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

[godai.livejournal.com]

just so i'm reading this right.

if a = 1 then a' = 0 right?

the primes are acting as nots?

[godai.livejournal.com]

truth table for me came out to be all 1's except for (a = 1 b,c,d = 0) and (b = 1 and a,c,d = 0)

[godai.livejournal.com]

dammit. all 0's is a zero as well.

Karnaugh Maps of DOOM!

[captainspam.livejournal.com]

All righty... /me cracks out Magic Pencil, his scrappy notebook, and his mad Discrete Mathematics and Computer Logic skillz.

(I'm assuming that's a NOT bubble at the end of the AND right before Z. So, we're trying to kill that AND gate at the end. Knowledge of this fact makes this much easier)

I'll try this by going the way of the lovely Karnaugh Map. Your friend and mine. I get a map that looks like:

ab\cd  00  01  11  10
 
 00     0   1   1   1
 01     0   1   1   1
 11     1   1   1   1
 10     0   1   1   1

That reduces down to c + ab + c'd. If you want, I'll try to type up a hideously long post on how I got there, but my twisted logic says that's right. Somehow. Extra opinions would be nice, but that sounds right to me.

The answer you got forms ALMOST the same Karnaugh Map, except with a 0 for ab/cd 01/01 (a'bc'd). Problem is, that DOES make a 1 due to the top part of the thing; ab would wind up as 0, OR that with d' to make another 0, AND that with c' to make another 0, which, in the final AND, comes out with a 0, which is NOTted into a 1 for Z.

Hope that helps... do ask if you want to know how I got there (It involves a lot of looking at that Karnaugh Map to save time)...

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.

[ex-ford479.livejournal.com]

*does it programmer style*

Z=!((((A && B) || !D) && !C) && ((((!C || D) && !B) || !A)))

*whips out PHP to do the logic for me*


AB/CD 00 01 10 11
00 0 1 1 1
01 0 1 1 1
10 0 1 1 1
11 1 1 1 1


Works out to Z = A && B || C || D

Dunno if that was the actual assignment, but that's what worked out.
*checks his initial function*
Seems to be right. Tell me if I'm a fuckup.