Self doubt- Min no of NOR gates

Question

What is the minimum number of 2 input NOR gates required to realise:

F(A,B,C) = (A'+B')(B'+C')(C'+A')

Case 1- Compliments of A,B,C are not available

Case 2- Compliments of A,B,C are available

Comments

Input size$?$

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Since arity is not given I am taking it greater than equal to 2, then I am getting 5 NOR GATES. And if arity is 2 then it should be 7 NOR Gates.

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Sorry for incomplete question, I have added arity, 2 input nor gates we need.

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If compliments are available with 2 arity nor gates,then 4 NOR gates are needed.

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case 1.- > 9

((AB+BC)''+AC)'

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@raahul can you show how you have simplified and got 4 gates?

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please correct ...i made a misake. Acc. to me 6.

Answer 1

 

Here, points to remember are:

1.a)For NOR gates, OR-AND logic is suitable to construct with.

1.b)For NAND gates, AND-OR logic is suitable to construct with. 

So, don't try to construct the circuit directly, better start with above points to avoid errors.

2.Bubbling is just like 44 = -(-44), which is next step after starting with point 1.Here, bubble means putting a NOT gate, you all know.

3.Replace the bubbled gates with NOR or NAND accordingly.

Bubbled AND = NOR, Bubbled OR = NAND.  

So, With compliments: 6 NOR gates, Without Compliments: 9 NOR are needed in our question.