Ray-casting algorithm
You are encouraged to solve this task according to the task description, using any language you may know.
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Given a point and a polygon, check if the point is inside or outside the polygon using the ray-casting algorithm.
A pseudocode can be simply:
count ← 0
foreach side in polygon:
if ray_intersects_segment(P,side) then
count ← count + 1
if is_odd(count) then
return inside
else
return outside
Where the function ray_intersects_segment return true if the horizontal ray starting from the point P intersects the side (segment), false otherwise.
An intuitive explanation of why it works is that every time we cross a border, we change "country" (inside-outside, or outside-inside), but the last "country" we land on is surely outside (since the inside of the polygon is finite, while the ray continues towards infinity). So, if we crossed an odd number of borders we was surely inside, otherwise we was outside; we can follow the ray backward to see it better: starting from outside, only an odd number of crossing can give an inside: outside-inside, outside-inside-outside-inside, and so on (the - represents the crossing of a border).
So the main part of the algorithm is how we determine if a ray intersects a segment.
Looking at the image on the right, we can easily be convinced of the fact that rays starting from points in the hatched area (like P1 and P2) surely do not intersect the segment AB. We also can easily see that rays starting from points in the greenish area surely intersect the segment AB (like point P3).
So the problematic points are those inside the white area (the box delimited by the points A and B), like P4.
Let us take into account a segment AB (the point A having y coordinate always smaller than B's y coordinate, i.e. point A is always below point B) and a point P. Let us use the cumbersome notation PAX to denote the angle between segment AP and AX, where X is always a point on the horizontal line passing by A with x coordinate bigger than the maximum between the x coordinate of A and the x coordinate of B. As explained graphically by the figures on the right, if PAX is greater than the angle BAX, then the ray starting from P intersects the segment AB. (In the images, the ray starting from PA does not intersect the segment, while the ray starting from PB in the second picture, intersects the segment).
Points on the boundary or "on" a vertex are someway special and through this approach we do not obtain coherent results. They could be treated apart, but it is not necessary to do so.
An algorithm for the previous speech could be (if P is a point, Px is its x coordinate):
ray_intersects_segment:
P : the point from which the ray starts
A : the end-point of the segment with the smallest y coordinate
(A must be "below" B)
B : the end-point of the segment with the greatest y coordinate
(B must be "above" A)
if Py ≥ Ay or Py ≤ B then
return false
else if Px > max(Ax, Bx) then
return false
else
if Px < min(Ax, Bx) then
return true
else
if Ax ≠ Bx then
m_red ← (By - Ay)/(Bx - Ax)
else
m_red ← ∞
end if
if Ax ≠ Px then
m_blue ← (Py - Ay)/(Px - Ax)
else
m_blue ← ∞
end if
if m_blue ≥ m_red then
return true
else
return false
end if
end if
end if
(As it can be seen, if the ray passes through a vertex, it is not considered intersection; there exists a set of "situations" where this behaviour brings to wrong results; to avoid this, we should "move" the point up or down by a small quantity)
C
Required includes and definitions:
<lang c>#include <stdio.h>
- include <stdlib.h>
- include <stdbool.h>
- include <math.h>
typedef struct {
double x, y;
} point_t;
typedef struct {
point_t *vertices; int edges[];
} polygon_t;</lang>
Polygons for testing:
<lang c>point_t square_v[] = {
{0,0}, {10,0}, {10,10}, {0,10},
{2.5,2.5}, {7.5,0.1}, {7.5,7.5}, {2.5,7.5}
};
polygon_t square = {
square_v,
{ 0,1, 1,2, 2,3, 3,0, -1 }
};
polygon_t squarehole = {
square_v,
{ 0,1, 1,2, 2,3, 3,0,
4,5, 5,6, 6,7, 7,4, -1 }
};
polygon_t strange = {
square_v,
{ 0,4, 4,3, 3,7, 7,6, 6,2, 2,1, 1,5, 5,0, -1 }
};</lang>
Check for intersection code:
<lang c>#define MAX(A,B) (((A)>(B))?(A):(B))
- define MIN(A,B) (((A)>(B))?(B):(A))
- define minP(A,B,C) ( (((A)->C) > ((B)->C)) ? (B) : (A) )
- define coeff_ang(PA,PB) ( ((PB)->y - (PA)->y) / ((PB)->x - (PA)->x) )
inline bool hseg_intersect_seg(point_t *s, point_t *a, point_t *b) {
if ( (s->y >= MAX(a->y, b->y) ||
s->y <= MIN(a->y, b->y)) ||
(s->x > MAX(a->x, b->x) ) ) return false; if ( s->x <= MIN(a->x, b->x) ) return true; double ca = (a->x != b->x) ? coeff_ang(a,b) : HUGE_VAL; // * ( (b->y - a->y) > 0 ? 1 : -1 ); point_t *my = minP(a,b,y); double cp = (s->x - my->x) ? coeff_ang(my, s) : HUGE_VAL; // * ( (s->y - my->y) > 0 ? 1 : -1); if ( cp >= ca ) return true; return false;
}</lang>
The ray-casting algorithm:
<lang c>bool point_is_inside(polygon_t *poly, point_t *pt) {
int cross = 0, i;
for(i=0; poly->edges[i] != -1 ; i+=2) {
if ( hseg_intersect_seg(pt,
&poly->vertices[ poly->edges[i] ], &poly->vertices[ poly->edges[i+1] ]) )
cross++; } return !(cross%2 == 0);
}</lang>
Testing:
<lang c>#define MAKE_TEST(S) do { \
printf("point (%.5f,%.5f) is ", test_points[i].x, test_points[i].y); \
if ( point_is_inside(&S, &test_points[i]) ) \
printf("INSIDE " #S "\n"); \
else \
printf("OUTSIDE " #S "\n"); \
} while(0); \
int main() {
point_t test_points[] = { {5,5}, {5,8}, {2,2},
{0,0}, {10,10}, {2.5,2.5}, {0.01,5}, {2.2,7.4}, {0,5}, {10,5}, {-4,10}};
int i;
for(i=0; i < sizeof(test_points)/sizeof(point_t); i++) {
MAKE_TEST(square);
MAKE_TEST(squarehole);
MAKE_TEST(strange);
printf("\n");
}
return EXIT_SUCCESS;
}</lang>
The test's output reveals the meaning of coherent results: a point on the leftmost vertical side of the square (coordinate 0,5) is considered outside; while a point on the rightmost vertical side of the square (coordinate 10,5) is considered inside, but on the top-right vertex (coordinate 10,10), it is considered outside again.