Art Gallery Problem
The Art Gallery Problem is formulated in geometry as the minimum number of guards that need to be placed in an n-vertex simple polygon such that all points of the interior are visible. A simple polygon is a connected closed region whose boundary is defined by a finite number of line segments. It is a family of related visibility problems in computational geometry. Visibility is defined such that two points u and v are mutually visible if the line segment joining them lies inside the polygon. Here, we discuss several variants. The common terms used in the variants discussed below:
- A simple (not necessarily convex) polygon P to describe the art gallery.
- A set of points S to describe the guards where each guard is represented by a point in P.
- A rule that a point A ∈ S can guard another point B ∈ P if and only if A ∈ S, B ∈ P and line segment AB is contained in P.
- A question on whether all points in polygon P are guarded by S.
Many variants of this Art Gallery Problem are classified as NP-hard problems. Here, we will focus on the ones that admit polynomial solutions:
- Variant 1: Determine the upper bound of the smallest size of set S.
- Variant 2: Determine if ∃ a critical point C in polygon P and ∃ another point D ∈ P such that if the guard is at position C, the guard cannot protect point D.
- Variant 3: Determine if polygon P can be guarded with just one guard.
- Variant 4: Determine the smallest size of set S if the guards can only be placed at the vertices of polygon P and only the vertices need to be guarded.
Note that there are many more variants and at least one book has been written for it.
- The solution for variant 1 is a theoretical work of the Art Gallery theorem by V´aclav Chv´atal. He states that [N/3] guards are always sufficient and sometimes necessary to guard a simple polygon with n vertices.
- The solution for variant 2 involves testing if polygon P is concave (and thus has a critical point). We can use the negation of isConvex function.
- The solution for variant 3 can be hard if one has not seen the solution before. We can use the cutPolygon function. We cut polygon P with all lines formed by the edges in P in a counter-clockwise fashion and retain the left side at all times. If we still have a non-empty polygon at the end, one guard can be placed in that non-empty polygon which can protect the entire polygon P.
- The solution for variant 4 involves the computation of Minimum Vertex Cover of the ‘visibility graph’ of polygon P. In general this is another NP-hard problem.
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