Abstract For positive numbers j and k, an L(j, k)-labeling f of G is an assignment of numbers to
vertices of G such that |f(u) − f(v)| ≥ j if uv ∈ E(G), and |f(u) − f(v)| ≥ k if d(u, v) = 2. Then
the span of f is the difference between the maximum and the minimum numbers assigned by f . The
L(j, k)-number of G, denoted by λj,k(G), is the minimum span over all L(j, k)-labelings of G. In this
paper, we give some results about the L(j, k)-number of the direct product of a path and a cycle for
j ≤ k.
Keywords L(j, k)-labeling, product of a path and a cycle
MR(2010) Subject Classification 05C78, 05C15
1 Introduction
The ever-growing wireless networks of computers and the need to let remote computers exchangedata all over the world, lead to increasing scarcity of available codes. The optimal assignmentof a limited code to increasing stations (including transmitters and receivers) became essentialand the code assignment problem (CAP) evolved. One major variation of CAP requires thespread of the codes between “very close” stations to be narrow to avoid the direct collision, andthat between two “close” stations to be wide to avoid the secondary collision.
In order to avoid secondary collision, Bertossi and Bonuccelli [2] introduced a kind of codeassignment, that is, two at distance two stations transmit on different codes. Thus, the sec-ondary collision avoidance problem can be formulated as follows: assign disparate codes toeach pair of vertices at distance two and use the minimum number of different codes. By corre-sponding codes to labels, the above problem is equivalent to an L(0, 1)-labeling problem, wherevertices at distance two must be assigned different labels.
In general, it seems reasonable to consider the direct collision. Basing on this premise,Jin and Yeh [20] generalized the code assignment problem to L(j, k)-labeling problem, wherej ≤ k. For instance, if j = 1, this means that any two adjacent stations are required to beassigned different codes to avoid direct interference, then to avoid secondary interference, aswell as to avoid direct interference, any two stations at distance two need to be assigned lagercode differences, that is, k = 2. Recently, the L(j, k)-labeling problem for j ≤ k has been paidattention to study, see [2, 5, 12, 20].
Received January 17, 2012, revised August 17, 2012, accepted October 19, 2012
Supported by Faculty Research Grant, Hong Kong Baptist University
1438 Shiu W. C. and Wu Q.
We define the code assignment problem as following. Let G = (V, E) be a graph. For anyu, v ∈ V , let dG(u, v) denote the distance (length of a shortest path) between u and v in G. Ifthere is no ambiguity, then we simply denote dG(u, v) by d(u, v). For positive numbers j and k,an L(j, k)-labeling f of G is an assignment of numbers to vertices of G such that |f(u)−f(v)| ≥ j
if uv ∈ E(G), and |f(u)− f(v)| ≥ k if d(u, v) = 2. Then the span of f is the difference betweenthe maximum and the minimum numbers assigned by f . The L(j, k)-number of G, denoted byλj,k(G), is the minimum span over all L(j, k)-labelings of G.
L(j, k)-numbers of graphs for j ≥ k have been studied in many literatures. Interestedreaders are referred to the articles [13–16, 21, 28, 29, 31] or the surveys [3, 35].
By now, about the L(j, k)-numbers of direct product of paths and cycles, there alreadyexist some results. Jha et al. [16] found that the L(2, 1)-labeling number of direct of paths andcycles is 6 for some special cycles. Also, they gave an upper bound 2k +2d−2 of direct of cyclesfor some special cycles in the paper [17]. Shao et al. [29, 31] gave an upper bound for directproduct of graphs. Moreover, graphs of grids, Cartesian product of paths and cycles, strongproduct of paths and cycles are similar to direct product of paths and cycles, some results onL(j, k)-numbers of above graphs are referred to [1, 4, 6–11, 15–19, 21–26, 30–33].
Remark 1.1 In some classical literatures, j and k were assumed to be integers. Resultsin those papers are easily extended when j and k are positive real numbers. So researchersconsider j and k being positive numbers on the L(j, k)-labeling problems recently.
For j ≤ k, Wu et al. [34] recently determined the circular L(j, k)-number of direct productof path and cycle. In this paper, we are only concerned the L(j, k)-labeling problems of thedirect product of a path and a cycle for j ≤ k.
