Convex Hull using Graham Scan
Prerequisite:
A convex hull is the smallest convex polygon that contains a given set of points. It is a useful concept in computational geometry and has applications in various fields such as computer graphics, image processing, and collision detection.
A convex polygon is a polygon in which all interior angles are less than 180 degrees. A convex hull can be constructed for any set of points, regardless of their arrangement.
Graham Scan Algorithm:
The Graham scan algorithm is a simple and efficient algorithm for computing the convex hull of a set of points. It works by iteratively adding points to the convex hull until all points have been added.
- The algorithm starts by finding the point with the smallest y-coordinate. This point is always on the convex hull. The algorithm then sorts the remaining points by their polar angle with respect to the starting point.
- The algorithm then iteratively adds points to the convex hull. At each step, the algorithm checks whether the last two points added to the convex hull form a right turn. If they do, then the last point is removed from the convex hull. Otherwise, the next point in the sorted list is added to the convex hull.
- The algorithm terminates when all points have been added to the convex hull.
Implementation of Graham Scan:
Phase 1 (Sort points):
- The first step in implementing the Graham scan algorithm is to sort the points by their polar angle with respect to the starting point.
- Once the points have been sorted, the starting point is added to the convex hull. Once the points are sorted, they form a simple closed path (See the following diagram).
Phase 2 (Accept or Reject Points):
Once we have the closed path, the next step is to traverse the path and remove concave points on this path. How to decide which point to remove and which to keep? Again, orientation helps here. The first two points in sorted array are always part of Convex Hull. For remaining points, we keep track of recent three points, and find the angle formed by them. Let the three points be prev(p), curr(c) and next(n). If orientation of these points (considering them in same order) is not counterclockwise, we discard c, otherwise we keep it. Following diagram shows step by step process of this phase.
Below are the implementation of the above algorithm.
CPP
// A C++ program to find convex hull of a set of points. Refer // for explanation of orientation() #include <iostream> #include <stack> #include <stdlib.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 find next to top in a stack Point nextToTop(stack<Point> &S) { Point p = S.top(); S.pop(); Point res = S.top(); S.push(p); return res; } // A utility function to swap two points void 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 distSq(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; // clock 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 (distSq(p0, *p2) >= distSq(p0, *p1))? -1 : 1; return (o == 2)? -1: 1; } // Prints convex hull of a set of n points. void convexHull(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 or choose the left // most point in case of tie 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); // If two or more points make same angle with p0, // Remove all but the one that is farthest from p0 // Remember that, in above sorting, our criteria was // to keep the farthest point at the end when more than // one points have same angle. int m = 1; // Initialize size of modified array for ( int i=1; i<n; i++) { // Keep removing i while angle of i and i+1 is same // with respect to p0 while (i < n-1 && orientation(p0, points[i], points[i+1]) == 0) i++; points[m] = points[i]; m++; // Update size of modified array } // If modified array of points has less than 3 points, // convex hull is not possible if (m < 3) return ; // Create an empty stack and push first three points // to it. stack<Point> S; S.push(points[0]); S.push(points[1]); S.push(points[2]); // Process remaining n-3 points for ( int i = 3; i < m; i++) { // Keep removing top while the angle formed by // points next-to-top, top, and points[i] makes // a non-left turn while (S.size()>1 && orientation(nextToTop(S), S.top(), points[i]) != 2) S.pop(); S.push(points[i]); } // Now stack has the output points, print contents of stack while (!S.empty()) { Point p = S.top(); cout << "(" << p.x << ", " << p.y << ")" << endl; S.pop(); } } // 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]); convexHull(points, n); return 0; } |
Java
// Java equivalent of the above code import java.util.*; // A Java program to find convex hull of a set of points. Refer // for explanation of orientation() public class ConvexHull { // Define Infinite (Using INT_MAX // caused overflow problems) static final int INF = 10000 ; // 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 static int orientation(Point p, Point q, Point r) { // for details of below formula. 