← 2025 Paper 2

UPSC Maths 2025 Paper 2 Q6b — Solution

15 marks · Section B

Question

Simplify the Boolean function $F(x, y, z) = xyz + x'yz + xy'z + xyz'$ and draw the corresponding GATE network.

Technique

Simplify by the absorption/adjacency law $AB + AB' = A$ (equivalently a Karnaugh map): pairs of minterms differing in exactly one variable combine, and the minterm $xyz$ may be reused by idempotence.

Solution

The four product terms are minterms (in variables $x,y,z$): $xyz = m_7,\quad x'yz = m_3,\quad xy'z = m_5,\quad xyz' = m_6.$ So $F = \sum m(3,5,6,7)$.

Combine adjacent minterms

Use $AB + AB' = A$. The term $xyz$ shares a face with each of the other three:

• $xyz + x'yz = yz(x + x') = yz.$
• $xyz + xy'z = xz(y + y') = xz.$
• $xyz + xyz' = xy(z + z') = xy.$

The minterm $xyz$ may be reused freely (idempotent law $xyz + xyz = xyz$). Hence $F = (xyz + x'yz) + (xyz + xy'z) + (xyz + xyz') = yz + xz + xy.$

So $\boxed{F(x,y,z) = xy + yz + zx.}$

This is the majority function: $F=1$ iff at least two of $x,y,z$ are $1$ (minterms $3,5,6,7$).

K-map confirmation

Three 2-cell groupings: the $yz$ column ($yz=11$), the $xz$ pair ($x=1,z=1$), and the $xy$ pair ($x=1,y=1$) → $F = yz + xz + xy$.

GATE network

        x ---+----------\
        y ---+--[ AND ]---\
                            \
        y ---+----------\    \
        z ---+--[ AND ]---+--[ OR ]--- F
                          /
        x ---+----------/
        z ---+--[ AND ]-/

Three 2-input AND gates compute $xy$, $yz$, $zx$; their outputs feed one 3-input OR gate giving $F = xy + yz + zx$.

Answer

$F(x,y,z) = xy + yz + zx$ realised by three 2-input AND gates feeding a single 3-input OR gate (the majority gate).