I've been trying to attempt a particular question that I need to translate truth table into boolean expression but I'm completely stuck on one point now.

First, I worked it out by using Sums of product by getting X first.

(I did this by taking the inputs with X = 1)

A'B'C'D' + A'B'C'D + A'B'CD + A'BC'D' + A'BCD' + A'BCD +AB'C'D + AB'CD + ABC'D' + ABC'D

= A'B'C'(D'+D) + A'CD(B'+B) + A'BC(D+D') + AB'D(C'+C) + ABC'(D'+D) (Distributive Law)

= A'B'C' + A'CD + A'BC + AB'D + ABC' (Double Complement Law)

[Currently I'm stuck here and don't know how to proceed]


1 Answer 1


If you want to obtain a minimal DNF in this way, it is generally a good idea to follow the Quine-McCluskey algorithm (https://en.wikipedia.org/wiki/Quine-McCluskey_algorithm). Otherwise, it is easy to get stuck when you use distributivity to combine each term with just one of several possible partners (if you look at your kmap, this would correspond to arranging the $1$ into pairs in such a way that none of them combine into larger blocks).

One other problem with your computation is that you lost the term $A'BC'D'$ when forming pairs (the reason why you still ended up with five pairs os that $A'BCD$ is covered twice).

  • $\begingroup$ Oh I didn't notice that. Thank you very much! For the record, I was able to obtain a new set of equation A'B'C' + A'BD' + A'CD + AB'D + ABC' I understand that using laws of logic is hard to obtain the minimal expression. However, I would like to double confirm that, this expression can almost not be simplified any further right? I've compared it once again with the laws to no avail. $\endgroup$
    – Lippy
    Commented Mar 27, 2016 at 11:43
  • $\begingroup$ It cannot be simplified further without some kind of backtracking. If you draw these terms into your Karnaugh map, you'll see that they form a different partition of the $1$s in the middle rows (i.e. the ones for $A'$), essentially trapping you in a local minimum. To get the minimal DFA from there, you would have to tear apart $A'B'C', A'BD', A'CD$ and re-combine the terms into $A'BC,A'B'D,A'C'D'$ instead.. $\endgroup$ Commented Mar 27, 2016 at 12:23
  • $\begingroup$ Thank you once again, cleared and helped me out a lot! I'll try to figure out the rest for myself from here on out. I wish that I could rep you but I'm still new to this website. $\endgroup$
    – Lippy
    Commented Mar 27, 2016 at 12:47

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