Prerequisites : Bitwise operators in C, Bitwise Hacks for Competitive Programming, Bit Tricks for Competitive Programming

- Compute XOR from 1 to n (direct method) :
`// Direct XOR of all numbers from 1 to n`

`int`

`computeXOR(`

`int`

`n)`

`{`

`if`

`(n % 4 == 0)`

`return`

`n;`

`if`

`(n % 4 == 1)`

`return`

`1;`

`if`

`(n % 4 == 2)`

`return`

`n + 1;`

`else`

`return`

`0;`

`}`

Input: 6 Output: 7

Refer Compute XOR from 1 to n for details.

- We can quickly calculate the total number of combinations with numbers smaller than or equal to with a number whose sum and XOR are equal. Instead of using looping (Brute force method), we can directly find it by a mathematical trick i.e.
// Refer Equal Sum and XOR for details. Answer = pow(2, count of zero bits)

- How to know if a number is a power of 2?
`// Function to check if x is power of 2`

`bool`

`isPowerOfTwo(`

`int`

`x)`

`{`

`// First x in the below expression is`

`// for the case when x is 0`

`return`

`x && (!(x & (x - 1)));`

`}`

Refer check if a number is power of two for details.

/li> - Find XOR of all subsets of a set. We can do it in O(1) time. The answer is always 0 if given set has more than one elements. For set with single element, the answer is value of single element. Refer XOR of the XOR’s of all subsets for details.
- We can quickly find number of leading, trailing zeroes and number of 1’s in a binary code of an integer in C++ using GCC. It can be done by using inbuilt function i.e.
Number of leading zeroes: builtin_clz(x) Number of trailing zeroes : builtin_ctz(x) Number of 1-bits: __builtin_popcount(x)

Refer GCC inbuilt functions for details.

- Convert binary code directly into an integer in C++.
`// Conversion into Binary code//`

`#include <iostream>`

`using`

`namespace`

`std;`

`int`

`main()`

`{`

`auto`

`number = 0b011;`

`cout << number;`

`return`

`0;`

`}`

Output: 3

- The Quickest way to swap two numbers:
a ^= b; b ^= a; a ^= b;

Refer swap two numbers for details.

- Simple approach to flip the bits of a number: It can be done by a simple way, just simply subtract the number from the value obtained when all the bits are equal to 1 .

For example:Number : Given Number Value : A number with all bits set in given number. Flipped number = Value – Number. Example : Number = 23, Binary form: 10111; After flipping digits number will be: 01000; Value: 11111 = 31;

- We can find the most significant set bit in O(1) time for a fixed size integer. For example below cocde is for 32 bit integer.
`int`

`setBitNumber(`

`int`

`n)`

`{`

`// Below steps set bits after`

`// MSB (including MSB)`

`// Suppose n is 273 (binary`

`// is 100010001). It does following`

`// 100010001 | 010001000 = 110011001`

`n |= n>>1;`

`// This makes sure 4 bits`

`// (From MSB and including MSB)`

`// are set. It does following`

`// 110011001 | 001100110 = 111111111`

`n |= n>>2;`

`n |= n>>4;`

`n |= n>>8;`

`n |= n>>16;`

`// Increment n by 1 so that`

`// there is only one set bit`

`// which is just before original`

`// MSB. n now becomes 1000000000`

`n = n + 1;`

`// Return original MSB after shifting.`

`// n now becomes 100000000`

`return`

`(n >> 1);`

`}`

Refer Find most significant set bit of a number for details.

- We can quickly check if bits in a number are in alternate pattern (like 101010). We compute n ^ (n >> 1). If n has an alternate pattern, then n ^ (n >> 1) operation will produce a number having set bits only. ‘^’ is a bitwise XOR operation. Refer check if a number has bits in alternate pattern for details.

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