Check whether four points make a parallelogram

Given four points in a 2-dimensional space we need to find out whether they make a parallelogram or not.  

A parallelogram has four sides. Two opposite sides are parallel and are of same lengths.

Examples:

Points = [(0, 0),  (4, 0),  (1, 3),  (5, 3)]
Above points make a parallelogram.

Points = [(0, 0), (2, 0), (4, 0), (2, 2)]
Above points does not make a parallelogram 
as first three points itself are linear.

Problems for checking square and rectangle can be read from Square checking and Rectangle checking but in this problem, we need to check for the parallelogram. The main properties of the parallelogram are that opposite sides of parallelogram are parallel and of equal length and diagonals of parallelogram bisect each other. We use the second property to solve this problem. As there are four points, we can get total 6 midpoints by considering each pair. Now for four points to make a parallelogram, 2 of the midpoints should be equal and rest of them should be different. In below code, we have created a map, which stores pairs corresponding to each midpoint. After calculating all midpoints, we have iterated over the map and check the occurrence of each midpoint, If exactly one midpoint occurred twice and other have occurred once, then given four points make a parallelogram otherwise not. 

C++

// C++ code to test whether four points make a
// parallelogram or not
#include <bits/stdc++.h>
using namespace std;

// structure to represent a point
struct point {
    double x, y;
    point() { }
    point(double x, double y)
        : x(x), y(y) { }

    // defining operator < to compare two points
    bool operator<(const point& other) const
    {
        if (x < other.x) {
            return true;
        } else if (x == other.x) {
            if (y < other.y) {
                return true;
            }
        }
        return false;
    }
};

// Utility method to return mid point of two points
point getMidPoint(point points[], int i, int j)
{
    return point((points[i].x + points[j].x) / 2.0,
                (points[i].y + points[j].y) / 2.0);
}

// method returns true if point of points array form
// a parallelogram
bool isParallelogram(point points[])
{
    map<point, vector<point> > midPointMap;

    // looping over all pairs of point to store their
    // mid points
    int P = 4;
    for (int i = 0; i < P; i++) {
        for (int j = i + 1; j < P; j++) {
            point temp = getMidPoint(points, i, j);

            // storing point pair, corresponding to
            // the mid point
            midPointMap[temp].push_back(point(i, j));
        }
    }

    int two = 0, one = 0;

    // looping over (midpoint, (corresponding pairs))
    // map to check the occurrence of each midpoint
    for (auto x : midPointMap) {

        // updating midpoint count which occurs twice
        if (x.second.size() == 2)
            two++;

        // updating midpoing count which occurs once
        else if (x.second.size() == 1)
            one++;

        // if midpoint count is more than 2, then
        // parallelogram is not possible
        else
            return false;
    }

    // for parallelogram, one mid point should come
    // twice and other mid points should come once
    if (two == 1 && one == 4)
        return true;

    return false;
}

// Driver code to test above methods
int main()
{
    point points[4];

    points[0] = point(0, 0);
    points[1] = point(4, 0);
    points[2] = point(1, 3);
    points[3] = point(5, 3);

    if (isParallelogram(points))
        cout << "Given points form a parallelogram";
    else
        cout << "Given points does not form a "
                "parallelogram";
    return 0;
}

Java

import java.util.*;

public class Main {
    // structure to represent a point
    static class Point {
        double x, y;
        Point() { }
        Point(double x, double y) {
            this.x = x;
            this.y = y;
        }

        // defining equals and hashCode method to compare two points
        @Override
        public boolean equals(Object obj) {
            if (obj == this) return true;
            if (!(obj instanceof Point)) return false;
            Point other = (Point) obj;
            return Double.compare(x, other.x) == 0
                    && Double.compare(y, other.y) == 0;
        }

        @Override
        public int hashCode() {
            return Objects.hash(x, y);
        }
    }

    // Utility method to return mid point of two points
    static Point getMidPoint(Point[] points, int i, int j) {
        return new Point((points[i].x + points[j].x) / 2.0,
                         (points[i].y + points[j].y) / 2.0);
    }

