Distance two edge labelings of lattices

Abstract

Suppose G is a graph. Two edges e and e′ in G are said to be adjacent if they share a common end vertex, and distance two apart if they are nonadjacent but both are adjacent to a common edge. Let j and k be two positive integers. An L(j,k)-edge-labeling of a graph G is an assignment of nonnegative integers, called labels, to the edges of G such that the difference between labels of any two adjacent edges is at least j, and the difference between labels of any two edges that are distance two apart is at least k. The minimum range of labels over all L(j,k)-edge-labelings of a graph G is called the L(j,k)-edge-labeling number of G, denoted by $\lambda_{j,k}'(G)$. Let m, j and k be positive integers. An m-circular-L(j,k)-edge-labeling of a graph G is an assignment f from {0,1,…,m−1} to the edges of G such that, for any two edges e and e′, |f(e)−f(e′)| _m ≥j if e and e′ are adjacent, and |f(e)−f(e′)| _m ≥k if e and e′ are distance two apart, where |a| _m =min{a,m−a}. The minimum m such that G has an m-circular-L(j,k)-edge-labeling is called the circular-L(j,k)-edge-labeling number of G, denoted by $\sigma_{j,k}'(G)$. This paper investigates the L(1,1)-edge-labeling numbers, the L(2,1)-edge-labeling numbers and the circular-L(2,1)-edge-labeling numbers of the hexagonal lattice, the square lattice, the triangular lattice and the strong product of two infinite paths.

Appendix: The proof of Theorem 3.1

Proof

We only prove the lower bound here. Suppose $\sigma_{2,1}'(\varGamma_{4})\geq11$ is not true, then $\sigma_{2,1}'(\varGamma_{4})\leq10$. Let f: E(Γ ₄)→{0,1,…,9} be a 10-circular-L(2,1)-edge-labeling of Γ ₄. We first prove the following claim.

Claim A

For any two integers i and j,

$$\bigl|f(e_{i,j})-f(e_{i,j+1})\bigr|_{10}=1, \quad \mbox{\textit{and}}\quad\bigl|f(h_{i,j})-f(h_{i+1,j})\bigr|_{10}=1. $$

Proof

By the symmetry of Γ ₄, we only need to show that |f(e _i,j )−f(e _i,j+1)|₁₀=1 for any two integers i and j. Again by the symmetry of Γ ₄, we may assume i=j=1, and we only prove |f(e _1,1)−f(e _1,2)|₁₀=1. With no loss of generality, assume f(e _1,1)=0. Our goal now is to prove that f(e _1,2)∉{2,3,4,5,6,7,8}. By the symmetry of the labels, we actually only need to prove that f(e _1,2)∉{2,3,4,5}. The proof is distinguished into the following four cases.

Case 1. f(e _1,2)=2.

In this case, we have

$$\bigl\{f(h_{1,0}),f(e_{0,1}),f(h_{1,1}), f(h_{2,1}),f(e_{2,1}),f(h_{2,0})\bigr\}=\{3,4,5,6,7,8 \}. $$

Due to the distance one condition, we must have

$$\bigl\{f(h_{1,0}),f(e_{0,1}),f(h_{1,1})\bigr\}= \{3,5,7\},\qquad \bigl\{ f(h_{2,1}),f(e_{2,1}),f(h_{2,0}) \bigr\}=\{4,6,8\} $$

or

$$\bigl\{f(h_{1,0}),f(e_{0,1}),f(h_{1,1})\bigr\}= \{4,6,8\},\qquad \bigl\{ f(h_{2,1}),f(e_{2,1}),f(h_{2,0}) \bigr\}=\{3,5,7\}. $$

With no loss of generality, assume

$$\bigl\{f(h_{1,0}),f(e_{0,1}),f(h_{1,1})\bigr\}= \{3,5,7\},\qquad \bigl\{ f(h_{2,1}),f(e_{2,1}),f(h_{2,0}) \bigr\}=\{4,6,8\}. $$

If f(h _1,1)=5, then the only labels left for e _0,2 and h _1,2 are 8,9, violating the distance one condition. Thus f(h _1,1)=7. But then {f(e _0,2),f(h _1,2)}={4,9}, implying f(e _2,2)=f(h _2,2)=5, a contradiction. Therefore Case 1 can not happen.

