Convex Hull | Monotone chain algorithm
Given a set of points, the task is to find the convex hull of the given points. The convex hull is the smallest convex polygon that contains all the points.
Please check this article first: Convex Hull | Set 1 (Jarvis’s Algorithm or Wrapping)
Examples:
Input: Points[] = {{0, 3}, {2, 2}, {1, 1}, {2, 1}, {3, 0}, {0, 0}, {3, 3}}
Output:
(0, 0)
(3, 0)
(3, 3)
(0, 3)
Approach: Monotone chain algorithm constructs the convex hull in O(n * log(n)) time. We have to sort the points first and then calculate the upper and lower hulls in O(n) time. The points will be sorted with respect to x-coordinates (with respect to y-coordinates in case of a tie in x-coordinates), we will then find the left most point and then try to rotate in clockwise direction and find the next point and then repeat the step until we reach the rightmost point and then again rotate in the clockwise direction and find the lower hull.
Below is the implementation of the above approach:
CPP
// C++ implementation of the approach #include <bits/stdc++.h> #define llu long long int using namespace std; struct Point { llu x, y; bool operator<(Point p) { return x < p.x || (x == p.x && y < p.y); } }; // Cross product of two vectors OA and OB // returns positive for counter clockwise // turn and negative for clockwise turn llu cross_product(Point O, Point A, Point B) { return (A.x - O.x) * (B.y - O.y) - (A.y - O.y) * (B.x - O.x); } // Returns a list of points on the convex hull // in counter-clockwise order vector<Point> convex_hull(vector<Point> A) { int n = A.size(), k = 0; if (n <= 3) return A; vector<Point> ans(2 * n); // Sort points lexicographically sort(A.begin(), A.end()); // Build lower hull for ( int i = 0; i < n; ++i) { // If the point at K-1 position is not a part // of hull as vector from ans[k-2] to ans[k-1] // and ans[k-2] to A[i] has a clockwise turn while ( k >= 2 && cross_product(ans[k - 2], ans[k - 1], A[i]) <= 0) k--; ans[k++] = A[i]; } // Build upper hull for ( size_t i = n - 1, t = k + 1; i > 0; --i) { // If the point at K-1 position is not a part // of hull as vector from ans[k-2] to ans[k-1] // and ans[k-2] to A[i] has a clockwise turn while (k >= t && cross_product(ans[k - 2], ans[k - 1], A[i - 1]) <= 0) k--; ans[k++] = A[i - 1]; } // Resize the array to desired size ans.resize(k - 1); return ans; } // Driver code int main() { vector<Point> points; // Add points points.push_back({ 0, 3 }); points.push_back({ 2, 2 }); points.push_back({ 1, 1 }); points.push_back({ 2, 1 }); points.push_back({ 3, 0 }); points.push_back({ 0, 0 }); points.push_back({ 3, 3 }); // Find the convex hull vector<Point> ans = convex_hull(points); // Print the convex hull for ( int i = 0; i < ans.size(); i++) cout << "(" << ans[i].x << ", " << ans[i].y << ")" << endl; return 0; } |
Python3
class Point( object ): def __init__( self , x, y): self .x = x self .y = y # A utility function to find next # to top in a stack def nextToTop(S): a = S.pop() b = S.pop() S.append(a) return b # A utility function to swap two # points def swap(p1, p2): return p2, p1 # A utility function to return # square of distance between # two points def distSq(p1, p2): return (p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y) # Prints convex hull of a set of n # points. def convexHull(points, n): # There must be at least 3 points if (n < 3 ): return # Initialize Result hull = [] # Find the leftmost point l = 0 for i in range ( 1 , n): 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. p = l q = 0 while ( True ): # Add current point to result hull.append(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 i in range ( 0 , n): # 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 we don't come to first point if (p = = l): break # Print Result printHull(hull) # 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 or wise else : return 2 # counterclock or wise # Prints convex hull of a set of n points. def printHull(hull): print ( "The points in Convex Hull are:" ) for i in range ( len (hull)): print ( "(" , hull[i].x, ", " , hull[i].y, ")" ) # Driver Code if __name__ = = "__main__" : points = [] points.append(Point( 0 , 3 )) points.append(Point( 2 , 2 )) points.append(Point( 1 , 1 )) points.append(Point( 2 , 1 )) points.append(Point( 3 , 0 )) points.append(Point( 0 , 0 )) points.append(Point( 3 , 3 )) n = len (points) convexHull(points, n) # This code is contributed by ishankhandelwals. |
Javascript
