Problem:
Problem E: Symmetry

Open Binary and Object Group organizes a programming contest every year. Mr. Hex belongs to this
group and joins the judge team of the contest. This year, he created a geometric problem with its solution
for the contest. The problem required a set of points forming a line-symmetric polygon for the input.
Preparing the input for this problem was also his task. The input was expected to cover all edge cases, so
he spent much time and attention to make them satisfactory.

However, since he worked with lots of care and for a long time, he got tired before he finished. So He
might have made mistakes - there might be polygons not meeting the condition. It was not reasonable to
prepare the input again from scratch. The judge team thus decided to find all line-asymmetric polygons
in his input and fix them as soon as possible. They asked a programmer, just you, to write a program to
find incorrect polygons.

You can assume the following:

Edges of the polygon must not cross or touch each other except for the end points of adjacent
edges.

It is acceptable for the polygon to have adjacent three vertexes on a line, but in such a case, there
must be the vertex symmetric to each of them.

Input

The input consists of a set of points in the following format.

N

x1 y1

x2 y2

...

xN yN

The first line of the input contains an integer N (3 ≤ N ≤ 1000), which denotes the number of points.
The following N lines describe each point. The i-th line contains two integers x1, y1 (-10000 ≤ xi, yi ≤ 10000), which denote the coordinates of the i-th point.

Note that, although the points are the vertexes of a polygon, they are given in an artibrary order, not
necessarily clockwise or counterclockwise.

Output

Output "Yes" in a line if the points can form a line-symmetric polygon, otherwise output "No".

Sample Input 1

4
0 1
1 0
0 0
1 1

Sample Output 1

Yes

Sample Input 2

4
0 1
1 -1
0 0
1 1

Sample Output 2

No

Sample Input 3

9
-1 1
0 1
1 1
-1 0
0 0
1 0
-1 -1
0 -1
1 -1

Sample Output 3

No

Sample Input 4

3
-1 -1
0 0
1 1

Sample Output 4

No

Sample Input 5

4
0 2
0 0
-1 0
1 0

Sample Output 5

Yes

Solution in JAVA (judge verdict: Wrong Answer)

import java.awt.geom.Line2D;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class Main{
	public static void main(String[] args) {
		new Main().solver();

	}

	double EPS = 1e-2;

	void solver() {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		double gx = 0, gy = 0;
		double[] x = new double[N];
		double[] y = new double[N];
		for (int i = 0; i < N; i++) {
			x[i] = sc.nextInt();
			y[i] = sc.nextInt();
			gx += x[i];
			gy += y[i];
		}
		gx /= N;
		gy /= N;
		int[] ord = convexHull(x, y);
		boolean[] vertex_of_convexHull = new boolean[N];
		for (int i = 0; i < ord.length; i++) {
			vertex_of_convexHull[ord[i]] = true;
		}

		double[] angle = new double[ord.length];

		for (int i = 0; i < ord.length; i++) {
			int n0 = ord[i];
			int n1 = ord[(i - 1 + ord.length) % ord.length];
			int n2 = ord[(i + 1) % ord.length];
			angle[i] = angle((x[n1] - x[n0]), (y[n1] - y[n0]), (x[n2] - x[n0]), (y[n2] - y[n0]));
		}

		for (int i = 0; i < angle.length; i++) {

			// the middle point between p_(i+1) and p_(i-1)
			double[] p = { (x[ord[(i + 1) % angle.length]] + x[ord[(i - 1 + angle.length) % angle.length]]) / 2.0,
					(y[ord[(i + 1) % angle.length]] + y[ord[(i - 1 + angle.length) % angle.length]]) / 2.0 };
			// the formula for the pseudo symmetric line
			double dy = (p[1] - y[ord[(i + 1) % ord.length]]);
			double dx = p[0] - x[ord[(i + 1) % ord.length]];
			double a, b, c;
			// ax+b-cy=0;
			// ax+b-cy=0;
			if (dy == 0) {
				a = 1;
				b = -p[0];
				c = 0;
			} else {
				a = -dx / dy;
				b = -a * p[0] + p[1];
				c = 1;
			}
			// tr(c * gy - a * gx - b, 0);
			if (!eq(c * gy - a * gx - b, 0))
				continue;
			boolean flag = true;
			for (int j = 1; j < angle.length / 2; j++) {
				if (!eq(angle[(i + j) % angle.length], angle[(i - j + angle.length) % angle.length])) {
					flag = false;
					break;
				}
			}
			if (flag) {
				ArrayList<Pair> list = new ArrayList<>();
				for (int j = 0; j < N; j++) {
					if (!vertex_of_convexHull[j] || (angle.length % 2 == 0 && ord[angle.length / 2] == j)) {
						list.add(new Pair(angle(x[j] - x[ord[i]], y[j] - y[ord[i]], 1, -a),
								Math.abs(a * x[j] + b - c * y[j]) / (Math.sqrt(a * a + c * c))));
					}
				}
				list.sort(null);
				for (int j = 0; j < list.size(); j++) {
					while (list.size() > 0 && eq(list.get(j).dist, 0)) {
						list.remove(j);
					}
					while (j + 1 < list.size() && eq(list.get(j).angle, list.get(j + 1).angle)
							&& eq(list.get(j).dist, list.get(j + 1).dist)) {
						list.remove(j);
						list.remove(j);
					}
				}
				if (list.isEmpty()) {
					System.out.println("Yes");
					return;
				}
			}
		}

