How many m-input NAND gates will be required to construct a n-input NAND gate, where (m <= n)?

Question

How many $m-$ input NAND gates are needed to construct a $n-$ input NAND gate, where $m \leq n$

 

Examples:

> constructing $3-$input NAND gate using $2-$input NAND gate

> constructing $4-$input NAND gate using $2-$input NAND gate

> constructing $4-$input NAND gate using $4-$input NAND gate

 

Is there a general solution to this problem, or is there NO general solution available (i.e., a localized solution may exist)?

Comments

I don't think so, there is no such relational solution to it.. you need to draw and check the particular gate results !!

Answer 1

I did for AND gates, I belive the same logic will work for NAND gates too :

 

one every missing input we just provide 1 since $A\cdot 1 == A$, so the final result will have no change.

but in case of NAND gate, the output is $\bar A + \bar B$ so, putting $1$ in the missing input with $A$ will yield $\bar A$.
 

That can be worked around by finding how many NAND gates will have missing input, because for each, we'll need an extra NAND to invert the output.

Comments

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