On the super edge-magicness of graphs of equal order and size

Label the central vertex of one component $LK_{1,1}$ with $3$ and the other vertex of this component with $6$. Then label the central vertex of the other component isomorphic to $LK_{1,1}$ with $5$ and the other vertex vertex of this component with $2$. Finally, label the central vertex of the component isomorphic to $LK_{1,n}$ with $4$, and the vertices of degree 1 in $LK_{1,n}$ with the remaining labels in $[1,n+5]$. Then it is not hard to see that the resulting labeling is super edge-magic.∎

In order to label $2LK_{1,m}\cup LK_{1,n}$, consider the labeling of $2LK_{1,1}\cup LK_{1,n}$ obtained in the proof of Theorem 2.1. Add $m-1$ new edges with their corresponding vertices of degree 1 attached to the central vertex of each of the two components that are originally isomorphic to $LK_{1,1}$. Label the new vertices of the component that has the central vertex labeled 5 with the numbers $-1,-3,-5,\ldots,-(2m-3)$. Label the remaining vertices (that is to say the new vertices in the component with the central vertex labeled 3) with the numbers $0,-2,-4,-6,\ldots,-(2m-4)$. Then it is easy to check that by adding $(2m-2)$ to each of the original labels, we obtain a super edge-magic labeling of the graph $2LK_{1,m}\cup LK_{1,n}$. ∎

It is easy check that any bijective function $f:V(LK_{1,n})\to[1,n+1]$ is super edge-magic, for all $n\in\mathbb{N}$ (see for instance, [13]). Hence, we can assume that $s\in\mathbb{N}$. Let us define the graph $(2s+1)LK_{1,n}$ as follows: $V((2s+1)LK_{1,n})=\{v_{i}\}^{2s+1}_{i=1}\cup\{v_{i}^{j}\}^{2s+1}_{i=1},j=1,2,\ldots,n$ and $E((2s+1)LK_{1,n})=\{v_{i}v_{i}\}^{2s+1}_{i=1}\cup\{v_{i}v_{i}^{j}\}^{2s+1}_{i=1},j=1,2,\ldots,n$. Next, we show that $(2s+1)LK_{1,n}$ is super edge-magic. Consider the labeling $f:V((2s+1)LK_{1,n})\to[1,(n+1)(2s+1)]$ defined by

Then, $f$ is a super edge-magic labeling of $(2s+1)LK_{1,n}$ . ∎

Let us define the graph $G=LK_{1,m}\cup 2LK_{1,n}\cup(2s)LK_{1,1},\ s\in\mathbb{N}$ as follows: $V(G)=\{v_{i}\}^{2s+3}_{i=1}\cup\{v^{j}_{1}\}^{n}_{j=1}\cup\{v^{k}_{2}\}^{m}_{k=1}\cup\{v^{j}_{3}\}^{n}_{j=1}\cup\{v^{1}_{i}\}^{2s+3}_{i=4}$ and $E(G)=\{v_{i}v_{i}\}^{2s+3}_{i=1}\cup\{v_{1}v^{j}_{1}\}^{n}_{j=1}\cup\{v_{2}v^{k}_{2}\}^{m}_{k=1}\cup\{v_{3}v^{j}_{3}\}^{n}_{j=1}\cup\{v_{i}v^{1}_{i}\}^{2s+3}_{i=4}$. Consider the labeling $f:V(G)\to[-2n+3,4s+m+5]$ defined by

By adding $(2n-2)$ to each of the original labels, we obtain a super edge-magic labeling of the graph $LK_{1,m}\cup 2LK_{1,n}\cup(2s)LK_{1,1},\ s\in\mathbb{N}$. ∎

It suffices to label the vertices of the caterpillar in a traditional super edge-magic way. ∎

