Read coordinates of 4 points. Do they make a squar

Question

I calculate the distance between the points and if the distances are equal the point make a square, else no. My code works only if I read the coordinates in the following order A(x, y), B(x, y), C(x, y), D(x, y) or reverse. But if I read like this for example A(x, y), B(x, y), D(x, y), C(x, y) it won't work because the dist method will calculate the square's diagonal length. How can I solve this problem?

#include <iostream>
using namespace std;

struct {
    int x;
    int y;
}a[10];

int dist(int x1, int y1, int x2, int y2)
{
    int c1, c2;
    c1 = x2-x1;
    c2 = y2-y1;
    return (c1*c1)+(c2*c2);
}

int main()
{
    int d1, d2, d3, d4;
    for (int i=1; i<=4; i++)
    {
        cout << 'X' << i << '='; cin >> a[i].x;
        cout << 'Y' << i << '='; cin >> a[i].y;
    }
    d1 = dist(a[1].x, a[1].y, a[2].x, a[2].y);
    d2 = dist(a[2].x, a[2].y, a[3].x, a[3].y);
    d3 = dist(a[3].x, a[3].y, a[4].x, a[4].y);
    d4 = dist(a[4].x, a[4].y, a[1].x, a[1].y);
    if(d1==d2 && d1==d3 && d1==d4)
        cout << "Is a square";
    else
        cout << "Is not a square";
    return 0;
}

Answer 1

Checking the distances is not enough, you'll need to check at least an angle, as the shape could be a rhombus.

Checking only angles is also not enough, because you could end up with a rectangle.

There are a total of 6 distances between the points. Calculate all of them. Out of those 6, four should be equal (call their length x). - this guarantees a rhombus

The other two should be equal between themselves (call their length y). this guarantees a rectangle

Put a rhombus and a rectangle together and BAM! - square.

Answer 2

Choose one vertex of the square (without loss of generality say A) and consider this the origin. Get 3 vectors formed from the origin to each other corner (AB, AC, AD). Vectorially these are given by B-A, C-A and D-A. Calculate the inner product of each vector with the other. If the vertices form a rectangle, one inner product will be zero (the perpendicular edge vectors). If they form a square then the other 2 inner products also must be equal to each other due to the common angle of 45 degrees between them. Thus if one inner product is zero and the other 2 are equal to eachother and the 4 distances are the same you have a square.

Answer 3

With four points, you have a total of six distances. If you compute the distance squared (which is easier, of course), then for any pair of edges with length L1, L2, the following is true

L1 == L2 || 2*L1 == L2 || L1 = 2*L2

You can create the following piece of code to implement this:

int isSquare (point *p1, point *p2, point *p3, point *p4) {
double dx, dy;
double dd[6];
point *pp[4];

pp[0]=p1; pp[1]=p2; pp[2]=p3; pp[3]=p4;
int ii, jj, kk, nn;
kk = 0;
// loop over all combinations of first and second point
// six in all
for(ii=0; ii<3; ii++) {
  for(jj=ii+1; jj<4; jj++) {
    dx = pp[ii]->x - pp[jj]->x;
    dy = pp[ii]->y - pp[jj]->y;
    dd[kk]= dx*dx + dy*dy;
    if (dd[kk]==0) return 0; // two identical points: not a square
    if(kk>1) {
      for(nn= 0; nn < kk-1; nn++) {
        // if both are "sides", we expect their length to be the same;
        // if one is a diagonal and the other a side, their ratio is 2
        // since we are working with the square of the number
        if (!(((2*dd[nn] == dd[kk] ) || \
               (dd[nn] == dd[kk])    || \
               (2*dd[kk] == dd[nn] )))) return 0;
      }
    }
     kk++;
  }
}
return 1; // got here: all combinations tested OK, we have a square

}

Answer 4

You can actually do it with only angles using inner products. Each vertex A should have two other vertices B and C such that AB and AC are at a right angle (0 inner product) as well as one vertex D such that AB and AD as well AC and AD are both at exactly 45 degrees (normalized dot product = acos(45 degrees), i.e. ~ 0.6675). If that is true for all four vertices you have a square.