Isometric path numbers of graphs

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Isometric path numbers of graphs

夡

Jun-Jie Pan

, Gerard J. Chang

a_{Department of Applied Mathematics, National Chiao Tung University, Hsinchu 30050, Taiwan} b_{Department of Mathematics, National Taiwan University, Taipei 10617, Taiwan}

c_{National Center for Theoretical Sciences, Taipei Office, Taiwan}

Received 10 November 2003; received in revised form 28 March 2006; accepted 12 April 2006 Available online 9 June 2006

Abstract

An isometric path between two vertices in a graph G is a shortest path joining them. The isometric path number of G, denoted by ip(G), is the minimum number of isometric paths needed to cover all vertices of G. In this paper, we determine exact values of isometric path numbers of complete r-partite graphs and Cartesian products of 2 or 3 complete graphs.

© 2006 Elsevier B.V. All rights reserved.

Keywords: Isometric path; Complete r-partite graph; Hamming graphs

1. Introduction

An isometric path between two vertices in a graph G is a shortest path joining them. The isometric path number of G, denoted by ip(G), is the minimum number of isometric paths required to cover all vertices of G. This concept has a

close relationship with the game of cops and robbers described as follows.

The game is played by two players, the cop and the robber, on a graph. The two players move alternately, starting with the cop. Each player’s first move consists of choosing a vertex at which to start. At each subsequent move, a player may choose either to stay at the same vertex or to move to an adjacent vertex. The object for the cop is to catch the robber, and for the robber is to prevent this from happening. Nowakowski and Winkler[7]and Quilliot [9]independently proved that the cop wins if and only if the graph can be reduced to a single vertex by successively removing pitfalls, where a pitfall is a vertex whose closed neighborhood is a subset of the closed neighborhood of another vertex.

As not all graphs are cop-win graphs, Aigner and Fromme[1]introduced the concept of cop-number of a general graph G, denoted by c(G), which is the minimum number of cops needed to put into the graph in order to catch the robber. On the way to giving an upper bound for the cop-numbers of planar graphs, they showed that a single cop moving on an isometric path P guarantees that after a finite number of moves the robber will be immediately caught if he moves onto P. Observing this fact, Fitzpatrick[3]then introduced the concept of isometric path cover and pointed out that c(G)
ip(G).

The isometric path number of the Cartesian product Pn1Pn2 . . . Pnrhas been studied in the literature. Fitzpatrick [4]gave bounds for the case when n1= n2= · · · = nr. Fisher and Fitzpatrick[2]gave exact values for the case r = 2.

夡_{Supported in part by the National Science Council under Grant NSC92-2115-M-002-015.} E-mail address:gjchang@math.ntu.edu.tw(G.J. Chang).

0012-365X/$ - see front matter © 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2006.04.003

Fitzpatrick et al.[5]gave a lower bound, which is in fact the exact value if r + 1 is a power of 2, for the case when

n1= n2= · · · = nr= 2. Pan and Chang[8]gave a linear-time algorithm to solve the isometric path problem on block

graphs.

In this paper, we determine exact values of isometric path numbers of all complete r-partite graphs and Cartesian products of 2 or 3 complete graphs. Recall that a complete r-partite graph is a graph whose vertex set can be partitioned into disjoint union of r nonempty parts, and two vertices are adjacent if and only if they are in different parts. We use Kn1,n2,...,nr to denote a complete r-partite graph whose parts are of sizes n1, n2, . . . , nr, respectively. A Hamming

graph is the Cartesian product of complete graphs, which is the graph Kn1Kn2 . . . Knr with vertex set

V (Kn1Kn2 . . . Knr) = {(x1, x2, . . . , xr) : 0
xi< ni for 1
i
r} and edge set E(Kn1Kn2 . . . Knr) is

{(x1, x2, . . . , xr)(y1, y2, . . . , yr) : xi = yi ∈ V (Ki) for all i except just one xj = yj}.

2. Completer-partite graphs

The purpose of this section is to determine exact values of the isometric path numbers of all complete r-partite graphs.

