GATE CS 2015 Set 3 Q.53 (CS Q.43): Boolean Algebra and Logic Minimization

The total number of prime implicants of the function $f(w, x, y, z) = \sum(0, 2, 4, 5, 6, 10)$ is _____.

Your answer

Official answer and solution

Official answer: 3
From: Official GATE 2015 answer key

Solution

Plot minterms on a 4-variable Karnaugh map and identify all maximal rectangular implicants.

The minterms of $f(w, x, y, z)$ are: $m_0(0000), m_2(0010), m_4(0100), m_5(0101), m_6(0110), m_{10}(1010)$ .

Examine possible group formations:

Minterms $\{m_0, m_2, m_4, m_6\}$ ${m_{0}, m_{2}, m_{4}, m_{6}}$ :
- Binary: $0000, 0010, 0100, 0110$ .
- They form a 4-cell group covering all cells with $w = 0$ and $z = 0$ .
- Product term: $w' z'$ . This cannot be expanded further, so it is a Prime Implicant.

Minterms $\{m_4, m_5\}$ ${m_{4}, m_{5}}$ :
- Binary: $0100, 0101$ .
- They differ only in $z$ , giving product term $w' x y'$ .
- Since $m_5$ shares no other adjacent minterms in the function, this 2-cell group is maximal and is a Prime Implicant.

Minterms $\{m_2, m_{10}\}$ ${m_{2}, m_{10}}$ :
- Binary: $0010, 1010$ .
- They differ only in $w$ , giving product term $x' y z'$ .
- Since $m_8$ is not a minterm, this 2-cell group cannot expand into a 4-cell corner group. Thus, $x' y z'$ is maximal and is a Prime Implicant.

All 6 minterms are covered by these 3 prime implicants ( $w'z', w'xy', x'yz'$ ), and no other maximal groups exist.
Hence, the total number of prime implicants is 3.

Final answer

Common trap

Counting essential prime implicants instead of total prime implicants, or missing the vertical 2-cell wrap-around group {m2, m10}.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 3
Question: 53 of 65, printed as CS Q.43
Section: Computer Science and Information Technology

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