degree of graph, which extends the result 𝜓3 (𝐺) ≤ (2𝑛 + 𝑚)/6 of [4]. Theorem 2.1. Let 𝐺 be a graph of order 𝑛, size 𝑚 and average degree 𝑑(𝐺), and 𝑘 be a positive integer such that 2𝑘 ≤ 𝑑(𝐺) ≤ 2𝑘 + 2. Then 𝜓3 (𝐺) ≤

𝑘𝑛 𝑚 + 𝑘 + 2 (𝑘 + 1)(𝑘 + 2)

Proof. Proof by induction on 𝑘 (the basis of induction 𝑘 = 1 coincides with the bound from [4]). Assume (𝑘 − 1)𝑛 𝑚 𝜓3 (𝐺) ≤ + 𝑘+1 𝑘(𝑘 + 1) for all graphs with average degree 𝑑, 2𝑘 − 2 ≤ 𝑑 ≤ 2𝑘. We aim to prove that for a graph 𝐺 with average degree 𝑑, 2𝑘 ≤ 𝑑 ≤ 2𝑘 + 2 the following holds 𝑘𝑛 𝑚 𝜓3 (𝐺) ≤ + . 𝑘 + 2 (𝑘 + 1)(𝑘 + 2) Systematically remove from 𝐺 vertices of degree at least 2(𝑘 + 1). We get a graph 𝐺′ of Δ(𝐺′ ) ≤ 2(𝑘 + 1) − 1. For 𝐺′ we now look at two cases. In the first case we do not use the induction step. If 𝑘𝑛′ ≤ 𝑚′ , then by Lemma 2.1 𝜓3 (𝐺′ ) ≤ = = ≤

𝑘 𝑛′ 𝑘+1 𝑘(𝑘 + 2) 𝑛′ (𝑘 + 1)(𝑘 + 2) 𝑘(𝑘 + 1) + 𝑘 ′ 𝑛 (𝑘 + 1)(𝑘 + 2) 𝑘 𝑚′ 𝑛′ + 𝑘+2 (𝑘 + 1)(𝑘 + 2)

In the second case, if 𝑘𝑛′ ≥ 𝑚′ we use the induction step. 𝜓3 (𝐺′ ) ≤ = ≤ =

𝑘−1 ′ 𝑛 + 𝑘+1 𝑘−1 ′ 𝑛 + 𝑘+1 𝑘−1 ′ 𝑛 + 𝑘+1 𝑘 𝑛′ + 𝑘+2

𝑚′ 𝑘(𝑘 + 1) 2𝑚′ 𝑚′ + 𝑘(𝑘 + 1)(𝑘 + 2) (𝑘 + 1)(𝑘 + 2) 2𝑛′ 𝑚′ + (𝑘 + 1)(𝑘 + 2) (𝑘 + 1)(𝑘 + 2) 𝑚′ (𝑘 + 1)(𝑘 + 2) 3

Next we prove a lower bound for 𝜓3 (𝐺), showing that the upper bound of Theorem 2.1 is tight. Theorem 2.2. For every positive rational number 𝑎𝑏 > 1, 2𝑘 ≤ there exists a graph 𝐺 with average degree 𝑑(𝐺) = 𝑎𝑏 and 𝜓3 (𝐺) ≥

𝑎 𝑏

≤ 2𝑘 + 2,

𝑘𝑛 𝑚 + . 𝑘 + 2 (𝑘 + 1)(𝑘 + 2)

Proof. Denote by 𝐻𝑛 a complete graph on 𝑛 vertices minus one perfect matching (assume 𝑛 is even). Clearly ∣𝑉 (𝐻𝑛 )∣ = 𝑛 and ∣𝐸(𝐻𝑛 )∣ =

𝑛(𝑛 − 1) 𝑛 𝑛(𝑛 − 2) − = . 2 2 2

Clearly 𝑑𝑖𝑠𝑠(𝐻𝑛 ) ≤ 2, because arbitrary three vertices of 𝐻𝑛 forms a path of order three. Therefore 𝜓3 (𝐻𝑛 ) = 𝑛 − 2, for 𝑛 ≥ 2. We construct the graph 𝐺 as the disjoint union of ∙ 𝑥 components 𝐻2𝑘+2 , ∙ 𝑦 components 𝐻2𝑘+4 . We let 𝑥 = (2𝑏 − 𝑎 + 2𝑘𝑏)(𝑘 + 2) and 𝑦 = (𝑎 − 2𝑘𝑏)(𝑘 + 1). First, we verify the average degree of the graph 𝐺. There are 𝑥(2𝑘 + 2) vertices of degree 2𝑘 and 𝑦(2𝑘 + 4) vertices of degree 2𝑘 + 2, hence we get 𝑑(𝐺) = = = = =

