← 2021 Paper 2

UPSC 2021 Maths Optional Paper 2 Q6b — Step-by-Step Solution

15 marks · Section B

Boolean algebra · Numerical Analysis · asked 15× in 13 yrs · Read the full method →

Question

Obtain Boolean function $F(x,y,z)$ from given truth table. Simplify and draw GATE network.

Technique

K-map simplification; recognise majority function pattern.

Solution

Given truth table:

Step 1 — Identify rows where $F=1$

Rows: $(1,1,1),(1,1,0),(1,0,1),(0,1,1)$.

Step 2 — Sum of minterms (DNF)

$F=xyz+xy\bar z+x\bar y z+\bar x y z$.

Step 3 — Simplify

Group: $xy(z+\bar z)+\bar y z(\bar x\to$… let me try Karnaugh map.

K-map (3 variables $x,y,z$):

Group of 4: cells $(x,y\bar z), (x,yz), (\bar x,yz)$ — wait, let me re-look.

Cells where $F=1$:

• $(\bar x, yz)$: row 0, col 3.
• $(x, \bar y z)$: row 1, col 2.
• $(x, yz)$: row 1, col 3.
• $(x, y\bar z)$: row 1, col 4.

Group 1: row $x$, all four cells: cells $(x,\bar y z), (x,yz), (x,y\bar z)$ — only 3 of 4 in row $x$ are 1 (not $\bar y\bar z$). So can’t take whole row.

Group 1: $(x,yz)$ and $(x,y\bar z)$ — gives $xy$. Group 2: $(x,yz)$ and $(x,\bar y z)$ — gives $xz$. Group 3: $(\bar x,yz)$ and $(x,yz)$ — gives $yz$.

So $F=xy+xz+yz$ (with overlap at $xyz$, but that’s fine for SOP).

Verify:

• $(1,1,1)$: $1+1+1=1$ ✓.
• $(1,1,0)$: $1+0+0=1$ ✓.
• $(1,0,1)$: $0+1+0=1$ ✓.
• $(1,0,0)$: $0+0+0=0$ ✓.
• $(0,1,1)$: $0+0+1=1$ ✓.
• $(0,1,0)$: $0+0+0=0$ ✓.
• $(0,0,1)$: $0+0+0=0$ ✓.
• $(0,0,0)$: $0+0+0=0$ ✓.

All match.

Answer