    // method returns true if point of points array form
    // a parallelogram
    static boolean isParallelogram(Point[] points) {
        Map<Point, List<Point>> midPointMap = new HashMap<>();

        // looping over all pairs of point to store their
        // mid points
        int P = 4;
        for (int i = 0; i < P; i++) {
            for (int j = i + 1; j < P; j++) {
                Point temp = getMidPoint(points, i, j);

                // storing point pair, corresponding to
                // the mid point
                if (!midPointMap.containsKey(temp)) {
                    midPointMap.put(temp, new ArrayList<>());
                }
                midPointMap.get(temp).add(new Point(i, j));
            }
        }

        int two = 0, one = 0;

        // looping over (midpoint, (corresponding pairs))
        // map to check the occurrence of each midpoint
        for (List<Point> pointsList : midPointMap.values()) {
            int size = pointsList.size();
            // updating midpoint count which occurs twice
            if (size == 2) {
                two++;
            }
            // updating midpoing count which occurs once
            else if (size == 1) {
                one++;
            }
            // if midpoint count is more than 2, then
            // parallelogram is not possible
            else {
                return false;
            }
        }

        // for parallelogram, one mid point should come
        // twice and other mid points should come once
        if (two == 1 && one == 4) {
            return true;
        }
        return false;
    }

    // Driver code to test above methods
    public static void main(String[] args) {
        Point[] points = new Point[4];

        points[0] = new Point(0, 0);
        points[1] = new Point(4, 0);
        points[2] = new Point(1, 3);
        points[3] = new Point(5, 3);

        if (isParallelogram(points)) {
            System.out.println("Given points form a parallelogram");
        } else {
            System.out.println("Given points do not form a parallelogram");
        }
    }
}

// This code is contributed by Prince Kumar

Python3

# Python program for the above approach

from typing import List
from collections import defaultdict
import math

# structure to represent a point
class Point:
    def __init__(self, x=0, y=0):
        self.x = x
        self.y = y

    # defining equals and hashCode method to compare two points
    def __eq__(self, other):
        return (self.x == other.x) and (self.y == other.y)

    def __hash__(self):
        return hash((self.x, self.y))

# Utility method to return mid point of two points
def get_mid_point(points: List[Point], i: int, j: int) -> Point:
    return Point((points[i].x + points[j].x) / 2.0, (points[i].y + points[j].y) / 2.0)

# method returns true if point of points array form a parallelogram
def is_parallelogram(points: List[Point]) -> bool:
    mid_point_map = defaultdict(list)

    # looping over all pairs of point to store their mid points
    P = 4
    for i in range(P):
        for j in range(i + 1, P):
            temp = get_mid_point(points, i, j)

            # storing point pair, corresponding to the mid point
            mid_point_map[temp].append((i, j))

    two = 0
    one = 0

    # looping over (midpoint, (corresponding pairs)) map to check the occurrence of each midpoint
    for points_list in mid_point_map.values():
        size = len(points_list)
        # updating midpoint count which occurs twice
        if size == 2:
            two += 1
        # updating midpoing count which occurs once
        elif size == 1:
            one += 1
        # if midpoint count is more than 2, then parallelogram is not possible
        else:
            return False

    # for parallelogram, one mid point should come twice and other mid points should come once
    if two == 1 and one == 4:
        return True
    return False

# Driver code to test above methods
if __name__ == "__main__":
    points = [Point(0, 0), Point(4, 0), Point(1, 3), Point(5, 3)]

    if is_parallelogram(points):
        print("Given points form a parallelogram")
    else:
        print("Given points do not form a parallelogram")

# This code is contributed by rishabmaldfijo

C#

// C# code to test whether four points make a
// parallelogram or not
using System;
using System.Collections.Generic;

// structure to represent a point
public struct Point
{
    public double x, y;
    public Point(double x, double y)
    {
        this.x = x;
        this.y = y;
    }

    // defining operator < to compare two points
    public static bool operator < (Point p1, Point p2)
    {
        if (p1.x < p2.x)
            return true;
        else if (p1.x == p2.x) {
            if (p1.y < p2.y)
                return true;
        }
        return false;
    }