𝑥(2𝑘 + 2)(2𝑘) + 𝑦(2𝑘 + 4)(2𝑘 + 2) 𝑥(2𝑘 + 2) + 𝑦(2𝑘 + 4) (2𝑏 − 𝑎 + 2𝑘𝑏)(𝑘 + 2)(2𝑘 + 2)(2𝑘) + (𝑎 − 2𝑘𝑏)(𝑘 + 1)(2𝑘 + 4)(2𝑘 + 2) (2𝑏 − 𝑎 + 2𝑘𝑏)(𝑘 + 2)(2𝑘 + 2) + (𝑎 − 2𝑘𝑏)(𝑘 + 1)(2𝑘 + 4) (2𝑏 − 𝑎 + 2𝑘𝑏)(2𝑘) + (𝑎 − 2𝑘𝑏)(2𝑘 + 2) (2𝑏 − 𝑎 + 2𝑘𝑏) + (𝑎 − 2𝑘𝑏) 2𝑘𝑏 − (𝑎 + 2𝑘𝑏)(2𝑘) + (𝑎 − 2𝑘𝑏)(2𝑘) + 2(𝑎 − 2𝑘𝑏) 2𝑏 𝑎 . 𝑏

Now, let us verify the value of 𝜓3 of 𝐺. Clearly, 𝜓3 (𝐺) = 𝑥 ⋅ 𝜓3 (𝐻2𝑘+2 ) + 𝑦 ⋅ 𝜓3 (𝐻2𝑘+4 ) = 𝑥(2𝑘) + 𝑦(2𝑘 + 2).

4

And this exactly matches the value of the formula 𝜓3 (𝐺) = = = = =

3

𝑘𝑛 𝑚 + 𝑘 + 2 (𝑘 + 1)(𝑘 + 2) 𝑘(𝑥(2𝑘 + 2) + 𝑦(2𝑘 + 4)) 𝑥(2𝑘)(2𝑘 + 2) + 𝑦(2𝑘 + 2)(2𝑘 + 4) + 𝑘+2 2(𝑘 + 1)(𝑘 + 2) 𝑘𝑥(2𝑘 + 2) 𝑥(2𝑘) + 𝑦(2𝑘 + 4) + 2𝑘𝑦 + 𝑘+2 (𝑘 + 2) 𝑘𝑥(2𝑘 + 2) 𝑥(2𝑘) + 2𝑘𝑦 + + 2𝑦 𝑘+2 (𝑘 + 2) 2𝑘𝑥 + 2𝑘𝑦 + 2𝑦.

Regular graphs

In the main theorem of this section we shall use the following result of Erd˝os and Gallai [7]. Theorem 3.1 ([7]). If 𝐺 is a graph on 𝑛 vertices that does not contain a path of order 𝑘 then it cannot have more than 𝑛(𝑘−2) edges. Moreover, the 2 bound is achieved when the graph consists of disjoint complete graphs on 𝑘 − 1 vertices. Theorem 3.2. Let 𝑘 ≥ 2 and 𝑑 ≥ 𝑘 − 1 be positive integers. Then for any 𝑑-regular graph 𝐺 the following holds: 𝜓𝑘 (𝐺) ≥

𝑑−𝑘+2 ∣𝑉 (𝐺)∣. 2𝑑 − 𝑘 + 2

Proof. Let 𝑆 ⊆ 𝑉 (𝐺) be a vertex 𝑘-path cover and 𝑇 = 𝑉 (𝐺) ∖ 𝑆. Let 𝐸𝑆 , 𝐸𝑇 be the number of edges with both endvertices in 𝑆 and 𝑇 respectively. Let 𝐸𝑆𝑇 be the number of edges with one endvertex in 𝑆 and the second vertex in 𝑇 . Then obviously ∣𝐸(𝐺)∣ = 21 𝑑∣𝑉 (𝐺)∣ = ∣𝐸𝑆 ∣ + ∣𝐸𝑆𝑇 ∣ + ∣𝐸𝑇 ∣. Since 𝐺 is 𝑑-regular, 𝑑∣𝑆∣ = 2∣𝐸𝑆 ∣ + ∣𝐸𝑆𝑇 ∣. Therefore ∣𝑆∣ ≥ 𝑑1 ∣𝐸𝑆𝑇 ∣. Similarly, ∣𝐸𝑆𝑇 ∣ + 2∣𝐸𝑇 ∣ = 𝑑∣𝑇 ∣. Since, 𝐺[𝐸𝑇 ] does not contain the path of order 𝑘, according to Theorem 3.1, we have ∣𝐸𝑇 ∣ ≤ ∣𝑇 ∣(𝑘−2) . Combining all 2 the previous formulas we immediately have ∣𝑆∣ ≥

1 1 1 𝑑−𝑘+2 ∣𝐸𝑆𝑇 ∣ = (𝑑∣𝑇 ∣ − 2∣𝐸𝑇 ∣) ≥ (𝑑∣𝑇 ∣ − ∣𝑇 ∣(𝑘 − 2)) = ∣𝑇 ∣. 𝑑 𝑑 𝑑 𝑑

Then ∣𝑆∣ + ∣𝑇 ∣ ∣𝑇 ∣ 𝑑 2𝑑 − 𝑘 + 2 =1+ ≤1+ = ∣𝑆∣ ∣𝑆∣ 𝑑−𝑘+2 𝑑−𝑘+2 5

and ∣𝑆∣ ≥

𝑑−𝑘+2 ∣𝑉 (𝐺)∣. 2𝑑 − 𝑘 + 2

Corollary 3.1. If 𝐺 is a cubic graph then 𝜓𝑘 (𝐺) ≥ 𝜓3 (𝐺) ≥ 25 ∣𝑉 (𝐺)∣.

