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Antimagic Labeling of Central Digraphs Constructed from Loops
Raja Rawat
TL;DR
The paper examines antimagic labeling for central digraphs arising from loops, connecting graph-labeling questions with the algebraic framework of central digraphs. It gives a lexicographic consecutive arc-labeling and proves it is antimagic for every loop of order k ≥ 3, with the property depending only on k.
Problem
The paper studies antimagic labeling in central digraphs constructed from loops, bringing graph-labeling questions together with the algebraic framework of central digraphs.
Source: §1 · pp. 1–2Method
It orders the loopless central digraph’s arcs lexicographically by tail and head, labels them consecutively, and proves the resulting vertex weights are pairwise distinct.
Source: §1 · p. 2, §3 · pp. 7–8Results
Takeaways & Limitations
Non-isomorphic loops of the same order yield loopless central digraphs that are antimagic under the same explicit labeling.
Source: §4 · p. 9The stated result covers loops of order k ≥ 3; the setup specifies identity 0.
Source: §2 · p. 2
Abstract
from arXiv · showhide
Let $(T, \circ)$ be a loop of order $k \geq 3$ and $D_S^*$ the loopless central digraph obtained from the associated central digraph $D_S$ by removing the $k$ self-loops at the diagonal vertices. We construct an explicit arc-labeling $f$ of $D_S^*$ ordering all arcs lexicographically by $(\text{tail}, \text{head})$ and assigning labels $1, \ldots, m$ in that order and prove that $f$ is an antimagic labeling. Consequently, $D_S$ is antimagic for every loop of order $k \geq 3$, a property that depends only on $k$ and is invariant across all non-isomorphic loops of the same order.
1 Introduction
The paper brings antimagic labeling to central digraphs constructed from loops, drawing together graph-labeling and algebraic frameworks. It proves that an explicit consecutive labeling makes the loopless digraph antimagic for every loop of order k ≥ 3.
1 Introduction
- Antimagic labeling assigns labels 1 through m bijectively so that the vertex sums are pairwise distinct.The paper describes earlier results for graphs with minimum degree at least c log n, regular graphs of odd degree, and graphs with sufficiently large average degree.Source: §1 · p. 1
- A central groupoid satisfies (x∗y)∗(y∗z) = y, and its associated digraph has adjacency matrix A satisfying A^2 = J_n.Finite central groupoids have order k^2, and every vertex in the associated digraph has in-degree and out-degree k.Source: §1 · p. 2
- The loopless central digraph is obtained by deleting the k diagonal self-loops; non-diagonal vertices retain in-degree and out-degree k, while diagonal vertices have degree k − 1.The paper uses the associated left quasigroup with left identity and right zero to construct the digraph from a loop.Source: §1 · p. 2
- For every loop of order k ≥ 3, lexicographically labeling the loopless central digraph’s arcs consecutively gives an antimagic labeling.The digraph has k^2 vertices and k^3 − k arcs.Source: §1 · p. 2
2 Setup and Notation
The setup fixes a loop of order k ≥ 3 and its associated left quasigroup, then labels arcs in lexicographic order. The proof uses the digraph’s degree structure and vertex sums to derive the oriented weights.
2 Setup and Notation
- The setup takes a loop of order k ≥ 3 with identity 0 and its associated left quasigroup with left identity and right zero.The digraph is loopless after its k diagonal self-loops are removed.Source: §2 · pp. 2–3
- Vertices are encoded as u = l1k + l2, and arcs are ordered first by tail and then by head, assigning consecutive labels 1 through m.Because this is a total order on the arc set, the labeling is bijective; arcs leaving a given tail receive consecutive labels.Source: §2 · pp. 3–4
- The loopless digraph has k^3 − k arcs after removing k self-loops from k^3 arcs; non-diagonal vertices retain in- and out-degree k, while diagonal vertices have in- and out-degree k − 1.Source: §2 · pp. 3–4
- The labeling’s tail offset counts earlier outgoing arcs, accounting for each earlier diagonal vertex’s one-arc degree reduction.The diagonal count before u is used to express the offset.Source: §2 · pp. 3–4
- For a diagonal vertex (a, a), the actual oriented weight differs from its corresponding non-diagonal formula by k − 1 − a.The section derives oriented weights by subtracting the in-sum from the out-sum.Source: §2 · pp. 6–7
3 Proof of Main Theorem 1.1
The proof establishes distinct oriented vertex weights by analyzing consecutive vertices in the row-by-row ordering. It first proves that out-sums are pairwise distinct, then compares weight changes across transition cases.
