How do I convert CC-NOT gates to OR operators?

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SUMMARY

The discussion focuses on converting CC-NOT gates to OR operators using Fredkin Gates. The user successfully derived AND and XOR gates but struggled with the OR operation. The solution involves using the expression NOT((NOT(INPUT1)) AND (NOT(INPUT2))) to achieve the OR functionality. This method is confirmed as a straightforward approach to recreate OR gates from existing logic gates.

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  • Understanding of Fredkin Gates and their properties
  • Knowledge of basic logic gate operations (AND, OR, NOT)
  • Familiarity with XOR gate construction
  • Basic understanding of truth tables
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  • Research the implementation of Fredkin Gates in quantum computing
  • Study the Feynman Lectures in Computing for deeper insights
  • Explore advanced logic gate constructions using CC-NOT gates
  • Learn about the applications of universal gates in digital circuits
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This discussion is beneficial for computer scientists, electrical engineers, and anyone interested in quantum computing and logic gate design.

khkwang
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Fredkin Gates are supposed to be universal. So far I've gotten AND, OR and NOT out of them but I can't figure out XOR. Any help?

I know that A XOR B = (A AND NOT B) OR (B AND NOT A), but trying to recreate that with Fredkin Gates is not very elegant... is that the only way?

Edit: I guess I can't change the title of my thread...

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nvm this bottom part of this post, it's been solved already.

According to Feynman Lectures in Computing, by using only C-NOT, CC-NOT and NOT gates we can recreate AND, OR and XOR gates.

I understand how to create AND and XOR, but I can't work out the OR. I spent a good few hours pondering and trying out different truth tables but I just don't get it. Can anyone demonstrate a way?
 
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If you have AND and NOT, then NOT ( (NOT (INPUT1)) AND (NOT (INPUT2)) ) = ((INPUT1) OR (INPUT2))
 
Ahh so simple. Can't believe I forgot something so basic. Thanks, you actually helped me clear up a whole other question as well.
 
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