Perform XNOR of two numbers
The problem at hand involves performing an XNOR operation on two given numbers. An XNOR operation (also known as equivalence operation) takes two binary inputs and produces a single binary output. It returns 1 if both inputs are the same (either both 0 or both 1), and 0 if the inputs are different. In this article, we'll explore the problem statement, understand the concept behind the XNOR operation, provide a step-by-step approach to solving the problem, and analyze the time complexity of the solution.
Problem Statement and Description
The task is to perform the XNOR operation on two given integers,
y. The goal is to
compute the result of
x ⊙ y, where
⊙ represents the XNOR operation.
For example, if
x is 12 (binary 1100) and
y is 6 (binary 0110), then:
- The XNOR of the least significant bits is 0 (0 ⊙ 0 = 1).
- The XNOR of the next bits is 1 (1 ⊙ 1 = 1).
- The XNOR of the next bits is 1 (0 ⊙ 0 = 1).
- The XNOR of the most significant bits is 1 (0 ⊙ 0 = 1). So, the result is 0101, which is 5 in decimal.
7 in binary is 00111 18 in binary is 10010. Now we can perform the XNOR operation on each corresponding bit:
7 00111 18 10010 ------------ 1 XNOR 0 = 0 1 XNOR 1 = 1 1 XNOR 0 = 0 0 XNOR 0 = 1 0 XNOR 1 = 0
Therefore, the XNOR of 7 and 18 is 01010 in binary, which is 10 in decimal.
Idea to Solve the Problem
To perform the XNOR operation on two numbers
y, we can break down the problem into
XNOR operations on individual bits of the two numbers. We'll compare the corresponding bits of
y and compute the XNOR for each pair of bits, resulting in the corresponding bits of the final output.
Here's a high-level pseudocode for solving the problem:
function setAllBit(num): n = num n = n | n >> 1 n = n | n >> 2 n = n | n >> 4 n = n | n >> 8 n = n | n >> 16 return n function xnor(x, y): result = 0 if x > y: result = (setAllBit(x) ^ x) ^ y else: result = (setAllBit(y) ^ y) ^ x return result function main(): task = create ExclusiveNOR object task.xnor(12, 6) task.xnor(7, 18) task.xnor(45, 32)
setAllBitfunction takes an integer
numand activates all the bits after the most significant bit, effectively creating a number with all bits set to 1 up to the position of the most significant bit.
xnorfunction takes two integers,
y, and performs the XNOR operation using the
xis greater than
y, the function computes
(setAllBit(x) ^ x) ^ y.
setAllBit(x)ensures that all bits up to the most significant bit are set to 1.
- The first XNOR (
^ x) compares
- The second XNOR (
^ y) compares the result of the first XNOR with
yis greater than
x, the function computes
(setAllBit(y) ^ y) ^ xfollowing a similar logic.
mainfunction creates an
ExclusiveNORobject and calls the
xnorfunction for three sets of inputs.
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setAllBit function runs in O(1) time because it operates on a fixed number of bits, regardless of
the input value. The
xnor function has a constant number of operations for each input pair, and its
time complexity is also O(1). The
main function runs the
xnor function three times, so its
time complexity is still O(1).