GATE CSE 1988 | Question: 3a-b

Question

   

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.

Comments

Learn the "Compact Truth Table" concept Here: https://youtu.be/Hn9MhOJs6F0

Both Parts A, B can be Easily Solved after watching the above lecture, with Many More Variations.

Answer 1

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.

Comments

Thank You Sir :-)

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Learn the "Compact Truth Table" concept Here: https://youtu.be/Hn9MhOJs6F0

Both Parts A, B can be Easily Solved after watching the above lecture, with Many More Variations.

Answer 2

 

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

********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...

Comments

No need of BC....as both 1s (011 and 111) are already covered

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First we can list the K-map in terms of SSOP.
If two SSOP terms can be minimized on the reduced literal (C / C’) then we should do so. 

After that, we can treat the reduced literal (C / C’) as a constant.
For terms not containing C / C’, const 1 can be considered (since we are working with SOP here).

Answer 3

On solving the first k-map we get the following expression:
AC+A'C'+BC

in the reduced k-map:
 

if C=1, we get,  AC+BC

if C=0, we get, A'C'

on combining both we get AC+A'C'+BC

Answer 4

The question is different but the concept is similar . 

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