following

For context: a mux chip/gate takes in three inputs (a, b, sel), then outputs either input a or b depending on what the value of sel (selector) is. If sel=0, then the output should be a, but if sel=1, the output should be b.

I believe my first thought was to use the nand gate with a and sel- I can't remember exactly why- I think I wanted to return a 1 if the selector was choosing a since nand returns 1 as long as its two inputs aren't both 1 (so if a=1 and sel=1 output would be 0), but if a=1 and sel=0, it would return 1, so I had to find a way around that. I think I chose to do nand(a, sel) and not(sel) so that this would only be true if sel=0. I had no idea what to do if it was selecting for b though (or what to do after I decided the value was a or b), so I scrapped that. I might've been on the right track, though?

I think my next ideas were to try and look at it from a more theoretical way- I brute-forced the not, and, or, and xor gates by comparing the (expected) truth tables to nand's, but mux is more complex- there's three inputs! I figured the mux chip's logic had two functions- one, to decide whether the input to pass on was a or b, and two, to figure out whether to pass on 0 or 1. I couldn't figure out the logic for this though, so I tried looking at it from another approach- the output could only be a 0 or 1 regardless of what sel was, so maybe I could see what cases there'd be (e.g. out=0 when a=0 and sel=0 (regardless of b) or b=0 and sel=1 (regardless of a)), but I couldn't figure out how to "pick" which case was suitable... alas.

I almost looked up the answer, but I decided to go through the Nand2Tetris course's slides to see if there was any helpful info in there. I saw an implementation example for an "AndMuxOr" gate, and for a second, thought I got spoiled for the answer (in my defense, the truth tables looked very similar). I then saw a hint for implementing the mux gate- that it only needed the and, or, and not gates- here I was thinking it needed nand or maybe xor gates. I also realized that maybe, I didn't need to figure out whether I was picking a or b- that's the purpose of the sel input after all, isn't it?

So I went back to trying the mux implementation again- I skimmed through the book chapter again, and saw something about arithmetically representing Boolean functions, and something clicked in my head. (Well, I read it before and I preferred the arithmetic representation as opposed to prefix/word notations, but figured I should get used to the latter...) Basically, + represents or, * represents and, and an overline represents not. Since sel was 0 or 1, I could "cancel" the input that wasn't being selected, just like with multiplication. I think it's easier to explain with the equation, so here: asel + bsel
So when sel=0, we get a*0 + b*0 = a*1 + b*0 = a, and for sel=1, a*1 + b*1 = a*0 + b*1 = b

I haven't tried doing demux yet, but since demux has two outputs, I assume I'd have to break the equation for mux into two pieces...

Quick addition: I only tested my mux implementation after writing this entire section. Thankfully it worked- if it didn't after I wrote this entire section, I would've cried