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The minimum number of NAND gates requires to implement A ⊕ B ⊕ C is
  • a)
    8
  • b)
    10
  • c)
    9
  • d)
    6
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The minimum number of NAND gates requires to implement A⊕ B&oplu...
The Sum output bit of a full adder is given by:
Looking at the Full-Adder circuit, we can see that,

S = A ⊕ B ⊕ C
From full adder circuit,
Cout =  AB + Cin (A ⊕ B)
Now, logic gate for above boolean expression can be drawn as,

Connecting them to a NAND gate, we now have the Full-Adder NAND Equivalent.

Hence, 9 minimum number of NAND gates requires to implement A ⊕ B ⊕ C or Full-Adder circuit.
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Community Answer
The minimum number of NAND gates requires to implement A⊕ B&oplu...
The minimum number of NAND gates required to implement A depends on the specific logic function or circuit being implemented.

If A is a single input and single output logic gate, such as an inverter or NOT gate, then only one NAND gate is required.

If A is a two-input logic gate, such as an AND gate or OR gate, then two NAND gates are required. One NAND gate can be used to implement the AND or OR function, and another NAND gate can be used as an inverter if necessary.

If A is a more complex logic circuit with multiple inputs and outputs, the minimum number of NAND gates required will depend on the specific circuit design and logic function being implemented. In general, the number of NAND gates required will be determined by the number of inputs, outputs, and the desired logic function.
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The minimum number of NAND gates requires to implement A⊕ B⊕ C isa)8b)10c)9d)6Correct answer is option 'C'. Can you explain this answer?
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