Can't understand the Fredkin gate well

[James LeBron]

I'm having trouble understanding why the Fredkin gate is universal. I am getting nowhere when I am trying to build a NOR gate, which would prove that the Fredkin is universal.

(There's 3 inputs, C, I1, and I2)

[Bill Simpson]

I believe you can implement a NOR gate using only 2 Fredkin gates.

Hint: use one of them as an inverter. You implement that by doing ...

And then the other one must be a ... And you implement that by doing ...

[James LeBron]

Thanks for the response. I actually figured it out earlier, since I was confused as to what the inputs meant at first. I thought that inputs A and B that were to be NOR-ed together were the bottom 2 inputs (aka I1 and I2 that I said on the first post) on the gate. But it's actually better to have input A be on the top of the first gate, and then input B on top of the second gate.

Then, for inputs I1 and I2 on the first bottom gate, set them as 1 and 0, respectively. That way, we pass on "A-not" to the second gate (If A = 0, it passes 1; If A = 1, the Fredkin gate switches I1 and I2 to pass on 0 to the next gate) as I1 (for the second gate), with I2 (for the second gate) being 0 and the second "true" input B being the top of the second gate. This way, if B = 1, then the second gate's I1 and I2 switch to pass on 0. If B = 0, then the output depends on "A-not". If A = 1, then it passes 0, which it should. If A = 0, then it passes 1, which it should, since a NOR is only true if both A and B are 0.

And you are right, only 2 Fredkin gates are necessary.