Suppose G is the complete r-partite graph Kn1,n2,···,nr of n vertices, where r 2, n1n2 · · · nr and n = n1+ n2+ · · · + nr. Suppose G have
parts of odd sizes. We note that every isometric path in G has at most three vertices. Consequently,

ip(G)n/3.

Also, for any path of three vertices in an isometric path coverC, two end vertices of the path are in one part of G and the center vertex is in another part. In case when two paths of three vertices inC have a common end vertex, we may replace one by a path of two vertices. And, a path of one vertex can be replaced by a path of two vertices. So, without loss of generality, we may only consider isometric path covers in which every path is of two or three vertices, and two 3-vertex paths have different end vertices.

Lemma 1. If 3n1> 2n, then ip(G) = n1/2.

Proof. First, ip(G)n1/2 since every isometric path contains at most two vertices in the first part.

On the other hand, we use induction on n−n1to prove that ip(G)
n1/2. When n−n1=1, we have G=Kn−1,1. In this case, it is clear that ip(G)
n1/2. Suppose n − n12 and the claim holds for n− n1< n − n1. Then we remove

two vertices from the first part and one vertex from the second part to form an isometric 3-path P. Since 3n1> 2n,

we have n1− 2 > 2(n − n1− 1) > 0 and so n1− 2 > n2. Then, the remaining graph Ghas r2, n1= n1− 2 and

n= n − 3. It then still satisfies 3n1> 2n. As n− n1= n − n1− 1, by the induction hypothesis, ip(G)
n1/2 and

so ip(G)
n1/2 + 1 = n1/2. 

Lemma 2. If 3
> n, then ip(G) = (n +
)/4.

Proof. SupposeC is an optimum isometric path cover with p2paths of two vertices and p3paths of three vertices.

Then

2p2+ 3p3n.

Note that at most n −
vertices in G can be paired up as the end vertices of the 3-paths in P. Hence p3
(n −
)/2

and so

2p2+ 2p3n − (n −
)/2 = (n +
)/2 or ip(G) = p2+ p3(n +
)/4.

On the other hand, we use induction on n −
to prove that ip(G)
(n +
)/4. When n −

1, we have n =
and G is the complete graph of order n. So, ip(G) = n/2 = (n +
)/4. Suppose n −
2 and the claim holds

for n−
< n −
. In this case, 3
> n
+ 2 which implies
> 1 and n > 3. Then we may remove two vertices

from the first part and one vertex from an odd part other than the first part to form an isometric 3-path P of G. The remaining graph G has n= n − 3 and
=
− 1. It then satisfies 3
> n. Note that r2 unless G = K2,1,1in

which n = 4 and
= 2 imply ip(G) = 2 = (n +
)/4. By the induction hypothesis, ip(G)
(n+
)/4 and so

ip(G)
(n+
)/4 + 1 = (n +
)/4. 

Lemma 3. If 3n1
2n and 3

n, then ip(G) = n/3.

Proof. Since every isometric path in G has at most three vertices, ip(G)n/3.

On the other hand, we use induction on n to prove that ip(G)
n/3. When n
8, by the assumptions that 3n1
2n

and 3

n we have G ∈ {K2,1, K2,2, K3,2, K2,2,1, K4,2, K4,1,1, K3,3, K3,2,1, K2,2,2, K2,2,1,1, K4,3, K4,2,1, K3,2,2,

K2,2,2,1, K5,3, K5,2,1, K4,4, K4,3,1, K4,2,2, K4,2,1,1, K3,3,2, K3,2,2,1, K2,2,2,2, K2,2,2,1,1}. It is straightforward to check

that ip(G)
n/3.

Suppose n9 and the claim holds for n< n. We remove two vertices from the first part and one vertex from the jth

part to form an isometric 3-path P for G, where j is the largest index such that j 2 and nj is odd (when ni are even for all i 2, we choose j = r). Then, the remaining subgraph Ghas n= n − 3 and
=
− 1 or

2. Therefore, 3

n and n9 imply that 3

nin any case. We shall prove that 3n1
2naccording to the following cases.

Case 1: n1n2+ 2.

In this case, n1− 2n2nifor all i 2 and so n1= n1− 2. Therefore, 3n1= 3(n1− 2)
2(n − 3) = 2n.

