Given a set of points in the plane. the convex hull of the set is the smallest convex polygon that contains all the points of it.
We strongly recommend to see the following post first. How to check if two given line segments intersect? The idea of Jarvis’s Algorithm is simple, we start from the leftmost point (or point with minimum x coordinate value) and we keep wrapping points in counterclockwise direction.
The big question is, given a point p as current point, how to find the next point in output?
The idea is to use orientation() here. Next point is selected as the point that beats all other points at counterclockwise orientation, i.e., next point is q if for any other point r, we have “orientation(p, q, r) = counterclockwise”.
Algorithm: Step 1) Initialize p as leftmost point. Step 2) Do following while we don’t come back to the first (or leftmost) point. 2.1) The next point q is the point, such that the triplet (p, q, r) is counter clockwise for any other point r.
To find this, we simply initialize q as next point, then we traverse through all points.
For any point i, if i is more counter clockwise, i.e., orientation(p, i, q) is counter clockwise, then we update q as i.
Our final value of q is going to be the most counter clockwise point. 2.2) next[p] = q (Store q as next of p in the output convex hull). 2.3) p = q (Set p as q for next iteration).
// 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
functionorientation(p, q, r)
let val = (q.y - p.y) * (r.x - q.x) -
(q.x - p.x) * (r.y - q.y);
if(val == 0) return0; // collinear
return(val > 0)? 1: 2; // clock or counterclock wise
// Prints convex hull of a set of n points.
// There must be at least 3 points
if(n < 3) return;
// Initialize Result
let hull = ;
// Find the leftmost point
let l = 0;
for(let 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.
let p = l, q;
// Add current point to result
// Search for a point 'q' such that
// orientation(p, q, x) 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(let i = 0; i < n; i++)
// If i is more counterclockwise than
// current q, then update q
if(orientation(points[p], points[i], points[q])
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
// Print Result
for(let temp of hull.values())
document.write("("+ temp.x + ", "+
temp.y + ")<br>");
/* Driver program to test above function */
let points = newArray(7);
points = newPoint(0, 3);
points = newPoint(2, 3);
points = newPoint(1, 1);
points = newPoint(2, 1);
points = newPoint(3, 0);
points = newPoint(0, 0);
points = newPoint(3, 3);
let n = points.length;
// This code is contributed by avanitrachhadiya2155
Output: The output is points of the convex hull.
Time Complexity:O(m * n), where n is number of input points and m is number of output or hull points (m <= n). For every point on the hull we examine all the other points to determine the next point.
Worst case, Time complexity:O(n2). The worst case occurs when all the points are on the hull (m = n).
Auxiliary Space: O(n), since n extra space has been taken. Set 2- Convex Hull (Graham Scan) Note : The above code may produce different results for different order of inputs, when there are collinear points in the convex hull. For example, it produces output as (0, 3) (0, 0) (3, 0) (3, 3) for input (0, 3), (0, 0), (0, 1), (3, 0), (3, 3) and output as (0, 3) (0, 1) (0, 0) (3, 0) (3, 3) for input as (0, 3), (0, 1), (0, 0), (3, 0), (3, 3). We generally need the farthest next point in case of collinear, we can get the desired result in case of collinear points by adding one more if condition. Please refer this modified code. Sources: http://www.cs.uiuc.edu/~jeffe/teaching/373/notes/x05-convexhull.pdf http://www.dcs.gla.ac.uk/~pat/52233/slides/Hull1x1.pdf Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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