    // defining operator > to compare two points
    public static bool operator > (Point p1, Point p2)
    {
        if (p1.x > p2.x)
            return true;
        else if (p1.x == p2.x) {
            if (p1.y > p2.y)
                return true;
        }
        return false;
    }
}

public class GFG {
    // Utility method to return mid point of two points
    static Point getMidPoint(Point[] points, int i, int j)
    {
        return new Point((points[i].x + points[j].x) / 2.0,
                         (points[i].y + points[j].y) / 2.0);
    }

    // method returns true if point of points array form
    // a parallelogram
    static bool isParallelogram(Point[] points)
    {
        Dictionary<Point, List<Point> > midPointMap
            = new Dictionary<Point, List<Point> >();

        // looping over all pairs of point to store their
        // mid points
        int P = 4;
        for (int i = 0; i < P; i++) {
            for (int j = i + 1; j < P; j++) {
                Point temp = getMidPoint(points, i, j);

                // storing point pair, corresponding to
                // the mid point
                if (midPointMap.ContainsKey(temp)) {
                    midPointMap[temp].Add(new Point(i, j));
                }
                else {
                    midPointMap[temp] = new List<Point>();
                    midPointMap[temp].Add(new Point(i, j));
                }
            }
        }

        int two = 0, one = 0;

        // looping over (midpoint, (corresponding pairs))
        // map to check the occurrence of each midpoint
        foreach(var x in midPointMap)
        {

            // updating midpoint count which occurs twice
            if (x.Value.Count == 2)
                two++;

            // updating midpoing count which occurs once
            else if (x.Value.Count == 1)
                one++;

            // if midpoint count is more than 2, then
            // parallelogram is not possible
            else
                return false;
        }

        // for parallelogram, one mid point should come
        // twice and other mid points should come once
        if (two == 1 && one == 4)
            return true;

        return false;
    }

    // Driver code to test above methods
    static public void Main(string[] args)
    {
        Point[] points = new Point[4];

        points[0] = new Point(0, 0);
        points[1] = new Point(4, 0);
        points[2] = new Point(1, 3);
        points[3] = new Point(5, 3);

        if (isParallelogram(points)) {
            Console.WriteLine(
                "Given points form a parallelogram");
        }
        else {
            Console.WriteLine(
                "Given points do not form a parallelogram");
        }
    }
}
// This code is contributed by prasad264

Javascript

// JavaScript code to test whether four points make a
// parallelogram or not

// structure to represent a point
class Point {
    constructor(x, y) {
        this.x = x;
        this.y = y;
    }
    // defining operator < to compare two points
    static compare(a, b) {
        if (a.x < b.x) {
            return -1;
        } else if (a.x == b.x) {
            if (a.y < b.y) {
                return -1;
            }
        }
        return 1;
    }
}

// Utility method to return mid point of two points
function getMidPoint(points, i, j) {
    return new Point(
        (points[i].x + points[j].x) / 2.0,
        (points[i].y + points[j].y) / 2.0
    );
}

// method returns true if point of points array form
// a parallelogram
function isParallelogram(points) {
    const midPointMap = new Map();

    // looping over all pairs of point to store their
    // mid points
    const P = 4;
    for (let i = 0; i < P; i++) {
        for (let j = i + 1; j < P; j++) {
            const temp = getMidPoint(points, i, j);

            // storing point pair, corresponding to
            // the mid point
            const pair = [i, j];
            if (!midPointMap.has(temp)) {
                midPointMap.set(temp, []);
            }
            midPointMap.get(temp).push(pair);
        }
    }

    let two = 0, one = 0;

    // looping over (midpoint, (corresponding pairs))
    // map to check the occurrence of each midpoint
    for (const [midPoint, pairs] of midPointMap) {
        // updating midpoint count which occurs twice
        if (pairs.length == 2) {
            two++;
        }
        // updating midpoing count which occurs once
        else if (pairs.length == 1) {
            one++;
        }
        // if midpoint count is more than 2, then
        // parallelogram is not possible
        else {
            return false;
        }
    }