Case 2: n1
n2+ 1 and n2
4.

In this case, n1
n2
4 and n6. Then, 3n1
12
2n.

Case 3: n1
n2+ 1 and n25 and r = 2.

In this case, n1
n2− 1 and n= n − 3 = n1+ n2− 32n2− 3. Then, 3n1
3n2− 3
4n2− 8 < 2n.

Case 4: n1
n2+ 1 and n25 and r 3.

In this case, n1
n2and n= n − 3n1+ n2+ 1 − 32n2− 2. Then, 3n1
3n2
4n2− 5 < 2n. 

According to Lemmas 1–2, we have the following theorem.

Theorem 4. Suppose G is the complete r-partite graph Kn1,n2,...,nr on n vertices with r 2, n1n2 · · · nr and n = n1+ n2+ · · · + nr. If there are exactly
indices i with niodd, then

ip(G) =

n1/2 if 3n1> 2n;

(n +
)/4 if 3
> n;

n/3 if 3

n and 3n1
2n.

In the proofs of the lemmas above, the essential point for the arguments is not the fact that each partitioning set of the complete r-partite graph is trivial. If we add some edges into the graph but still keep that each partite set can be partitioned inton_i/2 pairs of two nonadjacent vertices and ni− 2ni/2 vertices, then the same result still holds.

Corollary 5. Suppose G is the graph obtained from the complete r-partite graph Kn1,n2,...,nr of n vertices by adding edges such that each ith part can be partitioned inton_i/2 pairs of two nonadjacent vertices and ni− 2ni/2 vertex,

where r 2, n1n2 · · · nrand n = n1+ n2+ · · · + nr. If there are exactly
indices i with niodd, then

ip(G) =

n1/2 if 3n1> 2n;

(n +
)/4 if 3
> n;

n/3 if 3

n and 3n1
2n.

3. Hamming graphs

This section establishes isometric path numbers of Cartesian products of two or three complete graphs.

Suppose G is the Hamming graph Kn1Kn2 . . . Knr of n vertices, where n = n1n2. . . nr and ni2 for 1
i
r. We note that every isometric path in G has at most r + 1 vertices. Consequently,

Recall that the vertex set of Kn1Kn2 . . . Knr is

V (Kn1Kn2 . . . Knr) = {(x1, x2, . . . , xr) : 0
xi< ni for 1
i
r} and edge set E(Kn1Kn2 . . . Knr) is

{(x1, x2, . . . , xr)(y1, y2, . . . , yr) : xi = yi ∈ V (Ki) for all i except just one xj = yj}. We first consider the case when r = 2.

Theorem 6. If n12 and n22, then ip(Kn1Kn2) = n1n2/3.

Proof. We only need to prove that ip(Kn1Kn2)
n1n2/3. We shall prove this assertion by induction on n1+ n2.

For the case when n1+ n2
6, the isometric path covers

C2,2= {(0, 0)(0, 1), (1, 0)(1, 1)},

C2,3= {(0, 0)(0, 1)(1, 1), (0, 2)(1, 2)(1, 0)},

C2,4= {(0, 0)(0, 1)(1, 1), (0, 2)(1, 2)(1, 0), (0, 3)(1, 3)} and

C3,3= {(0, 0)(2, 0)(2, 2), (0, 1)(0, 2)(1, 2), (1, 0)(1, 1)(2, 1)}

for K2K2, K2K3, K2K4and K3K3, respectively, give the assertion.

Suppose n1+ n27 and the assertion holds for n1+ n2< n1+ n2. For the case when all ni
4, without loss of generality we may assume that n1= 4 and 3
n2
4. As we can partition the vertex set of K_n1K_n2 into the vertex

sets of two copies of distance invariant induced subgraphs K2K_n2,

ip(Kn1Kn2)
2ip(K2Kn2)
22n2/3 = n1n2/3.

For the case when there is at least one ni5, say n15, again we can partition the vertex set of Kn1Kn2 into the vertex sets of two distance invariant induced subgraphs K3Kn2 and Kn1−3Kn2. Then,

ip(Kn1Kn2)
ip(K3Kn2) + ip(Kn1−3Kn2)
3n2/3 + (n1− 3)n2/3 = n1n2/3. 

