Hyperfactorial in Java

Hyperfactorial of a number is the outcome of multiplying a given number of continuous numbers from 1 to that number, where each number is raised to its power.

Mathematically,

Thus,

H(1) = 1 ^ 1 = 1

H(2) = 1 ^ 1 × 2 ^ 2 = 1 × 4 = 4

H(3) = 1 ^ 1 × 2 ^ 2 × 3 ^ 3 = 1 × 4 × 27 = 108

H(4) = 1 ^ 1 × 2 ^ 2 × 3 ^ 3 × 4 ^ 4 = 1 × 4 × 27 × 256 = 27648

H(5) = 1 ^ 1 × 2 ^ 2 × 3 ^ 3 × 4 ^ 4 × 5 ^ 5 = 1 × 4 × 27 × 256 × 3125 = 86400000

Java Program to Find the Hyperfactorial Number

The following program uses the mathematical formula to find the Hyperfactorial numbers.

FileName: HyperFactorialNumber.java

  // A program that finds the first 5 hyperfactorial numbers     public class HyperFactorialNumber   {  // the method finds the hyperFactorial number using the number no  public int findHyperFactorialNo(int no)  {  int ans = 1;    for(int i = 1; i <= no; i++)  {  int hyperFactorialNo = i;    for(int j = 2; j <= i; j++)  {  hyperFactorialNo = hyperFactorialNo * i;  }    ans = ans * hyperFactorialNo;    }  return ans;  }  // main method  public static void main(String argvs[])  {  // creating an object of the class HyperFactorialNumber  HyperFactorialNumber obj = new HyperFactorialNumber();  System.out.println(“The first 5 HyperFactorial numbers are: “);  // loop to iterate till 5 to find  // the first 5 hyperFactorial numbers  for(int i = 1; i <= 5; i++)  {  int hyperFactorialNo = obj.findHyperFactorialNo(i);  System.out.print(hyperFactorialNo + ” “);  }  }  }  

Output:

The first 5 HyperFactorial numbers are:   1 4 108 27648 86400000

To find each hyperfactorial number, the above program is taking O(n ^ 2) time. We can do the optimization to reduce the time complexity. Consider the following program.

FileName: HyperFactorialNumber1.java

  // A program that finds the first 5 hyperfactorial numbers     public class HyperFactorialNumber1   {  // the method finds the value of x ^ y  public int findPow(int x, int y)  {  int ans = 1;  while(true)  {  if( y == 0)  {  break;  }  if(y % 2 == 1)  {  ans = ans * x;  y = y – 1;  }    else  {  x = x * x;  y = y / 2;  }  }    return ans;  }    // the method finds the hyperFactorial number using the number no  public int findHyperFactorialNo(int no)  {  int ans = 1;    for(int i = 1; i <= no; i++)  {  ans = ans * findPow(i , i);  }  return ans;  }  // main method  public static void main(String argvs[])  {  // creating an object of the class HyperFactorialNumber1  HyperFactorialNumber1 obj = new HyperFactorialNumber1();    System.out.println(“The first 5 HyperFactorial numbers are: “);  // loop to iterate till 5 to find  // the first 5 hyperFactorial numbers  for(int i = 1; i <= 5; i++)  {  int hyperFactorialNo = obj.findHyperFactorialNo(i);  System.out.print(hyperFactorialNo + ” “);  }  }  }  

Output:

The first 5 HyperFactorial numbers are:   1 4 108 27648 86400000

The findPow() method has the time complexity of O(log(n)), and this method has been called by a for-loop that runs with the time complexity of O(n). Thus, the total time complexity is O(nlog(n)), which is better than the previous code.

Finding Hyperfactorial Number Using Recursion

Using recursion also, one can find the HyperFactorial numbers. The recursive formula to find the hyperfactorial number is:

H(1) = 1

H(p) = H(p – 1) × p ^ p, where p >= 2

Thus,

H(2) = H(2 – 1) × 2 ^ 2 = H(1) × 4 = 1 × 4 = 4

H(3) = H(3 – 1) × 3 ^ 3 = H(2) × 27 = 4 × 27 = 108

H(4) = H(4 – 1) × 4 ^ 4 = H(3) × 256 = 108 × 256 = 27648

H(5) = H(5 – 1) × 5 ^ 5 = H(4) × 15625 = 27648 × 15625 = 86400000

Let’s translate the above recursive formula into the Java code.

FileName: HyperFactorialNumber2.java

  // A program that finds the first 5 hyperfactorial numbers recursively    public class HyperFactorialNumber2  {        // the method finds the value of x ^ y  public int findPow(int x, int y)  {  int ans = 1;  while(true)  {  if( y == 0)  {  break;  }  if(y % 2 == 1)  {  ans = ans * x;  y = y – 1;  }    else  {  x = x * x;  y = y / 2;  }  }    return ans;  }      // the method finds the hyperfactorial number recursively  public int findHyperFactorialNo(int no)  {  // handling the base case  if(no == 1)  {  return 1;  }    else  {  // recursively find the hyperfactorial numbers  return findHyperFactorialNo(no – 1) * findPow(no, no);  }    }      // main method  public static void main(String argvs[])  {    // creating an object of the class HyperFactorialNumber2  HyperFactorialNumber2 obj = new HyperFactorialNumber2();    System.out.println(“The first 5 HyperFactorial numbers are: “);      // loop to iterate till 5 to find  // the first 5 hyperFactorial numbers  for(int i = 1; i <= 5; i++)  {  int hyperFactorialNo = obj.findHyperFactorialNo(i);    System.out.print(hyperFactorialNo + ” “);    }    }  }  

Output:

The first 5 HyperFactorial numbers are:   1 4 108 27648 86400000

Next TopicIdentifier Expected Error in Java

Hyperfactorial in Java

Hyperfactorial in Java

Java Program to Find the Hyperfactorial Number

Finding Hyperfactorial Number Using Recursion

How to Take Multiple String Input in Java Using Scanner

Java constructor returns a value, but what

You may also like