5−𝑘 8−𝑘 ∣𝑉

(𝐺)∣, in particular

Corollary 3.2. Let 𝑘 ≥ 2 be a fixed positive integer and 𝑄𝑑 the 𝑑-dimensional hypercube. Then 1 𝜓𝑘 (𝑄𝑑 ) = . lim 𝑑→∞ ∣𝑉 (𝑄𝑑 )∣ 2 𝑑−𝑘+2 Proof. By Theorem 3.2 we know that 2𝑑−𝑘+2 ∣𝑉 (𝑄𝑑 )∣ ≤ 𝜓𝑘 (𝑄𝑑 ) ≤ 𝜓2 (𝑄𝑑 ) ≤ 1 2 ∣𝑉 (𝑄𝑑 )∣, and the result immediately follows.

Theorem 3.3. Let 𝑘 ≥ 2 and 𝑑 ≥ 𝑘 − 1 be positive integers. There exists a 𝑑-regular graph 𝐺 with 𝜓𝑘 (𝐺) ≤

𝑑−𝑘+2 ∣𝑉 (𝐺)∣. 2𝑑 − 𝑘 + 2

Proof. Given values of 𝑘 and 𝑑, a graph 𝐺 may be constructed as follows. The vertex set is 𝑉 (𝐺) = 𝐴 ∪ 𝐵 with ∣𝐴∣ = (𝑘 − 1)(𝑑 − 𝑘 + 2) ∣𝐵∣ = 𝑑(𝑘 − 1). Let the set 𝐴 be an independent set in 𝐺 and let 𝐵 consist of 𝑑 components of 𝐾𝑘−1 in 𝐺. The remaining edges between 𝐴 and 𝐵 may be defined arbitrarily to fill 𝑑-regularity of 𝐺. Clearly, 𝐴 is a 𝑘-path vertex cover of 𝐺 since 𝐵 does not contain a path on 𝑘 vertices and ∣𝐴∣ (𝑘 − 1)(𝑑 − 𝑘 + 2) 𝑑−𝑘+2 𝜓𝑘 (𝐺) ≤ = = . ∣𝑉 (𝐺)∣ ∣𝐴∣ + ∣𝐵∣ (𝑘 − 1)(𝑑 − 𝑘 + 2) + 𝑑(𝑘 − 1) 2𝑑 − 𝑘 + 2

4

Cartesian products

Recall that the Cartesian product 𝐺□𝐻 of graphs 𝐺 = (𝑉 (𝐺), 𝐸(𝐺)) and 𝐻 = (𝑉 (𝐻), 𝐸(𝐻)) has the vertex set 𝑉 (𝐺)×𝑉 (𝐻), and vertices (𝑢, 𝑣), (𝑥, 𝑦) are adjacent whenever 𝑢 = 𝑥 and 𝑣𝑦 ∈ 𝐸(𝐻), or 𝑢𝑥 ∈ 𝐸(𝐺) and 𝑣 = 𝑦. By 𝑝𝐺 and 𝑝𝐻 we denote the natural projections to the factors 𝐺 and 𝐻, respectively. For a fixed vertex 𝑢 ∈ 𝑉 (𝐺), the 𝐻-fiber 𝑢 𝐻 is the subgraph of 𝐺□𝐻 induced by the set of vertices {(𝑢, 𝑣), 𝑣 ∈ 𝑉 (𝐻)}; analogously, for a fixed vertex 𝑣 ∈ 𝑉 (𝐻), the 𝐺-fiber 𝐺𝑣 is the subgraph of 𝐺□𝐻 induced by the set of vertices {(𝑢, 𝑣), 𝑢 ∈ 𝑉 (𝐺)}. Note that each 𝐻-fiber is isomorphic to 𝐻, and is an induced subgraph of 𝐺□𝐻. Hence any vertex 𝑘-path cover 𝑆 of 𝐺□𝐻 must contains at least 6

𝜓𝑘 (𝐻) vertices in every 𝐻-fiber. Since there are ∣𝑉 (𝐺)∣ such fibers, we get ∣𝑆∣ ≥ ∣𝑉 (𝐺)∣ ⋅ 𝜓𝑘 (𝐻). By reversing the roles of 𝐺 and 𝐻 we immediately get the following result. Proposition 4.1. Let 𝐺 and 𝐻 be graphs and 𝑘 ≥ 2 be a positive integer. Then 𝜓𝑘 (𝐺□𝐻) ≥ max{𝜓𝑘 (𝐺) ⋅ ∣𝑉 (𝐻)∣, ∣𝑉 (𝐺)∣ ⋅ 𝜓𝑘 (𝐻)}. For 𝑘 = 2 this bound was observed already by Vizing [16], who expressed it in terms of independence numbers as follows 𝛼(𝐺□𝐻) ≤ min{𝛼(𝐺) ∣𝑉 (𝐻)∣ , 𝛼(𝐻) ∣𝑉 (𝐺)∣}.

