Minimum side of square embedded in Regular polygon with N sides
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 |
Python3
# Python 3 program to find the minimum # side of the square in which # a regular polygon with even sides # can completely embed import math # PI value in Python 3 using # acos function 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 def nGon(N): # Projection angle variation # from axes proAngleVar = 0 # 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 negX = 1.0e + 99 posX = - 1.0e + 99 negY = 1.0e + 99 posY = - 1.0e + 99 for j in range (N): # Projection from all N points # on X-axis px = math.cos( 2 * pi * j / N + proAngleVar) # Projection from all N points # on Y-axis py = math.sin( 2 * pi * j / N + proAngleVar) negX = min (negX, px) posX = max (posX, px) negY = min (negY, py) posY = max (posY, py) # Maximum side opt2 = max (posX - negX, posY - negY) # Return the portion of side # forming the square return (opt2 / math.sin(pi / N) / 2 ) # Driver code if __name__ = = "__main__" : N = 10 print ( '%.5f' % nGon(N)) # This code is contributed by ukasp. |
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 |
Javascript
<script> // Javascript program to find the minimum // side of the square in which // a regular polygon with even sides // can completely embed // PI value in Java using // acos function let 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 function nGon( N) { // Projection angle variation // from axes let 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 let negX = 1.0e+99, posX = -1.0e+99, negY = 1.0e+99, posY = -1.0e+99; for ( let j = 0; j < N; ++j) { // Projection from all N points // on X-axis let px = Math.cos(2 * pi * j / N + proAngleVar); // Projection from all N points // on Y-axis let 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 let opt2 = Math.max(posX - negX, posY - negY); // Return the portion of side // forming the square return opt2 / Math.sin(pi / N) / 2; } // Driver code let N = 10; document.write(nGon(N).toFixed(5)); // This code is contributed by Princi Singh </script> |
3.19623
Time Complexity: O(N), where N is the number of sides of the polygon.
Auxiliary Space: O(1)