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The Karnaugh map of a function of $(A, B, C)$ is shown on the left hand side of the above figure.

The reduced form of the same map is shown on the right hand side, in which the variable $C$ is entered in the map itself. Discuss,

  1. The methodology by which the reduced map has been derived and
  2. the rules (or steps) by which the boolean function can be derived from the entries in the reduced map.
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3 Answers

Best answer
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We can get the truth table as $$\begin{array}{|c|c|c|c|}\hline
A&B&C&F\\\hline
0&0&0&1\\
0&0&1&0\\
0&1&0&1\\
0&1&1&1\\
1&0&0&0\\
1&0&1&1\\
1&1&0&0\\
1&1&1&1\\ \hline
\end{array}$$ Now, to reduce the K-map to a Variable Entrant Map we can write the function $F$ in terms of $C$. i.e., wherever $F$ is becoming $1$ dependent on $C$ (i.e. when $C$ complements $F$ must become $0)$, we replace $1$ with $C$ or $\bar C$ based on whichever is giving output $1.$ So, we can rewrite the truth table as $$\begin{array}{|c|c|c|c|}\hline
A&B&F\\\hline
0&0&\bar C\\
0&1&1\\
1&0&C\\
1&1&C\\
 \hline
\end{array}$$ In the above truth table $F=1$ for second row, because when $A = 0,B=1, F= 1$ for both $C$ and $\bar C$ making $F$ independent of $C.$ Now, if we draw the $K$-map for the above truth table we get the reduced Variant Entrant map given.

edited by
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$f(A,B,C)$ = $\overline{A}\overline{C} + BC + AC + \overline{A}B$

********This method is known as variable entrant map*******

Step 1:

SOP= $\overline{A}B$

Step 2:  $"\overline{C}"$

SOP = $\overline{A} = \overline{A}\overline{C}$

Step 3: "C"

SOP = B + A  = BC + AC

Merge all these terms $F(A,B,C) =    \overline{A}\overline{C} + AC + BC +\overline{A}B$

 

 

lemme me know if i m wrong...

edited by
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0 votes

The question is different but the concept is similar . 

step 4 ) SOP of VEM is obtained by ORing the previous SOP expressions .

 

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