(1)

A characterization of pairs of graphs that achieve the equality in (1) has been obtained in [9], and independently also in [5]. The characterization involves the existence of homomorphims of one factor to the so-called independence graph of the other factor. We will extend this result, by presenting the corresponding characterization of pairs of graphs that achieve the equality in 4.1. Recall that homomorphism from a graph 𝐺 to a graph 𝐻 is a mapping ℎ : 𝑉 (𝐺) → 𝑉 (𝐻) such that for any pair 𝑢, 𝑣 ∈ 𝑉 (𝐺): if 𝑢𝑣 ∈ 𝐸(𝐺) then ℎ(𝑢)ℎ(𝑣) ∈ 𝐸(𝐻). If there exists a homomorphism from a graph 𝐺 to a graph 𝐻 we write 𝐺 → 𝐻. Let 𝐺 be a graph, 𝑘 ≥ 2 an integer, and let 𝐴,𝐵 be two subsets of 𝑉 (𝐺). We say that 𝐴, 𝐵 form a good 𝑘-partition in 𝐺 if there is no path on 𝑘 vertices in 𝐺 of the form 𝑎1 , . . . , 𝑎𝑙1 = 𝑏𝑙1 , . . . , 𝑏𝑙2 = 𝑎𝑙2 , . . . , 𝑎𝑘 such that 𝑎𝑖 ∈ 𝐴, 𝑏𝑗 ∈ 𝐵 (for all applicable indices 𝑖 and 𝑗); in addition, we allow that the path ends with 𝑏𝑘 and/or begins with 𝑏1 . In other words, there is no 𝑘 path in 𝐴 ∪ 𝐵. In particular, any two 𝜓𝑘 -sets 𝐴, 𝐵 such that 𝐴 ∩ 𝐵 = ∅ form a good 𝑘-partition of 𝐺. On the other hand, if one of the sets 𝐴, 𝐵 is not a vertex 𝑘-path cover then 𝐴, 𝐵 will clearly not form a good 𝑘-partition of 𝐺; indeed, if say 𝐴 is not a vertex 𝑘-path cover then there is already a path of length 𝑘 in 𝑉 (𝐺) − 𝐴 that begins with 𝑏1 and ends with 𝑏𝑘 . We say that a vertex 𝑘-path cover of minimum size 𝜓𝑘 (𝐺) is a 𝜓𝑘 -set of 𝐺, and let 𝒮 be the set of all 𝜓𝑘 -sets of 𝐺. Then the 𝜓𝑘 -graph of 𝐺, denoted Ψ𝑘 (𝐺) is the graph with 𝑉 (Ψ𝑘 (𝐺)) = 𝒮, and 𝐴, 𝐵 ∈ 𝒮 are adjacent whenever they form a good 𝑘-partition in 𝐺. Theorem 4.1. Let 𝐺, 𝐻 be arbitrary graphs. If 𝜓𝑘 (𝐺□𝐻) = max{𝜓𝑘 (𝐺) ⋅ ∣𝑉 (𝐻)∣, ∣𝑉 (𝐺)∣ ⋅ 𝜓𝑘 (𝐻)} then 𝐻 → Ψ𝑘 (𝐺) or 𝐺 → Ψ𝑘 (𝐻). Proof. Suppose first that 𝐺 and 𝐻 are graphs such that 𝜓𝑘 (𝐺)∣𝑉 (𝐻)∣ ≥ 𝜓𝑘 (𝐻)∣𝑉 (𝐺)∣, and that the equality holds, thus 𝜓𝑘 (𝐺□𝐻) = 𝜓𝑘 (𝐺)∣𝑉 (𝐻)∣. 7

