Argh, I say. Argh.

[napoleonherself]

I have this circuit. I have to simplify this circuit using boolean logic. For about two and a half hours I repeatedly reworked it and got different answers. Then I finally got to the point where reworking it kept getting me the same answer. However, now I keep getting the TRUTH TABLES turning out differently. Basically, I just want to know if I'm on the right track or not.


I get that reducing to a'd + b'd + c + ab . Anyone who knows this stuff and feels like helping, am I right? Or do I need to work it again?

Comments

Re: Karnaugh Maps of DOOM!

[apm.livejournal.com]

Actually, that particular K-map reduces a little further to ab + c + d. Always overlap if you can.

Re: Karnaugh Maps of DOOM!

[captainspam.livejournal.com]

/me looks over it again

Oh, right... the large 3x4 block of 1's reduces to two 2x4 blocks. My mistake. Yeah, ab + c + d.

Does the map at least look right to anyone else?

Re: Karnaugh Maps of DOOM!

[apm.livejournal.com]

I'm pretty sure the original ab + a'd + b'd + c equates to that, since (ab + a'd + b'd) -> (ab + d(a' + b')). If d=1, then the result must be true, because:

a b d | a' + b'  ab |
0 0 1 |   1      0  | 1
0 1 1 |   1      0  | 1
1 0 1 |   1      0  | 1
1 1 1 |   0      1  | 1

Therefore, (ab + d(a' + b')) -> (ab + d).
I tried for a while to figure that out purely using boolean logic reduction rules, but failed because I am rusty.

So I assume that because your K-map corresponds to Jenny's answer, it and Jenny's answer are both probably correct.