Lemma 1.2 Let j and k be two positive numbers with j ≤ k. Suppose G is a graph and H isan induced subgraph of G. Then
λj,k(G) ≥ λj,k(H).
Georges and Mauro [9] determined the L(j, k)-number for cycles with j ≥ k. Niu [27]studied the L(j, k)-number of cycles for j ≤ k and got the following result:
Theorem 1.3 ([27]) Let Cn be a cycle with n vertices (n ≥ 3).
1. For k ≥ 2j, λj,k(Cn) =
⎧⎪⎪⎨
⎪⎪⎩
2j, if n = 3,
j + k, if n ≡ 0 (mod 4),
2k, otherwise.
2. For 32j < k < 2j, λj,k(Cn) =
⎧⎪⎪⎨
⎪⎪⎩
4j, if n = 5,
3j, if n ≡ 0 (mod 4),
2k, otherwise.
3. For j ≤ k ≤ 32j, λj,k(Cn) =
⎧⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎩
2j, if n = 3,
4j, if n = 5,
2k, if n ≡ 0 (mod 3) and n �= 3,
3j, otherwise.
L(j, k)-number of Direct Product of Path and Cycle 1439
2 L(j, k)-numbers
A set of labels is said to be k-separated if the distance of any two labels from this set is greaterthan or equal to k.
Given two graphs G and H, the direct product of G and H is the graph G×H with vertexset V (G)×V (H) in which two vertices (x, y) and (x′, y′) are adjacent if and only if xx′ ∈ E(G)and yy′ ∈ E(H).
Let the path Pn = x1x2 · · ·xn and the cycle Cm = y1y2 · · · ymy1. For convenience, we denotethe vertex (xs, yt) ∈ V (Pn × Cm) by vs,t throughout this paper.
For example, direct product of P3 and C3 is a graph as Figure 1 shows.
Figure 1 Graph P3 × C3
Note that P2 × Cm is C2m when m is odd, and P2 × Cm is the union of two disjoint cyclesCm when m is even. Hence, we can use Theorem 1.3 to obtain the following theorem directly.
Theorem 2.1 Let j and k be positive numbers and m ≥ 3.1. For k ≥ 2j,
λj,k(P2 × Cm) =
⎧⎨
⎩
j + k, m ≡ 0 (mod 4),
2k, otherwise.
2. For 32j < k < 2j,
λj,k(P2 × Cm) =
⎧⎨
⎩
3j, if m ≡ 0 (mod 4),
2k, otherwise.
3. For j ≤ k ≤ 32j,
λj,k(P2 × Cm) =
⎧⎨
⎩
2k, if m ≡ 0 (mod 3),
3j, otherwise.
We shall discuss L(j, k)-number of the graph Pn×Cm for n ≥ 3 and m ≥ 3. For convenience,we define a class of graphs first. For n ≥ 2 and m ≥ 2, the palisade graph P (n, m) is a graphwith vertex set {u2i,j | 0 ≤ i ≤ �n−1
such that u2i+1,j is adjacent with u2i,j , u2i,j+1, u2i+2,j and u2i+2,j+1 for 0 ≤ i ≤ �n−22 � and
0 ≤ j ≤ m − 1. For a fixed i, the sequence of vertices ui,j according to the nature order is
called the i-th row of the graph. The sequence of vertices {u2i,0}�n−1
2 �i=0 is called the first column
and the sequence of vertices {u2i,m}�n−1
2 �i=0 is called the last column of the graph. The circular
palisade graph CP(n, m) is the graph obtained from P (n, m) by identifying the first and thelast columns. For example, Figure 2 is the graph CP(3, 3).
1440 Shiu W. C. and Wu Q.
Figure 2 Graph CP(3, 3)
Suppose m ≥ 3 is odd. We rewrite the second index of vs,t ∈ V (Pn ×Cm), where t ∈ Zm ={0, 1, . . . , m − 1}. Since 2 is a generator of the additive group Zm, t = 2i ∈ Zm for a uniquei ∈ Zm. Then d(vs,2i, vs′,2l) = 2 if and only if |i − l|m = 1 and s = s′, or i = l and |s − s′| = 2,where |x − y|k = min{|x − y|, k − |x − y|}.Remark 2.2 When m is even, Pn × Cm is isomorphic to the union of two CP(n, m
2 ). Whenm is odd, Pn ×Cm is isomorphic to CP(n, m). Namely, the corresponding is u2i,j ↔ v2i,2j andu2i+1,j ↔ v2i+1,2j+1.