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 ; // clock or counterclock wise } // Prints convex hull of a set of n points. static void convexHull(Point points[], int n) { // There must be at least 3 points if (n < 3 ) return ; // Initialize Result Vector<Point> hull = new Vector<Point>(); // Find the leftmost point int l = 0 ; for ( int i = 1 ; i < n; i++) if (points[i].x < points[l].x) l = i; // Start from leftmost point, keep moving // counterclockwise until reach the start point // again. This loop runs O(h) times where h is // number of points in result or output. int p = l, q; do { // Add current point to result hull.add(points[p]); // Search for a point 'q' such that // orientation(p, x, q) is counterclockwise // for all points 'x'. The idea is to keep // track of last visited most counterclock- // wise point in q. If any point 'i' is more // counterclock-wise than q, then update q. q = (p + 1 ) % n; for ( int i = 0 ; i < n; i++) { // If i is more counterclockwise than // current q, then update q if (orientation(points[p], points[i], points[q]) == 2 ) q = i; } // Now q is the most counterclockwise with // respect to p. Set p as q for next iteration, // so that q is added to result 'hull' p = q; } while (p != l); // While we don't come to first // point // Print Result for ( int i = 0 ; i < hull.size(); i++) System.out.println( "(" + hull.get(i).x + ", " + hull.get(i).y + ")" ); } // Driver program to test above function public static void main(String[] args) { Point points[] = new Point[ 8 ]; points[ 0 ] = new Point( 0 , 3 ); points[ 1 ] = new Point( 1 , 1 ); points[ 2 ] = new Point( 2 , 2 ); points[ 3 ] = new Point( 4 , 4 ); points[ 4 ] = new Point( 0 , 0 ); points[ 5 ] = new Point( 1 , 2 ); points[ 6 ] = new Point( 3 , 1 ); points[ 7 ] = new Point( 3 , 3 ); int n = points.length; convexHull(points, n); } } //Point class to store points class Point { int x, y; Point() { x = 0 ; y = 0 ; } Point( int a, int b) { x = a; y = b; } }; |
C#
// C# equivalent of the above code using System; using System.Collections.Generic; public class ConvexHul { // Define Infinite (Using INT_MAX caused overflow problems) const int INF = 10000; // 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 static int Orientation(Point p, Point q, Point r) { // See https://www.w3wiki.net/orientation-3-ordered-points/ for details of below formula. 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; // clock or counterclock wise } // Prints convex hull of a set of n points. static void ConvexHull(Point[] points, int n) { // There must be at least 3 points if (n < 3) return ; // Initialize Result List<Point> hull = new List<Point>(); // Find the leftmost point int l = 0; for ( int i = 1; i < n; i++) if (points[i].x < points[l].x) l = i; // Start from leftmost point, keep moving counterclockwise until reach the start point again. This loop runs O(h) times where h is number of points in result or output. int p = l, q; do { // Add current point to result hull.Add(points[p]); // Search for a point 'q' such that orientation(p, x, q) is counterclockwise for all points 'x'. The idea is to keep track of last visited most counterclock- wise point in q. If any point 'i' is more counterclockwise than q, then update q. q = (p + 1) % n; for ( int i = 0; i < n; i++) { // If i is more counterclockwise than current q, then update q if (Orientation(points[p], points[i], points[q]) == 2) q = i; } // Now q is the most counterclockwise with respect to p. Set p as q for next iteration, so that q is added to result 'hull' p = q; } while (p != l); // While we don't come to first // point // Print Result foreach (Point i in hull) Console.WriteLine( "(" + i.x + ", " + i.y + ")" ); } // Driver program to test above function public static void Main(String[] args) { Point[] points = new Point[8]; points[0] = new Point(0, 3); points[1] = new Point(1, 1); points[2] = new Point(2, 2); points[3] = new Point(4, 4); points[4] = new Point(0, 0); points[5] = new Point(1, 2); points[6] = new Point(3, 1); points[7] = new Point(3, 3); int n = points.Length; ConvexHull(points, n); } } // Point class to store points public class Point { public int x, y; public Point() { x = 0; y = 0; } public Point( int a, int b) { x = a; y = b; } }; |
Javascript
// JavaScript program to find convex hull of a set of // points. Refer // for explanation of orientation() // A class used to store the x and y coordinates of points class Point { constructor(x = null , y = null ) { this .x = x; this .y = y; } } // A global point needed for sorting points with reference // to the first point let p0 = new Point(0, 0); // A utility function to find next to top in a stack function nextToTop(S) { return S[S.length - 2]; } // A utility function to return square of distance // between p1 and p2 function distSq(p1, 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 function orientation(p, q, r) { let val = ((q.y - p.y) * (r.x - q.x) - (q.x - p.x) * (r.y - q.y)); if (val == 0) return 0; // collinear else if (val > 0) return 1; // clock wise else return 2; // counterclock wise } // A function used by cmp_to_key function to sort an array // of points with respect to the first point function compare(p1, p2) { // Find orientation let o = orientation(p0, p1, p2); if (o == 0) { if (distSq(p0, p2) >= distSq(p0, p1)) return -1; else return 1; } else { if (o == 2) return -1; else return 1; } } // Prints convex hull of a set of n points. function convexHull(points, n) { // Find the bottommost point let ymin = points[0].y; let min = 0; for ( var i = 1; i < n; i++) { let y = points[i].y; // Pick the bottom-most or choose the left // most point in case of tie 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 points[0], points[min] = points[min], points[0]; // 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 let p0 = points[0]; points.sort(compare); // If two or more points make same angle with p0, // Remove all but the one that is farthest from p0 // Remember that, in above sorting, our criteria was // to keep the farthest point at the end when more than // one points have same angle. let