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For a set of distances $D$, a graph $G$ on $n$ vertices is said to be $D$-magic if there exists a bijection $f:V\rightarrow \{1,2, \ldots , n\}$ and a constant $k$ such that for any vertex $x$, $\sum_{y\in N_D(x)} f(y) = k$, where…

Combinatorics · Mathematics 2019-09-10 Rinovia Simanjuntak , Palton Anuwiksa

A graph $G$ is $k$-$weighted-list-antimagic$ if for any vertex weighting $\omega\colon V(G)\to\mathbb{R}$ and any list assignment $L\colon E(G)\to2^{\mathbb{R}}$ with $|L(e)|\geq |E(G)|+k$ there exists an edge labeling $f$ such that…

Combinatorics · Mathematics 2023-06-22 Zhanar Berikkyzy , Axel Brandt , Sogol Jahanbekam , Victor Larsen , Danny Rorabaugh

An antimagic labeling of a graph $G(V,E)$ is a bijection $f: E \to \{1,2, \dots, |E|\}$ so that $\sum_{e \in E(u)} f(e) \neq \sum_{e \in E(v)} f(e)$ holds for all $u, v \in V(G)$ with $u \neq v$, where $E(v)$ is the set of edges incident to…

Combinatorics · Mathematics 2023-08-01 Johnny Sierra , Daphne Der-Fen Liu , Jessica Toy

A graph labeling assigns values to the components of a graph (vertices, edges, etc.). In particular, distance magic labelings have been widely studied in undirected graphs. In such a labeling, the vertices are labeled with unique values…

A graph $G=(V,E)$ is said to be distance magic if there is a bijection $f$ from a vertex set of $G$ to the first $|V(G)|$ natural numbers such that for each vertex $v$, its weight given by $\sum_{u \in N(v)}f(u)$ is constant, where $N(v)$…

Combinatorics · Mathematics 2024-02-09 Himadri Mukherjee , Ravindra Pawar , Tarkeshwar Singh

A labeling of a graph is a bijection from $E(G)$ to the set $\{1, 2,..., |E(G)|\}$. A labeling is \textit{antimagic} if for any distinct vertices $u$ and $v$, the sum of the labels on edges incident to $u$ is different from the sum of the…

Combinatorics · Mathematics 2011-10-12 Daniel W. Cranston

For a graph on $m$ edges, a bijective function between the edge set of the graph and $\{1,2,\ldots,m\}$ is an antimagic labeling provided that when adding the labels of the edges incident to the same vertex, the sums are pairwise distinct.…

Combinatorics · Mathematics 2025-03-20 Wei-Tian Li , Po-Wen Yang

An antimagic labelling of a graph $G = (V,E)$ is a bijection from $E$ to $\{1,2, \ldots, |E|\}$, such that all vertex-sums are pairwise distinct, where the vertex-sum of each vertex is the sum of labels over edges incident to this vertex. A…

Combinatorics · Mathematics 2026-03-04 Grégoire Beaudoire , Cédric Bentz , Christophe Picouleau

Motivated by the conjecture of Hartsfield and Ringel on antimagic labelings of undirected graphs, Hefetz, M\"{u}tze, and Schwartz initiated the study of antimagic labelings of digraphs in 2010. Very recently, it has been conjectured in…

Combinatorics · Mathematics 2019-06-17 Donglei Yang

In $1990$, Hartsfield and Ringel introduced antimagic graphs. Hartsfield and Ringel conjectured that every connected graph (and in particular, a tree) except $K_2$ is antimagic. In $2010$, Hefetz et al.\ raised two questions: Is every…

Combinatorics · Mathematics 2025-06-19 Dr A. N. Bhavale

The concept of antimagic labelings of a graph is to produce distinct vertex sums by labeling edges through consecutive numbers starting from one. A long-standing conjecture is that every connected graph, except a single edge, is antimagic.…

Combinatorics · Mathematics 2019-05-21 Fei-Huang Chang , Hong-Bin Chen , Wei-Tian Li , Zhishi Pan

An antimagic labeling of a graph $G$ with $m$ edges is a bijection from $E(G)$ to $\{1,2,\ldots,m\}$ such that for all vertices $u$ and $v$, the sum of labels on edges incident to $u$ differs from that for edges incident to $v$. Hartsfield…

Combinatorics · Mathematics 2015-08-06 Daniel W. Cranston

An antimagic labelling of a graph is a bijection from the set of edges to $\{1, 2, \ldots , m\}$, such that all vertex-sums are pairwise distinct, where the vertex-sum of a vertex is the sum of labels on the edges incident to it. We say a…

Combinatorics · Mathematics 2025-03-13 Grégoire Beaudoire , Cédric Bentz , Christophe Picouleau

A graph is distance magic if it admits a bijective labeling of its vertices by integers from $1$ up to the order of the graph in such a way that the sum of the labels of all the neighbors of a vertex is independent of a given vertex. We…

Combinatorics · Mathematics 2026-03-10 Petr Kovář , Ksenija Rozman , Primož Šparl

Given a graph $G$ with $n$ vertices and an Abelian group $A$ of order $n$, an $A$-distance antimagic labelling of $G$ is a bijection from $V(G)$ to $A$ such that the vertices of $G$ have pairwise distinct weights, where the weight of a…

Combinatorics · Mathematics 2016-10-05 S. Cichacz , D. Froncek , K. Sugeng , Sanming Zhou

An antimagic labeling of a graph $G$ is a one-to-one correspondence between the edge set $E(G)$ and $\lbrace 1,2,...,|E(G)|\rbrace$ in which the sum of the edge labels incident on the distinct vertices are distinct. Let…

Combinatorics · Mathematics 2023-11-20 Nivedha D , Devi Yamini S

An antimagic labeling of a graph $G$ is a $1-1$ correspondence between the edge set $E(G)$ and $\lbrace 1,2,...,|E(G)|\rbrace$ in which the sum of the labels of edges incident to the distinct vertices are different. The edge corona of any…

Combinatorics · Mathematics 2022-12-01 D. Nivedha , S. Devi Yamini

A simple graph $G$ is said to admit an antimagic orientation if there exist an orientation on the edges of $G$ and a bijection from $E(G)$ to $\{1,2,\ldots,|E(G)|\}$ such that the vertex sums of vertices are pairwise distinct, where the…

Combinatorics · Mathematics 2022-07-19 Eranda Dhananjaya , Wei-Tian Li

An {\em antimagic labeling} of a graph with $m$ edges and $n$ vertices is a bijection from the set of edges to the integers $1,...,m$ such that all $n$ vertex sums are pairwise distinct, where a vertex sum is the sum of labels of all edges…

Combinatorics · Mathematics 2007-05-23 N. Alon , G. Kaplan , A. Lev , Y. Roditty , R. Yuster

Let $G=(V,E)$ be an $n$-vertex graph with $m$ edges. A function $f : V \cup E \rightarrow \{1, \ldots, n+m\}$ is an edge-magic labeling of $G$ if $f$ is bijective and, for some integer $k$, we have $f(u)+f(v)+f(uv) = k$ for every edge $uv…