Can a Circuit with 2 Not Gates Negate 3 Inputs?

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SUMMARY

The discussion centers on designing a circuit that negates three inputs (A, B, C) using only two NOT gates alongside an unlimited number of AND and OR gates. The consensus is that while three NOT gates can easily achieve this, the challenge lies in proving whether two NOT gates can suffice. The proposed solution involves using a majority circuit to determine the number of 1 bits among the inputs and subsequently deriving the negated outputs based on the input combinations. The discussion also touches on the limitations of embedding the circuit for additional negations.

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  • Understanding of digital logic gates, specifically NOT, AND, and OR gates.
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implet
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Hi,

Homework Statement


You are trying to design a circuit with 3 inputs and 3 outputs. The circuit should negate each of its inputs. e.g. A-> not A, B-> not B, C-> not C. Clearly we can do this using 3 'not' gate. Is it possible to do this using 2 'not' gates and an unlimited number of 'and' and 'or' gates?


Homework Equations


None I know of.


The Attempt at a Solution


I'm not sure whether this is true.
To prove it's true: Find a circuit. I've tried many with no success.
To prove it's false: Use an information based argument? Use properties of the different gates?

Thanks :)

P.S. This is an OCW question from 6.080
 
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Someone previously asked the same question on this forum: link
 
try to make a two bit signal that represents the number of 1 bits in the 3 inputs.
these are the two bits you have to invert.

the high bit of this is easy, that is a 3-way majority circuit.

you can use the inverted high bit to compute the low bit.

Once you have that, you can make a decoder that produces 4 outputs
that are 1 if 0,1,2 or 3 of the input bits are equal to 1.

finally to calculate a' you can comtine the cases for 0,1 and 2 1-bits in the input.

if there are 0 1-bits in the input then a must be 0, so a' = 1
if there is 1 1-bit in the input then a is 0 if b and c are 1.
if there are 2 1-bits in the input then ...

and then you can do the same for b' and c'

Finally. What is the problem you run into, if you try to embed this circuit into another copy of it to get more than 2 nots from 2 not gates?
 

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