m = 1; // Initialize size of modified array for ( var i = 1; i < n; i++) { // Keep removing i while angle of i and i+1 is same // with respect to p0 while ((i < n - 1) && (orientation(p0, points[i], points[i + 1]) == 0)) i += 1; points[m] = points[i]; m += 1; // Update size of modified array } // If modified array of points has less than 3 points, // convex hull is not possible if (m < 3) return ; // Create an empty stack and push first three points // to it. let S = []; S.push(points[0]); S.push(points[1]); S.push(points[2]); // Process remaining n-3 points for ( var i = 3; i < m; i++) { // Keep removing top while the angle formed by // points next-to-top, top, and points[i] makes // a non-left turn while ( true ) { if (S.length < 2) break ; if (orientation(nextToTop(S), S[S.length - 1], points[i]) >= 2) break ; S.pop(); } S.push(points[i]); } // Now stack has the output points, // print contents of stack while (S.length > 0) { let p = S[S.length - 1]; console.log( "(" + p.x + ", " + p.y + ")" ); S.pop(); } } // Driver Code let 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) ]; let n = points.length; convexHull(points, n); // This code is contributed by phasing17 |
Python3
# A Python3 program to find convex hull of a set of points. Refer # for explanation of orientation() from functools import cmp_to_key # A class used to store the x and y coordinates of points class Point: def __init__( self , x = None , y = None ): self .x = x self .y = y # A global point needed for sorting points with reference # to the first point p0 = Point( 0 , 0 ) # A utility function to find next to top in a stack def nextToTop(S): return S[ - 2 ] # A utility function to return square of distance # between p1 and p2 def distSq(p1, 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 def orientation(p, q, r): val = ((q.y - p.y) * (r.x - q.x) - (q.x - p.x) * (r.y - q.y)) if val = = 0 : return 0 # collinear elif val > 0 : return 1 # clock wise else : return 2 # counterclock wise # A function used by cmp_to_key function to sort an array of # points with respect to the first point def compare(p1, p2): # Find orientation o = orientation(p0, p1, p2) if o = = 0 : if distSq(p0, p2) > = distSq(p0, p1): return - 1 else : return 1 else : if o = = 2 : return - 1 else : return 1 # Prints convex hull of a set of n points. def convexHull(points, n): # Find the bottommost point ymin = points[ 0 ].y min = 0 for i in range ( 1 , n): y = points[i].y # Pick the bottom-most or choose the left # most point in case of tie if ((y < ymin) or (ymin = = y and points[i].x < points[ min ].x)): ymin = points[i].y min = i # Place the bottom-most point at first position points[ 0 ], points[ min ] = points[ min ], points[ 0 ] # 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 = sorted (points, key = cmp_to_key(compare)) # If two or more points make same angle with p0, # Remove all but the one that is farthest from p0 # Remember that, in above sorting, our criteria was # to keep the farthest point at the end when more than # one points have same angle. m = 1 # Initialize size of modified array for i in range ( 1 , n): # Keep removing i while angle of i and i+1 is same # with respect to p0 while ((i < n - 1 ) and (orientation(p0, points[i], points[i + 1 ]) = = 0 )): i + = 1 points[m] = points[i] m + = 1 # Update size of modified array # If modified array of points has less than 3 points, # convex hull is not possible if m < 3 : return # Create an empty stack and push first three points # to it. S = [] S.append(points[ 0 ]) S.append(points[ 1 ]) S.append(points[ 2 ]) # Process remaining n-3 points for i in range ( 3 , m): # Keep removing top while the angle formed by # points next-to-top, top, and points[i] makes # a non-left turn while (( len (S) > 1 ) and (orientation(nextToTop(S), S[ - 1 ], points[i]) ! = 2 )): S.pop() S.append(points[i]) # Now stack has the output points, # print contents of stack while S: p = S[ - 1 ] print ( "(" + str (p.x) + ", " + str (p.y) + ")" ) S.pop() # Driver Code input_points = [( 0 , 3 ), ( 1 , 1 ), ( 2 , 2 ), ( 4 , 4 ), ( 0 , 0 ), ( 1 , 2 ), ( 3 , 1 ), ( 3 , 3 )] points = [] for point in input_points: points.append(Point(point[ 0 ], point[ 1 ])) n = len (points) convexHull(points, n) # This code is contributed by Kevin Joshi |
Output:
(0, 3)
(4, 4)
(3, 1)
(0, 0)
Time Complexity: O(nLogn), where n be the number of input points.
The first step (finding the bottom-most point) takes O(n) time. The second step (sorting points) takes O(nLogn) time. The third step takes O(n) time. In the third step, every element is pushed and popped at most one time. So the sixth step to process points one by one takes O(n) time, assuming that the stack operations take O(1) time. Overall complexity is O(n) + O(nLogn) + O(n) + O(n) which is O(nLogn).
Auxiliary Space: O(n), as explicit stack is used, since no extra space has been taken.
Applications of Convex Hulls:
Convex hulls have a wide range of applications, including:
- Collision detection: Convex hulls can be used to quickly determine whether two objects are colliding. This is useful in computer graphics and physics simulations.
- Image processing: Convex hulls can be used to segment objects in images. This is useful for tasks such as object recognition and tracking.
- Computational geometry: Convex hulls are used in a variety of computational geometry algorithms, such as finding the closest pair of points and computing the diameter of a set of points.
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