(So far, 208 people got it right out of 414 for a success rate of 50%)

(Thank you to Clive "Max" Maxfield for submitting this question.  Make sure to have a look at his "Cool Beans" blog and to read everything by Max imagining a strong British accent.)


This is a digital logic problem. Let's assume we have a "black box" with three inputs called A, B, and C, and three outputs called not_A, not_B, and not_C. We want each of the outputs to be the logical negation of its corresponding input (e.g., not_A = !A).

The question is, is this possible using only AND and OR gates along with just two (2) NOT gates, but no NAND, NOR, XOR, or XNOR gates?

Pick one:
100% Yes
66% Maybe
33% Maybe
100% No

[ - ]
Comment by SharpenNovember 12, 2019

Can you show a diagram of the gates please. I don’t understand what you’ve writte

[ - ]
Comment by anon12November 14, 2019

The gate picture for the proposed solution is quite messy, as it involves several layers of gates. You can do it yourself of course from the equations---I tried and it was a complete spaghetti, and the equations actually explain the idea behind the solution a little better.

Having said that, I came up with a different, simpler design:

Aout = not Ain

Bout = not(Bin and Cin) and Bin

Cout = not(Bin and Cin) and Cin

which is fairly easy to draw, as it involves only five total gates.

It works because not(Bin and Cin) is notBin or notCin by de Morgan laws, and when you OR it with e.g. Bin you can cancel out Bin OR notBin, so notCin is what's left.

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