
A simple graph <em>G</em> admits a <em>K</em><sub>1,n</sub>-covering if every edge in <em>E</em>(<em>G</em>) belongs to a subgraph of <em>G</em> isomorphic to <em>K</em><sub>1,n</sub>. The graph <em>G</em> is <em>K</em><sub>1,n</sub>-supermagic if there exists a bijection <em>f</em> : <em>V</em>(<em>G</em>) ∪ <em>E</em>(<em>G</em>) → {1, 2, 3,..., |<em>V</em>(<em>G</em>) ∪ <em>E</em>(<em>G</em>)|} such that for every subgraph <em>H</em>' of <em>G</em> isomorphic to <em>K</em><sub>1,n</sub>, ∑v<sub> ∈ V(H') </sub> f(v) + ∑<sub>e ∈ E(H')</sub> f(e) is a constant and <em>f</em>(<em>V</em>(<em>G</em>)) = {1, 2, 3,..., |<em>V</em>(<em>G</em>)|}. In such a case, <em>f</em> is called a <em>K</em><sub>1,n</sub>-supermagic labeling of <em>G</em>. In this paper, we give a method how to construct <em>K</em><sub>1,n</sub>-supermagic graphs from the old ones.
QA1-939, $k_{1, n}$-covering \sep $k_{1, n}$-supermagic labeling \sep $k_{1, n}$-supermagic graph, Mathematics
QA1-939, $k_{1, n}$-covering \sep $k_{1, n}$-supermagic labeling \sep $k_{1, n}$-supermagic graph, Mathematics
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