What is the minimum number of 2-input NOR gates required to implement a 4-variable function expressed in sum of-min-terms form as f = ∑(0, 2, 5, 7, 8, 10, 13, 15)?

Assume that all the inputs and their complements are available.

This question was previously asked in
GATE CS 2019 Official Paper
View all GATE CS Papers >

Answer (Detailed Solution Below) 3

Free
GATE CS Full Mock Test
5.3 K Users
65 Questions 100 Marks 180 Mins

Detailed Solution

Download Solution PDF

Let us assume that A, B, C, and D are the 4 variables.

K- map for given function:

F1 R.S M.P 19.08.19 D1.1

f(A, B, C, D) = BD + B̅ D̅

Implementing the above function by using XOR gates with complements of variables:

\(f\left( {A,\;B,\;C,\;D} \right) = \overline {\overline {B + \bar D} + \overline {\bar B + D} }\)

\(f\left( {A,\;B,\;C,\;D} \right) = \left( {B + \bar D} \right).\left( {\bar B + D} \right)\)

f(A, B, C, D) = BD + B̅ D̅

Therefore, the minimum number of 2-input NOR gates needed to implement the function is 3.
Latest GATE CS Updates

Last updated on Jan 8, 2025

-> GATE CS 2025 Admit Card has been released on 7th January 2025.

-> The exam will be conducted on 1st February 2025 in 2 shifts.

-> Candidates applying for the GATE CE must satisfy the GATE Eligibility Criteria.

-> The candidates should have BTech (Computer Science). Candidates preparing for the exam can refer to the GATE CS Important Questions to improve their preparation.

-> Candidates must check their performance with the help of the GATE CS mock tests and GATE CS previous year papers for the GATE 2025 Exam.

Hot Links: teen patti club teen patti game paisa wala lucky teen patti teen patti all games