......
)( 0,0v )( 2,0 −mv)( 4,0 −mv
)( 1,1v )( 3,1v
)( 2,0v )( 4,0v
)( 1,1 −mv)( 3,1 −mv
0,0u 1,0u 2,0u 2,0 −mu 1,0 −mu)( 0,0v
0,0u
0,1u 1,1u
)( 5,1v
2,1u 2,1 −mu1,1 −mu
)( 0,2v0,2u
)( 2,2v1,2u
)( 4,2v2,2u
)( 4,2 −mv2,2 −mu
)( 2,2 −mv1,2 −mu
)( 0,2v0,2u
Figure 3 The correspondence of vertices of P3 × Cm onto CP(3,m) for odd m
Remark 2.3 From Remark 2.2, finding a λ-L(j, k)-labeling of Pn×Cm is equivalent to findinga λ-L(j, k)-labeling of CP(n, m) for m ≥ 2 and n ≥ 3. Suppose we have a λ-L(j, k)-labeling ofP (n, m) such that the first and the last columns receive the same labels. We call such a labelingas a restricted λ-L(j, k)-labeling of P (n, m). Then after identifying those vertices in the firstand the last columns accordingly, we get a λ-L(j, k)-labeling of CP(n, m). Thus, finding aλ-L(j, k)-labeling of CP(n, m) is equivalent to finding a restricted λ-L(j, k)-labeling of P (n, m).
For any r, x ∈ R, [x]r ∈ [0, r) denotes the remainder of x upon division of r.
Theorem 2.4 Let j and k be positive numbers and m ≥ 3. For k ≥ 2j,
λj,k(P3 × Cm) =
⎧⎪⎪⎨
⎪⎪⎩
3k, if m ≡ 0 (mod 4),
5k, if m = 3, 6,
4k, otherwise.
Proof Case 1 m ≡ 0 (mod 4). It is easy to obtain λj,k(K1,4) = 3k. Then, by Lemma 1.2,
λj,k(P3 × Cm) ≥ λj,k(K1,4) = 3k.
Now we are going to prove λj,k(P3 × Cm) ≤ 3k. From Remark 2.3, we only need to find arestricted 3k-L(j, k)-labeling of P (3, m
2 ). The labeling f is defined as follows.If s = 0, 2, then f(us,t) = (s + [t]2)k, where 0 ≤ t ≤ m
2 .If s = 1, then f(us,t) = ( 1
2 + [t]2)k, where t ∈ Z m2.
It can be checked that f is a 3k-L(j, k)-labeling of CP(3, m2 ). Thus, λj,k(P3 × Cm) = 3k.
L(j, k)-number of Direct Product of Path and Cycle 1441
Case 2 Suppose m = 3. Vertices v0,0, v0,1, v0,2, v2,0, v2,1, v2,2 are at distance 2 from eachother. So the labels of these vertices under any L(j, k)-labeling are k-separated. Hence, λj,k(P3×C3) ≥ 5k.
On the other hand, we give a 5k-L(j, k)-labeling as follows. f(vs,t) = ( [s]22 + 3� s
2�+ t)k, fors = 0, 1, 2; t = 0, 1, 2. It can be verified that f satisfies the constraints of the L(j, k)-labelingproblem. Hence, λj,k(P3 × C3) = 5k.
Suppose m = 6. The graph P3×C6 is composed of two disjoint P3×C3’s. Hence, λj,k(P3×C6) = λj,k(P3 × C3) = 5k.
Case 3 m �≡ 0 (mod 4) and m is neither 3 nor 6. That is, m = 4n + i, where i = 1, 2, 3, n =1, 2, . . . . If m = 4n+2, then the graph P3 ×Cm is composed of two disjoint P3 ×C2n+1’s (thatis two disjoint CP(3, 2n+1)’s). Then we just need to consider two kinds of graph CP(3, 4n+1)and CP(3, 4n + 3). That means we only need to consider CP(3, m) for odd m ≥ 5.