Case 2. f(e _1,2)=3.

In this case, we have

$$\bigl\{f(h_{1,0}),f(e_{0,1}),f(h_{1,1}), f(h_{2,1}),f(e_{2,1}),f(h_{2,0})\bigr\}=\{2,4,5,6,7,8 \}. $$

Similar as in Case 1, we may assume that

$$\bigl\{f(h_{1,0}),f(e_{0,1}),f(h_{1,1})\bigr\}= \{2,5,7\},\qquad \bigl\{ f(h_{2,1}),f(e_{2,1}),f(h_{2,0}) \bigr\}=\{4,6,8\}. $$

Clearly $f(h_{2,1})\not= 4$. If f(h _2,1)=6, then {f(e _2,2),f(h _2,2)}⊆{1,9}, implying f(e _0,2)=f(h _1,2)=8, a contradiction. If f(h _2,1)=8, then {f(e _2,2),f(h _2,2)}={1,5}, and so {f(e _0,2),f(h _1,2)}={6,9}. This is a contradiction since f(h _1,1)=5 or 7. Thus Case 2 will not happen.

Case 3. f(e _1,2)=4.

In this case, we have

$$\bigl\{f(h_{1,0}),f(e_{0,1}),f(h_{1,1}), f(h_{2,1}),f(e_{2,1}),f(h_{2,0})\bigr\}=\{2,3,5,6,7,8 \}. $$

Similar to Case 1, we may assume that

$$\bigl\{f(h_{1,0}),f(e_{0,1}),f(h_{1,1})\bigr\}= \{2,5,7\},\qquad \bigl\{ f(h_{2,1}),f(e_{2,1}),f(h_{2,0}) \bigr\}=\{3,6,8\}, $$

or

$$\bigl\{f(h_{1,0}),f(e_{0,1}),f(h_{1,1})\bigr\}= \{3,5,7\},\qquad \bigl\{ f(h_{2,1}),f(e_{2,1}),f(h_{2,0}) \bigr\}=\{2,6,8\}. $$

In the former case, we clearly have $f(h_{2,1})\not= 3$. If f(h _2,1)=8, then {f(e _2,2),f(h _2,2)}={1,2}, a contradiction. Thus we assume f(h _2,1)=6. Note that f(h _1,1)∈{2,7}. If f(h _1,1)=2, then {f(e _0,2),f(h _1,2)}={8,9}, a contradiction. If f(h _1,1)=7, then {f(e _0,2),f(h _1,2)}⊆{1,9}. But then the two edges e _2,2 and h _2,2 can not be labeled properly, a contradiction.

In the latter case, we must have f(h _1,1)=7. It follows that {f(e _2,2),f(h _2,2)}={1,9}, implying {f(e _0,2),f(h _1,2)}={1,2}, a contradiction. Thus Case 3 will not happen.

Case 4. f(e _1,2)=5.

In this case, we have

$$\bigl\{f(h_{1,0}),f(e_{0,1}),f(h_{1,1}), f(h_{2,1}),f(e_{2,1}),f(h_{2,0})\bigr\}=\{2,3,4,6,7,8 \}. $$

Similar to Case 1, we may assume

$$\bigl\{f(h_{1,0}),f(e_{0,1}),f(h_{1,1})\bigr\}= \{2,4,7\}, \qquad \bigl\{ f(h_{2,1}),f(e_{2,1}),f(h_{2,0}) \bigr\}=\{3,6,8\}. $$

Clearly $f(h_{1,1})\not=4$. If f(h _1,1)=2, then {f(e _0,2),f(h _1,2)}={8,9}, a contradiction. Thus f(h _1,1)=7. It is clear that f(h _2,1)=8 or 3. If f(h _2,1)=8, then {f(e _2,2),f(h _2,2)}={1,2}, a contradiction. If f(h _2,1)=3, then {f(e _2,2),f(h _2,2),f(e _0,2),f(h _1,2)}⊆{1,9}, a contradiction. Therefore Case 4 will not happen.