We just prove (i), the other items can be proved similarly. By Lemma 2.1, $C_{k}\otimes_{h}(2LK_{1,1}\cup LK_{1,n})\cong 2C_{k}\odot\overline{K}_{1}\cup C_{k}\odot\overline{K}_{n}$. Combining this with Theorem 2.1 and Theorem 1.1, $2C_{k}\odot\overline{K}_{1}\cup C_{k}\odot\overline{K}_{n}$ is edge-magic. In particular, if $k$ is odd, $C_{k}$ is super edge-magic and hence the graph $2C_{k}\odot\overline{K}_{1}\cup C_{k}\odot\overline{K}_{n}$ is super edge-magic. ∎

Assume to the contrary that $LK_{1,m}\cup LK_{1,n}$ is a super edge-magic graph and let $D$ be the digraph obtained from it by orienting its edges in such a way that all vertices in $D$ have indegree 1. By definition, $D$ is also super edge-magic. Let $f$ be a super edge-magic labeling of $D$, and let $A(D)=(a_{ij})$ be the adjacency matrix induced by $f$ where the vertices take the name of their labels in the super edge-magic labeling. By Lemma 3.1, all the entries 1 are located in exactly two rows of $A(D)$. Let these two row be row $i$ and row $j$ and assume that $i<j$. If $a_{i1}=a_{i2}=\ldots=a_{il}=1$ and $a_{il+1}=0$, for some $l\geq 1$, then there is no $1$ in the diagonal with induced sum $i+l+1$, contradicting Lemma 1.2. Thus, we only have two possible generic forms for the adjacency matrix, either row $i$ is of the form $(0\ldots 01\ldots 10\ldots\ldots 10\ldots 0)$, with one more block of zeros than blocks of ones, or, $(0\ldots 01\ldots 10\ldots\ldots 01\ldots 1)$, with exactly the same number of blocks of zeros and ones. Notice that, the first possibility is forbidden by Lemma 1.3, since otherwise, the adjacency matrix induced by the complementary labeling of $f^{c}$ would have exactly two rows with $1$ entries, namely $k,l$ with $k<l$, and row $k$ of the form $(1\ldots 10\ldots)$, which is a contradiction, as we have shown above. Hence, in what follows assume that row $i$ and row $j$ are of the form

respectively, where the $T_{l}$ and $B_{l}$ are blocks of $1$’s. By Lemma 1.3, we can assume that $\sum_{i=1}^{k}|T_{i}|=m+1$ and $\sum_{i=1}^{k}|B_{i}|=n+1$.

Notice that the block $B_{1}$ must be used to cover the zeros between $T_{1}$ and $T_{2}$, $B_{2}$ must be used to cover the zeros between $T_{2}$ and $T_{3}$ and so on. Since the length of $B_{1}$ must be equal to the number of zeros between $T_{1}$ and $T_{2}$ which is equal to the length of $B_{2}$ and so on,we get $\mid B_{i}\mid=(n+1)/k\ \hbox{for}\ i=1,2,\ldots,k$. A similar argument shows that $\mid T_{i}\mid=(m+1)/k\ \hbox{for }i=1,2,\ldots,k$. More over, since the first zero in the top row after $T_{1}$ appears in column $((m+n+2)/k)+1$, this can only be covered if the bottom row is $((m+n+2)/k)+i)$. This implies that

On the other hand, the possible positions of $1$’s in the top row and bottom row are $(l^{\prime}(m+n+2)-(m+1))/k+y$, $l^{\prime}=1,2,\ldots,k-1$, $y\in[1,(m+1)/k]$ and $(l-1)(m+n+2))/k+w$, $l=1,2,\ldots,k-1$, $w\in[1,(n+1)/k],$ respectively. Since we assume that the graph is super edge-magic, each of the two rows contains a $1$ in the main diagonal. Thus, there exist $l,l^{\prime},y$ and $w$ such that, $i=(l^{\prime}(m+n+2)-(m+1))/k+y$ and $j=(l-1)(m+n+2))/k+w$. Hence, by(1) we obtain that $w=(2-l+l^{\prime})(m+n+2)/k-(m+1)/k+y$. Since $l^{\prime}\leq l$, $w$ is either negative or greater than $(n+1)/k$ which contradicts that $w\in[1,(n+1)/k]$. ∎

