# Find Simple Closed Path for a given set of points

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

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.

- Consider the remaining n-1 points and sort them by polor angle in counterclockwise order around points[0]. If polor 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.

**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 colinear` `// 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; ` `// colinear` ` ` `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, chose 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;` `}` |

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.

Source:

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

This article is contributed by **Rajeev Agrawal**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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