n-SuperHyperGraph

3.28. Optimal Sets and Numbers For Family of n-SuperHyperGraph

Proof. Suppose n-SuperHyperGraph n-S H G = (Gn ⊆ P n(V ), En ⊆ P n(V )). If there’s no super-vertex, then all super-vertices are super-resolved. Hence if I choose Gn, then there’s no super-vertex such that super-vertex is super-resolved. It implies Gn is super-resolving set but Gn isn’t optimal-super-resolving set. Since if I construct one set from Gn such that only one super-vertex is out of S, then S is super-resolving set. It implies Gn isn’t optimal-super-resolving set. Thus Gn is super-resolving set. Proposition 3.27.6. Assume n-SuperHyperGraph n-S H G = (Gn ⊆ P n(V ), En ⊆ P n(V )). Then Gn is super-coloring set.

Proof. Suppose n-SuperHyperGraph n-S H G = (Gn ⊆ P n(V ), En ⊆ P n(V )). All super-vertices belong to a super-edge have to color differently. If Gn is chosen, then all super-vertices have different colors. It induces that t colors are used where t is the number of super-vertices. Every super-vertex has unique color. Thus Gn is super-coloring set.

3.28 Optimal Sets and Numbers For Family of n-SuperHyperGraph

Proposition 3.28.1. Assume G is a family of n-SuperHyperGraph. Then Gn is super-dominating set for al l members of G, simultaneously.

Proof. Suppose G is a family of n-SuperHyperGraph. Thus Gn is superdominating set for every given n-SuperHyperGraph of G. It implies Gn is super-dominating set for all members of G, simultaneously. Proposition 3.28.2. Assume G is a family of n-SuperHyperGraph. Then Gn is super-resolving set for al l members of G, simultaneously.

Proof. Suppose G is a family of n-SuperHyperGraph. Thus Gn is super-resolving set for every given n-SuperHyperGraph of G. It implies Gn is super-resolving set for all members of G, simultaneously. Proposition 3.28.3. Assume G is a family of n-SuperHyperGraph. Then Gn is super-coloring set for al l members of G, simultaneously.

Proof. Suppose G is a family of n-SuperHyperGraph. Thus Gn is super-coloring set for every given n-SuperHyperGraph of G. It implies Gn is super-coloring set for all members of G, simultaneously. Proposition 3.28.4. Assume G is a family of n-SuperHyperGraph. Then Gn \ {Xn} is super-dominating set for al l members of G, simultaneously.

Proof. Suppose G is a family of n-SuperHyperGraph. Thus Gn \ {Xn} is superdominating set for every given n-SuperHyperGraph of G. One super-vertex is out of Gn \ {Xn}. It’s super-dominated from any super-vertex in Gn \ {Xn}. Hence every given two super-vertices are super-dominated from any super-vertex in Gn \ {Xn}. It implies Gn \ {Xn} is super-dominating set for all members of G, simultaneously. 147