Lemma 7. If n1, n2and n3are positive even integers, then

ip(Kn1Kn2Kn3) = n1n2n3/4.

Proof. We only need to prove that ip(Kn1Kn2Kn3)
n1n2n3/4. First, the isometric path cover C2,2,2= {(0, 0, 0)

(0, 0, 1) (0, 1, 1)(1, 1, 1), (1, 0, 1)(1, 0, 0)(1, 1, 0)(0, 1, 0)} for K2K2K2proves the assertion for the case when

n1= n2= n3= 2. For the general case, as the vertex set of Kn1Kn2Kn3 can be partitioned into the vertex sets of

n1n2n3/8 copies of distance invariant induced subgraphs K2K2K2,

ip(Kn1Kn2Kn3)
(n1n2n3/8)ip(K2K2K2)
n1n2n3/4. 

Lemma 8. If n33 is odd, then ip(K2K2Kn3) = n3+ 1.

Proof. First, we prove that ip(K2K2Kn3)n3+ 1. Suppose to the contrary that the graph can be covered by n3

isometric paths

Pi : (xi1, xi2, xi3)(yi1, yi2, yi3)(zi1, zi2, zi3)(wi1, wi2, wi3),

i = 1, 2, . . . , n3. These paths are in fact vertex-disjoint paths of four vertices, each contains exactly one type-j edge

for j = 1, 2, 3, where an edge (x1, x2, x3)(y1, y2, y3) is type-j if x_j = y_j. For each P_i we then have x_i1= 1 − w_i1

and xi2= 1 − wi2, which imply that xi1+ xi2has the same parity as wi1+ wi2. As Pi has just one type-3 edge, by symmetry, we may assume either xi3 = yi3= zi3= wi3or xi3= yi3 = zi3= wi3, for which we call Pi type 1-3 or type 2-2, respectively. For a type 2-2 path Piwe may further assume that xi1= yi1= zi1= wi1.

For 0
x3< n3, the x3-square is the set S(x3) ={(0, 0, x3), (0, 1, x3), (1, 0, x3), (1, 1, x3)}. Note that a type 1-3 path

and two vertices in S(wi3). We call a type 1-3 path Piadjacent to another type 1-3 path Pj if the last three vertices of

Pi and the first vertex of Pj form a square. This defines a digraph D whose vertices are all type 1-3 paths, in which each vertex has out-degree one and in-degree at most one. In fact, each vertex then has in-degree one. In other words, each type 1-3 path Pi corresponds to exactly one type 1-3 path Pj such that the last three vertices of Pi and the first vertex of Pjform a square. Consequently, the vertices of all type 1-3 paths together form p squares; and so the vertices of all type 2-2 paths form the other n3− p squares.

Since xi1= yi1= zi1= wi1for a type 2-2 path Pi, the first two vertices of a type 2-2 path together with the first two vertices of another type 2-2 path form a square. This shows that there is an even number of type 2-2 paths. Therefore, there is an odd number of type 1-3 paths.

On the other hand, in a type 1-3 path Piwe have that xi1+ xi2= yi1+ yi2has the different parity as zi1+ zi3, and the same parity as wi1+ wi2. We call the path Pieven or odd when xi1+ xi2is even or odd, respectively. So Pi is adjacent to a type 1-3 path whose parity is the same as zi1+ zi2. That is, a type 1-3 path is adjacent to a type 1-3 path of different parity. Therefore, the digraph D is the union of some even directed cycles. This is a contradiction to the fact that there is an odd number of type 1-3 paths.

The arguments above prove that ip(K2K2K_n3)n3+ 1. On the other hand, since the vertex set of K2K2K_n3

is the union of the vertex sets of (n3+ 1)/2 copies of K2K2K2, by the coverC2,2,2in the proof of Lemma 7, we

have ip(K2K2Kn3)
n3+ 1. 

Theorem 9. If all ni2, then ip(Kn1Kn2Kn3) = n1n2n3/4 except for the case when two ni are 2 and the third

is odd. In the exceptional case, ip(Kn1Kn2Kn3) = n1n2n3/4 + 1.