		for (int i = 0; i < angle.length; i++) {
			// the middle point between p_(i+1) and p_(i-1)
			double[] p = { (x[ord[i]] + x[ord[(i - 1 + angle.length) % angle.length]]) / 2.0,
					(y[ord[i]] + y[ord[(i - 1 + angle.length) % angle.length]]) / 2.0 };
			// The symmetric line must go on the center of gravity
			double dy = p[1] - y[ord[i]];
			double dx = p[0] - x[ord[i]];
			double a, b, c;
			// ax+b-cy=0;
			if (dy == 0) {
				a = 1;
				b = -p[0];
				c = 0;
			} else {
				a = -dx / dy;
				b = -a * p[0] + p[1];
				c = 1;
			}
			if (!eq(c * gy - a * gx - b, 0))
				continue;
			boolean flag = true;
			for (int j = 0; j < angle.length / 2; j++) {
				if (!eq(angle[(i + j) % angle.length], angle[((i - 1) - j + angle.length) % angle.length])) {
					flag = false;
					break;
				}
			}
			if (flag) {
				ArrayList<Pair> list = new ArrayList<>();
				for (int j = 0; j < N; j++) {
					if (!vertex_of_convexHull[j] || (angle.length % 2 == 1 && ord[angle.length / 2 + 1] == j)) {
						list.add(new Pair(angle(x[j] - p[0], y[j] - p[1], 1, a),
								Math.abs(a * x[j] + b - c * y[j]) / (Math.sqrt(a * a + c * c))));
					}
				}
				list.sort(null);
				for (int j = 0; j < list.size(); j++) {
					while (list.size() > 0 && eq(list.get(j).dist, 0))
						list.remove(j);
					while (j + 1 < list.size() && eq(list.get(j).angle, list.get(j + 1).angle)
							&& eq(list.get(j).dist, list.get(j + 1).dist)) {
						list.remove(j);
						list.remove(j);
					}
				}
				if (list.isEmpty()) {
					System.out.println("Yes");
					return;
				}
			}
		}
		System.out.println("No");

	}

	double angle(double x1, double y1, double x2, double y2) {
		double rad = Math
				.acos(dot_product(x1, y1, x2, y2) / (Math.sqrt(x1 * x1 + y1 * y1) * Math.sqrt(x2 * x2 + y2 * y2)));
		// return Math.toDegrees(rad);
		return rad;
	}

	double dot_product(double x1, double y1, double x2, double y2) {
		return x1 * x2 + y1 * y2;
	}

	public static int[] convexHull(final double[] x, final double[] y) {
		int n = x.length;
		Integer[] ord = new Integer[n];
		for (int i = 0; i < n; i++)
			ord[i] = i;
		Arrays.sort(ord, new Comparator<Integer>() {
			@Override
			public int compare(Integer a, Integer b) {
				if (x[a] != x[b])
					return x[a] < x[b] ? -1 : 1;
				if (y[a] != y[b])
					return y[a] < y[b] ? -1 : 1;
				return 0;
			}
		});

		int[] ret = new int[n + 1];
		int p = 0;
		for (int i = 0; i < n; i++) {
			if (p >= 1 && x[ret[p - 1]] == x[ord[i]] && y[ret[p - 1]] == y[ord[i]])
				continue;
			while (p >= 2 && Line2D.relativeCCW(x[ret[p - 2]], y[ret[p - 2]], x[ret[p - 1]], y[ret[p - 1]], x[ord[i]],
					y[ord[i]]) == 1)
				p--;
			ret[p++] = ord[i];
		}

		int inf = p + 1;
		for (int i = n - 2; i >= 0; i--) {
			if (x[ret[p - 1]] == x[ord[i]] && y[ret[p - 1]] == y[ord[i]])
				continue;
			while (p >= inf && Line2D.relativeCCW(x[ret[p - 2]], y[ret[p - 2]], x[ret[p - 1]], y[ret[p - 1]], x[ord[i]],
					y[ord[i]]) == 1)
				p--;
			ret[p++] = ord[i];
		}

		return Arrays.copyOf(ret, p - 1);
	}

	void tr(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

	class Pair implements Comparable<Pair> {
		double angle;
		double dist;

		Pair(double angle, double dist) {
			this.angle = angle;
			this.dist = dist;
		}

		@Override
		public int compareTo(Pair o) {
			int ret = Double.compare(this.dist, o.dist);
			return (ret != 0) ? ret : Double.compare(this.angle, o.angle);
		}
	}

	boolean eq(double a, double b) {
		return Math.abs(a - b) < EPS;
	}

}