Thus Claim A holds. □

By Claim A and the cyclic property, we may assume that f(e _1,2)=0 and f(e _1,1)=1. Then f(h _1,1),f(h _2,1)∈{3,4,5,6,7,8}. According to Claim A, we must have {f(h _1,1),f(h _2,1)}={3,4}, or {4,5}, or {5,6}, or {6,7}, or {7,8}. Next we shall derive contradictions in each of these cases and thus prove the theorem. Due to the symmetry of the labels, we only need to deal with the first three cases. Please see Figs. 10, 11, 12, 13 for illustrations of these cases.

Fig. 10

Subcase A.1

Fig. 11

Subcases A.2 and A.3

Fig. 12

Case B

Fig. 13

Case C

Case A. {f(h _1,1),f(h _2,1)}={3,4}.

By the symmetry of Γ ₄, we may assume that f(h _1,1)=3 and f(h _2,1)=4. Then

$$\begin{array}{@{}ll} \bigl\{f(e_{0,2}),f(h_{1,2})\bigr\}\subseteq\{5,6,7,8\},\cr\noalign {\vspace{3pt}} \bigl\{f(e_{2,1}),f(h_{2,0})\bigr\}\subseteq\{6,7,8,9\},\cr\noalign {\vspace{3pt}} \bigl\{f(e_{2,2}),f(h_{2,2})\bigr\}\subseteq\{2,6,7,8\}. \end{array} $$

Since e _0,2 is adjacent to h _1,2, {f(e _0,2),f(h _1,2)}={5,7}, {5,8}, or {6,8}. We consider these three subcases individually.

Subcase A.1. {f(e _0,2),f(h _1,2)}={5,7}.

We first suppose f(e _0,2)=5 and f(h _1,2)=7. By Claim A, we have f(e _0,1)=6 and f(h _2,2)∈{6,8}. Since f(e _2,1)∉{1,3}, by Claim A, f(e _2,2)∈{6,8}. It follows that {f(e _2,1),f(h _2,0)}={7,9}. By Claim A, the case f(e _2,2)=6 and f(e _2,1)=9 will not happen. Thus we only need to consider the following three situations.

(1) If f(h _2,0)=9, f(e _2,1)=7, f(h _2,2)=6, and f(e _2,2)=8, then by Claim A, f(h _3,2)=5, f(h _3,0)=0, and f(h _3,1)=3. It follows that f(e _3,2)=1, implying by Claim A that f(e _3,1)=2. But this violates the distance one condition.

(2) If f(h _2,0)=9, f(e _2,1)=7, f(h _2,2)=8, and f(e _2,2)=6, then by Claim A, f(h _3,1)=3 and f(h _3,0)=0, implying f(e _3,1)=5. By Claim A, f(e _3,2)∈{4,6}, contradicting the fact that f(e _2,2)=6 and f(h _3,1)=3.

(3) If f(h _2,0)=7, f(e _2,1)=9, f(h _2,2)=6, and f(e _2,2)=8, then by Claim A, f(h _3,2)=5 and f(h _3,0)=6, implying f(h _3,1)=3. But then the edge e _3,1 can not be labeled properly, a contradiction.

Now suppose f(e _0,2)=7 and f(h _1,2)=5. Then by Claim A, f(h _2,2)=6, implying f(e _2,2)∈{2,8}. It follows that f(h _2,0),f(e _2,1)∈{7,8,9}. By Claim A, we have f(e _2,2)=8, and so f(h _2,0),f(e _2,1)∈{7,9}. We deal with the following two situations.