3 Proof of Main Theorem 1.1
- Pairwise distinct out-sums follow from checking consecutive differences across the three degree-transition cases.When both consecutive vertices are non-diagonal, the out-sum difference is k^2 > 0.Source: §3 · pp. 7–8
- Each vertex’s in-sum uses exactly d_u labels, one from each in-arc’s label block.The in-arcs are determined by the edge rules for vertices with second coordinate zero or nonzero.Source: §3 · p. 8
- The proof checks four types of consecutive-vertex transitions to establish distinct oriented weights.The cases are determined by diagonal status and whether the sequence crosses a row boundary.Source: §3 · p. 8
- For non-diagonal-to-diagonal and diagonal-to-nondiagonal transitions, the oriented-weight change is strictly negative in the stated ranges.The respective bounds are at most −5 and at most −2.Source: §3 · p. 9
4 Consequences
The antimagic property is uniform across loops of a fixed order: it depends on k, not the loop’s isomorphism class. The proof applies the same labeling to every loop of that order.
4 Consequences
- The antimagic property depends only on k, so non-isomorphic loops of the same order have antimagic associated digraphs under the same explicit labeling.The out-sums are the same for each loop of order k, and the distinctness argument applies uniformly.Source: §4 · p. 9
- The uniformity follows because the proof’s out-sums agree across loops of order k, while its distinctness and weight arguments apply to all of them.Source: §4 · p. 9
Worked Example: Antimagic Labeling for k = 3
A worked example uses k = 3 to illustrate the theorem’s steps with concrete numbers before the general argument.
Worked Example: Antimagic Labeling for k = 3
- The example uses the smallest nontrivial case, k = 3.Source: Worked Example: Antimagic Labeling for… · p. 10
- It illustrates every step of Theorem 1.1 with concrete numbers so readers can check the formulas.Source: Worked Example: Antimagic Labeling for… · p. 10
- The numerical illustration is presented before examination of the general argument.Source: Worked Example: Antimagic Labeling for… · p. 10
S and its arc set
For k = 3, vertices are encoded as u = 3a + b, and the loopless central digraph deletes self-loops at diagonal vertices. The verification computes out-sums, in-sums, and weights, showing strict weight decrease and loop-independent weights for order 3.
S and its arc set
- For k = 3, the nine vertices are encoded by u = 3a + b, and the loopless digraph deletes diagonal self-loops from the lexicographically ordered arc list.The table marks diagonal vertices and labels the 24 arcs.Source: S and its arc set · p. 10, Table 1 · p. 10
Step 1: Computing the out-sums S−(u)
- The nine out-sums are pairwise distinct but not monotone, with dips at diagonal vertices.Their values sum to 300, matching 24·25/2.Source: Step 1: Computing the out-sums S−(u) · p. 10
Step 3: Computing the in-sums S+(u)
- In-sums are obtained from arcs entering each vertex, excluding the diagonal source when the target is diagonal.For k = 3, the examples give S+(0) = 26 and S+(1) = 29.Source: Step 3: Computing the in-sums S+(u) · p. 11
Verification of the in-sum formulas
- For k = 3, the closed-form in-sum formulas reproduce the column-offset values C(0) = 24, C(1) = 32, and C(2) = 40.Examples verify S+(1, 0) = 35 and S+(1, 1) = 25.Source: Verification of the in-sum formulas · pp. 11–12
Step 4: Computing the weights w(u) and verifying strict decrease
- For k = 3, every consecutive weight difference is negative, and the listed transition cases match the predicted differences.The example also checks non-diagonal and diagonal weight formulas, including w(2, 1) = −17 and w(1, 1) = 0.Source: Table 4 · p. 12, Step 4: Computing the weights w(u) and… · p. 13
Observations
- The weights are strictly decreasing and antisymmetric about the midpoint; for k = 3, all nine are distinct and the same sequence occurs for every loop of order 3.The out-sums need not be monotone, while the weights are; the unique zero weight is at the middle vertex.Source: Observations · p. 13