Algorithm - Geometry


http://www.geeksforgeeks.org/check-four-segments-form-rectangle/


We can solve this problem by using properties of a rectangle. First, we check total unique end points of segments, if count of these points is not equal to 4 then the line segment can’t make a rectangle. Then we check distances between all pair of points, there should be at most 3 different distances, one for diagonal and two for sides and at the end we will check the relation among these three distances, for line segments to make a rectangle these distance should satisfy Pythagorean relation because sides and diagonal of rectangle makes a right angle triangle. If they satisfy mentioned conditions then we will flag polygon made by line segment as rectangle otherwise not.
struct Segment
{
    int ax, ay;
    int bx, by;
 
    Segment(int ax, int ay, int bx, int by) :
              ax(ax), ay(ay), bx(bx), by(by)
    {}
};
 
// Utility method to return square of distance
// between two points
int getDis(pair<int, int> a, pair<int, int> b)
{
    return (a.first - b.first)*(a.first - b.first) +
           (a.second - b.second)*(a.second - b.second);
}
 
// method returns true if line Segments make
// a rectangle
bool isPossibleRectangle(Segment segments[])
{
    set< pair<int, int> > st;
 
    // putiing all end points in a set to
    // count total unique points
    for (int i = 0; i < N; i++)
    {
        st.insert(make_pair(segments[i].ax, segments[i].ay));
        st.insert(make_pair(segments[i].bx, segments[i].by));
    }
 
    // If total unique points are not 4, then
    // they can't make a rectangle
    if (st.size() != 4)
        return false;
 
    //  dist will store unique 'square of distances'
    set<int> dist;
 
    //  calculating distance between all pair of
    // end points of line segments
    for (auto it1=st.begin(); it1!=st.end(); it1++)
        for (auto it2=st.begin(); it2!=st.end(); it2++)
            if (*it1 != *it2)
                dist.insert(getDis(*it1, *it2));
 
    // if total unique distance are more than 3,
    // then line segment can't make a rectangle
    if (dist.size() > 3)
        return false;
 
    // copying distance into array. Note that set maintains
    // sorted order.
    int distance[3];
    int i = 0;
    for (auto it = dist.begin(); it != dist.end(); it++)
        distance[i++] = *it;
 
    // If line seqments form a square
    if (dist.size() == 2)
      return (2*distance[0] == distance[1]);
 
    //  distance of sides should satisfy pythagorean
    // theorem
    return (distance[0] + distance[1] == distance[2]);
}

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