c++ : check if two triangles intersect or not

Question

Problem :

Given the vertices of two triangles, check whether both of them have any common interior point. No points on the edges or vertices are considered interior to a triangle.
I think this is a problem to find whether two triangles intersect or not.

My idea is :

1. Make three line segments for each of the two triangles
2. For each pair of line segment (A,B) (where A is in triangle #1 and B is in Triangle #2) check whether they intersect or not.
3. If any pair of segments intersect, then the two triangles have a interior point. Otherwise not .

My code :

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>

#define READ freopen("in.txt","r",stdin)
#define ui64 unsigned long long int
#define i64 long long int

using namespace std;

class Point{
public:
    i64 x, y;
    Point() : x(0), y(0) {}
    Point(int a, i64 b) : x(a), y(b) {}
};

class Segment{
public:
    Point a, b;
    Segment() : a(0,0), b(0,0) {}
    Segment(Point e, Point r) : a(e), b(r) {}
    Segment(const Segment &s){
        a = s.a;
        b = s.b;
    }
    i64 Direction(const Point &p){
        return (p.x - a.x) * (b.y - a.y) - (b.x - a.x) * (p.y - a.y);
    }
    inline bool inside(int a,int b,int c){
        return a<=c && c<=b;
    }
    bool onSegment(const Point &p){
        return inside(min(a.x,b.x), max(a.x,b.x), p.x) && inside(min(a.y,b.y), max(a.y,b.y), p.y);
    }
    bool Intersect(Segment &s){
        int d1 = this->Direction(s.a);
        int d2 = this->Direction(s.b);

        int d3 = s.Direction(a);
        int d4 = s.Direction(b);

        if(((d1>0 && d2<0) || (d1<0 && d2>0)) && ((d3>0 && d4<0) || (d3<0 && d4>0))) return true;

        //check if the two segments just touch each other but don't cross  
        //i ignored this because as the problem said ...  
        //No points on the edges or vertices are considered interior to a triangle.
        /*if(!d1 && this->onSegment(s.a)) return true;
        if(!d2 && this->onSegment(s.b)) return true;
        if(!d3 && s.onSegment(a)) return true;
        if(!d4 && s.onSegment(b)) return true;*/
        return false;
    }
};

Point p1[3], p2[3];
Segment s1[3], s2[3];

bool check()
{
    for(int i=0;i<3;i++)
        for(int j=0;j<3;j++)
            if(s1[i].Intersect(s2[j]))
                return true;
    return false;
}

int main()
{
    //READ;

    int t, ts=0;

    scanf("%d",&t);//number of test cases
    while(t--){
        //points for first triangle
        for(int i=0;i<3;i++){
            scanf("%lld %lld",&p1[i].x, &p1[i].y);
        }
        //points for second triangle
        for(int i=0;i<3;i++){
            scanf("%lld %lld",&p2[i].x, &p2[i].y);
        }
        for(int i=0;i<3;i++){
            s1[i] = Segment(p1[i], p1[(i+1)%3]);
            s2[i] = Segment(p2[i], p2[(i+1)%3]);
        }
        printf("pair %d: %s\n",++ts, check() ? "yes" : "no");
    }

    return 0;
}  

However for this input ..

1

0 0 5 0 2 4
4 0 5 0 -4 16

My program gives output

no

but the correct answer is

yes

and there are many cases for which my program doesn't work.
I checked my Segment class for other program and it worked fine.
But in this , i can't find the mistake.
Please help.

Thank you.

Answer 1

I think there is a bug in the line

if(((d1>0 && d2<0) || (d1<0 && d2>0)) && ((d3>0 && d4<0) || (d3<0 && d4>0))) return true;

If I understand it correctly, the Direction function tells you whether a point is to the left or right of the vector representing your line segment. The statement above assumes that both your other segment's points are either to the left or right, but doesn't account for the case when one point of e.g. segment B lies on e.g. segment A but the other doesn't.

So I would change it so that, for example, instead of strictly greater than (>), you check for greater than or equal to (>=).

This may explain the problem with your example. In your example, it just happens that the point (4,0) lies on the segment (0,0),(5,0) and the point (2,4) lies on the segment (-4,16),(4,0).

Answer 2

If the number of rectangles is more than two, you should consider solving this by line sweep. If it is just two rectangles, you can do a couple of checks that give you the answer if they have common interior point.

The observation is that either the segments of the triangles intersect or if one rectangle is entirely contained in the other one. We break it down to a couple of checks if lines intersect or a rectangle contains a point.

The first condition can be checked by pair-wise comparison of the appropriate segments if they intersect. The second condition can be checked by testing if an endpoint of any segment is contained in the other rectangle. Not a beautiful solution but easily codeable.