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I got some homework for Boolean Simplification, could someone please help me and solve this because I'm literally going to rip my head apart :D I've tried this multiple times but I keep ending up doing it wrong...

A'BC'D'+ AB'C'D' + AB'CD' + ABC'D + ABCD'

Any help would be greatly appreciated!

amWhy
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Andrius Zapolskis
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  • How can you tell you're doing it wrong? Do you have an answer key? If so, please edit your post to include the answer you're aiming for. (It actually happens, though not very often, that a given solution might contain a typo). Why don't you show work from one of those multiple tries you made, so we can help steer you in the right direction. – amWhy Dec 03 '16 at 16:46
  • I destroyed my answers from stress and anger, I'm counting this to be my last resort. I'm sorry I can't update it with my last answer, but it was wrong.

    I started from this -> C'.D' (BC' + B'C') A (B'CD + BC'D)

     – Andrius Zapolskis Dec 03 '16 at 16:55
  • I'm aiming to simplify that expression as much as possible. There's no particular answer – Andrius Zapolskis Dec 03 '16 at 17:08
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    Then how can you tell your attempts ended "up doing it wrong", if there is not some answer/objective you aim to seek? Seems that your claim of "multiple" attempts, to the point that you're "literally going to rip my head apart" is an overly dramatic way of trying to make users here feel sorry for you. – amWhy Dec 03 '16 at 17:22
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    Perhaps you've encountered the expression "Be a man!" At the very least, Andrius, stop acting like a child. – amWhy Dec 03 '16 at 17:27
  • I came here to get help, not get criticised. – Andrius Zapolskis Dec 03 '16 at 17:30
  • We offer help for those who have made an effort to help themselves, and/or who work along with us in answering a question. This site is not meant to be a drive-through-answer-upon-demand site. – amWhy Dec 03 '16 at 17:32
  • Just help me get started, I didn't think that this site was a drive-through-answer-upon-demand site because I've looked all other related topics and everyone is taking part, yes.. I can't because I'm just stuck and would appreciate some help. – Andrius Zapolskis Dec 03 '16 at 17:42

1 Answers1

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$A'BC'D'+ AB'C'D' + AB'CD' + ABC'D + ABCD'$

$=A'BC'D'+ AB'D'(C'+C) + ABC'D + ABCD'$

$=A'BC'D'+ AB'D' + ABC'D + ABCD'$

I don't think it can be simplified further unless we use other constructs like $\implies$ and XOR.

Meet Taraviya
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  • Is that the final answer? – Andrius Zapolskis Dec 03 '16 at 16:52
  • I have this question, and I have never done Computer Architecture before, this really has me baffled. And if ==> and XOR are part of Boolean Algebra, than we can try and simplify this further, can't we? – Andrius Zapolskis Dec 03 '16 at 17:02
  • This boils down to the meaning of simplification. Adding more constructs can surely make the expression smaller,but does it get simplified? Smaller is not always more simple. – Meet Taraviya Dec 03 '16 at 17:09
  • My question was "Using Boolean Algebra simplify the minterm as far as you can, showing each major step that you take. The rule you are applying at each step should also be made clear" – Andrius Zapolskis Dec 03 '16 at 17:13
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    Andrius, your not in any kind of position to demand Meet "tailor" the answer to your liking. I actually, in another answer field, just completed a proof using XOR. But given your last comment, and your lack of any sort of demonstrated effort on your part, I refuse to post it now. – amWhy Dec 03 '16 at 17:18
  • I'm sorry that I'm not smart enough to understand Computer Architecture, I'm a first year University student in United Kingdom, Sheffield. I have never came across Computer Architecture, and having given a homework that includes simplifying the expression that i have posted, I was lost and I didn't even know how to start, @amWhy I just came here to get some help, maybe I did over do it with the "Going to rip my head apart" but that's how I literally feel, sorry for expressing my emotions, didn't know that was not allowed here. I appreciate all the help that was given to me by you and Taraviya – Andrius Zapolskis Dec 03 '16 at 17:36
  • This has nothing to do with "smarts". Nothing I've said applies only to geniuses. It has to do with maturity and taking owndership of your question. It has to do with being respectful to those from whom you "want an answer." And it has to do with being willing to be up front, from the get-go...No drama..."What did you try? and/or "What, specifically, do you not understand? What's keeping you from attempting. Where did the problem come from; etc etc What boolean operators have you learned? Which can you use. (i.e., many ways to provide us with the context from which we can help you the most.) – amWhy Dec 03 '16 at 18:11
  • I simplified the expression up to A+B (The + Contains a circle around it) + C'D' + D + ABC'D + ABC'D what stopping me from attempting further is, what do you do about the C'D? For example A + A' = 1 can D + D' also be = 1? – Andrius Zapolskis Dec 03 '16 at 18:16
  • And when it's 1, you dont show that in the expression, you would just take A + A' out of the expression, right? – Andrius Zapolskis Dec 03 '16 at 18:20
  • A'BC'D + AB'D' + ABC'D + ABCD' AB = (C'D + CD') AB = (1.D + 1.D') AB = (D + D') AB = (1) = AB A'BC'D' + AB'D' + AB D' (A'BC' + AB') = D (A⊕B + C') = D + C' + A⊕B + AB – Andrius Zapolskis Dec 03 '16 at 19:01
  • $(C'D+CD')(AB) = (C\oplus D)AB\neq AB$ – amWhy Dec 03 '16 at 19:57