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Number of quadrilaterals possible from the given points

  • Difficulty Level : Hard
  • Last Updated : 09 Jul, 2021

Given four points (x, y) in the cartesian coordinate. Find the possible no of quadrilaterals than can be formed by joining all the four points. 
Examples: 
 

Input: A=(0, 9), B=(-1, 0), C=(5, -1), D=(5, 9)
Output: Only one quadrilateral is possible (ABCD) in any orientation 

Input: A=(0, 9), B=(-1, 0), C=(5, -1), D=(0, 3)
Output: 3 quadrilaterals are possible (ABCD), (ADBC), (ABDC)

Figure for 2nd example: 
 

Case 2 description

Approach: 
 

  1. We need to check whether any of the given points are same. If yes, no of quadrilateral = 0
  2. Then we need to check whether any of the 3 points of the given 4 points are collinear or not. If yes, no of quadrilateral=0.Check Program to check if three points are collinear link to check collinearity of 3 points.
    1. if its a convex quadrilateral then there is only one possible quadrilateral.
    2. if its a concave quadrilateral then there are 3 possible quadrilaterals.

This can be determined by How to check if two given line segments intersect? of diagonals. 
In case of a convex quadrilateral, the diagonals will intersect whereas in case of a concave quadrilateral the diagonals won’t intersect. 
since we don’t know the orientation of the points, we can’t specifically determine the diagonals so all the distinct line segments(no common points in the two line segments) of the quadrilateral and determine whether they intersect or not.
Refer to the figure to understand how to determine the type of quadrilateral : 
 



Intersection of line segments

Convex quadrilateral:
 

line AB and line DC do not intersect 
line AD and line BC do not intersect 
line AC and line BD intersect 
so total no of intersection= 1

Concave quadrilateral 
 

line AB and line DC do not intersect 
line AD and line BC do not intersect 
line AC and line BD do not intersect 
so total no of intersection= 0 

if no of intersection = 1, its a convex quadrilateral so no of possibility= 1 
if no of intersection = 0, its a concave quadrilateral so no of possibilities = 3 
 

C++




// C++ implementation of above approach
#include <iostream>
using namespace std;
 
struct Point // points
{
    int x;
    int y;
};
 
// determines the orientation of points
int orientation(Point p, Point q, Point r)
{
    int val = (q.y - p.y) * (r.x - q.x) - (q.x - p.x) * (r.y - q.y);
 
    if (val == 0)
        return 0;
    return (val > 0) ? 1 : 2;
}
 
// check whether the distinct line segments intersect
bool doIntersect(Point p1, Point q1, Point p2, Point q2)
{
    int o1 = orientation(p1, q1, p2);
    int o2 = orientation(p1, q1, q2);
    int o3 = orientation(p2, q2, p1);
    int o4 = orientation(p2, q2, q1);
 
    if (o1 != o2 && o3 != o4)
        return true;
 
    return false;
}
 
// check if points overlap(similar)
bool similar(Point p1, Point p2)
{
 
    // it is same, we are returning false because
    // quadrilateral is not possible in this case
    if (p1.x == p2.x && p1.y == p2.y)
        return false;
 
    // it is not same, So there is a
    // possibility of a quadrilateral
    return true;
}
 
// check for collinearity
bool collinear(Point p1, Point p2, Point p3)
{
    int x1 = p1.x, y1 = p1.y;
    int x2 = p2.x, y2 = p2.y;
    int x3 = p3.x, y3 = p3.y;
 
    // it is collinear, we are returning false
    // because quadrilateral is not possible in this case
    if ((y3 - y2) * (x2 - x1) == (y2 - y1) * (x3 - x2))
        return false;
 
    // it is not collinear, So there
    // is a possibility of a quadrilateral
    else
        return true;
}
 
int no_of_quads(Point p1, Point p2, Point p3, Point p4)
{
    // ** Checking for cases where no quadrilateral = 0 **
 
