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Convex Hull Jarvis’s Algorithm or Wrapping in C++
In this tutorial, we will be discussing a program to find the convex hull of a given set of points using Jarvis’s Algorithm.
Convex hull is the smallest polygon convex figure containing all the given points either on the boundary on inside the figure.
In Jarvis’s algorithm, we select the leftmost point and keep wrapping points moving in the clockwise direction.
Example
#include <bits/stdc++.h> using namespace std; //structure of the point struct Point{ int x, y; }; //calculating the position of the points int cal_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 counterclockwise } //printing convex hull void convexHull(Point points[], int n){ if (n < 3) return; vector<Point> hull; //calculating the leftmost point int l = 0; for (int i = 1; i < n; i++) if (points[i].x < points[l].x) l = i; //moving in the clockwise direction int p = l, q; do{ //adding current point to result hull.push_back(points[p]); q = (p+1)%n; for (int i = 0; i < n; i++){ if (cal_orientation(points[p], points[i], points[q]) == 2) q = i; } p = q; } while (p != l); //if didn't reached the first point for (int i = 0; i < hull.size(); i++) cout << "(" << hull[i].x << ", " << hull[i].y << ")\n"; } int main(){ Point points[] = {{0, 3}, {2, 2}, {1, 1}, {2, 1}, {3, 0}, {0, 0}, {3, 3}}; int n = sizeof(points)/sizeof(points[0]); convexHull(points, n); return 0; }
Output
(0, 3) (0, 0) (3, 0) (3, 3)
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