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54 votes

The literal count of a Boolean expression is the sum of the number of times each literal appears in the expression. For example, the literal count of $\left(xy+xz'\right)$ is $4.$ What are the minimum possible literal counts of the product-of-sum and sum-of-product representations respectively of the function given by the following Karnaugh map? Here, $X$ denotes "don't care"

 

 

  1. $(11, 9)$
  2. $(9, 13)$
  3. $(9, 10)$
  4. $(11,11)$
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6 Answers

Best answer
96 votes
96 votes


We will be getting  two different grouping..

Grouping $1: (9,8)$ 

 

$\text{SOP}: 2 + 3 + 3 = 8$   
  $\text{POS}: 2 + 2 + 2 + 3 = 9$

 

 

 

GROUPING $2: (9,10)$

$\text{SOP}: 2 + 2 + 3 + 3 = 10$ 
  $\text{POS}: 2 + 2 + 2 + 3 = 9$

 

Both the grouping are correct representation of the function  $f(wxyz)$

PS: Some wrong beliefs about don't cares

  1. "once you have assumed a don't care as '1' u can't use the same don't care for grouping zeros and vice versa"
  2. "if don't care has been used in POS than can't be used in SOP"

Both these statements are wrong. Don't care simply means just don't care -- say we use don't care $d3$ for grouping $1$ in SOP we can use $d3$ for grouping 0 in POS. (The literals in SOP and POS may not be the same)

K-Map grouping is not unique. And the question says about minimal literals. So, the best answer would be (9,8) Since there is no option in GATE we can go with $(9,10)$ (the question setter might have missed Grouping 1)

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7 votes
7 votes

I am getting (9,8), 9 for POS and 8 for SOP.

5 votes
5 votes
POS

(z'+w')(y'+z')(x'+w')(x+y'+w)

9 literals

SOP:

y'w'+yw+xyz'+x'y'z'

10 literals

ans is c
Answer:

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