## Algorithms for Placing Guards

### Vertex Guards

First, it is important to notice that the proof of the floor(n/2) bound on the required number of vertex guards for a convex terrain depends on four coloring the surface's vertices. As such, we cannot directly apply this idea to place guards, since four coloring is NP-complete. However, five coloring is not, and in this section we will see how to use this algorithm in order to place vertex guards.

One simple algorithm that we can use to place vertex guards is as follows. It is based on the idea that if we five-color the vertices of a terrain, then each face to be covered is a triangle and therefore is seen by three differently colored vertices. Then if we take any set of three colors, since there are only five colors in the vertex coloring, at least one of them must be used in each face's vertices. This gives the following simple algorithm:

1. Five-color the vertices of the terrain
2. Select the three colors which are least used
3. Place one guard at each vertex of these colors
Work has been previously done to show that we can five-color the planar graph of the terrain (and therefore its vertices) in O(n) [5]. Obviously selecting the color set and placing the guards is also O(n), therefore the time complexity of this algorithm is O(n). As noted above, all faces must be covered by the three colors set, and we get at most floor(3n/5) guards by the pigeon hole principle concerning the size of the color set.

### Edge Guards

The algorithm used to place edge guards is slightly more complex than the one above, but still relies on the idea of five-coloring the vertices of the terrain. Let us look at the planar graph corresponding to the terrain. Then the algorithm is:
1. Five-color the vertices of the graph
2. For each combination of three colors (there are 10 of them), find the maximal matching of the graph for vertices colored with these three colors
• If the sum of all the edges of all the above sets is greater than 2n, we can find a combination of one matching and of a set of vertices such that guarding each edge in the matching and vertices from the remaining unguarded triangles requires at most floor(2n/5) guards
• Otherwise, we can find two matchings such that guarding the edges of both requires at most floor(2n/5) and together they cover the whole terrain

The proof of the above algorithm goes as follows. First observe that if we take a maximal matching for some color combination, then not all triangles might be guarded by this matching. However, if we add one edge guard for each of the remaining unguarded triangles, then we will cover the whole terrain. If a,b,c are the three colors in question, G(a,b,c) the total size of the set of guards, S(a,b,c) the set of vertices of one of these colors and M(a,b,c) the size of the matching, then we have

##### G(a,b,c) = S(a,b,c) - M(a,b,c)

Simply because for each edge that we guard, we effectively guard two different colors. Therefore, if we were to use only vertex guards, we would require S(a,b,c) guards, but for each edge guard used, we can take away two vertex guards.

Notice that , as we have 10 combinations of colors from our five coloring. Indeed, each color appears exactly 6 times in the list of combinations. Now, if , then we have that the average G(a,b,c) is

Since there must be at least one set G(a,b,c) such that G(a,b,c) is smaller than the average G(a',b',c') then we know that picking this set will yield a set of guards covering the entire graph (and by extension, the terrain) with size at most floor(2n/5).

If , then we claim that if the colors are named 1..5, then a pair of matchings from the following list is sufficient to guard the terrain: { M(1,2,3) and M(1,4,5), M(1,2,5) and M(2,3,4), M(1,2,4) and M(3,4,5), M(1,3,4) and M(2,3,5), M(1,3,5) and M(2,4,5) }. Without loss of generality, suppose we pick the first pair, M(1,2,3) and M(1,4,5). Then suppose there is a triangle that is not guarded. This triangle cannot contain an edge colored (1,2), (1,3), (2,3), (1,4), (1,5) or (4,5), since it would be part of the matching as none of the vertices of the triangle are guarded. So one of the edges must be (2,4), (2,5), (3,4) or (3,5). Therefore the third vertex of the triangle must be 1, 3, or 5 depending on this edge. But in all cases, we find that this implies we have two unmatched vertices that should be in M(1,2,3) or M(1,4,5)! Hence all triangles must be guarded. By contradiction, we have shown that one of the above pairs of matchings is sufficient to cover the terrain.

Now, the average size of these pairs is , and thus as noted above, there must be a pair that has size at most floor(2n/5). It follows that our algorithm can always find a set of edges (or edges and vertices) such that floor(2n/5) edge guards are sufficient to cover the entire terrain. Th algorithm takes O(n) time, since there is a fixed number of matchings, each taking O(n) time to be found, and five-coloring is O(n).