We first want to show that λj,k(P3×Cm) ≤ 4k. By Remark 2.3, it suffices to find a restricted4k-L(j, k)-labeling of P (3, m).
Firstly, Figure 4 shows a restricted 4k-L(j, k)-labeling of P (3, 5) and Figure 5 shows arestricted 3k-L(j, k)-labeling of P (3, 2).
k4
2
3k
0 0k
k k2
k2
k3
k3
k2
2
k2
k
2
5k2
3k
k4 0
Figure 4 A restricted 4k-L(j, k)-labeling of P (3, 5)
k 0
2
3k
2
k
0
k2k2 k3
Figure 5 A restricted 3k-L(j, k)-labeling of P (3, 2)
Suppose there is a restricted 4k-L(j, k)-labeling of P (3, m) with the first and the last columnslabeled by (0, 2k). Since the first and the last columns of P (3, 2) are also labeled by (0, 2k), wemay extend the labelings of P (3, m) and P (3, 2) to P (3, m + 2) by identifying the last columnof P (3, 2) (the dotted line box) with the first column of P (3, m) (the solid line box). Clearly,it is still a 4k-L(j, k)-labeling with the first and the last columns labeled by (0, 2k).
By the way, we can describe the 4k-L(j, k)-labeling of P3 ×Cm defined above as an explicitform for odd m ≥ 7. When s = 0 or 2,
f(vs,2i) =
⎧⎨
⎩
{[i]2 + s
}k, 0 ≤ i ≤ m − 6;
[i − m + s]5k, m − 5 ≤ i ≤ m − 1.
1442 Shiu W. C. and Wu Q.
When s = 1,
f(v1,2i+1) =
⎧⎪⎨
⎪⎩
{
[i]2 +12
}
k, 0 ≤ i ≤ m − 2;
5k
2, i = m − 1.
Hence, we have λj,k(P3 × Cm) ≤ 4k.Next, we prove that λj,k(P3 × Cm) ≥ 4k. Suppose λ = λj,k(P3 × Cm) < 4k. Assume g is a
λ-L(j, k)-labeling of P3 × Cm. Let g(vs,2i) = gs,2i, where i ∈ Zm. Without loss of generality,we let g0,0 = 0. Then g0,2, g2,0, g2,2 ∈ [k, 4k).
If g2,0 ∈ [2k, 3k), then g0,2, g2,2 ∈ [k, 2k)∪[3k, 4k), g0,4, g2,4 ∈ [0, k)∪[2k, 3k). Repeating thisargument, we have g0,2j , g2,2j ∈ [k, 2k)∪ [3k, 4k) when j is odd; and g0,2j , g2,2j ∈ [0, k)∪ [2k, 3k)when j is even. Now since m − 1 is even, g0,m−2 = g0,2(m−1) ∈ [0, k) ∪ [2k, 3k). It contradictsthat g0,m−2, g0,0 and g2,0 are k-separated.
If g2,0 ∈ [k, 2k) or [3k, 4k), then we will also obtain a similar contradiction. Hence, λj,k(P3×Cm) = 4k. �
Example 2.1 Applying the method described in the proof above, we have restricted 4k-L(j, k)-labelings of P (3, 7) (see Figure 6) and P (3, 9) (see Figure 7), respectively.
0 0 0
0
k
k
k k2
k2 k2k2
k3
k3k3
k4
2
k
k4
2
k2
k
2
3k2
3k
2
3k
2
5k
Figure 6 A restricted 4k-L(j, k)-labeling of P (3, 7)
0 k
0
000 k2
k
kk
k2k2 k2k2
k3
k3k3 k3 k4
2
k
k4
2
3k2
k2
k
2
k
2
5k
2
3k
2
3k
2
3k
Figure 7 A restricted 4k-L(j, k)-labeling of P (3, 9)
Lemma 2.5 Let G be a graph indicated in Figure 8. If k ≥ j, then λj,k(G) = 3k + j.
a3a0
b1
b0 b3
a2 b2
a1
Figure 8 A subgraph of Pn × Cm
Proof Let labels of vertices of G be indicated in the figure above under a λ-L(j, k) labeling.Hence, {a0, a1, a2, a3, b0, b1, b2, b3} ⊂ [0, λ] and at least one of these labels is 0. Without loss ofgenerality, we may assume that 0 ∈ {a0, a1, a2, b0}.
