Find Simple Closed Path for a given set of points (original) (raw)

Last Updated : 23 Jul, 2025

Given a set of points, connect the dots without crossing.

Simple Closed Path for a given set of points 1

Example:

Input: points[] = {(0, 3), (1, 1), (2, 2), (4, 4), (0, 0), (1, 2), (3, 1}, {3, 3}};

Output: Connecting points in following order would not cause any crossing {(0, 0), (3, 1), (1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (0, 3)}

We strongly recommend you to minimize your browser and try this yourself first.
The idea is to use sorting.

Find the bottom-most point by comparing y coordinate of all points. If there are two points with same y value, then the point with smaller x coordinate value is considered. Put the bottom-most point at first position.

find the bottom-most point by comparing y coordinate of all points

Consider the remaining n-1 points and sort them by polar angle in counterclockwise order around points[0]. If polar angle of two points is same, then put the nearest point first.
Traversing the sorted array (sorted in increasing order of angle) yields simple closed path.

traversing the sorted array

How to compute angles?
One solution is to use trigonometric functions.
Observation: We don’t care about the actual values of the angles. We just want to sort by angle.
Idea: Use the orientation to compare angles without actually computing them!

Below is C++ implementation of above idea.

C++ `

// A C++ program to find simple closed path for n points // for explanation of orientation() #include <bits/stdc++.h> using namespace std;

struct Point { int x, y; };

// A global point needed for sorting points with reference // to the first point. Used in compare function of qsort() Point p0;

// A utility function to swap two points int swap(Point &p1, Point &p2) { Point temp = p1; p1 = p2; p2 = temp; }

// A utility function to return square of distance between // p1 and p2 int dist(Point p1, Point p2) { return (p1.x - p2.x)(p1.x - p2.x) + (p1.y - p2.y)(p1.y - p2.y); }

// To find orientation of ordered triplet (p, q, r). // The function returns following values // 0 --> p, q and r are collinear // 1 --> Clockwise // 2 --> Counterclockwise 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;  // collinear
return (val > 0)? 1: 2; // clockwise or counterclock wise

}

// A function used by library function qsort() to sort // an array of points with respect to the first point int compare(const void *vp1, const void *vp2) { Point *p1 = (Point *)vp1; Point *p2 = (Point *)vp2;

// Find orientation int o = orientation(p0, *p1, *p2); if (o == 0) return (dist(p0, *p2) >= dist(p0, *p1))? -1 : 1;

return (o == 2)? -1: 1; }

// Prints simple closed path for a set of n points. void printClosedPath(Point points[], int n) { // Find the bottommost point int ymin = points[0].y, min = 0; for (int i = 1; i < n; i++) { int y = points[i].y;

 // Pick the bottom-most. In case of tie, choose the
 // left most point
 if ((y < ymin) || (ymin == y &&
     points[i].x < points[min].x))
    ymin = points[i].y, min = i;

}

// Place the bottom-most point at first position swap(points[0], points[min]);

// Sort n-1 points with respect to the first point. // A point p1 comes before p2 in sorted output if p2 // has larger polar angle (in counterclockwise // direction) than p1 p0 = points[0]; qsort(&points[1], n-1, sizeof(Point), compare);

// Now stack has the output points, print contents // of stack for (int i=0; i<n; i++) cout << "(" << points[i].x << ", " << points[i].y <<"), "; }

// Driver program to test above functions int main() { Point points[] = {{0, 3}, {1, 1}, {2, 2}, {4, 4}, {0, 0}, {1, 2}, {3, 1}, {3, 3}}; int n = sizeof(points)/sizeof(points[0]); printClosedPath(points, n); return 0; }

Java

import java.util.*;

class Point { int x, y; Point(int x, int y) { this.x = x; this.y = y; } }

class ConvexHull { static Point p0;

static void swap(Point p1, Point p2) {
    Point temp = p1;
    p1 = p2;
    p2 = temp;
}

static int dist(Point p1, Point p2) {
    return (int)Math.pow(p1.x - p2.x, 2) +
           (int)Math.pow(p1.y - p2.y, 2);
}

