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)}
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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 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.
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; } |
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:
http://www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf
This article is contributed by Aarti_Rathi and 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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