no. of gates

Question

Comments

can u plz explain sir how?

---

yup! 4 is only possible if we are allowed to use more than 2 input nor gate.

---

after solving it I am getting f =  AB+CD+AD

For AB,CD,AD 1 NOR gate is required as variables are present in complimented form also than to OR them we will require two NOR gates in this way i am getting 5. plz explain how 4 is possible?

Answer 1

 After solving with K- Map , we get expression as (A+C) . (A+D) . (B+D)

 which can be written as (A+ CD). (B+D)

                                        =  $ \overline{ \overline{(A+ CD).(B+D)} } $

                                        =  $ \overline{\overline{A+CD}+\overline{B+D}} $

                                        =  $ \overline{\overline{A+\overline{C'+D'}}+\overline{B+D}} $

As, variable are available both primed and Unprimed form. We are having 4 , 2-Input Nor Gates..