Let 𝐴 be a 𝜓𝑘 -set of 𝐺□𝐻. Note that in each 𝐺-fiber of 𝐺□𝐻 at least 𝜓𝑘 (𝐺) vertices are from 𝐴. Combining this with ∣𝐴∣ = 𝜓𝑘 (𝐺)∣𝑉 (𝐻)∣ we find that in each 𝐺-fiber there are exactly 𝜓𝑘 (𝐺) vertices from 𝐴. In other words, vertices of 𝐴 in each 𝐺-fiber form a 𝜓𝑘 -set of this 𝐺-fiber. Let 𝑢, 𝑣 ∈ 𝑉 (𝐻) be adjacent vertices. Then we claim that the sets 𝑝𝐺 (𝐴∩𝐺𝑢 ) and 𝑝𝐺 (𝐴∩𝐺𝑣 ) are two 𝜓𝑘 -sets of that form a good 𝑘-partition of 𝐺. Indeed, otherwise there would exist a path of length 𝑘 in 𝐺 of the form 𝑎1 , . . . , 𝑎𝑙1 = 𝑏𝑙1 , . . . , 𝑏𝑙2 = 𝑎𝑙2 , . . . , 𝑎𝑘 as described above, and hence in 𝐺□𝐻 −𝐴 there would be a path (𝑎1 , 𝑢), . . . , (𝑎𝑙1 , 𝑢), (𝑏𝑙1 , 𝑣), . . . , (𝑏𝑙2 , 𝑣), (𝑎𝑙2 , 𝑢), . . . , (𝑎𝑘 , 𝑢) of the same length 𝑘. This is a contradiction with 𝐴 being a vertex 𝑘path cover of 𝐺. Hence the mapping ℎ : 𝐻 → Ψ𝑘 (𝐺), defined by ℎ(𝑢) = 𝑝𝐺 (𝐴 ∩ 𝐺𝑢 ) is a homomorphism, since for any 𝑢𝑣 ∈ 𝐸(𝐻), also ℎ(𝑢) and ℎ(𝑣) are adjacent. Note that for 𝑘 = 2 the converse of the statement of Theorem 4.1 is also true. We get nicer results if we simplify the structure of one of the factors. If say, 𝐻 is isomorphic to the complete graph 𝐾𝑛 , then Ψ𝑘 (𝐻) has a nice structure; its vertices consist of all 𝑘-combinations of the set {1, . . . , 𝑛}, and two such 𝑘-combinations are adjacent whenever their union is {1, . . . , 𝑛}. Clearly Ψ𝑘 (𝐾𝑛 ) is nontrivial when 2𝑘 ≥ 𝑛. The part from Theorem 4.1, i.e. 𝐻 → Ψ𝑘 (𝐾𝑛 ), thus translates to the colorings of a graph 𝐻 with 𝑘combinations. The following theorem, containing the formulas for 𝜓𝑘 of grids (Cartesian products of paths), is the main result of this section. Theorem 4.2.

(i) 𝜓3 (𝑃2𝑛+1 □ 𝑃2𝑘 ) = 2𝑛𝑘 + ⌊ 2𝑘 3 ⌋, where 𝑛, 𝑘 ≥ 1,

(ii) 𝜓3 (𝑃2𝑛 □ 𝑃2𝑘 ) = 2𝑛𝑘, where 𝑛, 𝑘 ≥ 1, (iii) 𝜓3 (𝑃2𝑛+1 □ 𝑃2𝑘+1 ) = 𝑛(2𝑘 + 1) + ⌊ 2𝑘+1 3 ⌋, where 1 ≤ 𝑛 ≤ 𝑘. Proof. (i) Let us label the vertices of 𝑃2𝑛+1 consecutively by 𝑢1 , 𝑢2 , . . . , 𝑢2𝑛+1 and the vertices of 𝑃2𝑘 by 𝑣1 , 𝑣2 , . . . , 𝑣2𝑘 . With this labeling, the vertex 𝑣𝑗 (𝑢𝑖 , 𝑣𝑗 ) ∈ 𝑉 (𝑃2𝑛+1 □ 𝑃2𝑘 ) is in 𝑖th fiber 𝑢𝑖 𝑃2𝑘 and in 𝑗 th fiber 𝑃2𝑛+1 . 2𝑘 Let us first prove that 𝜓3 (𝑃2𝑛+1 □ 𝑃2𝑘 ) ≤ 2𝑛𝑘 + ⌊ 3 ⌋. We construct a vertex 3-path cover set 𝑆 in the following way: 𝑆 = {(𝑢2𝑖+1 , 𝑣3𝑗+3 ), (𝑢2𝑖 , 𝑣3𝑗+1 ), (𝑢2𝑖 , 𝑣3𝑗+2 )∣ for applicable indices 𝑖 and 𝑗} . It is not difficult to verify that 𝑆 is a vertex 3-path cover set (Fig. 1) and gives the desired upper bond.

8

v1 v2 v3 v4 v5 v6 v 3k 2 v 3k 1 v 3k u1 u2 u3 u4 u5

u2n 1 u2n u2n 1

Figure 1: A vertex 3-path covering of 𝑃2𝑛+1 □ 𝑃2𝑘 Now let us prove that 𝜓3 (𝑃2𝑛+1 □ 𝑃2𝑘 ) ≥ 2𝑛𝑘 + ⌊ 2𝑘 3 ⌋. Let 𝑆 be the 𝑢 𝑖 optimal vertex 3-path covering. Consider the fiber 𝑃2𝑘 , for some 𝑖 ∈ {1, 2, . . . , 2𝑛}. Suppose (𝑢𝑖 , 𝑣𝑗 ) ∕∈ 𝑆, for some 𝑗 ∈ {1, 2, . . . , 2𝑘 −1}, and all its neighbors 𝑢 in 𝑖 𝑃2𝑘 are in 𝑆. Then in the fiber 𝑢𝑖+1 𝑃2𝑘 either the vertex (𝑢𝑖+1 , 𝑣𝑗 ) is in 𝑆 (left-hand side of Fig. 2) or all its neighbors in this fiber are in 𝑆 (right-hand side of Fig. 2).

