GATE CS 2016 Set 1 Q.16 (CS Q.6): Boolean Algebra and Logic Minimization

Consider the Boolean operator $\#$ with the following properties:
$x \# 0 = x$ , $x \# 1 = \bar{x}$ , $x \# x = 0$ and $x \# \bar{x} = 1$ . Then $x \# y$ is equivalent to

$x\bar{y} + \bar{x}y$

$x\bar{y} + \bar{x}\bar{y}$

$\bar{x}y + xy$

$xy + \bar{x}\bar{y}$

Official answer and solution

Official answer: A
From: Official GATE 2016 answer key

Solution

Match the given operational properties to the fundamental Boolean XOR operator.

Examine the properties given for the binary operator $\#$ :

$x \# 0 = x$
$x \# 1 = \bar{x}$
$x \# x = 0$
$x \# \bar{x} = 1$

Compare these properties with standard Boolean operations:

For the exclusive OR (XOR) operator $\oplus$ $\oplus$ :
- $x \oplus 0 = x \cdot 1 + \bar{x} \cdot 0 = x$
- $x \oplus 1 = x \cdot 0 + \bar{x} \cdot 1 = \bar{x}$
- $x \oplus x = x\bar{x} + \bar{x}x = 0$
- $x \oplus \bar{x} = x \cdot x + \bar{x} \cdot \bar{x} = x + \bar{x} = 1$
  All four axioms uniquely identify $\#$ as the XOR operator.

The sum-of-products (SOP) boolean expression for XOR is:

x \oplus y = x\bar{y} + \bar{x}y

Final answer

Common trap

Confusing XOR ( $x\bar{y} + \bar{x}y$ ) with XNOR ( $xy + \bar{x}\bar{y}$ ), which yields $x \odot 0 = \bar{x}$ and $x \odot x = 1$ .

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2016 Computer Science and Information Technology (CS) Session 1
Question: 16 of 65, printed as CS Q.6
Section: Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.

Question

Solution

Source