L(j, k)-number of Direct Product of Path and Cycle 1443
Suppose b0 = 0. Then {a0, a1, a2, a3} ⊂ [j, λ]. Since a0, a1, a2 and a3 are k-separate,λ − j ≥ 3k. Hence, λ ≥ 3k + j.
Suppose ai = 0 for some i ∈ {0, 1, 2}. Then b0 ≥ j.
1. Suppose b0 < k. We have {b1, b2, b3} ⊂ [b0 + k, λ] by the definition of L(j, k)-labeling.Then λ − b0 − k ≥ 2k. Hence, λ ≥ 3k + b0 ≥ 3k + j.
2. Suppose λ − b0 < k. We have {b1, b2, b3} ⊂ [0, b0 − k]. This implies that b0 ≥ 3k. Ifλ − b0 ≥ j, then λ ≥ b0 + j ≥ 3k + j. If λ − b0 < j, then {a0, a1, a2, a3} ⊂ [0, b0 − j]. We haveb0 − j ≥ 3k and hence λ ≥ b0 ≥ 3k + j.
3. Suppose k ≤ b0 ≤ λ − k.
a. Suppose b0 − j < k. We have ({a0, a1, a2, a3}\{ai}) ⊂ [b0 + j, λ]. This implies thatλ − b0 − j ≥ 2k. Hence, λ ≥ 2k + b0 + j ≥ 3k + j.
b. Suppose λ − (b0 + j) < k. We have that [b0 + j, λ] contains at most one ar for somer. Since ai = 0, [k, b0 − j] contains at least two aq’s. This implies that b0 − j − k ≥ k. Sinceλ − b0 ≥ k, λ ≥ k + b0 ≥ 3k + j.
c. Suppose k + j ≤ b0 ≤ λ−k− j. If b0 −k ≥ k, then λ ≥ b0 +k + j ≥ 3k + j. If b0 −k < k,then similar to Case b, we have that [b0 + k, λ] contains at least two bq’s. This implies thatλ − b0 − k ≥ k. Hence, λ ≥ 2k + b0 ≥ 3k + j.
Combining the above cases, we have λ ≥ 3k + j.
On the other hand, if we let a0 = j, a1 = 2k + j, a2 = 3k + j, a3 = k + j, b0 = 0, b1 = k,b2 = 2k + j and b3 = 3k + j, then we have a 3k + j-L(j, k)-labeling of G. Hence, we proved thelemma. �
Theorem 2.6 Let j and k be positive numbers, m ≥ 3 and n ≥ 4. For k ≥ 2j,
λj,k(Pn × Cm) =
⎧⎪⎪⎨
⎪⎪⎩
3k + j, if m ≡ 0 (mod 4),
5k, if m = 3, 6,
4k, otherwise.
Proof Case 1 m ≡ 0 (mod 4). Since Pn×Cm is composed by two CP(n, m2 )’s, we only need
to find a restricted (3k + j)-L(j, k)-labeling of P (n, m2 ) which is defined as follows. f(us,t) =
j[s]2 + k[2� s2� + t]4, where 0 ≤ s ≤ n − 1; 0 ≤ t ≤ m
2 − 1 for s odd and 0 ≤ t ≤ m2 for s even.
It can be verified that f is a restricted (3k + j)-L(j, k)-labeling of P (n, m2 ). Hence, we have a
(3k + j)-L(j, k)-labeling of Pn × Cm.
On the other hand, the graph G in Figure 8 is a subgraph of Pn × Cm. By Lemma 1.2, wehave λj,k(Pn × Cm) ≥ λj,k(G) = 3k + j. Thus, λj,k(Pn × Cm) = 3k + j.
Case 2 Suppose m = 3. We define a 5k-L(j, k)-labeling f as follows. f(vs,t) = k[3� s2� + t]6,
where 0 ≤ s ≤ n − 1, t = 0, 1, 2. It can be checked that f is a 5k-L(j, k)-labeling function ofPn × C3.