    // for parallelogram, one mid point should come
    // twice and other mid points should come once
    if (two == 1 && one == 4) {
        return false;
    }
    return true;
}

// Driver code to test above methods
const points = [
    new Point(0, 0),
    new Point(4, 0),
    new Point(1, 3),
    new Point(5, 3),
];

if (isParallelogram(points)) {
    console.log("Given points form a parallelogram");
} else {
    console.log("Given points do not form a parallelogram");
}

// Contributed by adityasha4x71

Output:

Given points form a parallelogram

Time Complexity: O(p²logp) , where p is number of points
Auxiliary Space: O(p²), where p is number of points

Approach 2 : Using Vectors:

• Another approach to check if four points form a parallelogram is to use vector operations. We can calculate the vectors formed by the pairs of points and check if they satisfy the properties of a parallelogram.
• Here is the algorithm for this approach:
• Take four points A, B, C, and D as input.
• Calculate vectors AB and CD using the formula (B – A) and (D – C).
• Calculate vectors AC and BD using the formula (C – A) and (D – B).
• Check if AB and CD are parallel by taking their cross product. If the cross product is zero, then they are parallel.
• Check if AC and BD are parallel by taking their cross product. If the cross product is zero, then they are parallel.
• If AB and CD are parallel and AC and BD are parallel, then the points form a parallelogram.

Here is the C++ code implementation of this approach:

C++

#include <bits/stdc++.h>
using namespace std;

// structure to represent a point
struct point {
    double x, y;
    point() { }
    point(double x, double y)
        : x(x), y(y) { }
};

// method to calculate cross product of two vectors
double crossProduct(point a, point b) {
    return a.x * b.y - a.y * b.x;
}

// method to check if four points form a parallelogram
bool isParallelogram(point A, point B, point C, point D) {
    // calculate vectors AB, CD, AC, and BD
    point AB = point(B.x - A.x, B.y - A.y);
    point CD = point(D.x - C.x, D.y - C.y);
    point AC = point(C.x - A.x, C.y - A.y);
    point BD = point(D.x - B.x, D.y - B.y);

    // check if AB and CD are parallel
    if (crossProduct(AB, CD) == 0) {
        // check if AC and BD are parallel
        if (crossProduct(AC, BD) == 0) {
            // points form a parallelogram
            return true;
        }
    }

    // points do not form a parallelogram
    return false;
}

// Driver code to test above method
int main() {
    point A = point(0, 0);
    point B = point(4, 0);
    point C = point(1, 3);
    point D = point(5, 3);

    if (isParallelogram(A, B, C, D))
        cout << "Given points form a parallelogram";
    else
        cout << "Given points do not form a parallelogram";
    return 0;
}

Javascript

// structure to represent a point
class point {
    constructor(x, y) {
        this.x = x;
        this.y = y;
    }
}

// method to calculate cross product of two vectors
function crossProduct(a, b) {
    return a.x * b.y - a.y * b.x;
}

// method to check if four points form a parallelogram
function isParallelogram(A, B, C, D) {
    // calculate vectors AB, CD, AC, and BD
    let AB = new point(B.x - A.x, B.y - A.y);
    let CD = new point(D.x - C.x, D.y - C.y);
    let AC = new point(C.x - A.x, C.y - A.y);
    let BD = new point(D.x - B.x, D.y - B.y);

    // check if AB and CD are parallel
    if (crossProduct(AB, CD) == 0) {
        // check if AC and BD are parallel
        if (crossProduct(AC, BD) == 0) {
            // points form a parallelogram
            return true;
        }
    }

    // points do not form a parallelogram
    return false;
}

// Driver code to test above method
let A = new point(0, 0);
let B = new point(4, 0);
let C = new point(1, 3);
let D = new point(5, 3);

if (isParallelogram(A, B, C, D))
    console.log("Given points form a parallelogram");
else
    console.log("Given points do not form a parallelogram");

Output:

Given points form a parallelogram

Time Complexity:  O(n^2logn) where n is the number of points
Auxiliary Space:  O(n^2) since the map data structure is used to store the midpoints, and the worst-case number of midpoints that can be stored is n^2.

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