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Find the minimum product of sums of the following expression

$f=ABC + \bar{A}\bar{B}\bar{C}$

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Minimal POS $f=(\bar{A} + B) (A+\bar{C}) (\bar{B}+C)$
by Boss (11.4k points)
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There are many such pos possible given is one of the Possible POS.
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sir i am getting (A+B') (B+C') (B+A') ..i dont what is going wrong with me please help sir ..!
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you put AB on the top of the k-map and c is below..and here is opposite in above answer.
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please change 1 to 0
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@air1ankit in pos we will take 1 in complement and 0 as normal while in sop we take 1 as normal and 0 as a complemented

check are you correct here.

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@Anu007

Given is the minimal POS!
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Yes it is minimal POS
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there is a typo in the solution, it's 0 at A'BC, but it is 1 in the above K-map.

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Here there are no essential prime implicants. So many POS are possible.
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@Lokesh .

when we give include red circle too to give the answer as @Lokesh . used the only black circle to give the answer ??? 0

I think we can do using the prime implicant chart (Branching method), we got the same answer as well as it depends upon the choice. So more than one answer is possible. In $1990$ time paper are descriptive, that time solution matters.  f

We get $\overline{f}(A,B,C)=A\overline{B}+\overline{A}C+B\overline{C}$             $\text{SOP}$

We can also write $f(A,B,C)=(\overline{A}+B)\cdot(A+\overline{C})\cdot(\overline{B}+C)$         $\text{POS}$

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@Lakshman Patel RJIT Thank you for your explanation and tell me one thing more ,when we combine all possible combination (red circle) and combine all combination (black circle) is as similar as only (black combination)??

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@ There are more than one POS possible but it may or may not be minimal. It depends upon the choices.

Here i take $A\overline{B}$ first then i apply the procedure and then take $\overline{A}C$ and after that remaining $B\overline{C}.$

It is always better to use the prime implicant chart method if there is no essential prime implicant.

I show you another answer also possible when i take differently,it depends what we choose first.  I hope you understand.

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thankyou.. for your explanation
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@Lakshman Patel RJIT why there is no 0 in A=0, B=0, C=0 position?

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I make $K$ map for $\overline{f}(A,B,C)$
In this minimization given expression is in SOP (Sum of product form) but in question POS form minimization asked so

Firstly We have to make K-map according to SOP form for SOP form entry is 1 but for POS it is 0 ,we can make remaining cells zero in this way we can do minimization easily
by Boss (11k points)
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