On the other hand, P3 ×C3 is a subgraph of Pn ×C3. By Theorem 2.4 and Lemma 1.2, wehave λj,k(Pn × C3) ≥ λj,k(P3 × C3) = 5k. Hence, λj,k(Pn × C3) = 5k.
Suppose m = 6. The graph Pn × C6 is composed of two disjoint Pn × C3’s. Hence,λj,k(Pn × C6) = λj,k(Pn × C3) = 5k.
1444 Shiu W. C. and Wu Q.
Case 3 m �≡ 0 (mod 4) and m is neither 3 nor 6. Similar to Case 3 in the proof of Theo-rem 2.4, we only need to deal with Pn × Cm when m is odd and m ≥ 5. We will use a similarmethod as Case 3 in the proof of Theorem 2.4 to prove the result.
1) Suppose m ≡ 1 (mod 4).1.1) We label P (11, 5) as following.
k
k4
k3
k2
k
k k20
k3k2
k2
k0
k3
k3
k0
0
0
0
k4
k4
k4
k4 k2 k4
k2 k3
kk4
k k2 k3 0
0
k4
k2k3
k
k30k4
k2 kk4
k k2 k3 0
0
k4
k3
k3k
0
k2 k4
k
0
k2k3
Figure 9 A restricted 4k-L(j, k)-labeling of P (11, 5)
If we delete the last seven rows from P (11, 5) one by one, then we have restricted 4k-L(j, k)-labelings of P (10, 5), P (9, 5), P (8, 5), P (7, 5), P (6, 5), P (5, 5), and P (4, 5). When n ≥ 11, thelabels of the s-th row are defined to be the same as the labels of i-th row of P (11, 5), wheres ≡ i (mod 11), 0 ≤ i ≤ 10 and 0 ≤ s ≤ n − 1. Hence, we have a restricted 4k-L(j, k)-labelingof P (n, 5) for any n ≥ 4.
1.2) Suppose m ≥ 9. We label P (5, 9) and P (5, 4) as Figures 10 and 11. Note that thesequences of labels of the 0-th and the 5-th rows are the same.
Now suppose we have a restricted 4k-L(j, k)-labeling of P (5, m), where m ≡ 1 (mod 4),such that the labels of the first and the last columns are (0, 2k, 0). Since the last column ofP (5, 4) is labeled by (0, 2k, 0), we identify the last column of P (5, 4) with the first column ofP (5, m). Therefore, we have a restricted 4k-L(j, k)-labeling of P (5, m + 4) with the first andthe last columns are labeled by (0, 2k, 0). By induction, we have a restricted 4k-L(j, k)-labelingof P (5, m) for m ≡ 1 (mod 4) and m ≥ 9.
2
k25k
2
3k
2
7k0 k2 k3 k4 0
2
k
25k
2
3k
2
7k 0 k2 k3 k4
k
k
2
k25k
2
3k
2
7kk2 k3 k4 0 k2k
2
k
2
3k0 k2 k3 k4k
25k
2
7k
2
k25k
2
3k
2
7k0 k2 k3 k4 0k
Figure 10 A restricted 4k-L(j, k)-labeling of P (5, 9)
L(j, k)-number of Direct Product of Path and Cycle 1445
0
0
00
0k
k
k
k2 k3
k2
k2
2
k
k3
k3k2
2
3k
2
5k
2
7k
2
5k
2
7k2
k
2
3k
Figure 11 A restricted ( 7k2
)-L(j, k)-labeling of P (5, 4)
Let the labeling of P (5, m) defined above be f1. When n ≥ 5. P (5, m) is a subgraph ofP (n, m). We extend f1 to the whole graph by f1(us,t) = f1(ui,t) if s ≡ i (mod 4) for each t
and s ≥ 5, where i ∈ {0, 1, 2, 3}. Hence, we have a restricted 4k-L(j, k)-labeling of P (n, m) forodd m ≥ 9 with m ≡ 1 (mod 4) and n ≥ 5.
2) Suppose m ≡ 3 (mod 4).2.1) For m = 7, we give a restricted 4k-L(j, k)-labeling f2 of P (5, 7) as follows.
