# For which graphs we have domination number of (n-1)/2

Here is a theorem from a textbook:

For even number of n , G is a graph with n vertices and no isolated vertice ; the domination number is $$\frac{n}{2}$$ if and only if The components ofG are C4 or the corona HoK1 for any connected graph H (Payan and Xuong )

Here is a question: for which graphs we have $$\gamma(G)=\frac{n-1}{2}$$

In other words: for odd number n and same upper properties ,but $$\frac{n-1}{2}$$ domination number , what are the components of G ?

I would even appreciate linking Thank you

(For context for readers who may not be familiar with the corona product, $$H \circ K_1$$ means we take any connected graph $$H$$ and add a new degree-$$1$$ vertex - a leaf vertex - adjacent to every vertex of $$H$$.)
If we try to generalize to $$\gamma(G) = \frac{n-1}{2}$$, then we can definitely take any connected $$\frac{n+1}{2}$$-vertex graph $$H$$, and add a leaf to all but one vertex of $$H$$. To dominate each degree-$$1$$ vertex $$v$$, we need to either take $$v$$ or its neighbor, and taking all of their neighbors will give us a dominating set.
However, there are plenty of other examples, so we probably can't characterize them as easily as in the $$\frac n2$$ case where there's just $$C_4$$. Mathematica's GraphData command knows about the following:
Out of the $$15$$ graphs above, $$8$$ belong to a different infinite family. Take a graph $$H$$ on $$\frac{n+1}{2}$$ vertices; it need not be connected, but it should have at most two connected components. To $$\frac{n-3}{2}$$ of the vertices, add a leaf vertex, as before; add one last vertex (call it $$v$$) adjacent to both of the remaining vertices of $$H$$ (call them $$w_1, w_2$$). If $$H$$ was not connected, then to make the result connected, the $$w_1, w_2$$ should be in different components of $$H$$.
The $$\frac{n-1}{2}$$ vertices we added are not adjacent and have no common neighbors, so the resulting graph needs a dominating set of size at least $$\frac{n-1}{2}$$. We can get one by taking either $$V(H)-\{w_1\}$$ or $$V(H) - \{w_2\}$$.