arXiv:math/0310332v1 [math.CO] 21 Oct 2003
Isometric path numbers of graphs
∗
Jun-Jie Pan
Department of Applied Mathematics
National Chiao Tung University
Hsinchu 300, Taiwan
Gerard J. Chang
Department of Mathematics
National Taiwan University
Taipei 106, Taiwan
E-mail: gjchang@math.ntu.edu.tw
October 3, 2003
Abstract
An isometric path between two vertices in a graph G is a shortest path join-ing 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.
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 iso-metric 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 alternatively, 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 give an upper bound for the cop-numbers of planar graphs, they showed that a single cop moving on an isometric path P guarantee 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 . . . Pnr has
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.
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 com-plete r-partite graphs and Cartesian products of 2 or 3 comcom-plete 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
2
Complete r-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,
n1 ≥ n2 ≥ . . . ≥ nr and n = n1+ n2+ . . . + nr. Let G has α parts of odd sizes. We
notice that every isometric path in G has at most 3 vertices. Consequently, ip(G) ≥ ⌈n/3⌉.
Also, for any path of 3 vertices in an isometric path cover C, two end vertices of the path is in a part of G and the center vertex in another part. In case when two paths of 3 vertices in C have a common end vertex, we may replace one by a path of 2 vertices. And, a path of 1 vertex can be replaced by a path of 2 vertices. So, without loss of generality, we may only consider isometric path covers in which every path is of 2 or 3 vertices, and two 3-vertices 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 an induction on n − n1 to 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 − n1 ≥ 2 and the claim holds for n′ − n′1 < 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 G′ has r′ ≥ 2, n′
1 = n1 − 2 and n′ = n − 3. It then
still satisfies 3n′
1 > 2n′. As n′ − n′1 = n − n1 − 1, by the induction hypothesis,
ip(G′) ≤ ⌈n′
1/2⌉ and so ip(G) ≤ ⌈n′1/2⌉ + 1 = ⌈n1/2⌉.
Lemma 2 If 3α > n, then ip(G) = ⌈(n + α)/4⌉.
Proof. Suppose C is an optimum isometric path cover with p2 paths of 2 vertices
and p3 paths of 3 vertices. Then
2p2+ 3p3 ≥ n.
Notice that there are at most n − α vertices in G can be paired up as the end vertices of the 3-paths in P. Hence p3 ≤ (n − α)/2 and so
2p2+ 2p3 ≥ n − (n − α)/2 = (n + α)/2 or ip(G) = p2+ p3 ≥ ⌈(n + α)/4⌉.
On the other hand, we use an induction on n − α to prove that ip(G) ≤ ⌈(n + α)/4⌉. When n − α ≤ 1, we have n = α and G is the complete graph of order n. So, ip(G) = ⌈n/2⌉ = ⌈(n + α)/4⌉. Suppose n − α ≥ 2 and the claim holds for n′ − α′ < n − α. In this case, 3α > n ≥ α + 2 which implies α > 1 and n > 3. Then
we may remove two vertices from the first part of and one vertex form an odd part other than the first part to form a isometric 3-path P of G. The remaining graph G′
has n′ = n − 3 and α′ = α − 1. It then satisfies 3α′ > n′. Notice that r′ ≥ 2 unless
G = K2,1,1 in which n = 4 and α = 2 imply ip(G) = 2 = ⌈(n+α)/4⌉. By the induction
hypothesis, ip(G′) ≤ ⌈(n′+ α′)/4⌉ and so ip(G) ≤ ⌈(n′+ α′)/4⌉ + 1 = ⌈(n + α)/4⌉.
Lemma 3 If 3n1 ≤ 2n and 3α ≤ n, then ip(G) = ⌈n/3⌉.
Proof. Since every isometric path in G has at most 3 vertices, ip(G) ≥ ⌈n/3⌉. On the other hand, we use an induction on n to prove that ip(G) ≤ ⌈n/3⌉. When n ≤ 8, by the assumptions that 3n1 ≤ 2n and 3α ≤ 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 n ≥ 9 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 G′ has n′ = n − 3 and
α′ = α − 1 or α′ ≤ 2. Therefore, 3α ≤ n and n ≥ 9 imply that 3α′ ≤ n′ in any case.
We shall prove that 3n′
1 ≤ 2n′ according to the following cases.
Case 1. n1 ≥ n2+ 2.
In this case, n1 − 2 ≥ n2 ≥ ni for all i ≥ 2 and so n′1 = n1 − 2. Therefore,
3n′
1 = 3(n1− 2) ≤ 2(n − 3) = 2n′.
Case 2. n1 ≤ n2+ 1 and n2 ≤ 4.
In this case, n′
1 ≤ n2 ≤ 4 and n′ ≥ 6. Then, 3n′1 ≤ 12 ≤ 2n′.
Case 3. n1 ≤ n2+ 1 and n2 ≥ 5 and r = 2.
In this case, n′
1 ≤ n2 − 1 and n′ = n − 3 = n1 + n2 − 3 ≥ 2n2 − 3. Then,
3n′
1 ≤ 3n2 − 3 ≤ 4n2 − 8 < 2n′.
Case 4. n1 ≤ n2+ 1 and n2 ≥ 5 and r ≥ 3.
In this case, n′
1 ≤ n2 and n′ = n − 3 ≥ n1 + n2 + 1 − 3 ≥ 2n2 − 2. Then,
3n′
1 ≤ 3n2 ≤ 4n2− 5 < 2n′.
According to Lemma 1, 2 and 3, we have the following theorem.
