Deleting points from Convex Hull - GeeksforGeeks

22 Apr, 2022
Given a fixed set of points. We need to find convex hull of given set. We also need to find convex hull when a point is removed from the set.
Example: 
Initial Set of Points: (-2, 8) (-1, 2) (0, 1) (1, 0)
                        (-3, 0) (-1, -9) (2, -6) (3, 0) 
                        (5, 3) (2, 5) 
Initial convex hull:- (-2, 8) (-3, 0) (-1, -9) (2, -6)
                      (5, 3)
Point to remove from the set : (-2, 8)
Final convex hull: (2, 5) (-3, 0) (-1, -9) (2, -6) (5, 3)
Prerequisite : Convex Hull (Simple Divide and Conquer Algorithm)The algorithm for solving the above problem is very easy. We simply check whether the point to be removed is a part of the convex hull. If it is, then we have to remove that point from the initial set and then make the convex hull again (refer Convex hull (divide and conquer) ). 
And if not then we already have the solution (the convex hull will not change). 
C++
// C++ program to demonstrate delete operation// on Convex Hull.#include<bits/stdc++.h>using namespace std; // stores the center of polygon (It is made// global because it is used in compare function)pair<int, int> mid; // determines the quadrant of a point// (used in compare())int quad(pair<int, int> p){    if (p.first >= 0 && p.second >= 0)        return 1;    if (p.first <= 0 && p.second >= 0)        return 2;    if (p.first <= 0 && p.second <= 0)        return 3;    return 4;} // Checks whether the line is crossing the polygonint orientation(pair<int, int> a, pair<int, int> b,                pair<int, int> c){    int res = (b.second-a.second)*(c.first-b.first) -              (c.second-b.second)*(b.first-a.first);     if (res == 0)        return 0;    if (res > 0)        return 1;    return -1;} // compare function for sortingbool compare(pair<int, int> p1, pair<int, int> q1){    pair<int, int> p = make_pair(p1.first - mid.first,                                 p1.second - mid.second);    pair<int, int> q = make_pair(q1.first - mid.first,                                 q1.second - mid.second);     int one = quad(p);    int two = quad(q);     if (one != two)        return (one < two);    return (p.second*q.first < q.second*p.first);} // Finds upper tangent of two polygons 'a' and 'b'// represented as two vectors.vector<pair<int, int> > merger(vector<pair<int, int> > a,                               vector<pair<int, int> > b){    // n1 -> number of points in polygon a    // n2 -> number of points in polygon b    int n1 = a.size(), n2 = b.size();     int ia = 0, ib = 0;    for (int i=1; i<n1; i++)        if (a[i].first > a[ia].first)            ia = i;     // ib -> leftmost point of b    for (int i=1; i<n2; i++)        if (b[i].first < b[ib].first)            ib=i;     // finding the upper tangent    int inda = ia, indb = ib;    bool done = 0;    while (!done)    {        done = 1;        while (orientation(b[indb], a[inda], a[(inda+1)%n1]) >=0)            inda = (inda + 1) % n1;         while (orientation(a[inda], b[indb], b[(n2+indb-1)%n2]) <=0)        {            indb = (n2+indb-1)%n2;            done = 0;        }    }     int uppera = inda, upperb = indb;    inda = ia, indb=ib;    done = 0;    int g = 0;    while (!done)//finding the lower tangent    {        done = 1;        while (orientation(a[inda], b[indb], b[(indb+1)%n2])>=0)            indb=(indb+1)%n2;         while (orientation(b[indb], a[inda], a[(n1+inda-1)%n1])<=0)        {            inda=(n1+inda-1)%n1;            done=0;        }    }     int lowera = inda, lowerb = indb;    vector<pair<int, int> > ret;     //ret contains the convex hull after merging the two convex hulls    //with the points sorted in anti-clockwise order    int ind = uppera;    ret.push_back(a[uppera]);    while (ind != lowera)    {        ind = (ind+1)%n1;        ret.push_back(a[ind]);    }     ind = lowerb;    ret.push_back(b[lowerb]);    while (ind != upperb)    {        ind = (ind+1)%n2;        ret.push_back(b[ind]);    }    return ret; } // Brute force algorithm to find convex hull for a set// of less than 6 pointsvector<pair<int, int> > bruteHull(vector<pair<int, int> > a){    // Take any pair of points from the set and check    // whether it is the edge of the convex hull or not.    // if all the remaining points are on the same side    // of the line then the line is the edge of convex    // hull otherwise not    set<pair<int, int> >s;     for (int i=0; i<a.size(); i++)    {        for (int j=i+1; j<a.size(); j++)        {            int x1 = a[i].first, x2 = a[j].first;            int y1 = a[i].second, y2 = a[j].second;             int a1 = y1-y2;            int b1 = x2-x1;            int c1 = x1*y2-y1*x2;            int pos = 0, neg = 0;            for (int k=0; k<a.size(); k++)            {                if (a1*a[k].first+b1*a[k].second+c1 <= 0)                    neg++;                if (a1*a[k].first+b1*a[k].second+c1 >= 0)                    pos++;            }            if (pos == a.size() || neg == a.size())            {                s.insert(a[i]);                s.insert(a[j]);            }        }    }     vector<pair<int, int> >ret;    for (auto e : s)        ret.push_back(e);     // Sorting the points in the anti-clockwise order    mid = {0, 0};    int n = ret.size();    for (int i=0; i<n; i++)    {        mid.first += ret[i].first;        mid.second += ret[i].second;        ret[i].first *= n;        ret[i].second *= n;    }    sort(ret.begin(), ret.end(), compare);    for (int i=0; i<n; i++)        ret[i] = make_pair(ret[i].first/n, ret[i].second/n);     return ret;} // Returns the convex hull for the given set of pointsvector<pair<int, int>> findHull(vector<pair<int, int>> a){    // If the number of points is less than 6 then the    // function uses the brute algorithm to find the    // convex hull    if (a.size() <= 5)        return bruteHull(a);     // left contains the left half points    // right contains the right half points    vector<pair<int, int>>left, right;    for (int i=0; i<a.size()/2; i++)        left.push_back(a[i]);    for (int i=a.size()/2; i<a.size(); i++)        right.push_back(a[i]);     // convex hull for the left and right sets    vector<pair<int, int>>left_hull = findHull(left);    vector<pair<int, int>>right_hull = findHull(right);     // merging the convex hulls    return merger(left_hull, right_hull);} // Returns the convex hull for the given set of points after// remviubg a point p.vector<pair<int, int>> removePoint(vector<pair<int, int>> a,                                   vector<pair<int, int>> hull,                                   pair<int, int> p){    // checking whether the point is a part of the    // convex hull or not.    bool found = 0;    for (int i=0; i < hull.size() && !found; i++)        if (hull[i].first == p.first &&                hull[i].second == p.second)            found = 1;     // If point is not part of convex hull    if (found == 0)        return hull;     // if it is the part of the convex hull then    // we remove the point and again make the convex hull    // and if not, we print the same convex hull.    for (int i=0; i<a.size(); i++)    {        if (a[i].first==p.first && a[i].second==p.second)        {            a.erase(a.begin()+i);            break;        }    }     sort(a.begin(), a.end());    return findHull(a);} // Driver codeint main(){    vector<pair<int, int> > a;    a.push_back(make_pair(0, 0));    a.push_back(make_pair(1, -4));    a.push_back(make_pair(-1, -5));    a.push_back(make_pair(-5, -3));    a.push_back(make_pair(-3, -1));    a.push_back(make_pair(-1, -3));    a.push_back(make_pair(-2, -2));    a.push_back(make_pair(-1, -1));    a.push_back(make_pair(-2, -1));    a.push_back(make_pair(-1, 1));     int n = a.size();     // sorting the set of points according    // to the x-coordinate    sort(a.begin(), a.end());    vector<pair<int, int> >hull = findHull(a);     cout << "Convex hull:\n";    for (auto e : hull)        cout << e.first << " "             << e.second << endl;     pair<int, int> p = make_pair(-5, -3);    removePoint(a, hull, p);     cout << "\nModified Convex Hull:\n";    for (auto e:hull)        cout << e.first << " "             << e.second << endl;     return 0;}
Output: 
convex hull:
-3 0
-1 -9
2 -6
5 3
2 5
Time Complexity: It is simple to see that the maximum time taken per query is the time taken to construct the convex hull which is O(n*logn). So, the overall complexity is O(q*n*logn), where q is the number of points to be deleted.
This article is contributed by Amritya Vagmi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
khushboogoyal499
surinderdawra388
Geometric
Geometric
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Closest Pair of Points | O(nlogn) Implementation
Given n line segments, find if any two segments intersect
Check whether a given point lies inside a triangle or not
Polygon Clipping | Sutherland–Hodgman Algorithm
Program To Check whether a Triangle is Equilateral, Isosceles or Scalene
Check if two given circles touch or intersect each other
Window to Viewport Transformation in Computer Graphics with Implementation
Sum of Manhattan distances between all pairs of points
Optimum location of point to minimize total distance
Program to find area of a triangle