Suppose to the contrary that there exists $n\in\mathbb{N}$ such that $L\cup LK_{1,n}$ is super edge-magic and assume that each vertex is identified with the label assigned by a super edge-magic labeling. By definition, the digraph $D\cong\overrightarrow{L}\cup\overrightarrow{LK}_{1,n}$ obtained from $L\cup LK_{1,n}$ by orienting its edges in such a way that all vertices in $D$ have indegree $1$ in $D$ is also super edge-magic. The adjacency matrix of $D$ contains exactly two rows with $1$ entries, namely row $i$ and row $j$, $1\leq i<j\leq n+2$. If $a_{i1}=a_{i2}=\ldots=a_{il}=1$ and $a_{il+1}=0$, for some $l\geq 1$, then there is no $1$ in the diagonal with induced sum $i+l+1$, contradicting Lemma 1.2.

Thus, we only have three possible generic forms for the adjacency matrix, either row $i$, $1\leq i<n+2$, is of the form a) $(0\ldots 01)$ or, b) $(01\ldots 1)$ or, c) $(0\ldots 010\ldots 0)$. Notice that, a) and b) are not possible since each of the two rows should contain a $1$ in the main diagonal, which is not possible in any of the two configurations. Finally, (c) is forbidden by Lemma 1.3, since otherwise, the adjacency matrix induced by the complementary labeling of $f^{c}$ would have exactly two rows with $1$ entries, namely $k,l$ with $k<l$, and row $k$ of the form $(1\ldots 101\ldots 1)$, which is a contradiction, as we have shown above. ∎

Assume to the contrary that there exists a $n\in\mathbb{N}$ such that $2L\cup LK_{1,n}$ is super edge-magic. Following the same idea that we used in the two previous results, we consider the digraph $D\cong 2\overrightarrow{L}\cup\overrightarrow{LK}_{1,n}$ where the star is oriented as in the previous two proofs. By definition, $2L\cup LK_{1,n}$ is super edge-magic if and only if the digraph $D\cong 2\overrightarrow{L}\cup\overrightarrow{LK}_{1,n}$ is super edge-magic. Let $f$ be a super edge-magic labeling of D, and let $A(D)=(a_{ij})$ be the adjacency matrix induced by $f$ where the vertices take the name of their labels in the super edge-magic labeling. Since all vertices of $D$ have indegree $1$, we can represent the adjacency matrix of $D$ by a vector $(v_{1},\ldots,v_{n})$ such that $v_{k}=i$ if and only if $a_{ik}=1$. There are three possible generic forms.

Case 1: Let $v_{1}=v_{2}=\ldots=v_{k}=1$, $1\leq k\leq n+1$ and $v_{k+1}>1$. Then there is no $1$ in the diagonal with induced sum $k+2$, contradicting Lemma 1.2.

Case 2: Let $i$ be such that $a_{1i}=1$. By case 1, $i>1$. Let $v_{1}=v_{2}=\ldots=v_{k}=i$ and $v_{k+1}\neq i$ , $2\leq k\leq n+1$ and $2\leq i\leq n+2$. Since each row with entries being $1$ has a $1$ in the main diagonal and $i>1$, the other two rows of the adjacency matrix represent the two loops. Let $v_{k+1}$ and $v_{l},k+2\leq l\leq n+3$ be the components that represent the two loops. Then, $\min\{v_{k+1},v_{l}\}>v_{k}$. If $v_{l}>v_{k+1}$, then there is no $1$ in the diagonal with induced sum $i+k+1$. If $v_{l}<v_{k+1},v_{l}\neq i$, then $v_{k}<v_{l}<v_{k+1},k+2\leq l\leq n+2$. This implies that one of the two rows representing the loop components cannot have a 1 in the main diagonal. Hence we get a contradiction in each of the above possible scenarios.

Case 3: Let $v_{1}=v_{2}=\ldots=v_{k}=n+3$, $1\leq k\leq n+1$. Notice that this must be a part of the component $LK_{1,n}$ otherwise there is no $1$ in the main diagonal. In particular, this implies that $v_{n+3}=1$, which in view of Lemma 1.3 and case 1, is also a contradiction. ∎