Given an even number N which represents the number of sides of a regular polygon with N vertices, the task is to find the square of the minimum size such that given Polygon can completely embed in the square.
A Polygon is a convex figure and has equal sides and equal angles. All sides have length 1.
Embedding: Place Polygon in the square in such way that each point which lies inside or on a border of N should also lie inside or on a border of the square.
Examples:
Input: N = 4
Output: 1
Explanation:
Regular polygon with 4 Sides is square with side 1.
Given polygon can easily embed on the square with side 1.
Input: N = 6
Output: 1.931851653
Explanation :
Regular polygon with 6 Sides is Hexagon with side 1.
Given polygon can easily embed on the square with side 1.931851653.
Approach: The idea is to observe that on a 3-D plane, when a polygon is embedded in a square, it might be rotated. A similar approach has been discussed in Hexagon problem and Octagon problem . Therefore, we take the projection of each side on both the axis using the mathematical functions sin() and cos(). The overall sum of all the projections is the minimum side of the square required in this problem.
Below is the implementation of the above approach:
C++
// C++ program to find the minimum // side of the square in which // a regular polygon with even sides // can completely embed #include <bits/stdc++.h> using namespace std; // PI value in C++ using // acos function const double pi = acos(-1.0); // Function to find the minimum // side of the square in which // a regular polygon with even sides // can completely embed double nGon(int N) { // Projection angle variation // from axes double proAngleVar; // Projection angle variation // when the number of // sides are in multiple of 4 if (N % 4 == 0) { proAngleVar = pi * (180.0 / N) / 180; } else { proAngleVar = pi * (180.0 / (2 * N)) / 180; } // Distance between the end points double negX = 1.0e+99, posX = -1.0e+99, negY = 1.0e+99, posY = -1.0e+99; for (int j = 0; j < N; ++j) { // Projection from all N points // on X-axis double px = cos(2 * pi * j / N + proAngleVar); // Projection from all N points // on Y-axis double py = sin(2 * pi * j / N + proAngleVar); negX = min(negX, px); posX = max(posX, px); negY = min(negY, py); posY = max(posY, py); } // Maximum side double opt2 = max(posX - negX, posY - negY); // Return the portion of side // forming the square return (double)opt2 / sin(pi / N) / 2; } // Driver code int main() { int N = 10; cout << nGon(N); return 0; } |
Java
// Java program to find the minimum // side of the square in which // a regular polygon with even sides // can completely embed class GFG{ // PI value in Java using // acos function static double pi = Math.acos(-1.0); // Function to find the minimum // side of the square in which // a regular polygon with even sides // can completely embed static double nGon(int N) { // Projection angle variation // from axes double proAngleVar; // Projection angle variation // when the number of // sides are in multiple of 4 if (N % 4 == 0) { proAngleVar = pi * (180.0 / N) / 180; } else { proAngleVar = pi * (180.0 / (2 * N)) / 180; } // Distance between the end points double negX = 1.0e+99, posX = -1.0e+99, negY = 1.0e+99, posY = -1.0e+99; for (int j = 0; j < N; ++j) { // Projection from all N points // on X-axis double px = Math.cos(2 * pi * j / N + proAngleVar); // Projection from all N points // on Y-axis double py = Math.sin(2 * pi * j / N + proAngleVar); negX = Math.min(negX, px); posX = Math.max(posX, px); negY = Math.min(negY, py); posY = Math.max(posY, py); } // Maximum side double opt2 = Math.max(posX - negX, posY - negY); // Return the portion of side // forming the square return (double)opt2 / Math.sin(pi / N) / 2; } // Driver code public static void main(String[] args) { int N = 10; System.out.printf("%.5f",nGon(N)); } } // This code is contributed by 29AjayKumar |
C#
// C# program to find the minimum // side of the square in which // a regular polygon with even sides // can completely embed using System; class GFG{ // PI value in Java using // acos function static double pi = Math.Acos(-1.0); // Function to find the minimum // side of the square in which // a regular polygon with even sides // can completely embed static double nGon(int N) { // Projection angle variation // from axes double proAngleVar; // Projection angle variation // when the number of // sides are in multiple of 4 if (N % 4 == 0) { proAngleVar = pi * (180.0 / N) / 180; } else { proAngleVar = pi * (180.0 / (2 * N)) / 180; } // Distance between the end points double negX = 1.0e+99, posX = -1.0e+99, negY = 1.0e+99, posY = -1.0e+99; for (int j = 0; j < N; ++j) { // Projection from all N points // on X-axis double px = Math.Cos(2 * pi * j / N + proAngleVar); // Projection from all N points // on Y-axis double py = Math.Sin(2 * pi * j / N + proAngleVar); negX = Math.Min(negX, px); posX = Math.Max(posX, px); negY = Math.Min(negY, py); posY = Math.Max(posY, py); } // Maximum side double opt2 = Math.Max(posX - negX, posY - negY); // Return the portion of side // forming the square return (double)opt2 / Math.Sin(pi / N) / 2; } // Driver code public static void Main() { int N = 10; Console.Write(string.Format("{0:F5}", nGon(N))); } } // This code is contributed by Nidhi_biet |
3.19623
Time Complexity: O(N), where N is number of sides of polygon.