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; // collinear
    return (val > 0)? 1: 2; 
}

static int compare(Point p1, Point p2) {
    int o = orientation(p0, p1, p2);
    if (o == 0)
        return (dist(p0, p2) >= dist(p0, p1))? -1 : 1;
    return (o == 2)? -1: 1;
}

static void printClosedPath(Point points[], int n) {
    int ymin = points[0].y, min = 0;
    for (int i = 1; i < n; i++) {
        int y = points[i].y;
        if ((y < ymin) || (ymin == y &&
            points[i].x < points[min].x))
            ymin = points[i].y;
            min = i;
    }
    swap(points[0], points[min]);
    p0 = points[0];
    Arrays.sort(points, 1, n, (p1, p2) -> compare(p1, p2));
    for (int i=0; i<n; i++)
        System.out.println("(" + points[i].x + ", " + points[i].y + "), ");
}

public static void main(String[] args) {
    Point[] points = {new Point(0, 3), new Point(1, 1), new Point(2, 2), new Point(4, 4),
                      new Point(0, 0), new Point(1, 2), new Point(3, 1), new Point(3, 3)};
    int n = points.length;
    printClosedPath(points, n);
}

}

Python3

from functools import cmp_to_key

A Python program to find simple closed path for n points

for explanation of orientation()

A global point needed for sorting points with reference

to the first point. Used in compare function of qsort()

p0 = None

A utility function to return square of distance between

p1 and p2

def dist(p1, p2): return (p1[0] - p2[0])(p1[0] - p2[0]) + (p1[1] - p2[1])(p1[1] - p2[1])

To find orientation of ordered triplet (p, q, r).

The function returns following values

0 --> p, q and r are collinear

1 --> Clockwise

2 --> Counterclockwise

def orientation(p, q, r): val = (q[1] - p[1]) * (r[0] - q[0]) - (q[0] - p[0]) * (r[1] - q[1])

if val == 0: return 0  # collinear
return 1 if val > 0 else 2 # clockwise or counterclock wise

A function used by library function qsort() to sort

an array of points with respect to the first point

def compare(vp1, vp2): p1 = vp1 p2 = vp2

# Find orientation
o = orientation(p0, p1, p2)
if o == 0:
    return -1 if dist(p0, p2) >= dist(p0, p1) else 1

return -1 if o == 2 else 1

Prints simple closed path for a set of n points.

def printClosedPath(points, n): global p0 # Find the bottommost point ymin = points[0][1] min = 0 for i in range(1,n): y = points[i][1]

    # Pick the bottom-most. In case of tie, choose the
    # left most point
    if (y < ymin) or (ymin == y and points[i][0] < points[min][0]):
        ymin = points[i][1]
        min = i

# Place the bottom-most point at first position
temp = points[0]
points[0] = points[min]
points[min] = temp

# Sort n-1 points with respect to the first point.
# A point p1 comes before p2 in sorted output if p2
# has larger polar angle (in counterclockwise
# direction) than p1
p0 = points[0]
points.sort(key=cmp_to_key(compare))

# Now stack has the output points, print contents
# of stack
for i in range(n):
    print("(",points[i][0],",",points[i][1],"), ", end="")

Driver program to test above functions

points = [[0, 3], [1, 1], [2, 2], [4, 4], [0, 0], [1, 2], [3, 1], [3, 3]] n = len(points)

printClosedPath(points, n)

C#

using System; using System.Collections.Generic;

public class Point { public int x, y; public Point(int x, int y) { this.x = x; this.y = y; } }

public class ClosestPath { static Point p0;

static int dist(Point p1, Point p2)
{
    return (p1.x - p2.x) * (p1.x - p2.x)
        + (p1.y - p2.y) * (p1.y - p2.y);
}

