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What is the equivalent Boolean expression in product-of-sums form for the Karnaugh map given in Fig

1. $B\overline{D} + \overline{B}D$

2. $(B + \overline{C} +D) (\overline{B} + C + \overline{D})$

3. $(B + {D})(\overline{B} +\overline{ D})$

4. $(B + \overline{D})(\overline{B} + {D})$

### 1 comment

(C) is correct option

Following two K-Maps are equivalent and represent the same boolean function.
While the first K-Map gives us the boolean expression in Sum-of-Product form, and the second K-Map gives us the same boolean function in Product-of-Sum form:

(C) is correct option!

If all the empty cells are filled with '0' and then the POS expression is calculated as:

 CD\AB 00 01 11 10 00 0 0 01 0 0 11 0 0 10 0 0

Then the POS expression of f = (D+B)(D'+B') = BD' + B'D

You are right
So here, what will be correct answer?? A or C??
The one in POS form
First of all find the sop form then just complement it to get the corresponding pos form which comes out to be option c

Say , F = something in sop
taking complement,
F' = something in pos , yes that is right
but what we get in pos is F'
and what we need in pos is F.
So if we have  F = some minterms  i.e, sop
get F' = remaining minterms , that is also sop
that complement, we get F is pos
This method doesn’t works! We have to find POS explicitly, complementing SOP won’t give correct POS here.
no by complementing we get f complement which is in pos form
but we want f in pos form
so best method is to convert sop to pos using boolean laws
Or Use the K-map get the Min terms s we get  in this question.#USING MIN-TERMS and convert them to MAX term Check for Options
as in using K map we get
F= D'B + B'D (Is would be easy till this part)
Now apply de-morgan's law
F'= (D + B') (D'+B)
solve this  we get
F'= DD' +DB + D'B' BB'

Apply de-morgan's again
F'' = (D'+B')(D+B)

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