Proof. The claim for the exceptional case holds according to Lemma 8.

For the main case, by Lemma 7, we may assume that at least one ni is odd. Again, we only need to prove that ip(Kn1Kn2Kn3)
n1n2n3/4. We shall prove the assertion by induction on

3

i=1ni. For the case when 3

i=1ni
10, the following isometric path covers for K2K3K3, K2K3K4, K3K3K3 and K3K3K4,

respectively, prove the assertion:

C2,3,3= {(0, 1, 1)(0, 1, 0)(0, 0, 0)(1, 0, 0), (0, 2, 2)(0, 2, 0)(1, 2, 0)(1, 1, 0), (0, 2, 1)(1, 2, 1)(1, 1, 1), (0, 0, 2)(0, 1, 2)(1, 1, 2), (0, 0, 1)(1, 0, 1)(1, 0, 2)(1, 2, 2)}; C2,3,4= {(0, 1, 1)(0, 1, 0)(0, 0, 0)(1, 0, 0), (0, 2, 1)(0, 2, 0)(1, 2, 0)(1, 1, 0), (0, 2, 3)(0, 2, 2)(1, 2, 2)(1, 1, 2), (0, 1, 3)(0, 1, 2)(0, 0, 2)(1, 0, 2), (0, 0, 1)(1, 0, 1)(1, 1, 1)(1, 1, 3), (1, 2, 1)(1, 2, 3)(1, 0, 3)(0, 0, 3)}; C2,3,5= C∗2,3,3∪ {(0, 1, 4)(0, 1, 3)(0, 2, 3)(1, 2, 3), (0, 0, 3)(0, 0, 4)(0, 2, 4)(1, 2, 4), (1, 0, 3)(1, 0, 4)} where C∗_2,3,3= C2,3,3\{(0, 2, 1)(1, 2, 1)(1, 1, 1), (0, 0, 2)(0, 1, 2)(1, 1, 2)}∪ {(0, 2, 1)(1, 2, 1)(1, 1, 1)(1, 1, 3), (0, 0, 2)(0, 1, 2)(1, 1, 2)(1, 1, 4)}; C3,3,3= {(0, 0, 0)(0, 2, 0)(1, 2, 0)(1, 2, 1), (1, 1, 0)(2, 1, 0)(2, 2, 0)(2, 2, 1), (0, 2, 1)(0, 1, 1)(1, 1, 1)(1, 1, 2), (1, 0, 1)(2, 0, 1)(2, 1, 1)(2, 1, 2), (0, 1, 0)(0, 1, 2)(0, 2, 2)(1, 2, 2), (0, 0, 1)(0, 0, 2)(2, 0, 2)(2, 2, 2), (1, 0, 2)(1, 0, 0)(2, 0, 0)}; C3,3,4= {(0, 0, 0)(0, 2, 0)(1, 2, 0)(1, 2, 1), (1, 1, 0)(2, 1, 0)(2, 2, 0)(2, 2, 1), (0, 2, 1)(0, 1, 1)(1, 1, 1)(1, 1, 2), (1, 0, 1)(2, 0, 1)(2, 1, 1)(2, 1, 2), (0, 1, 0)(0, 1, 2)(0, 2, 2)(1, 2, 2), (0, 0, 2)(2, 0, 2)(2, 2, 2)(2, 2, 3), (0, 1, 3)(1, 1, 3)(1, 0, 3)(1, 0, 2), (1, 0, 0)(2, 0, 0)(2, 0, 3)(2, 1, 3), (0, 0, 1)(0, 0, 3)(0, 2, 3)(1, 2, 3)}.

Suppose3_i=1ni11 and the assertion holds for 3

i=1ni< 3

For the case when there is some i, say i = 3, such that n37 or n3= 6 with all n_j3, we have ip(K_n1K_n2K_n3)

ip(Kn1Kn2K4) + ip(Kn1Kn2Kn3−4)
n1n24/4 + n1n2(n3− 4)/4 = n1n2n3/4.

For the case when some ni, say n3, is equal to 4, we may assume n1n2 and so n14. Then ip(Kn1Kn2 K4)
ip(K2Kn2K4) + ip(Kn1−2 Kn2K4) = 2n24/4 + (n1− 2) n24/4 = n1n2n3/4.

