GATE 2026 EC Q.56: solution and answer

GATE 2026 · ECQ.56 · 2 marks · MSQ

Consider the four-variable Boolean function,

$f (w, x, y, z) = \sum m (0, 2, 5, 7, 8, 10, 13, 14, 15)$ with ' $w$ ' as MSB and ' $z$ ' as LSB.

Which of the following expressions is/are the valid form(s) of $f (w, x, y, z)$ ?

(Complement is printed as an overbar and is written here with \overline{}.)

(A)
$\overline{x} \overline{z} + x z + w x y$
(B)
$x z + w x y + w \overline{x} \overline{z} + \overline{w} x y \overline{z}$
(C)
$\overline{x} \overline{z} + w x y + w \overline{x} \overline{z} + \overline{w} x y \overline{z}$
(D)
$\overline{x} \overline{z} + x z + w y \overline{z}$

See the question in the paper

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Answer and solution

ANSWER

Options (A) and (D) are both valid, and each is a minimal SOP form of $f$ . The only difference between them is how $m_{14}$ gets covered: $w x y$ (pairing it with $m_{15}$ ) or $w y \overline{z}$ (pairing it with $m_{10}$ ).

Official answer: A, DSource: Official answer key, GATE 2026 EC (IIT Guwahati)

Write each minterm as $w x y z$ (with $w$ as MSB): $0 = 0000, 2 = 0010, 5 = 0101, 7 = 0111, 8 = 1000, 10 = 1010, 13 = 1101, 14 = 1110, 15 = 1111$ . So $f = 0$ at $1, 3, 4, 6, 9, 11, 12$ . Complement is written here with .
K-map grouping of the 1s, the four corners: $x = 0, z = 0$ gives ${0, 2, 8, 10}$ . This is a quad, so the term is $\overline{x} \overline{z}$ .
K-map grouping of the 1s, the centre quad: $x = 1, z = 1$ gives ${5, 7, 13, 15}$ , so the term is $x z$ .
Only $m_{14} = 1110$ is still uncovered. It can pair with $m_{15} = 1111$ (common literals $w = 1, x = 1, y = 1$ ), giving $w x y$ , or with $m_{10} = 1010$ (common literals $w = 1, y = 1, z = 0$ ), giving $w y \overline{z}$ . Both pairs contain only 1s, so there are two equally minimal SOP forms: $f = \overline{x} \overline{z} + x z + w x y or f = \overline{x} \overline{z} + x z + w y \overline{z}$
Check (A): $\overline{x} \overline{z} \to {0, 2, 8, 10}$ , $x z \to {5, 7, 13, 15}$ , $w x y \to {14, 15}$ . The union is exactly the nine given minterms and nothing else, so (A) is valid.
Check (D): $w y \overline{z} \to {10, 14}$ . Together with the two quads the union is again exactly ${0, 2, 5, 7, 8, 10, 13, 14, 15}$ , so (D) is valid.
Check (B) and (C): both contain $\overline{w} x y \overline{z}$ , which is $0110 = m_{6}$ , and $f (6) = 0$ . So both expressions output 1 where $f$ is 0. In addition, (B) misses $m_{0}$ and $m_{2}$ (its $w \overline{x} \overline{z}$ covers only $8, 10$ ), and (C) has no term covering $m_{5}, m_{7}, m_{13}$ . Neither is valid.
Valid forms: (A) and (D).

KEY IDEA

Draw the K-map. The four corners give $\overline{x} \overline{z}$ and the centre quad gives $x z$ . After that, $m_{14}$ is the only 1 left, and it has two adjacent 1s ( $m_{15}$ and $m_{10}$ ). That gives two equally minimal covers, $w x y$ or $w y \overline{z}$ . A function can have more than one minimal SOP. To test any option, list the minterms its terms cover and compare that set with the given one.

WATCH OUT FOR

Stopping at the first minimal cover you find ( $\overline{x} \overline{z} + x z + w x y$ ) and marking only (A), without noticing that $m_{14}$ can also pair with $m_{10}$ , which makes (D) valid too. Another mistake is to judge (B) and (C) only by whether they cover every 1. Both also include $\overline{w} x y \overline{z} = m_{6}$ , where $f = 0$ .

From the GATE 2026 EC paper. Every question of this paper · Download the full paper · All 22 Boolean logic questions, with notes

GATE 2026 EC Q.56: solution and answer

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