vj

vj

ui ui 1

ui ui 1

Figure 2: Vertex (𝑢𝑖 , 𝑣𝑗 ) is not in 𝑆 Suppose (𝑢𝑖 , 𝑣𝑗 ) and (𝑢𝑖 , 𝑣𝑗+1 ), for some 𝑗 ∈ {1, 2, . . . , 2𝑘 − 1}, are not in 𝑆. Then the vertices (𝑢𝑖+1 , 𝑣𝑗 ) and (𝑢𝑖+1 , 𝑣𝑗+1 ) must be in 𝑆, otherwise 𝑆 is not a vertex 3-path covering (see Fig. 3). Let 𝐴 be the set of all vertices in fiber 𝑢𝑖 𝑃2𝑘 that are not in 𝑆, and 𝐵 the set of all vertices in fiber 𝑢𝑖+1 𝑃2𝑘 that are in 𝑆. We shall prove that ∣𝐴∣ ≤ ∣𝐵∣ by finding a one to one function 𝑓 from 𝐴 to 𝐵. Let (𝑢𝑖 , 𝑣𝑗 ) be a vertex in 𝐴. Suppose that the vertex (𝑢𝑖 , 𝑣2𝑘 ) is not in 𝐴. By Fig. 2 and Fig. 3 we set 𝑓 ((𝑢𝑖 , 𝑣𝑗 )) = (𝑢𝑖+1 , 𝑣𝑗 ) if (𝑢𝑖+1 , 𝑣𝑗 ) is in 𝑆, otherwise we set 𝑓 ((𝑢𝑖 , 𝑣𝑗 )) = (𝑢𝑖+1 , 𝑣𝑗+1 ). This is obviously a one to one function. We take a different approach if the vertex (𝑢𝑖 , 𝑣2𝑘 ) is in 𝐴. If also the vertex (𝑢𝑖 , 𝑣2𝑘−1 ) is in 𝐴 we can use the same argument then above since 9

vj vj 1

ui ui 1 Figure 3: Vertices (𝑢𝑖 , 𝑣𝑗 ) and (𝑢𝑖 , 𝑣𝑗+1 ) are not in 𝑆 according to Fig. 3 we can set 𝑓 ((𝑢𝑖 , 𝑣2𝑘 )) = (𝑢𝑖+1 , 𝑣2𝑘 ). If the vertex (𝑢𝑖 , 𝑣2𝑘−1 ) ∈ / 𝐴 and (𝑢𝑖+1 , 𝑣2𝑘 ) ∈ 𝐵 then we can set 𝑓 ((𝑢𝑖 , 𝑣2𝑘 )) = (𝑢𝑖+1 , 𝑣2𝑘 ) otherwise we can set 𝑓 ((𝑢𝑖 , 𝑣2𝑘 )) = (𝑢𝑖+1 , 𝑣2𝑘−1 ). It can happen that the vertex (𝑢𝑖+1 , 𝑣2𝑘−1 ) is already an image of the vertex (𝑢𝑖 , 𝑣2𝑘−2 ). If this situation occurs we can set ether 𝑓 ((𝑢𝑖 , 𝑣2𝑘−2 )) = (𝑢𝑖+1 , 𝑣2𝑘−2 ) or 𝑓 ((𝑢𝑖 , 𝑣2𝑘−2 )) = (𝑢𝑖+1 , 𝑣2𝑘−3 ) since at least one of the vertices (𝑢𝑖+1 , 𝑣2𝑘−2 ) or (𝑢𝑖+1 , 𝑣2𝑘−3 ) is in 𝐵. Note that if vertex (𝑢𝑖 , 𝑣2𝑘−2 ) exists in the grid then so must vertex (𝑢𝑖+1 , 𝑣2𝑘−3 ) since every 𝑃2𝑘 -fiber has even number of vertices. It can occur that the vertex (𝑢𝑖+1 , 𝑣2𝑘−3 ) is also an image of a vertex in fiber 𝑢𝑖 𝑃2𝑘 . In this case we can repeat the above procedure. We can always find a one to one function from 𝐴 to 𝐵 and hence ∣𝐴∣ ≤ ∣𝐵∣. Therefore two consecutive fibers, 𝑢𝑖 𝑃2𝑘 and 𝑢𝑖+1 𝑃2𝑘 , must have at least 2𝑘 vertices in 𝑆. There are 𝑛 such paired fibers in 𝑃2𝑛+1 □ 𝑃2𝑘 . Together with the additional fiber, which is isomorphic to the path 𝑃2𝑘 , there must be at lest 2𝑛𝑘 + ⌊ 2𝑘 3 ⌋ vertices in 𝑆. (ii) We follow the same line of thought as in (i), with the exception that in this case there is no additional fiber.

