Not Frequent
0/9

# Convex Hull

Authors: Benjamin Qi, Neo Wang, Chuyang Wang

Smallest convex polygon containing a set of points on a grid.

## Introduction

The Convex Hull is the subset of points that forms the smallest convex polygon which encloses all points in the set. To visualize this, imagine that each point is a pole. Then, imagine what happens if you were to wrap a rope around the outside of all the poles, and then pull infinitely hard, such that the connections between any two points that lie on the edge of the rope are lines. The set of points that touch the rope is the convex hull.

Convex Hull Visualization

Focus Problem – try your best to solve this problem before continuing!

## With Graham Scan

### Solution

C++

The Graham Scan algorithm works in 3 steps. First, it sorts all of the $n$ points by their counterclockwise angle around a pivot $P_0$, breaking ties by distance. This algorithm uses the leftmost (and bottommost if there is a tie), point as $P_0$.

We maintain a stack containing the points such that the following invariant holds: every three consecutive points $a,b,c$ of the stack form a counterclockwise turn. In other words, $c$ lies to the left of the line from $a$ to $b$. This condition implies that the points of the stack form the vertices of a convex polygon.

To start the creation of the convex hull, we choose 2 points. The pivot (first point), and the second point based on our initial sorting. After that, we attempt to add each point in $\texttt{cand}$ to the stack.

Denote our stack as $\texttt{hull}$, the top element of $\texttt{hull}$ as $\texttt{hull}[i]$ and $\texttt{cand}[j]$ as the $j$-th sorted candidate point. Before adding $\texttt{cand}[j]$ to the stack, we check whether $\texttt{hull}[i-1] \rightarrow \texttt{hull}[i] \rightarrow \texttt{cand}[j]$ forms a counterclockwise turn.

• If so, then we add $\texttt{cand}[j]$ to the stack and the invariant holds. Continue with $\texttt{cand}[j+1]$.
• Otherwise, $\texttt{hull}[i]$ lies within the convex hull of the other points in the stack along with $\texttt{cand}[j]$, so we pop $\texttt{hull}[i]$ from the stack and continue with $\texttt{cand}[j]$.

Illustration

Worked Example

#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i < (b); i++)#define FORE(i, a, b) for (int i = (a); i <= (b); i++)#define F0R(i, a) for (int i = 0; i < (a); i++)#define trav(a, x) for (auto &a : x)
#define f first

## With Monotone Chain

### Solution

With the Monotone Chain algorithm, we start by sorting the given $n$ points in ascending order with respect to their $x$ coordinates. If two points have the same $x$ coordinate, then we will look at the $y$ coordinate.

Next, we will calculate the convex hull in two parts - the upper and the lower hull. Firstly, we observe that the starting and ending points of both upper and lower hulls are the same. They are the points with the lowest and highest $x$ value respectively, $P_0$ and $P_{n-1}$. We start by adding $P_0$ and $P_1$ to the hull. (Note that $P_1$ doesn't necessarily have to be on the convex hull at the end). Then, starting from $P_2$, we iterate through the sorted points and add them to the hull. Let's denote the current point being added as $P_{k}$ and the last point still on the hull as $P_{i}$. When adding new points, we want to make sure that there is no right turn among all segments of the hull, just like in the Graham Scan algorithm discussed above. To achieve this, the segment $P_{i-1}P_i$ should always be on the right side of the segment $P_{i-1}P_k$. This can be calculated by using a cross-product:

• If $(P_i - P_{i-1}) \times (P_k - P_{i-1}) < 0$, the point $P_i$ lies on the left side of the segment $P_{i-1}P_k$. In this case, we have to remove point $P_i$ from the hull and repeat this check.
• If $(P_i - P_{i-1}) \times (P_k - P_{i-1}) = 0$, the point $P_i$ lies on the segment $P_{i-1}P_k$. Whether to include multiple collinear points depends on the question, but for the given question above, we will remove the point $P_i$ as well and repeat the check.
• Otherwise, the point $P_i$ lies on the right side of the segment $P_{i-1}P_k$. In this case, we can add $P_k$ to the hull and process the next point $P_{k+1}$ from the given point list.

After all the points have been processed, we have found the lower hull and will begin to find the upper hull in the same manner. This time, we add point $P_{n-2}$ to the hull and iterate from the end of the points, $P_{n-3}$, to the starting point $P_0$. (The point $P_{n-2}$ also doesn't necessarily have to be the convex hull and could be removed if it causes a right turn).

At the end, we have got all points of the convex hull in the counterclockwise order. To do this in the clockwise order, one only has to change the condition for $(P_i - P_{i-1}) \times (P_k - P_{i-1})$ from more than 0 to less than 0. In this case, the upper hull will be found first and then the lower hull.

This algorithm takes $\mathcal{O}(n \log n)$ time to sort the points and $\mathcal{O}(n)$ time to calculate the hull, giving a final time complexity of $\mathcal{O}(n \log n)$.

An animation of how it works:

Worked Example

C++

#include <bits/stdc++.h>using namespace std;
using pii = pair<int, int>;
vector<pii> points;vector<pii> hull;
// cross product, the signed area of these there pointsint area(pii O, pii P, pii Q) {

Java

import java.io.*;import java.util.*;
public class ConvexHull {	public static void main(String[] args) throws IOException {		BufferedReader in =		    new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(in.readLine());		while (N > 0) {

## Rotating Calipers

Focus Problem – try your best to solve this problem before continuing!

### This section is not complete.

Any help would be appreciated! Just submit a Pull Request on Github.

## Problems

StatusSourceProblem NameDifficultyTags
CFEasy
Show TagsConvex
PlatNormal
Show TagsConvex
CFNormal
Show TagsConvex, PURS
Old GoldNormal
Show TagsConvex
KattisHard
Show TagsConvex
Balkan OIHard
Show TagsConvex, MST
ACVery Hard
Show TagsConvex

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