    // check if any of the points are same
    bool same = true;
    same = same & similar(p1, p2);
    same = same & similar(p1, p3);
    same = same & similar(p1, p4);
    same = same & similar(p2, p3);
    same = same & similar(p2, p4);
    same = same & similar(p3, p4);
 
    // similar points exist
    if (same == false)
        return 0;
 
    // check for collinearity
    bool coll = true;
    coll = coll & collinear(p1, p2, p3);
    coll = coll & collinear(p1, p2, p4);
    coll = coll & collinear(p1, p3, p4);
    coll = coll & collinear(p2, p3, p4);
 
    // points are collinear
    if (coll == false)
        return 0;
 
    //** Checking for cases where no of quadrilaterals= 1 or 3 **
 
    int check = 0;
 
    if (doIntersect(p1, p2, p3, p4))
        check = 1;
    if (doIntersect(p1, p3, p2, p4))
        check = 1;
    if (doIntersect(p1, p2, p4, p3))
        check = 1;
 
    if (check == 0)
        return 3;
    return 1;
}
 
// Driver code
int main()
{
    struct Point p1, p2, p3, p4;
    // A =(0, 9), B = (-1, 0), C = (5, -1), D=(5, 9)
    p1.x = 0, p1.y = 9;
    p2.x = -1, p2.y = 0;
    p3.x = 5, p3.y = -1;
    p4.x = 5, p4.y = 9;
    cout << no_of_quads(p1, p2, p3, p4) << endl;
 
    // A=(0, 9), B=(-1, 0), C=(5, -1), D=(0, 3)
    p1.x = 0, p1.y = 9;
    p2.x = -1, p2.y = 0;
    p3.x = 5, p3.y = -1;
    p4.x = 0, p4.y = 3;
    cout << no_of_quads(p1, p2, p3, p4) << endl;
 
    // A=(0, 9), B=(0, 10), C=(0, 11), D=(0, 12)
    p1.x = 0, p1.y = 9;
    p2.x = 0, p2.y = 10;
    p3.x = 0, p3.y = 11;
    p4.x = 0, p4.y = 12;
    cout << no_of_quads(p1, p2, p3, p4) << endl;
 
    // A=(0, 9), B=(0, 9), C=(5, -1), D=(0, 3)
    p1.x = 0, p1.y = 9;
    p2.x = 0, p2.y = 9;
    p3.x = 5, p3.y = -1;
    p4.x = 0, p4.y = 3;
    cout << no_of_quads(p1, p2, p3, p4) << endl;
 
    return 0;
}

Java




// Java implementation of above approach
class GFG
{
static class Point // points
{
    int x;
    int y;
}
 
// determines the orientation of points
static int orientation(Point p, Point q,
                                Point r)
{
    int val = (q.y - p.y) * (r.x - q.x) -
              (q.x - p.x) * (r.y - q.y);
 
    if (val == 0)
        return 0;
    return (val > 0) ? 1 : 2;
}
 
// check whether the distinct
// line segments intersect
static boolean doIntersect(Point p1, Point q1,
                           Point p2, Point q2)
{
    int o1 = orientation(p1, q1, p2);
    int o2 = orientation(p1, q1, q2);
    int o3 = orientation(p2, q2, p1);
    int o4 = orientation(p2, q2, q1);
 
    if (o1 != o2 && o3 != o4)
        return true;
 
    return false;
}
 
// check if points overlap(similar)
static boolean similar(Point p1, Point p2)
{
 
    // it is same, we are returning
    // false because quadrilateral is
    // not possible in this case
    if (p1.x == p2.x && p1.y == p2.y)
        return false;
 
    // it is not same, So there is a
    // possibility of a quadrilateral
    return true;
}
 
// check for collinearity
static boolean collinear(Point p1, Point p2,
                         Point p3)
{
    int x1 = p1.x, y1 = p1.y;
    int x2 = p2.x, y2 = p2.y;
    int x3 = p3.x, y3 = p3.y;
 
    // it is collinear, we are returning
    // false because quadrilateral is not
    // possible in this case
    if ((y3 - y2) *
        (x2 - x1) == (y2 - y1) *
                     (x3 - x2))
        return false;
 