(4) If f(h _2,0)=9, f(e _2,1)=7, f(h _2,2)=6, and f(e _2,2)=8, then by Claim A, f(h _3,2)=5 and f(h _3,0)=0, implying f(h _3,1)=3. But then the edge e _3,1 can not be labeled properly, a contradiction.

(5) If f(h _2,0)=7, f(e _2,1)=9, f(h _2,2)=6, and f(e _2,2)=8, then by Claim A, f(h _3,2)=5 and f(h _3,0)=6, implying f(h _3,1)=3. But then we can not label the edge e _3,1 properly, a contradiction.

Subcase A.2. {f(e _0,2),f(h _1,2)}={5,8}.

Since f(h _1,2)∈{5,8} and f(h _2,2)∉{4,9}, by Claim A, we have f(h _2,2)=6 or 7. This implies that f(e _2,2)=2. This is a contradiction to Claim A since f(e _2,1)∉{1,3}.

Subcase A.3. {f(e _0,2),f(h _1,2)}={6,8}.

Since f(h _1,2)∈{6,8} and f(h _2,2)∉{5,9}, by Claim A, we have f(h _2,2)=7, implying f(e _2,2)=2. This is a contradiction to Claim A since f(e _2,1)∉{1,3}.

Case B. {f(h _1,1),f(h _2,1)}={4,5}.

With no loss of generality, assume f(h _1,1)=4 and f(h _2,1)=5. Then

$$\begin{array}{@{}ll} \bigl\{f(e_{0,2}),f(h_{1,2})\bigr\}\subseteq\{2,6,7,8\},\cr\noalign {\vspace{3pt}} \bigl\{f(e_{0,1}),f(h_{1,0})\bigr\}\subseteq\{6,7,8,9\},\cr\noalign {\vspace{3pt}} \bigl\{f(e_{2,1}),f(h_{2,0})\bigr\}\subseteq\{3,7,8,9\},\cr\noalign {\vspace{3pt}} \bigl\{f(e_{2,2}),f(h_{2,2})\bigr\}\subseteq\{2,3,7,8\}. \end{array} $$

Since e _0,2 is adjacent to h _1,2, we must have {f(e _0,2),f(h _1,2)}={2,6}, {2,7}, {2,8}, or {6,8}. If 2∈{f(e _0,2),f(h _1,2)}, then since {f(e _0,1),f(h _1,0)}⊆{6,7,8,9}, by Claim A, f(h _1,2)=2.

Subcase B.1. {f(e _0,2),f(h _1,2)}={2,6}.

Clearly f(e _2,0)=6 and f(h _1,2)=2. By Claim A, f(h _2,2)=3 and f(e _0,1)=7, implying {f(h _2,0),f(e _2,1)}={8,9}, a contradiction.

Subcase B.2. {f(e _0,2),f(h _1,2)}={2,7}.

Clearly f(e _2,0)=7 and f(h _1,2)=2. By Claim A, f(h _2,2)=3, implying f(e _2,2)=8. It follows that {f(e _2,1),f(h _2,0)}={7,9}, and so {f(e _0,1),f(h _1,0)}={6,8}.

If f(e _0,1)=6 and f(h _1,0)=8, then by Claim A, f(h _0,0)=9 and f(h _0,1)=3, leaving no label for the edge e _−1,1, a contradiction.

If f(e _0,1)=8 and f(h _1,0)=6, then by Claim A, f(h _2,0)=7 and f(e _2,1)=9, implying f(h _3,0)=6. Then, by Claim A, f(h _3,1)=4, and so f(h _3,2)=2. Now no label is available for the edge e _3,2, a contradiction.

Subcase B.3. {f(e _0,2),f(h _1,2)}={2,8}.

