How to find out all palindromic numbers

Question

A palindromic number or numeral palindrome is a \"symmetrical\" number like 16461, that remains the same when its digits are reversed.

The term palindromic is derive

Answer 1

I wrote these methods in C# which may be of some help. The main method builds a definitive list of all palindromic numbers up to the given number of digits. Its fast and commented throughout to help explain the processes I have used.

I've also included some support methods including a fast palindromic test and its worth pointing out that pow10[x] is an array of the powers of 10 to further improve speed.

                    lhs *= 10;
                    rhs /= 10;
                }
                palindrome = MathExt.Concat( lhs * 10, MathExt.ReverseDigits( rhs ) );      // Multiplying the lhs by 10 is equivalent to adding b == 0
                result.Add( palindrome );       // Add numbers of the form aaa + 0 + aaa

                lhs = a;
                for ( ulong b = 1; b != 10; b++ )
                {
                    rhs = a * 10 + b; // Adding b before we reverse guarantees that there is no trailing 0s
                    palindrome = MathExt.Concat( lhs, MathExt.ReverseDigits( rhs ) );       // Works except when b == 0
                    result.Add( palindrome );       // Add numbers of the form aaa + b + aaa
                }
                a++;
            }
            pow *= 10;        // Each pass of the outer loop add an extra digit to aaa
        } 
        return (result);
    }


    /// <summary>
    ///  Reverses the digits in a number returning it as a new number. Trailing '0's will be lost.
    /// </summary>
    /// <param name="n">The number to reverse.</param>
    /// <param name="radix">The radix or base of the number to reverse.</param>
    /// <returns>The reversed number.</returns>
    static public ulong ReverseDigits( ulong n, uint radix = 10 )
    {
        // Reverse the number
        ulong result = 0;
        do
        {
            // Extract the least significant digit using standard modular arithmetric
            result *= radix;
            result += n % radix;
            n /= radix;
        } while ( n != 0 );
        return (result);
    }


/// <summary>
    /// Concaternates the specified numbers 'a' and 'b' forming a new number 'ab'.
    /// </summary>
    /// <example>If a = 1234 and b = 5678 then Concat(a,b) = 12345678.</example>
    /// <param name="a">The first number.</param>
    /// <param name="b">The second number.</param>
    /// <returns>The concaternated number 'ab'.</returns>
    public static ulong Concat( this ulong a, ulong b )
    {
        // Concaternate the two numbers by shifting 'a' to the left by the number of digits in 'b' and then adding 'b'
        return (a * pow10[NumberOfDigits( b )] + b);
    }

    /// <summary>
    /// Evaluate whether the passed integer is a palindrome in base 10 or not.
    /// </summary>
    /// <param name="n">Integer to test.</param>
    /// <returns>True - Palindrome, False - Non palindrome.</returns>
    static public bool IsPalindrome( this ulong n )
    {
        uint divisor = NumberOfDigits( n ) - 1;
        do
        {
            // Extract the most and least significant digits of (n)
            ulong msd = n / pow10[divisor];
            ulong lsd = n % 10;

            // Check they match!
            if ( msd != lsd )
                return (false);

            // Remove the msd and lsd from (n) and test the next most and least significant digits.
            n -= msd * pow10[divisor];  // Remove msd
            n /= 10;                    // Remove lsd
            divisor -= 2;               // Number has reduced in size by 2 digits 
        } while ( n != 0 );
        return (true);
    }

Answer 2

import Queue
import copy

def printPalindromesTillK(K):
    q = Queue.Queue(K);
    for i in range(0, 10):
        q.put(str(i));
        q.put(str(i) + str(i));
    while(not q.empty()):
        elem = q.get();
        print  elem;
        for i in range(1, 10):
            item = str(i) + elem + str(i);
            if int(item) <= K:
               q.put(item); 

print printPalindromesTillK(10000);