GATE CSE 2009 | Question: 6

Question

What is the minimum number of gates required to implement the Boolean function $\text{(AB+C)}$ if we have to use only $2\text{-input NOR}$ gates?

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Comments

@mohan123

Thanks

---

Can you just tell me how did you convert AB+C to (A+B)(B+C) ? Is it SOP to POS ?

---

firstly u write the wrong  AB+C to (A+B)(B+C)  

correct  AB+C to (A+C)(B+C)

Distributive Law states that the multiplication of two variables and adding the result with a variable will result in the same value as the multiplication of addition of the variable with individual variables. For example:
A + BC = (A + B) (A + C)

Answer 1

Given boolean function is 

$f = AB + C$

$\quad = (A+C) . (B +C)$ 

$\quad =((A+C)' +(B+C)')'$

Therefore, $3$ NOR gates required .

Correct Answer: $B$

Comments

beautiful explanation.

---

@Barney  Because OR distributed over AND.

---

The given function is in SOP form. You can find the POS form using K-map.

Answer 2

Here is a simple approach to these kinds of questions.

 

This can be applied to any questions and you'll get min number of gates always.

Comments

@Akash Papnai and @`JEET

thanks for the explanations!

---

Thanks for the method. Will definitely help many including me

---

Very nice method. Thanks @Akash Papnai

Answer 3

Answer should be 3 nor gates...A.B + C can be written as (A+C).(B+C) which is a OR-AND realization and can be made with 3 NOR gates as OR- AND is similar to a NOR-NOR realization...

Answer 4

Almost same solution,