Theorem 4 Suppose G is the complete r-partite graph Kn1,n2,...,nr of n vertices with
r ≥ 2, n1 ≥ n2 ≥ . . . ≥ nr and n = n1 + n2+ . . . + nr. If there are exactly α indices
i with ni odd, then
ip(G) = ⌈n1/2⌉, if 3n1 > 2n; ⌈(n + α)/4⌉, if 3α > n; ⌈n/3⌉, if 3α ≤ n and 3n1 ≤ 2n.
In the proofs of the lemmas above, the essential points for the arguments is not the fact that each partite 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 into ⌊ni/2⌋ pairs of two nonadjacent vertices and ni− 2⌊ni/2⌋ vertex, then the same result
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 i-th part can be partitioned
into ⌊ni/2⌋ pairs of two nonadjacent vertices and ni− 2⌊ni/2⌋ vertex, where r ≥ 2,
n1 ≥ n2 ≥ . . . ≥ nr and n = n1+ n2+ . . . + nr. If there are exactly α indices i with
ni odd, then ip(G) = ⌈n1/2⌉, if 3n1 > 2n; ⌈(n + α)/4⌉, if 3α > n; ⌈n/3⌉, if 3α ≤ n and 3n1 ≤ 2n.
3
Hamming graphs
This section establishes isometric path numbers of Cartesian products of 2 or 3 com-plete graphs.
Suppose G is the Hamming graph Kn1Kn2 . . . Knr of n vertices, where
n = n1n2. . . nr and ni ≥ 2 for 1 ≤ i ≤ r. We notice that every isometric path in G
has at most r + 1 vertices. Consequently,
ip(G) ≥ ⌈n/(r + 1)⌉. Recall that the vertex set of Kn1Kn2 . . . Knr is
V (Kn1Kn2 . . . Knr) = {(x1, x2, . . . , xr) : 0 ≤ xi < ni for 1 ≤ i ≤ r}.
We first consider the case when r = 2
Theorem 6 If n1 ≥ 2 and n2 ≥ 2, 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, K2K4 and K3K3 respectively, gives the assertion.
Suppose n1 + n2 ≥ 7 and the assertion holds for n′1 + n′2 < 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 Kn1Kn2 into the vertex sets of
two copies of distance invariant induced subgraphs K2Kn2,
For the case when there is at least one ni ≥ 5, say n1 ≥ 5, 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, n2 and n3 are positive even integers, then
ip(Kn1Kn2Kn3) = n1n2n3/4.
Proof. We only need to prove that ip(Kn1Kn2Kn3) ≤ n1n2n3/4. First, the
isomet-ric 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 K2K2K2 proves 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 n3 ≥ 3 is odd, then ip(K2K2Kn3) = n3+ 1.
Proof. First, we claim 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 4 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 xj 6= yj. For each Pi we then have xi1 = 1 − wi1 and xi2 = 1 − wi2, which
imply that xi1+ xi2 has the same parity with wi1+ wi2. We call the path Pi evenor
odd when xi1+ xi2 is even or odd, respectively. Also, as Pi has just one type-3 edge,
by symmetric, we may assume either xi3 6= yi3 = zi3 = wi3 or xi3 = yi3 6= zi3 = wi3,
for which we call Pi type 1-3 or type 2-2 respectively. For a type 2-2 path Pi we may
further assume that xi16= yi1 = zi1= wi1.
For 0 ≤ x3 < n3, the x3-squareis the set S(x3) = {(0, 0, x3), (0, 1, x3), (1, 0, x3),
(1, 1, x3)}. Notice that a type 1-3 path Pi contains 1 vertex in S(xi3) and 3 vertices
in S(wi3), while a type 2-2 path Pi contains 2 vertices in S(xi3) and 2 vertices in
S(wi3). We call a type 1-3 path Pi is adjacent to another type 1-3 path Pj if the last
3 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, the “adjacent to” is a bijection. Consequently, vertices of all type 1-3 paths together form p squares; and so vertices of all type 2-2 paths form the other n3 − p squares.
Since xi1 6= 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 Pi we have xi1 + xi2 = yi1 + yi2 has the
different parity with zi1 + zi3, and the same parity with wi1 + wi2. So it 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 cycle. This is a contradiction to the fact that there is an odd number of type 1-3 paths.
The arguments above prove that ip(K2K2Kn3) ≥ n3+1. On the other hand,
since the vertex set of K2K2Kn3 is the union of the vertex sets of (n3+1)/2 copies of
K2K2K2, by the cover C2,2,2 in the proof of Lemma 7, we have ip(K2K2Kn3) ≤
n3 + 1.
Theorem 9 If all ni ≥ 2, then ip(Kn1Kn2Kn3) = ⌈n1n2n3/4⌉ except for the case
when twoni are 2 and the third is odd. In the exceptional case, ip(Kn1Kn2Kn3) =
n1n2n3/4 + 1.
Proof. 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 onP3
i=1ni. For the case when
P3
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)}; Let 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)}. 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 = C2,3,3∗ ∪ {(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)}; 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)}. Suppose P3
i=1ni ≥ 11 and the assertion holds for
P3
i=1n′i <
P3
i=1ni. We shall
consider the following cases.
For the case when there is some i, say i = 3, such that n3 ≥ 7 or n3 = 6 with all
nj ≥ 3, we have ip(Kn1Kn2Kn3) ≤ 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 n1 ≥ n2 and
so n1 ≥ 4. Then ip(Kn1Kn2K4) ≤ ip(K2Kn2K4) + ip(Kn1−2Kn2K4) =
⌈2n24/4⌉ + ⌈(n1 − 2)n24/4⌉ = ⌈n1n2n3/4⌉.
There are 6 remaining cases. The following isometric path covers prove the assertion for K2K3K6, K2K5K5 and K3K5K5, respectively:
C2,3,6 = C2,3,3∗ ∪ {(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)}. The other 3 cases 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.
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