Decide If a Point is Inside a Simple Polygon

Problem: Given a point x,  try to decide if x is inside a simple polygon P. P is denoted by its vertices {p1, p2, ... , pn} .

Solution: There exists an O(N) solution based on one important observation: any ray from x will cross the boundary of P for odd number of times if and only if x is inside P. Here a ray is a line start from a point. Before designing an algorithm based on this observation, there are two cases that we need to be cautious. One is when the ray across some vertices of P. Under such case, we need another way to  define the number of times that the ray across the boundary of P. The second case is when x is actually on the boundary of P, but this is easy to tell. Therefore, the skeleton of the algorithm looks as follows:

1. Decide if x is on the boundary of P, if so return True, otherwise return False. This step takes O(N) since there are N edges.
2. Find a ray that starts from x (we can just use a vertical ray and the ray can ends at the lowest y coordinate of the polygon + 1) , for each edge of P, check if the ray intersect with that edge. Each such operation takes O(1) and the total takes O(N).  Here we need to handle the special case where the ray across some vertices of P.  Assume i1, i2, ... , ik are the points at which the ray intersect the boundary, if any such point is not a vertex of P, then we define the number of times that the ray across the boundary of P as k. If any such point is a vertex of P, we define the number of times that the ray across the boundary of P as 1. For the rest, we  define the number of times that the ray across the boundary of P as num_of_non_vetex_points.