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Palindrome Bit

To Find the bit k’th position from right :

                                                            “   x & (  1 << (  k - 1  )  )  

// This function returns true if k'th bit in x is set (or 1).
// For example if x (0010) is 2 and k is 2, then it returns true
bool isKthBitSet(unsigned int x, unsigned int k)
    return (x & (1 << (k-1)))? true: false;
// This function returns true if binary representation of x is
// palindrome. For example (1000...001) is paldindrome
bool isPalindrome(unsigned int x)
    int l = 1; // Initialize left position
    int r = sizeof(unsigned int)*8; // initialize right position
    // One by one compare bits
    while (l < r)
        if (isKthBitSet(x, l) != isKthBitSet(x, r))
            return false;
        l++;     r--;
    return true;

The number from right to left and generate another number using these bits from left to right.

int isBitPalindrome(int x) {
    int reversed = 0, aux = x;
    while (aux > 0) {
                       Taking LSB of aux and putting it as LSB of reversed
                       variable. Before doing that shifting reversed to 
                       right, to build it from left to right.
                     reversed = (reversed  << 1) | (aux & 1);

                      Loop is depend of number of bits in aux. Taking next bit into 
                      LSB position by shifting aux toward right once.
                     aux = aux >> 1;
    return (reversed  == x) ? 1 : 0;


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Given a City table, whose fields are described as +-------------+----------+ | Field       | Type     | +-------------+----------+ | ID          | int(11)  | | Name        | char(35) | | CountryCode | char(3)  | | District    | char(20) | | Population  | int(11)  | +-------------+----------+
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