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Code Bit Logic

# Add two numbers using bitwise operators

The problem at hand involves adding two integers using bitwise operators. Instead of using the conventional arithmetic addition, this approach utilizes bitwise operators to perform the addition operation. The goal is to develop a program that can add two numbers efficiently using bitwise operations.

## Problem Statement and Description

Given two integers `a` and `b`, the task is to add these numbers using bitwise operators without using the conventional `+` operator. This involves simulating the binary addition process using bitwise AND, XOR, and left shift operations.

## Example

Consider the numbers 8 and -2. In binary form, 8 is represented as 1000, and -2 is represented as 11111111111111111111111111111110 (32-bit representation). When we add these two numbers using the bitwise addition approach, we get the result 6 in binary form (0110).

## Idea to Solve the Problem

The key idea to solve this problem lies in simulating the binary addition process using bitwise operations. Specifically, we use bitwise AND (`&`), bitwise XOR (`^`), and left shift (`<<`) operations to compute the sum of two numbers.

Tip

The time complexity increases when the bits are added by the bitwise operator. Because we will need a loop to do the addition.

## Pseudocode

Here's the pseudocode for the algorithm:

``````function addition(a, b):
print "[(a) + (b)] ",
carry_bit = 0
while b is not 0:
carry_bit = a & b
a = a ^ b
b = carry_bit << 1
print " : ", a``````

## Algorithm Explanation

1. Initialize a variable `carry_bit` to store the carry during addition.
2. Iterate in a loop until `b` becomes 0.
3. In each iteration, calculate the carry bit using bitwise AND of `a` and `b`.
4. Update `a` using bitwise XOR of `a` and `b`.
5. Update `b` by shifting the carry bit one position to the left.
6. After the loop completes, the value of `a` will contain the sum of the two numbers.
7. Print the sum.

## Resultant Output Explanation

For the given code and the provided test cases, let's break down the output:

1. For the numbers 8 and -2:

• Binary representation: 1000 and 11111111111111111111111111111110
• The bitwise addition process results in: 0110
• The output is "[(8) + (-2)] : 6".
2. For the numbers 5 and 3:

• Binary representation: 101 and 11
• The bitwise addition process results in: 1000
• The output is "[(5) + (3)] : 8".
3. For the numbers -3 and -5:

• Binary representation: 11111111111111111111111111111101 and 11111111111111111111111111111011
• The bitwise addition process results in: 11111111111111111111111111111000
• The output is "[( -3) + (-5)] : -8".

## Time Complexity

The time complexity of the algorithm primarily depends on the number of bits in the binary representation of the input numbers. The loop iterates as long as the value of `b` is not zero, and each iteration involves bitwise operations that are generally constant time. Therefore, the time complexity can be approximated as O(N), where N is the number of bits in the larger input number.

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