(iii) Note that in the proof of (i), two consecutive fibers, 𝑢𝑖 𝑃2𝑘 and 𝑢𝑖+1 𝑃 , have at least 2𝑘 vertices in 𝑆. The fiber 𝑢𝑖 𝑃 2𝑘 2𝑘+1 in this case has odd number of vertices and it can occur that two consecutive fibers do not have at least 2𝑘 + 1 vertices in 𝑆. Nevertheless, two consecutive fibers, 𝑢𝑖 𝑃 𝑢𝑖+1 𝑃 2𝑘+1 and 2𝑘+1 , must still have at least 2𝑘 vertices in 𝑆. Moreover if they have exactly 2𝑘 vertices in 𝑆, then for fibers 𝑢𝑖 𝑃2𝑘+1 and 𝑢𝑖+1 𝑃2𝑘+1 the vertices (𝑢𝑖 , 𝑣2𝑗 ) and (𝑢𝑖+1 , 𝑣2𝑗 ), 𝑗 ∈ {1, . . . , 𝑘}, must all be in 𝑆 (see the structure of sets 𝐴 and 𝐵 in (i)). Hence, both fibers 𝑢𝑖 𝑃2𝑘+1 and 𝑢𝑖+1 𝑃2𝑘+1 each contribute at least 𝑘 vetices to 𝑆. Having done the basic observation let us proceed with the proof. For the upper bound the same construction as in (i) suffices. Now we want to prove that 𝜓3 (𝑃2𝑛+1 □ 𝑃2𝑘+1 ) ≥ 𝑛(2𝑘 + 1) + ⌊ 2𝑘+1 3 ⌋. Asumme that 0 ≤ 𝑛 ≤ 𝑘 and 𝑆 is the optimal vertex 3-path covering of the graph 𝑃2𝑛+1 □ 𝑃2𝑘+1 . We proceed with induction on 𝑛. For 𝑛 = 0, the graph 𝑃1 □ 𝑃2𝑘+1 is isomorphic to the path 𝑃2𝑘+1 , and hence 𝜓3 (𝑃2𝑘+1 ) = ⌊ 2𝑘+1 3 ⌋. In the induction step we 10

consider two cases. Suppose that the last two fibers, 𝑢2𝑛 𝑃2𝑘+1 and 𝑢2𝑛+1 𝑃2𝑘+1 , contribute at least 2𝑘 + 1 vertices to 𝑆. Assume that 𝜓3 (𝑃2𝑛−1 □ 𝑃2𝑘+1 ) = (𝑛 − 1)(2𝑘 + 2𝑘+1 1) + ⌊ 2𝑘+1 3 ⌋ and 𝜓3 (𝑃2𝑛+1 □ 𝑃2𝑘+1 ) < 𝑛(2𝑘 + 1) + ⌊ 3 ⌋. If we remove the 𝑢 𝑢 fibers, 2𝑛 𝑃2𝑘+1 and 2𝑛+1 𝑃2𝑘+1 , we remove at least 2𝑘 + 1 vertices from 𝑆 an therefore 𝜓3 (𝑃2𝑛−1 □ 𝑃2𝑘+1 ) < 𝑛(2𝑘 + 1) + ⌊ 2𝑘+1 3 ⌋ − 2𝑘 − 1. By induction assumption it follows that ⌊ ⌋ ⌊ ⌋ 2𝑘 + 1 2𝑘 + 1 (𝑛 − 1)(2𝑘 + 1) + < 𝑛(2𝑘 + 1) + − 2𝑘 − 1 . 3 3 By doing some basic calculation this leads us to a contradiction. For the second case assume that the last two fibers, 𝑢2𝑛 𝑃2𝑘+1 and 𝑢2𝑛+1 𝑃2𝑘+1 , contribute exactly 2𝑘 vertices to 𝑆. According to the observation above, if we remove the last fiber 𝑢2𝑛+1 𝑃2𝑘+1 from the graph 𝑃2𝑛+1 □ 𝑃2𝑘+1 we remove at least 𝑘 vertices from 𝑆. Suppose again that 𝜓3 (𝑃2𝑛+1 □ 𝑃2𝑘+1 ) < 2𝑘+1 𝑛(2𝑘 + 1) + ⌊ 2𝑘+1 3 ⌋. Then 𝜓3 (𝑃2𝑛 □ 𝑃2𝑘+1 ) < 𝑛(2𝑘 + 1) + ⌊ 3 ⌋ − 𝑘. But 2𝑛 according to (i) we know that 𝜓3 (𝑃2𝑛 □ 𝑃2𝑘+1 ) = 2𝑛𝑘 + ⌊ 3 ⌋ and hence ⌊ ⌋ ⌊ ⌋ 2𝑛 2𝑘 + 1 2𝑛𝑘 + < 𝑛(2𝑘 + 1) + −𝑘. 3 3 By doing some basic calculation we get that 𝑘 < 𝑛, which again is a contradiction since 𝑛 ≤ 𝑘. As seen in Theorem 4.2, it is already hard to determine the exact value of 𝜓3 for grids therefore it would be nice to have at least some lower or upper bounds for 𝜓𝑘 , 𝑘 ≥ 4. Lemma 4.1. For each 𝑙 ≥ 2, 𝜓𝑙2 (𝑃2𝑙 □ 𝑃3𝑙 ) ≥ 𝑙. Proof. Assume to the contrary that 𝑆 is a 𝑘-path vertex cover of 𝑃2𝑙 □ 𝑃3𝑙 , with ∣𝑆∣ ≤ 𝑙 − 1. Then 𝑃2𝑙 □ 𝑃3𝑙 has at least 𝑙 𝑃3𝑙 -layers not containing any vertex of 𝑆. Denote such layers by 𝑃𝑓1 , 𝑃𝑓2 , . . . , 𝑃𝑓𝑙 . Since ∣𝑆∣ ≤ 𝑙 − 1, then two consecutive layers can be connected vertically in first and last thirds without crossing 𝑆. Formally, for each 𝑖 ∈ {1, . . . , (𝑙 − 1)} there exists 𝑖𝑎 ∈ {1, . . . , 𝑙} and 𝑖𝑏 ∈ {2𝑙 + 1, . . . , 3𝑙} such that for each ℎ ∈ {𝑎𝑖 , . . . , 𝑎𝑖+1 } the vertices (𝑖𝑎 , ℎ), (𝑖𝑏 , ℎ) ∈ / 𝑆. Finally, we construct a path 𝑃 moving horizontally on 𝑃𝑓1 , 𝑃𝑓2 , . . . , 𝑃𝑓𝑙 layers and vertically on 𝑖𝑎 , 𝑖𝑏 columns: 𝑃 =(1, 𝑓1 ), . . . , (1𝑏 , 𝑓1 ), (1𝑏 , 𝑓2 ), . . . , (2𝑎 , 𝑓2 ), (2𝑎 , 𝑓3 ), . . . , (3𝑏 , 𝑓3 ), (3𝑏 , 𝑓4 ), . . . , (4𝑎 , 𝑓4 ), . . . . Such a path contains at least 𝑙 vertices in each layer 𝑃𝑓1 , 𝑃𝑓2 , . . . , 𝑃𝑓𝑙 , hence the total order of the path 𝑃 is at least 𝑙2 , a contradiction. 11