0 k k2 k3 k4 0 k 0
0 k k2 k3k42
5k2
7k
k2 k3 k42k
23k
25k
27k k2
0 k k2 k3 k42k
23k
0 k k2 k3 k4 0 k 0
Figure 12 A restricted 4k-L(j, k)-labeling P (5, 7)
When n ≥ 5. We extend the labeling f2 to P (n, 7) by defining f2(us,t) = f2(ui,t) if s ≡ i
(mod 4) for each t and s ≥ 5, where i ∈ {0, 1, 2, 3}.2.2) For m ≥ 11, we first give a restricted 4k-L(j, k)-labeling f3 of P (5, 11) described in
Figure 13.
0 k k2 k3 k42k
23k
25k
27k 0 k 0
2k
23k
25k
27k 0 k k2 k3 k4
2k
23k
k2 k3 k42k
23k
25k
27k 0 k k2 k3 k2
25k
27k 0 k k2 k3 k4
2k
23k
25k
27k
0 k k2 k3 k42k
23k
25k
27k 0 k 0
Figure 13 A restricted 4k-L(j, k)-labeling P (5, 11)
Similar to the proof of Case (1.2), suppose we have a restricted 4k-L(j, k)-labeling of P (5, m),where m ≡ 3 (mod 4) and m ≥ 11, such that the labels of the first and the last columns are(0, 2k, 0). Then we identify the last column of P (5, 4) (described in Figure 11) with the firstcolumn of P (5, m). Therefore, we have a restricted 4k-L(j, k)-labeling of P (5, m + 4) withthe first and the last columns are labeled by (0, 2k, 0). By induction, we have a restricted4k-L(j, k)-labeling of P (5, m) for m ≡ 3 (mod 4) and m ≥ 11.
1446 Shiu W. C. and Wu Q.
Same as the above cases, we can extend the labeling of P (5, m) to P (n, m) for n ≥ 5.For those labelings of P (5, m) described above, if we delete the last row from P (5, m), then
we obtain a restricted 4k-L(j, k)-labeling of P (4, m). Hence, we have 4k-L(j, k)-labeling ofPn × Cm, for m �≡ 0 (mod 4), m is neither 3 nor 6 and n ≥ 4.
By Theorem 2.4 and Lemma 1.2, we have λj,k(Pn ×Cm) ≥ λj,k(P3 ×Cm) = 4k. Hence, wehave λj,k(Pn × Cm) = 4k. �
Example 2.2 Applying the method described in the proof above, we have restricted 4k-L(j, k)-labelings of P (5, 13) (see Figure 14) and P (5, 15) (see Figure 15), respectively.
k2
0 k k2 k3 0 k k2 k3 k42k
23k
25k
27k
2k
23k
25k
27k
2k
23k
25k
27k 0 k k2 k3 k4
k2 k3 0 k k2 k3 k42k
23k
25k
27k 0 k k2
25k
27k
2k
23k
25k
27k 0 k k2 k3 k4
2k
23k
0 k k2 k3 0 k k2 k3 k42k
23k
25k
27k
0
0
Figure 14 A restricted 4k-L(j, k)-labeling of P (5, 13)
k2
0 k k2 k3
2k
23k
25k
27k
k2 k3 0 k
25k
27k
2k
23k
0 k k2 k3
0 k k2 k3 k42k
23k
25k
27k 0 k 0
2k
23k
25k
27k 0 k k2 k3 k4
2k
23k
k2 k3 k42k
23k
25k
27k 0 k k2 k3 k2
25k
27k 0 k k2 k3 k4
2k
23k
25k
27k
0 k k2 k3 k42k
23k
25k
27k 0 k 0
Figure 15 A restricted 4k-L(j, k)-labeling of P (5, 15)
By Theorems 2.4 and 2.6, we have
Corollary 2.7 Let k and j be positive numbers. Suppose k ≥ 2j, m ≥ 2 and n ≥ 3.
λj,k(CP(n, m)) =
⎧⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎩
3k, if n = 3 and m is even,
j + 3k, if n ≥ 4 and m is even,
4k, m is odd and m ≥ 5,
5k, m = 3.
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