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; // collinear
    return (val > 0)
        ? 1
        : 2; // clockwise or counterclockwise
}

static int compare(Point p1, Point p2)
{
    int o = orientation(p0, p1, p2);
    if (o == 0)
        return (dist(p0, p2) >= dist(p0, p1)) ? -1 : 1;
    return (o == 2) ? -1 : 1;
}

static void printClosedPath(List<Point> points, int n)
{
    // Find the bottommost point
    int ymin = points[0].y;
    int min = 0;
    for (int i = 1; i < n; i++) {
        int y = points[i].y;
        if ((y < ymin)
            || (ymin == y
                && points[i].x < points[min].x)) {
            ymin = points[i].y;
            min = i;
        }
    }

    // Place the bottom-most point at first position
    Point temp = points[0];
    points[0] = points[min];
    points[min] = temp;

    // Sort n-1 points with respect to the first point.
    // A point p1 comes before p2 in sorted output if p2
    // has larger polar angle (in counterclockwise
    // direction) than p1
    p0 = points[0];
    points.Sort(compare);

    // Now stack has the output points, print contents
    // of stack
    for (int i = 0; i < n; i++) {
        Console.Write("(" + points[i].x + ", "
                      + points[i].y + "), ");
    }
}

public static void Main()
{
    List<Point> points = new List<Point>() {
        new Point(0, 3), new Point(1, 1),
            new Point(2, 2), new Point(4, 4),
            new Point(0, 0), new Point(1, 2),
            new Point(3, 1), new Point(3, 3)
    };
    int n = points.Count;

    printClosedPath(points, n);
}

} // This code is contributed by user_dtewbxkn77n

JavaScript

// A javascript program to find simple closed path for n points // for explanation of orientation()

// A global point needed for sorting points with reference // to the first point. Used in compare function of qsort() let p0;

// A utility function to return square of distance between // p1 and p2 function dist(p1, p2) { return (p1[0] - p2[0])(p1[0] - p2[0]) + (p1[1] - p2[1])(p1[1] - p2[1]); }

// To find orientation of ordered triplet (p, q, r). // The function returns following values // 0 --> p, q and r are collinear // 1 --> Clockwise // 2 --> Counterclockwise function orientation(p, q, r) { let val = (q[1] - p[1]) * (r[0] - q[0]) - (q[0] - p[0]) * (r[1] - q[1]);

if (val == 0) return 0;  // collinear
return (val > 0)? 1: 2; // clockwise or counterclock wise

}

// A function used by library function qsort() to sort // an array of points with respect to the first point function compare(vp1, vp2) { let p1 = vp1; let p2 = vp2;

// Find orientation let o = orientation(p0, p1, p2); if (o == 0) return (dist(p0, p2) >= dist(p0, p1))? -1 : 1;

return (o == 2)? -1: 1; }

// Prints simple closed path for a set of n points. function printClosedPath(points, n) { // Find the bottommost point let ymin = points[0][1]; let min = 0; for (let i = 1; i < n; i++) { let y = points[i][1];

 // Pick the bottom-most. In case of tie, choose the
 // left most point
 if ((y < ymin) || (ymin == y && points[i][0] < points[min][0])){
    ymin = points[i][1];
    min = i;
 }

}

// Place the bottom-most point at first position let temp = points[0]; points[0] = points[min]; points[min] = temp;

// Now stack has the output points, print contents // of stack for (let i=0; i<n; i++) console.log("(" + points[i][0] + "," + points[i][1] + "), "); }

// Driver program to test above functions let points = [[0, 3], [1, 1], [2, 2], [4, 4], [0, 0], [1, 2], [3, 1], [3, 3]]; let n = points.length;

printClosedPath(points, n);

// The code is contributed by Nidhi goel.

Output:

(0, 0), (3, 1), (1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (0, 3),

Time complexity of above solution is O(n Log n) if we use a O(nLogn) sorting algorithm for sorting points.
Auxiliary Space: O(1), since no extra space has been taken.

Source:
https://www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf