determine whether point lies inside triangle

2019-01-21 02:26发布

问题:

The program needs to read the values of three coordinates

  • P1(x1,y1)
  • P2(x2,y2)
  • P3(x3,y3)

as well as another coordinate P(x,y) and determine whether this point is inside a triangle formed from the 3 point above.

回答1:

The proper way to do this is by calculating the barycentric coordinates of the fourth point given the three points of your triangle. The formula to compute them is given at the end of the section "Converting to barycentric coordinates", but I hope to provide a less mathematics-intense explanation here.

Assume, for simplicity, that you have a struct, point, that has values x and y. You defined your points as:

point p1(x1, y1);
point p2(x2, y2);
point p3(x3, y3);

point p(x,y); // <-- You are checking if this point lies in the triangle.

Now, the barycentric coordinates, generally called alpha, beta, and gamma, are calculated as follows:

float alpha = ((p2.y - p3.y)*(p.x - p3.x) + (p3.x - p2.x)*(p.y - p3.y)) /
        ((p2.y - p3.y)*(p1.x - p3.x) + (p3.x - p2.x)*(p1.y - p3.y));
float beta = ((p3.y - p1.y)*(p.x - p3.x) + (p1.x - p3.x)*(p.y - p3.y)) /
       ((p2.y - p3.y)*(p1.x - p3.x) + (p3.x - p2.x)*(p1.y - p3.y));
float gamma = 1.0f - alpha - beta;

If all of alpha, beta, and gamma are greater than 0, then the point p lies within the triangle made of points p1, p2, and p3.

The explanation behind this is that a point inside a triangle can be described using the points of the triangle, and three coefficients (one for each point, in the range [0,1]):

p = (alpha)*p1 + (beta)*p2 + (gamma)*p3

Rearranging this function gives you the formula to compute barycentric coordinates, but I feel like the steps to do so might be beyond the scope of the question. They are provided on the Wikipedia page that I linked up top.

It follows that each coefficient must be greater than 0 in order for the point p to lie within the area described by the three points.



回答2:

Instead of P1, P2 and P3, lets assume the points as A,B and C.

          A(10,30)
            / \
           /   \
          /     \
         /   P   \      P'
        /         \
B (0,0) ----------- C(20,0) 

Algorithm :

1) Calculate area of the given triangle, i.e., area of the triangle ABC in the above diagram. 

    Area A = [ x1(y2 - y3) + x2(y3 - y1) + x3(y1-y2)]/2

2) Calculate area of the triangle PAB. We can use the same formula for this. Let this area be A1.
3) Calculate area of the triangle PBC. Let this area be A2.
4) Calculate area of the triangle PAC. Let this area be A3.
5) If P lies inside the triangle, then A1 + A2 + A3 must be equal to A.

Given below is a program in C:

#include <stdio.h>
#include <stdlib.h>

/* A utility function to calculate area of triangle formed by (x1, y1), 
   (x2, y2) and (x3, y3) */
float area(int x1, int y1, int x2, int y2, int x3, int y3)
{
   return abs((x1*(y2-y3) + x2*(y3-y1)+ x3*(y1-y2))/2.0);
}

/* A function to check whether point P(x, y) lies inside the triangle formed 
   by A(x1, y1), B(x2, y2) and C(x3, y3) */
bool isInside(int x1, int y1, int x2, int y2, int x3, int y3, int x, int y)
{   
   /* Calculate area of triangle ABC */
   float A = area (x1, y1, x2, y2, x3, y3);

   /* Calculate area of triangle PBC */  
   float A1 = area (x, y, x2, y2, x3, y3);

   /* Calculate area of triangle PAC */  
   float A2 = area (x1, y1, x, y, x3, y3);

   /* Calculate area of triangle PAB */   
   float A3 = area (x1, y1, x2, y2, x, y);

   /* Check if sum of A1, A2 and A3 is same as A */
   return (A == A1 + A2 + A3);
}

/* Driver program to test above function */
int main()
{
   /* Let us check whether the point P(10, 15) lies inside the triangle 
      formed by A(0, 0), B(20, 0) and C(10, 30) */
   if (isInside(0, 0, 20, 0, 10, 30, 10, 15))
     printf ("Inside");
   else
     printf ("Not Inside");

   return 0;
}

Time : O(1)

Space: O(1)



回答3:

Take the average of the three given points. This new point P4 will always lie inside the triangle.

Now check if P and P4 lie on the same side of each of the three lines P1P2 P2P3 and P3P1. You can do this by checking the signs of the cross products (P -> P1) x (P -> P2) and (P4 -> P1) x (P4 -> P2) (where P->P1 is the vector from P to P1), and then the other two pairs.



标签: math logic hint