Lychrel Number Implementation

Lychrel Number is a natural number that cannot form a palindrome through the iterative process of repeatedly reversing its digits and adding the resulting numbers. The process is sometimes called the 196-algorithm, after the most famous number associated with the process.
The first few numbers not known to produce palindromes when applying the 196-algorithm (i.e., a reverse-then-add sequence) are sometimes known as Lychrel numbers.
Examples:

Input : 56
Output : 56 is lychrel  : false
Explanation : 56 becomes palindromic after one iteration :
56 + 65 = 121

Input : 196
Output : 196 is lychrel  : true
Explanation : 196 becomes palindromic after 19 iterations :
196 + 691 = 887
887 + 788 = 1675
1675 + 5761 = 7436
7436 + 6347 = 13783
13783 + 38731 = 52514
....
16403234045 + 54043230461
70446464506 + 60546464407

The task is to find if a given number is Lycheral with given limit on number of iterations.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

1. Iterate given number of times
1. Add number to it's reverse
2. If
the newly formed number is palindrome
then
return false  // Number is not lychrel.
2. return true         // Number is lychrel

C++

 // C++ Program to check whether the given number // is Lychrel Number or not with given limit // on number of iterations. #include using namespace std;    long reverse(long); bool isPalindrome(long);    // Max Iterations static int MAX_ITERATIONS = 20;    // Function to check whether number is  // Lychrel Number string isLychrel(long number) {     for (int i = 0; i < MAX_ITERATIONS; i++)     {         number = number + reverse(number);                    if (isPalindrome(number))             return "false";     }            return "true"; }    // Function to check whether the number is // Palindrome bool isPalindrome(long number) {     return number == reverse(number); }    // Function to reverse the number long reverse(long number) {     long reverse = 0;     while (number > 0)     {         long remainder = number % 10;         reverse = (reverse * 10) + remainder;         number = number / 10;     }            return reverse; }    // drier program int main() {     long number = 295;     cout<

Java

 // Java Program to check whether the given number // is Lychrel Number or not with given limit // on number of iterations. import java.io.*;    public class LychrelNumberTest {     // Max Iterations     private static int MAX_ITERATIONS = 20;        // Function to check whether number is Lychrel Number     private static boolean isLychrel(long number)     {         for (int i = 0; i < MAX_ITERATIONS; i++)         {             number = number + reverse(number);             if (isPalindrome(number))                 return false;            }         return true;     }        // Function to check whether the number is Palindrome     private static boolean isPalindrome(final long number)     {         return number == reverse(number);     }        // Function to reverse the number     private static long reverse(long number)     {         long reverse = 0;            while (number > 0)         {             long remainder = number % 10;             reverse = (reverse * 10) + remainder;             number = number / 10;         }         return reverse;     }        // driver program     public static void main(String[] args)     {         long number = 295;         System.out.println(number + " is lychrel? "                            + isLychrel(number));     } }

Python3

 # Python3 Program to check whether the given number  # is Lychrel Number or not with given limit  # on number of iterations.     # Max Iterations  MAX_ITERATIONS = 20;     # Function to check whether number is  # Lychrel Number  def isLychrel(number):            for i in range(MAX_ITERATIONS):          number = number + reverse(number);                     if (isPalindrome(number)):              return "false";             return "true";     # Function to check whether the number  # is Palindrome  def isPalindrome(number):          return number == reverse(number);     # Function to reverse the number  def reverse(number):          reverse = 0;      while (number > 0):                  remainder = number % 10;          reverse = (reverse * 10) + remainder;          number = int(number / 10);             return reverse;     # Driver Code number = 295;  print(number," is lychrel? ",isLychrel(number));     # This code is contributed by mits

C#

 // C# Program to check whether the given number // is Lychrel Number or not with given limit // on number of iterations. using System;    class GFG {     // Max Iterations     private static int MAX_ITERATIONS = 20;        // Function to check whether number is Lychrel Number     private static bool isLychrel(long number)     {         for (int i = 0; i < MAX_ITERATIONS; i++)         {             number = number + reverse(number);             if (isPalindrome(number))                 return false;            }         return true;     }        // Function to check whether the number is Palindrome     private static bool isPalindrome( long number)     {         return number == reverse(number);     }        // Function to reverse the number     private static long reverse(long number)     {         long reverse = 0;            while (number > 0)         {             long remainder = number % 10;             reverse = (reverse * 10) + remainder;             number = number / 10;         }         return reverse;     }        // Driver program     public static void Main()     {         long number = 295;         Console.Write(number + " is lychrel? "                         + isLychrel(number));     } }    // This code is contributed by vt_m.

PHP

 0)      {          \$remainder = \$number % 10;          \$reverse = (\$reverse * 10) + \$remainder;          \$number = (int)(\$number / 10);      }             return \$reverse;  }     // Driver Code \$number = 295;  echo \$number . " is lychrel? " .              isLychrel(\$number);     // This code is contributed by mits  ?>

Output:

295 is lychrel ? true

This article is contributed by Pramod Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes arrow_drop_up

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.