COMBINATORIAL ANALYSIS (MATRIX PROBLEMS, ORDER THEORY)

NEW

Problem: Reconstructed conjecture: for every even $m\ge4$, every $n\ge3$, and each $d\in\{0,2\}$, the generalized prism $C_m\square P_n$ has a super $(a,d)$-edge-antimagic total labeling for some $a$. Here $P_n$ has $n$ vertices, and a super $(a,d)$-EAT labeling of a graph $G$ with $p$ vertices and $q$ edges is a bijection

with $f(V(G))=\{1,\dots,p\}$ and edge-weights

equal to $\{a,a+d,\dots,a+(q-1)d\}$. The source explicitly identifies $d=0$ with the super edge-magic case.

Result: The conjecture is false. In fact, $C_4\square P_n$ is not super $(a,0)$-edge-antimagic total for any $n\ge3$.

Let $G=C_4\square P_n$. Then

Also $G$ is bipartite, hence triangle-free, and $\Delta(G)\le4$.

Assume $G$ had a super $(a,0)$-EAT labeling. Since all edge-weights are equal, if $s(uv)=f(u)+f(v)$, then

The edge labels are exactly $p+1,\dots,p+q$, so the $q$ endpoint-sums $s(uv)$ are $q$ distinct consecutive integers. Since vertex labels are $1,\dots,p$, every endpoint-sum lies between $3$ and $2p-1$. Because $q=2p-4$, the only possible sets of endpoint-sums are

First suppose the set is $\{3,\dots,2p-2\}$. Let $v_i$ be the vertex labeled $i$. Sum $3$ forces $v_1v_2\in E(G)$, and sum $4$ forces $v_1v_3\in E(G)$. Sum $5$ is either $1+4$ or $2+3$, but $v_2v_3$ would form a triangle with $v_1$, so $v_1v_4\in E(G)$. Similarly, sum $6$ forces $v_1v_5\in E(G)$. Thus $v_1$ already has four neighbors $v_2,v_3,v_4,v_5$. For sum $7$, the only possibilities are $1+6$, $2+5$, or $3+4$. The first would give $v_1$ degree at least $5$, and the other two create triangles with $v_1$. Contradiction.

The second possible endpoint-sum set, $\{4,\dots,2p-1\}$, is symmetric. Sum $2p-1$ forces $v_pv_{p-1}$, sum $2p-2$ forces $v_pv_{p-2}$, sum $2p-3$ forces $v_pv_{p-3}$, and sum $2p-4$ forces $v_pv_{p-4}$, since the alternative pairs would create triangles. Then $v_p$ already has four neighbors. Sum $2p-5$ must be one of

but the first violates $\Delta(G)\le4$, and the other two create triangles. Contradiction.

Therefore $C_4\square P_n$ is not super edge-magic, i.e. not super $(a,0)$-EAT. Since $m=4$ is even and $n\ge3$, this refutes the conjecture for $d=0\in\{0,2\}$.

Citation: No external citation is needed; the disproof is the elementary counterexample family above.