There are six remaining cases. The following isometric path covers prove the assertion for K2K3K6, K2K5K5

and K3K5K5, respectively: C2,3,6= C∗2,3,3∪ {(0, 0, 4)(0, 0, 3)(1, 0, 3)(1, 2, 3), (0, 1, 3)(0, 1, 4)(0, 2, 4)(1, 2, 4), (0, 2, 3)(0, 2, 5)(1, 2, 5)(1, 1, 5), (0, 1, 5)(0, 0, 5)(1, 0, 5)(1, 0, 4)}; C2,5,5= C2,3,5\{(1, 0, 3)(1, 0, 4)} ∪ {(0, 4, 1)(0, 4, 0)(0, 3, 0)(1, 3, 0), (1, 4, 0)(1, 4, 1)(1, 3, 1)(0, 3, 1), (0, 4, 3)(0, 4, 2)(0, 3, 2)(1, 3, 2), (1, 4, 2)(1, 4, 3)(1, 3, 3)(0, 3, 3), (1, 0, 3)(1, 0, 4)(1, 4, 4), (0, 4, 4)(0, 3, 4)(1, 3, 4)}; C3,5,5= C2,3,5\{(1, 0, 3)(1, 0, 4)} ∪ {(0, 4, 0)(2, 4, 0)(2, 0, 0)(2, 0, 1), (0, 3, 0)(2, 3, 0)(2, 1, 0)(2, 1, 1), (0, 4, 1)(0, 3, 1)(1, 3, 1)(1, 3, 0), (1, 4, 0)(1, 4, 1)(2, 4, 1)(2, 2, 1), (1, 0, 3)(2, 0, 3)(2, 2, 3)(2, 2, 0), (1, 0, 4)(2, 0, 4)(2, 3, 4)(2, 3, 1), (0, 3, 2)(2, 3, 2)(2, 1, 2)(2, 1, 3), (0, 4, 4)(0, 4, 2)(2, 4, 2)(2, 0, 2), (0, 4, 3)(1, 4, 3)(1, 3, 3)(1, 3, 2), (0, 3, 3)(2, 3, 3)(2, 4, 3)(2, 4, 4), (0, 3, 4)(1, 3, 4)(1, 4, 4)(1, 4, 2), (2, 2, 2)(2, 2, 4)(2, 1, 4)}.

In the three other cases the claim follows from the following inequalities: ip(K2K5K6)
ip(K2K3K6) + ip(K2K2K6)
9 + 6 = 15,

ip(K3K3K5)
ip(K3K3K2) + ip(K3K3K3)
5 + 7 = 12,

ip(K5K5K5)
ip(K5K5K3) + ip(K5K5K2)
19 + 13 = 32. 

Acknowledgments

The authors thank to the referees for their constructive suggestions.

References

[1]M. Aigner, M. Fromme, A game of cops and robbers, Discrete Appl. Math. 8 (1984) 1–12.

[2]D.C. Fisher, S.L. Fitzpatrick, The isometric number of a graph, J. Combin. Math. Combin. Comput. 38 (2001) 97–110.

[3]S.L. Fitzpatrick, Ph.D. Thesis, Dalhousie University, Nova Scotia, Canada, 1997.

[4]S.L. Fitzpatrick, The isometric path number of the Cartesian product of paths, Congr. Numer. 137 (1999) 109–119.

[5]S.L. Fitzpatrick, R.J. Nowakowski, D. Holton, I. Caines, Covering hypercubes by isometric paths, Discrete Math. 240 (2001) 253–260.

[7]R. Nowakowski, P. Winkler, Vertex to vertex pursuit in a graph, Discrete Math. 43 (1983) 235–239.

[8]J.-J. Pan, G.J. Chang, Isometric path numbers of block graphs, Inform. Process. Letters 93 (2005) 99–102.

[9]A. Quilliot, Thèse de 3e cycle, Université de Paris VI, 1978.

Further reading

[1]F. Harary, A.J. Schwenk, Evolution of the path number of a graph: covering and packing in graphs, II in: R.C. Reed (Ed.), Graph Theory Comput.,