Coder Posted May 23, 2017 Share Posted May 23, 2017 (edited) How can I minimize these two Boolean expressions in shorter form. I have tried quite hard to solve but unable to do so. I can't apply any of the basic postulates here. Please help me. Edited May 23, 2017 by Coder Link to comment Share on other sites More sharing options...
Endy0816 Posted May 23, 2017 Share Posted May 23, 2017 You can solve them via a truth table. Only one of the variables is determining the result. Link to comment Share on other sites More sharing options...
KipIngram Posted May 23, 2017 Share Posted May 23, 2017 Coder: Are you familiar with Karnaugh maps? Endy0816: I may be tired or something, but I'm not seeing those simplify to just one variable. Link to comment Share on other sites More sharing options...
Endy0816 Posted May 23, 2017 Share Posted May 23, 2017 Yeah, double checked you're right. Link to comment Share on other sites More sharing options...
KipIngram Posted May 23, 2017 Share Posted May 23, 2017 The thing is, both of these expressions are in two-logic-level form. Product of sums, and sum of products. So in that sense they're already "simplified." Those are the forms that would give you the fastest operation in a logic circuit. So knowing what the "cost" criterion is would help. Link to comment Share on other sites More sharing options...
Coder Posted May 24, 2017 Author Share Posted May 24, 2017 The thing is, both of these expressions are in two-logic-level form. Product of sums, and sum of products. So in that sense they're already "simplified." Those are the forms that would give you the fastest operation in a logic circuit. So knowing what the "cost" criterion is would help. For example I have simplified this equation. 1st step arranging the literals. 2nd step applying Distributive Law 3rd step summation of y and its complement is equals to 1. So, like this how can I solve those above equations. And yes I am familiar with Karnaugh map. Link to comment Share on other sites More sharing options...
KipIngram Posted May 24, 2017 Share Posted May 24, 2017 (edited) Good example, but I don't think every expression of the initial form will necessarily go to the form you got in the example. In the example you have a part that's of the form a.b + /a.b, which will always go down to b. I didn't see that happening in the first of your two initial problems, which is the only one I've played with much. Are these problems presented in a context that promises they're reducible? I did a Karnaugh map and got this for the first one, (x+y)(/x+z)(y+z) yz x=0 x=1 00 0 0 01 0 1 11 1 1 10 1 0 xz + /xy So why are we having such trouble getting that algebraically? (x+y)(/x+z)(y+z) (x./x + x.z + y./x + y.z) (y+z) (x.z + y./x + y.z) (y+z) x.z.y + y./x.y + y.z.y + x.z.z + y./x.z + y.z.z x.y.z + /x.y + y.z + x.z + /x.y.z + y.z (x.y.z + /x.y.z) + (y.z + y.z) + /x.y + x.z y.z + /x.y + x.z I checked that and it gives the same Karnaugh map. So there's some rule that can be applied here. Here's what we need to do formally: The first term contributes only if y and z are both true. In that case one or the other of the other two terms will be true. So we can see that it simplifies. We just need the formal rule. The rule more or less has to say exactly that reduction: ab + /ca + cb = /ca + cb By the way, your second problem is directly in this form, so when we find this rule we can immediately apply it to get xy + /xz + yz = xy + /xz I'm assuming this isn't a rule you've been given, but it's a good one. And given that I didn't immediately recognize it upon looking at your second problem I imagine it wasn't given to me as a "rule" either. Ok, here you go. y.z + /x.y + x.z Multiply through by "true," defined as xz + /x + /z: (y.z + /x.y + x.z) (x.z + /x + /z) y.z.x.z + /x.y.x.z + x.z.x.z + y.z./x + /x.y./x + /x.x.z + /z.y.z + /z./x.y + /z.x.z Obvious simplifications: x.y.z + x.z + /x.y.z + /x.y + /x.y./z Note that x.y.z + x.z = x.z: x.z + /x.y + /x.y.z + /x.y./z Now note that /x.y.z + /x.y./z = /x.y x.z + /x.y + /x.y And finally /x.y + /x.y = /x.y: x.z + /x.y Done. Edited May 24, 2017 by KipIngram 2 Link to comment Share on other sites More sharing options...
Coder Posted May 24, 2017 Author Share Posted May 24, 2017 @KipIngram Thanks Link to comment Share on other sites More sharing options...
KipIngram Posted May 24, 2017 Share Posted May 24, 2017 Sure thing. Digital logic is my "expertise by education," so your problem was nagging me quite a bit, especially after I did the Karnaugh map and saw that it would simplify. Link to comment Share on other sites More sharing options...
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