    // it is not collinear, So there
    // is a possibility of a quadrilateral
    else
        return true;
}
 
static int no_of_quads(Point p1, Point p2,
                       Point p3, Point p4)
{
    // Checking for cases where
    // no quadrilateral = 0
 
    // check if any of the
    // points are same
    boolean same = true;
    same = same & similar(p1, p2);
    same = same & similar(p1, p3);
    same = same & similar(p1, p4);
    same = same & similar(p2, p3);
    same = same & similar(p2, p4);
    same = same & similar(p3, p4);
 
    // similar points exist
    if (same == false)
        return 0;
 
    // check for collinearity
    boolean coll = true;
    coll = coll & collinear(p1, p2, p3);
    coll = coll & collinear(p1, p2, p4);
    coll = coll & collinear(p1, p3, p4);
    coll = coll & collinear(p2, p3, p4);
 
    // points are collinear
    if (coll == false)
        return 0;
 
    // Checking for cases where
    // no of quadrilaterals= 1 or 3
 
    int check = 0;
 
    if (doIntersect(p1, p2, p3, p4))
        check = 1;
    if (doIntersect(p1, p3, p2, p4))
        check = 1;
    if (doIntersect(p1, p2, p4, p3))
        check = 1;
 
    if (check == 0)
        return 3;
    return 1;
}
 
// Driver code
public static void main(String args[])
{
    Point p1, p2, p3, p4;
    p1 = new Point();
    p2 = new Point();
    p3 = new Point();
    p4 = new Point();
     
    // A =(0, 9), B = (-1, 0),
    // C = (5, -1), D=(5, 9)
    p1.x = 0; p1.y = 9;
    p2.x = -1; p2.y = 0;
    p3.x = 5; p3.y = -1;
    p4.x = 5; p4.y = 9;
    System.out.println(no_of_quads(p1, p2, p3, p4));
 
    // A=(0, 9), B=(-1, 0),
    // C=(5, -1), D=(0, 3)
    p1.x = 0; p1.y = 9;
    p2.x = -1; p2.y = 0;
    p3.x = 5; p3.y = -1;
    p4.x = 0; p4.y = 3;
    System.out.println(no_of_quads(p1, p2, p3, p4));
 
    // A=(0, 9), B=(0, 10),
    // C=(0, 11), D=(0, 12)
    p1.x = 0; p1.y = 9;
    p2.x = 0; p2.y = 10;
    p3.x = 0; p3.y = 11;
    p4.x = 0; p4.y = 12;
    System.out.println(no_of_quads(p1, p2, p3, p4));
 
    // A=(0, 9), B=(0, 9),
    // C=(5, -1), D=(0, 3)
    p1.x = 0; p1.y = 9;
    p2.x = 0; p2.y = 9;
    p3.x = 5; p3.y = -1;
    p4.x = 0; p4.y = 3;
    System.out.println(no_of_quads(p1, p2, p3, p4));
}
}
 
// This code is contributed
// by Arnab Kundu

C#




// C# implementation of above approach
using System;
class GFG
{
public class Point // points
{
    public int x;
    public int y;
}
 
// determines the orientation of points
static int orientation(Point p, Point q,
                                Point r)
{
    int val = (q.y - p.y) * (r.x - q.x) -
            (q.x - p.x) * (r.y - q.y);
 
    if (val == 0)
        return 0;
    return (val > 0) ? 1 : 2;
}
 
// check whether the distinct
// line segments intersect
static bool doIntersect(Point p1, Point q1,
                        Point p2, Point q2)
{
    int o1 = orientation(p1, q1, p2);
    int o2 = orientation(p1, q1, q2);
    int o3 = orientation(p2, q2, p1);
    int o4 = orientation(p2, q2, q1);
 
    if (o1 != o2 && o3 != o4)
        return true;
 
    return false;
}
 
// check if points overlap(similar)
static bool similar(Point p1, Point p2)
{
 
    // it is same, we are returning
    // false because quadrilateral is
    // not possible in this case
    if (p1.x == p2.x && p1.y == p2.y)
        return false;
 