// JS implementation of the approach function Point(x, y) { this .x = x; this .y = y; } function crossProduct(O, A, B) { return (A.x - O.x) * (B.y - O.y) - (A.y - O.y) * (B.x - O.x); } function convexHull(A) { let n = A.length; let k = 0; if (n <= 3) return A; let ans = new Array(2 * n); // Sort points lexicographically A.sort((a, b) => { return a.x < b.x || (a.x == b.x && a.y < b.y); }); // Build lower hull for (let i = 0; i < n; ++i) { // If the point at K-1 position is not a part // of hull as vector from ans[k-2] to ans[k-1] // and ans[k-2] to A[i] has a clockwise turn while (k >= 2 && crossProduct(ans[k - 2], ans[k - 1], A[i]) <= 0) k--; ans[k++] = A[i]; } // Build upper hull for (let i = n - 1, t = k + 1; i > 0; --i) { // If the point at K-1 position is not a part // of hull as vector from ans[k-2] to ans[k-1] // and ans[k-2] to A[i] has a clockwise turn while (k >= t && crossProduct(ans[k - 2], ans[k - 1], A[i - 1]) <= 0) k--; ans[k++] = A[i - 1]; } // Resize the array to desired size ans.length = k - 1; return ans; } // Driver code let points = []; // Add points points.push( new Point(0, 3)); points.push( new Point(2, 2)); points.push( new Point(1, 1)); points.push( new Point(2, 1)); points.push( new Point(3, 0)); points.push( new Point(0, 0)); points.push( new Point(3, 3)); // Find the convex hull let ans = convexHull(points); // Print the convex hull for (let i = 0; i < ans.length; i++) console.log( "(" + ans[i].x + ", " + ans[i].y + ")" ); // This code is contributed by ishankhandelwals. |
C#
using System; using System.Collections.Generic; using System.Collections; using System.Linq; class Point { public int x; public int y; public Point( int x, int y) { this .x = x; this .y = y; } } class HelloWorld { // A utility function to return square // of distance between p1 and p2 public static 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 public 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; else if (val > 0) return 1; else return 2; } // Prints convex hull of a set of n points. public static void printHull(List<Point> hull) { Console.WriteLine( "The points in Convex Hull are:" ); for ( int i = 0; i < hull.Count; i++) { Console.WriteLine( "(" + hull[i].x + ", " + hull[i].y + ")" ); } } // Prints convex hull of a set of n // points. public static void convexHull(List<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; int q = 0; while ( true ) { // 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 we don't come to first point if (p == l) { break ; } } // Print Result printHull(hull); } static void Main() { List<Point> points = new List<Point>(); points.Add( new Point(0, 3)); points.Add( new Point(2, 2)); points.Add( new Point(1, 1)); points.Add( new Point(2, 1)); points.Add( new Point(3, 0)); points.Add( new Point(0, 0)); points.Add( new Point(3, 3)); int n = points.Count; convexHull(points, n); } } // The code is contributed by Nidhi goel. |
Java
import java.util.*; class Point implements Comparable<Point> { long x, y; public Point( long x, long y) { this .x = x; this .y = y; } // Implement compareTo method for sorting @Override public int compareTo(Point p) { return Long.compare(x, p.x) != 0 ? Long.compare(x, p.x) : Long.compare(y, p.y); } } public class ConvexHull { // Cross product of two vectors OA and OB // returns positive for counter clockwise // turn and negative for clockwise turn static long crossProduct(Point O, Point A, Point B) { return (A.x - O.x) * (B.y - O.y) - (A.y - O.y) * (B.x - O.x); } // Returns a list of points on the convex hull // in counter-clockwise order static List<Point> convexHull(List<Point> A) { int n = A.size(), k = 0 ; if (n <= 3 ) return A; List<Point> ans = new ArrayList<>( 2 * n); // Sort points lexicographically Collections.sort(A); // Build lower hull for ( int i = 0 ; i < n; ++i) { // If the point at K-1 position is not a part // of hull as vector from ans[k-2] to ans[k-1] // and ans[k-2] to A[i] has a clockwise turn while (k >= 2 && crossProduct(ans.get(k - 2 ), ans.get(k - 1 ), A.get(i)) <= 0 ) ans.remove(--k); ans.add(A.get(i)); k++; } // Build upper hull for ( int i = n - 2 , t = k; i >= 0 ; --i) { // If the point at K-1 position is not a part // of hull as vector from ans[k-2] to ans[k-1] // and ans[k-2] to A[i] has a clockwise turn while (k > t && crossProduct(ans.get(k - 2 ), ans.get(k - 1 ), A.get(i)) <= 0 ) ans.remove(--k); ans.add(A.get(i)); k++; } // Resize the array to desired size ans.remove(ans.size() - 1 ); return ans; } // Driver code public static void main(String[] args) { List<Point> points = new ArrayList<>(); // Add points points.add( new Point( 0 , 3 )); points.add( new Point( 2 , 2 )); points.add( new Point( 1 , 1 )); points.add( new Point( 2 , 1 )); points.add( new Point( 3 , 0 )); points.add( new Point( 0 , 0 )); points.add( new Point( 3 , 3 )); // Find the convex hull List<Point> ans = convexHull(points); // Print the convex hull for (Point p : ans) System.out.println( "(" + p.x + ", " + p.y + ")" ); } } |
(0, 0) (3, 0) (3, 3) (0, 3)
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