CODEWITHSUNDEEP
c++mathvectorpositioncoordinates
Erick.1993
Erick.1993

Reputation: 63

Collinear Points C++

I have a script that asks you to enter a number n and after that asks you to enter n number of (x,y) points.

At the end the script sort (x,y) points with respect to x

What I need is to detect collinear points and remove them so script will have only points at general position.

For example:

Input

10
8   16
2   16
5   9
9   16
15  18
3   17
8   10
3   12
10  17
5   17

Output

2 16
3 12
5 9
5 17
8 10
9 16
10 17
15 18

My Script is this:

#include <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;
int n,m;
class punto{
private:
    int x;
    int y;
public:
    punto();
    ~punto(){}
    void setx(int xn){ x = xn;}
    void sety(int yn) { y = yn; }
    void printcoord();

    friend bool compare_x(const punto& lhs, const punto& rhs);
    friend bool compare_y(const punto& lhs, const punto& rhs);
    };
punto::punto()
{
    x=0;
    y=0;
}
void punto:: printcoord()
{
    cout << x << " " << y << endl;
}
bool compare_x(const punto& lhs, const punto& rhs) {
    if (lhs.x == rhs.x)
        return lhs.y < rhs.y;
    return lhs.x < rhs.x;
}

int main()
{
    vector<punto> list;
            int x;
        int y;
        punto *p1;
    cin >>n;
    for(int contador=0; contador<n; contador++){
        if(contador<n){
                cin >> x;
                cin >> y;
                p1=new punto;
                p1->setx(x);
                p1->sety(y);
                list.push_back(*p1);
                cin.get();
            }
        }
    vector<punto>::iterator it;
     std::sort(list.begin(), list.end(), compare_x);
     for ( it = list.begin(); it != list.end(); ++it ) {
            it->printcoord();
     }
    return 0;
}

As you can see I'm missing a function to determine collinear points and remove them.

I would really appreciate any help!

Upvotes: 1

Views: 1668

Answers (1)

RollerSimmer
RollerSimmer

Reputation: 119

Use dot product. Let a, b, c be three points. If all three points lie on the same line, then the vector b-a and the vector c-a will have a dot product of |b-a||c-a|.

This is due to the fact that cos(angle(bac))=dot(b-a,c-a)/(mag(b-a)*mag(c-a)). If they are collinear, then the angle is zero. The cosine for zero is one, so dot(b-a,c-a)==mag(b-a)*mag(c-a) means they are collinear.

If your linear algebra skills are rusty or non-existent, then here's a refresher

dot(w,v) = w.x*v.x + w.y*v.y
mag(v)=sqrt(v.x*v.x + v.y*v.y)

The trick is to remove the costly square root function somehow. Squaring both sides, you have 

mag^2(v)=v.x*v.x+v.y*v.y 
cos^2(0)=1^2=1
dot^2(w,v)=(w.x*v.x + w.y*v.y)*(w.x*v.x + w.y*v.y)

just do a check 

if( (v.x*v.x + v.y*v.y)*(w.x*w.x + w.y*w.y)) == (w.x*v.x + w.y*v.y)*(w.x*v.x + w.y*v.y) )

Upvotes: 1

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