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GATE | GATE-CS-2017 (Set 1) | Question 28
  • Last Updated : 20 Dec, 2017

Consider the Karnaugh map given below, where X represents “ don’t care” and blank represents 0.


Assume for all inputs (a, c, d) the respective complements (a’, b’, c’, d’)are also available. The above logic is implemented 2-input NOR gates only.
The minimum number of gates required is ____________.

Note: This questions appeared as Numerical Answer Type.
(A) 1
(B) 2
(C) 3
(D) 4

Answer: (A)

Explanation: g_set1_q18
Solving the above k- map,
F(a,b,c,d)=a’c (x implies don’t care and only one don’t care is needed.)
= ((a’c)’)’ = (a + c’)’
Since complement of variables are also available. So, this can be implemented using one NOR gate.

Therefore, option A is correct


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