A cycle Ct in G is said to be minimal (minimal cycle or mt-cycle) if Ct does not contain any cycle of order less than t. The mt-cycle derivative of G, denoted by mtG′ is the graph obtain from G by taking the mt-cycles of G as vertices in mtG′ and two vertices in mtG′ are adjacent if and only if the two mt-cycles in G corresponding to these vertices have an edge in common.
The following theorem establishes the tree covering number of the cycle derivative of the ladder Ln = Pn × K2.
Theorem 3.4 Let n be a positive integer. Then tc[(Pn × K2)′] = 1
Proof: Forn=1,P1 ×K2 ∼=K2,then(P1 ×K2)′ isnull. Also,forn=2, Pn ×K2 ∼= C4. Thus, (Pn ×K2)′ = K1. Thus, in either case, tc[(Pn ×K2)′] = 0. If n ≥ 3, let u,v ∈ V(K2) and xi ∈ V(Pn), i = {1,2,3,...,n}, then the edge ei+1 = {[(xi+1,u),(xi+1,v)]} is an edge in common of the cycles ci and ci+1 respectively, hence the vertices ci and ci+1 in mG′ are adjacent vertices, that is, for each i = {1,2,3,...,n}, mG′ is a path of order n − 1, therefore tc[(Pn × K2)′] = 1.
A cycle Ct in G is said to be minimal (minimal cycle or mt-cycle) if Ct does not contain any cycle of order less than t. The mt-cycle derivative of G, denoted by mtG′ is the graph obtain from G by taking the mt-cycles of G as vertices in mtG′ and two vertices in mtG′ are adjacent if and only if the two mt-cycles in G corresponding to these vertices have an edge in common.The following theorem establishes the tree covering number of the cycle derivative of the ladder Ln = Pn × K2.Theorem 3.4 Let n be a positive integer. Then tc[(Pn × K2)′] = 1Proof: Forn=1,P1 ×K2 ∼=K2,then(P1 ×K2)′ isnull. Also,forn=2, Pn ×K2 ∼= C4. Thus, (Pn ×K2)′ = K1. Thus, in either case, tc[(Pn ×K2)′] = 0. If n ≥ 3, let u,v ∈ V(K2) and xi ∈ V(Pn), i = {1,2,3,...,n}, then the edge ei+1 = {[(xi+1,u),(xi+1,v)]} is an edge in common of the cycles ci and ci+1 respectively, hence the vertices ci and ci+1 in mG′ are adjacent vertices, that is, for each i = {1,2,3,...,n}, mG′ is a path of order n − 1, therefore tc[(Pn × K2)′] = 1.
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