First we present a lower bound with the help of Lemma 4.1. √ √ Proposition 4.2. For 𝑘 ≥ 4, 𝑛 ≥ 2 𝑘, 𝑚 ≥ 3 𝑘, the following holds 𝑛𝑚 √ ≤ 𝜓𝑘 (𝑃𝑛 □ 𝑃𝑚 ). 24 𝑘 Proof. We split the whole graph 𝑃𝑛 √ □ 𝑃𝑚 √ into 𝑟 disjoint subgraphs isomorphic to 𝑃2√𝑘 □ 𝑃3√𝑘 such that 𝑟(2 𝑘)(3 𝑘) ≥ 41 𝑛𝑚. By Lemma 4.1 a √ 𝑘-path vertex cover must have at least 𝑘 vertices in each subgraph isomorphic to 𝑃2√𝑘 □ 𝑃3√𝑘 in 𝐺, hence: √ 𝑛𝑚 𝜓𝑘 (𝑃𝑛 □ 𝑃𝑚 ) ≥ 𝑟 𝑘 ≥ √ . 24 𝑘

We conclude the paper with the following upper bound. Proposition 4.3. For 𝑘 ≥ 4 the following holds 2𝑛𝑚 𝑛𝑚 . 𝜓𝑘 (𝑃𝑛 □ 𝑃𝑚 ) ≤ √ − 𝑘 𝑘 √ Proof. We will construct a 𝑘-path vertex cover with at most 2𝑛𝑚 − 𝑛𝑚 𝑘 𝑘 { } √ vertices. Let 𝑆1 = (𝑖, 𝑗) ∈ 𝑃𝑛 □ 𝑃𝑚 ∣ 𝑖 ≡ 0 (mod⌊ 𝑘⌋) (for all applicable { √ } indicies 𝑖 and 𝑗) and similarly 𝑆2 = (𝑖, 𝑗) ∈ 𝑃𝑛 □ 𝑃𝑚 ∣ 𝑗 ≡ 0 (mod⌊ 𝑘⌋) . It is easy to see that 𝑆 = (𝑆1 ∪ 𝑆2 )∖(𝑆1 ∩ 𝑆2 ) is a 𝑘-path vertex cover, since the largest subgraph of 𝑃𝑛 □ 𝑃𝑚 with all vertices uncovered is isomorphic to 𝑃⌊√𝑘⌋−1 □ 𝑃⌊√𝑘⌋−1 . √ In a 𝑃𝑛 -layer we cover each ⌊ 𝑘⌋-th vertex, since there 𝑚 layers the √ . Similarly, ∣𝑆2 ∣ ≤ 𝑛𝑚 √ . The vertices size of 𝑆1 is at most ∣𝑆1 ∣ ≤ ⌊𝑛𝑚 𝑘⌋ ⌊ 𝑘⌋

(𝑖, 𝑗) ∈ 𝑆1 ∩ 𝑆2 can be left uncovered, since all the vertices (𝑖 ± 1, 𝑗) and (𝑖, 𝑗 ± 1) are in 𝑆. Since the size of ∣𝑆1 ∩ 𝑆2 ∣ ≤ 𝑛𝑚 𝑘 , the assertion follows.

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