intersection with P. A set G of points inside P or on the boundary of P is said to guard the
polygon if every point inside P or on the boundary of P is seen by a point in G. An interior
guard can lie anywhere inside or on the boundary of the polygon. Using a reduction from
Monotone 3SAT, we prove that the decision version of this problem is NP-hard. Because
interior guards can be placed anywhere inside the polygon, a clever gadget is introduced …