    // it is not same, So there is a
    // possibility of a quadrilateral
    return true;
}
 
// check for collinearity
static bool collinear(Point p1, Point p2,
                        Point p3)
{
    int x1 = p1.x, y1 = p1.y;
    int x2 = p2.x, y2 = p2.y;
    int x3 = p3.x, y3 = p3.y;
 
    // it is collinear, we are returning
    // false because quadrilateral is not
    // possible in this case
    if ((y3 - y2) *
        (x2 - x1) == (y2 - y1) *
                    (x3 - x2))
        return false;
 
    // it is not collinear, So there
    // is a possibility of a quadrilateral
    else
        return true;
}
 
static int no_of_quads(Point p1, Point p2,
                    Point p3, Point p4)
{
    // Checking for cases where
    // no quadrilateral = 0
 
    // check if any of the
    // points are same
    bool same = true;
    same = same & similar(p1, p2);
    same = same & similar(p1, p3);
    same = same & similar(p1, p4);
    same = same & similar(p2, p3);
    same = same & similar(p2, p4);
    same = same & similar(p3, p4);
 
    // similar points exist
    if (same == false)
        return 0;
 
    // check for collinearity
    bool coll = true;
    coll = coll & collinear(p1, p2, p3);
    coll = coll & collinear(p1, p2, p4);
    coll = coll & collinear(p1, p3, p4);
    coll = coll & collinear(p2, p3, p4);
 
    // points are collinear
    if (coll == false)
        return 0;
 
    // Checking for cases where
    // no of quadrilaterals= 1 or 3
 
    int check = 0;
 
    if (doIntersect(p1, p2, p3, p4))
        check = 1;
    if (doIntersect(p1, p3, p2, p4))
        check = 1;
    if (doIntersect(p1, p2, p4, p3))
        check = 1;
 
    if (check == 0)
        return 3;
    return 1;
}
 
// Driver code
static void Main()
{
    Point p1, p2, p3, p4;
    p1 = new Point();
    p2 = new Point();
    p3 = new Point();
    p4 = new Point();
     
    // A =(0, 9), B = (-1, 0),
    // C = (5, -1), D=(5, 9)
    p1.x = 0; p1.y = 9;
    p2.x = -1; p2.y = 0;
    p3.x = 5; p3.y = -1;
    p4.x = 5; p4.y = 9;
    Console.WriteLine(no_of_quads(p1, p2, p3, p4));
 
    // A=(0, 9), B=(-1, 0),
    // C=(5, -1), D=(0, 3)
    p1.x = 0; p1.y = 9;
    p2.x = -1; p2.y = 0;
    p3.x = 5; p3.y = -1;
    p4.x = 0; p4.y = 3;
    Console.WriteLine(no_of_quads(p1, p2, p3, p4));
 
    // A=(0, 9), B=(0, 10),
    // C=(0, 11), D=(0, 12)
    p1.x = 0; p1.y = 9;
    p2.x = 0; p2.y = 10;
    p3.x = 0; p3.y = 11;
    p4.x = 0; p4.y = 12;
    Console.WriteLine(no_of_quads(p1, p2, p3, p4));
 
    // A=(0, 9), B=(0, 9),
    // C=(5, -1), D=(0, 3)
    p1.x = 0; p1.y = 9;
    p2.x = 0; p2.y = 9;
    p3.x = 5; p3.y = -1;
    p4.x = 0; p4.y = 3;
    Console.WriteLine(no_of_quads(p1, p2, p3, p4));
}
}
 
// This code is contributed by mits

Javascript




<script>
 
// JavaScript implementation of above approach
 
class Point // points
{
    constructor()
    {
        this.x=0;
        this.y=0;
    }
}
 
