By Jurisic A.

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**Sample text**

Geodetic) set of Gi for i = 1, 2, . . , m. 5 Forcing Geodomination X 27 Y Z Fig. 20 ([171]). Given any pair of nonnegative integers a, b with a + b ≥ 2, there exists a two-connected graph G such that fh (G) = a, fg (G) = b, h(G) = a + b + c, and g(G) = a + 2b + c. Sketch of proof. Take a collection {Xi }ai=1 of disjoint graphs isomorphic to the graph X of Fig. 14. Take a collection {Yi }bi=1 of disjoint graphs isomorphic to the graph Y of Fig. 14. Take a collection {XZ }ci=1 of disjoint graphs isomorphic to the graph Z of Fig.

Then there is a connected graph G such that γ (G) = a, g(G) = b, and γg (G) = c. Sketch of proof. Take integers a, b ≥ 2. We distinguish cases. Case 1: c = a + b. Consider the graph G1 shown in Fig. 17. Check that {x, y, y2 , . . , y3s−1 } and {y3s , x1 , . . , xr } are a minimum dominating set and a minimum geodetic set of G, respectively. Notice also that the union of these two sets produces a minimum geodetic dominating set. Hence, taking r = b − 1 and s = a − 2, we obtain the desired values.

If b = 2, take any even cycle. Suppose that b ≥ 3. Consider the graph G1 in Fig. 11. Notice that {u1 , . . , ub−2 , v3 , x} is a minimum geodetic set of G1 and thus g(G1 ) = b. To prove that fg (G1 ) = 1, notice first that X = {u1, . . , ub−2 , v3 , x} and Y = {u1 , . . , ub−2 , v3 , y} are two distinct minimum geodetic sets and second that X is the unique minimum geodetic set containing {x}. Case 2: a > 1 and b = a + 1. Consider the graph G2 in Fig. 11. Notice that {u2 , . . , ua+1 , v1 } is a minimum geodetic set of G2 and thus g(G2 ) = a + 1.