Number of possible Triangles in a Cartesian coordinate system

Given n points in a Cartesian coordinate system. Count the number of triangles formed.

Examples:

Input  : point[] = {(0, 0), (1, 1), (2, 0), (2, 2)
Output : 3
Three triangles can be formed from above points.

A simple solution is to check if the determinant of the three points selected is non-zero or not. The following determinant gives the area of a Triangle (Also known as Cramer’s rule).

Area of the triangle with corners at (x1, y1), (x2, y2) and (x3, y3) is given by:

{\displaystyle Area = \pm \frac{1}{2}\begin{bmatrix} x1 & y1 & 1\\  x2 & y2 & 1\\  x3 & y3 & 1 \end{bmatrix}}

We can solve this by taking all possible combination of 3 points and finding the determinant.

C++

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// C++ program to count number of triangles that can
// be formed with given points in 2D
#include <bits/stdc++.h>
using namespace std;
  
// A point in 2D
struct Point
{
   int x, y;
};
  
// Returns determinant value of three points in 2D
int det(int x1, int y1, int x2, int y2, int x3, int y3)
{
   return x1*(y2 - y3) - y1*(x2 - x3) + 1*(x2*y3 - y2*x3);
}
  
// Returns count of possible triangles with given array
// of points in 2D.
int countPoints(Point arr[], int n)
{
    int result = 0;  // Initialize result
  
    // Consider all triplets of points given in inputs
    // Increment the result when determinant of a triplet
    // is not 0.
    for (int i=0; i<n; i++)
        for (int j=i+1; j<n; j++)
            for (int k=j+1; k<n; k++)
                if (det(arr[i].x, arr[i].y, arr[j].x,
                        arr[j].y, arr[k].x, arr[k].y))
                    result++;
  
    return result;
}
  
// Driver code
int main()
{
    Point arr[] = {{0, 0}, {1, 1}, {2, 0}, {2, 2}};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << countPoints(arr, n);
    return 0;
}

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Java

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// Java program to count number 
// of triangles that can be formed 
// with given points in 2D
  
class GFG{
// Returns determinant value 
// of three points in 2D
static int det(int x1, int y1, int x2, int y2, int x3, int y3)
{
    return (x1 * (y2 - y3) - y1 *
        (x2 - x3) + 1 * (x2 *
            y3 - y2 * x3));
}
  
// Returns count of possible 
// triangles with given array
// of points in 2D.
static int countPoints(int [][]Point, int n)
{
    int result = 0; // Initialize result
  
    // Consider all triplets of 
    // points given in inputs
    // Increment the result when 
    // determinant of a triplet is not 0.
    for(int i = 0; i < n; i++)
        for(int j = i + 1; j < n; j++)
            for(int k = j + 1; k < n; k++)
                if(det(Point[i][0], Point[i][1], 
                    Point[j][0], Point[j][1], 
                    Point[k][0], Point[k][1])>=0)
                    result = result + 1;
  
    return result;
}
  
// Driver code
public static void main(String[] args)
{
int Point[][] = {{0, 0},{1, 1},{2, 0},{2, 2}};
int n = Point.length;
System.out.println(countPoints(Point, n));
}
}
// This code is contributed by
// mits

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Python

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# Python program to count number 
# of triangles that can be formed 
# with given points in 2D
  
# Returns determinant value 
# of three points in 2D
def det(x1, y1, x2, y2, x3, y3):
    return (x1 * (y2 - y3) - y1 * 
            (x2 - x3) + 1 * (x2 *
             y3 - y2 * x3))
  
# Returns count of possible 
# triangles with given array
# of points in 2D.
def countPoints(Point, n):
  
    result = 0 # Initialize result
  
    # Consider all triplets of 
    # points given in inputs
    # Increment the result when 
    # determinant of a triplet is not 0.
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if(det(Point[i][0], Point[i][1], 
                       Point[j][0], Point[j][1], 
                       Point[k][0], Point[k][1])):
                    result = result + 1
  
    return result
  
# Driver code
Point = [[0, 0], [1, 1], 
         [2, 0], [2, 2]]
n = len(Point)
print(countPoints(Point, n))
  
# This code is contributed by
# Sanjit_Prasad

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C#

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// C# program to count number 
// of triangles that can be formed 
// with given points in 2D
using System;
  
class GFG{
// Returns determinant value 
// of three points in 2D
static int det(int x1, int y1, int x2, int y2, int x3, int y3)
{
    return (x1 * (y2 - y3) - y1 *
        (x2 - x3) + 1 * (x2 *
            y3 - y2 * x3));
}
  
// Returns count of possible 
// triangles with given array
// of points in 2D.
static int countPoints(int[,] Point, int n)
{
    int result = 0; // Initialize result
  
    // Consider all triplets of 
    // points given in inputs
    // Increment the result when 
    // determinant of a triplet is not 0.
    for(int i = 0; i < n; i++)
        for(int j = i + 1; j < n; j++)
            for(int k = j + 1; k < n; k++)
                if(det(Point[i,0], Point[i,1], Point[j,0], Point[j,1],Point[k,0], Point[k,1])>=0)
                    result = result + 1;
  
    return result;
}
  
// Driver code
public static void Main()
{
int[,] Point = new int[,] { { 0, 0 }, { 1, 1 }, { 2, 0 }, { 2, 2 } };
int n = Point.Length/Point.Rank;
Console.WriteLine(countPoints(Point, n));
}
}
// This code is contributed by mits

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PHP

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<?php
// PHP program to count number 
// of triangles that can be formed 
// with given points in 2D
  
// Returns determinant value 
// of three points in 2D
function det($x1, $y1, $x2,
             $y2, $x3, $y3)
{
    return ($x1 * ($y2 - $y3) - $y1 *
           ($x2 - $x3) + 1 * ($x2 *
            $y3 - $y2 * $x3));
}
  
// Returns count of possible 
// triangles with given array
// of points in 2D.
function countPoints($Point, $n)
{
    $result = 0; // Initialize result
  
    // Consider all triplets of 
    // points given in inputs
    // Increment the result when 
    // determinant of a triplet is not 0.
    for($i = 0; $i < $n; $i++)
        for($j = $i + 1; $j < $n; $j++)
            for($k = $j + 1; $k < $n; $k++)
                if(det($Point[$i][0], $Point[$i][1], 
                       $Point[$j][0], $Point[$j][1], 
                       $Point[$k][0], $Point[$k][1]))
                       $result = $result + 1;
  
    return $result;
}
  
// Driver code
$Point = array(array(0, 0),
               array(1, 1), 
               array(2, 0), 
               array(2, 2));
$n = count($Point);
echo countPoints($Point, $n);
  
// This code is contributed by
// mits
?>

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Output:

3

Time Complexity: O(n^{3}).

Optimization :
We can optimize the above solution to work in O(n2) using the fact that three points cannot form a triangle if they are collinear. We can use hashing to store slopes of all pairs and find all triangles in O(n2) time.

This article is contributed by Vrushank Upadhyay. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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