// determines the orientation of points
function orientation(p,q,r)
{
    let val = (q.y - p.y) * (r.x - q.x) -
              (q.x - p.x) * (r.y - q.y);
   
    if (val == 0)
        return 0;
    return (val > 0) ? 1 : 2;
}
 
// check whether the distinct
// line segments intersect
function doIntersect(p1,q1,p2,q2)
{
    let o1 = orientation(p1, q1, p2);
    let o2 = orientation(p1, q1, q2);
    let o3 = orientation(p2, q2, p1);
    let o4 = orientation(p2, q2, q1);
   
    if (o1 != o2 && o3 != o4)
        return true;
   
    return false;
}
 
// check if points overlap(similar)
function similar(p1,p2)
{
    // it is same, we are returning
    // false because quadrilateral is
    // not possible in this case
    if (p1.x == p2.x && p1.y == p2.y)
        return false;
   
    // it is not same, So there is a
    // possibility of a quadrilateral
    return true;
}
 
// check for collinearity
function collinear(p1,p2,p3)
{
    let x1 = p1.x, y1 = p1.y;
    let x2 = p2.x, y2 = p2.y;
    let x3 = p3.x, y3 = p3.y;
   
    // it is collinear, we are returning
    // false because quadrilateral is not
    // possible in this case
    if ((y3 - y2) *
        (x2 - x1) == (y2 - y1) *
                     (x3 - x2))
        return false;
   
    // it is not collinear, So there
    // is a possibility of a quadrilateral
    else
        return true;
}
 
function no_of_quads(p1,p2,p3,p4)
{
    // Checking for cases where
    // no quadrilateral = 0
   
    // check if any of the
    // points are same
    let same = true;
    same = same & similar(p1, p2);
    same = same & similar(p1, p3);
    same = same & similar(p1, p4);
    same = same & similar(p2, p3);
    same = same & similar(p2, p4);
    same = same & similar(p3, p4);
   
    // similar points exist
    if (same == false)
        return 0;
   
    // check for collinearity
    let coll = true;
    coll = coll & collinear(p1, p2, p3);
    coll = coll & collinear(p1, p2, p4);
    coll = coll & collinear(p1, p3, p4);
    coll = coll & collinear(p2, p3, p4);
   
    // points are collinear
    if (coll == false)
        return 0;
   
    // Checking for cases where
    // no of quadrilaterals= 1 or 3
   
    let check = 0;
   
    if (doIntersect(p1, p2, p3, p4))
        check = 1;
    if (doIntersect(p1, p3, p2, p4))
        check = 1;
    if (doIntersect(p1, p2, p4, p3))
        check = 1;
   
    if (check == 0)
        return 3;
    return 1;
}
 
// Driver code
let p1, p2, p3, p4;
p1 = new Point();
p2 = new Point();
p3 = new Point();
p4 = new Point();
 
// A =(0, 9), B = (-1, 0),
// C = (5, -1), D=(5, 9)
p1.x = 0; p1.y = 9;
p2.x = -1; p2.y = 0;
p3.x = 5; p3.y = -1;
p4.x = 5; p4.y = 9;
document.write(no_of_quads(p1, p2, p3, p4)+"<br>");
 
// A=(0, 9), B=(-1, 0),
// C=(5, -1), D=(0, 3)
p1.x = 0; p1.y = 9;
p2.x = -1; p2.y = 0;
p3.x = 5; p3.y = -1;
p4.x = 0; p4.y = 3;
document.write(no_of_quads(p1, p2, p3, p4)+"<br>");
 
// A=(0, 9), B=(0, 10),
// C=(0, 11), D=(0, 12)
p1.x = 0; p1.y = 9;
p2.x = 0; p2.y = 10;
p3.x = 0; p3.y = 11;
p4.x = 0; p4.y = 12;
document.write(no_of_quads(p1, p2, p3, p4)+"<br>");
 
// A=(0, 9), B=(0, 9),
// C=(5, -1), D=(0, 3)
p1.x = 0; p1.y = 9;
p2.x = 0; p2.y = 9;
p3.x = 5; p3.y = -1;
p4.x = 0; p4.y = 3;
document.write(no_of_quads(p1, p2, p3, p4)+"<br>");
 
 
// This code is contributed by avanitrachhadiya2155
 
</script>
Output: 
1
3
0
0

 




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