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How to minimize this Boolean expression in shorter form?


Coder

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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.

 

 

post-129532-0-99658300-1495565605.png

post-129532-0-37185700-1495565991.png

Edited by Coder
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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.

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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.

 

example.png

Answer.png

 

 

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.

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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 by KipIngram
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