We have f(e _2,0)=8 and f(h _1,2)=2. By Claim A, f(h _2,2)=3 and f(e _0,1)∈{7,9}. If f(e _0,1)=7, then {f(e _2,1),f(h _2,0)}={8,9}, a contradiction. If f(e _0,1)=9, then {f(e _2,1),f(h _2,0)}={7,8}, again a contradiction.

Subcase B.4. {f(e _0,2),f(h _1,2)}={6,8}.

In this case, it is easy to see that {f(e _0,1),f(h _1,0)}={7,9}, {f(e _2,1),f(h _2,0)}={3,8}, and {f(e _2,2),f(h _2,2)}={2,7}. Since {f(e _0,2),f(h _1,2)}={6,8} and {f(e _2,2),f(h _2,2)}={2,7}, it follows from Claim A that f(h _2,2)=7 and so f(e _2,2)=2. By Claim A, f(e _2,1)=3 and so f(h _2,0)=8. Now it follows from Claim A that f(h _3,1)=6, implying f(h _3,0)=9 and f(h _3,2)=8. But then there is no label for the edge e _3,1, a contradiction.

Case C. {f(h _1,1),f(h _2,1)}={5,6}.

With no loss of generality, assume f(h _1,1)=5 and f(h _2,1)=6. Then

$$\begin{array}{@{}ll} \bigl\{f(e_{0,2}),f(h_{1,2})\bigr\}\subseteq\{2,3,7,8\},\cr\noalign {\vspace{3pt}} \bigl\{f(e_{0,1}),f(h_{1,0})\bigr\}\subseteq\{3,7,8,9\},\cr\noalign {\vspace{3pt}} \bigl\{f(e_{2,1}),f(h_{2,0})\bigr\}\subseteq\{3,4,8,9\},\cr\noalign {\vspace{3pt}} \bigl\{f(e_{2,2}),f(h_{2,2})\bigr\}\subseteq\{2,3,4,8\}. \end{array} $$

Since e _0,2 is adjacent to h _1,2, we must have {f(e _0,2),f(h _1,2)}={2,7}, {2,8}, {3,7}, or {3,8}.

Subcase C.1. {f(e _0,2),f(h _1,2)}={2,7}.

If f(e _0,2)=2 and f(h _1,2)=7, then by Claim A, we have f(h _2,2)=8 and f(e _1,3)=1. Due to the cyclic property of circular edge labeling, this subcase is reduced to Case A and we are done.

If f(e _0,2)=7 and f(h _1,2)=2, then by Claim A, we have f(e _0,1)=8 and so f(h _1,0)=3. It follows from Claim A that f(h _2,0)=4, implying f(e _1,0)=0. Now this subcase is reduced to Case A and we are done.

Subcase C.2. {f(e _0,2),f(h _1,2)}={2,8}.

If f(h _1,2)=8, then since f(h _2,2)∉{7,9}, this is a contradiction to Claim A. Thus we assume f(e _0,2)=8 and f(h _1,2)=2. By Claim A, f(h _2,2)=3. But then we can not label the edge e _2,2 properly, a contradiction.

Subcase C.3. {f(e _0,2),f(h _1,2)}={3,7}.

In this case, we have {f(e _0,1),f(h _1,0)}⊆{8,9}, a contradiction.

Subcase C.4. {f(e _0,2),f(h _1,2)}={3,8}.

It is clear that {f(e _0,1),f(h _1,0)}={7,9} and {f(e _2,2),f(h _2,2)}={2,4}. By Claim A, we must have f(e _0,2)=8 and f(h _1,2)=3.

If f(e _0,1)=7 and f(h _1,0)=9, then by Claim A, f(h _0,1)=4 and f(h _0,0)=0, implying f(h _0,2)=2. But then we can not label the edge e _−1,1 properly, a contradiction.

If f(e _0,1)=9 and f(h _1,0)=7, then by Claim A, f(h _0,0)=6 and so f(h _0,1)=4, implying f(h _0,2)=2. But then we can not label the edge e _−1,1 properly, a contradiction.

This completes the proof of the theorem.  □