Related Concepts: 01 Boolean Algebra Foundations & Duality Principle | 04 Mathematical Conversions & Expansion of SOP and POS | 05 Tabular Method (Quine-McCluskey) & Prime Implicants

3.06 Boolean Algebra Puzzles & Exam Proofs

Exam Edge Cases & Mathematical Puzzles

High-yield exam questions often include 9 to 11-mark mathematical “puzzles” testing reverse-engineering, factoring constraints, and multi-variable gate expansions.

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1. The 5-Variable “Reverse Don’t Care” Puzzle

Major PYQ Problem (2017, 2020 - 11 Marks)

Question: The expression $BE+B^{\prime }D{E}^{\prime }$ is a simplified version of $A^{\prime }BE+BCDE+B{C}^{\prime }{D}^{\prime }E+A^{\prime }{B}^{\prime }D{E}^{\prime }+B^{\prime }{C}^{\prime }D{E}^{\prime }$. Are there any don’t care conditions? If so, what are they?

Solution:

Step 1: Minterms of Unsimplified Expression ($F_{\text{un}}$)

Expand each 5-variable term ($ABCDE$):

• $A^{\prime }BE\to 01XX1⟹m_9,m_{11},m_{13},m_{15}$
• $BCDE\to X1111⟹m_{15},m_{31}$
• $B{C}^{\prime }{D}^{\prime }E\to X1001⟹m_9,m_{25}$
• $A^{\prime }{B}^{\prime }D{E}^{\prime }\to 00X10⟹m_2,m_6$
• $B^{\prime }{C}^{\prime }D{E}^{\prime }\to X0010⟹m_2,m_{18}$ $\text{Unsimplified Minterms: }\{2,6,9,11,13,15,18,25,31\}$

Step 2: Minterms of Simplified Expression ($F_{\text{simp}}$)

• $BE\to X1XX1⟹m_9,m_{11},m_{13},m_{15},m_{25},m_{27},m_{29},m_{31}$
• $B^{\prime }D{E}^{\prime }\to X0X10⟹m_2,m_6,m_{18},m_{22}$ $\text{Simplified Minterms: }\{2,6,9,11,13,15,18,22,25,27,29,31\}$

Step 3: Isolate Don’t Care Minterms

$\text{Don’t Cares }d=F_{\text{simp}}\setminus F_{\text{un}}=\{22,27,29\}$

• $m_{22}(10110_2)=\mathbf{A}{\mathbf{B}}^{\prime }\mathbf{CD}{\mathbf{E}}^{\prime }$
• $m_{27}(11011_2)=\mathbf{AB}{\mathbf{C}}^{\prime }\mathbf{DE}$
• $m_{29}(11101_2)=\mathbf{ABC}{\mathbf{D}}^{\prime }\mathbf{E}■$

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2. The 8-Literal Limit Factoring Puzzle

Major PYQ Problem (2022 - 9 Marks)

Question: Express $F=B^{\prime }{C}^{\prime }{D}^{\prime }+A{B}^{\prime }C{D}^{\prime }+B{C}^{\prime }D+A^{\prime }BCD$ in three different ways using 8 or fewer literals.

Solution:

Way 1: Group by $B^{\prime }{D}^{\prime }$ and $BD$

$F=B^{\prime }{D}^{\prime }(C^{\prime }+AC)+BD(C^{\prime }+A^{\prime }C)$ Using absorption identity $(X^{\prime }+XY=X^{\prime }+Y)$: $\mathbf{F}={\mathbf{B}}^{\prime }{\mathbf{D}}^{\prime }({\mathbf{C}}^{\prime }+\mathbf{A})+\mathbf{BD}({\mathbf{C}}^{\prime }+{\mathbf{A}}^{\prime })\text{(8 literals)}$

Way 2: Group by $C^{\prime }$ and $C$

Expand Way 1: $F=A{B}^{\prime }{D}^{\prime }+B^{\prime }{C}^{\prime }{D}^{\prime }+A^{\prime }BD+B{C}^{\prime }D$ $\mathbf{F}={\mathbf{C}}^{\prime }({\mathbf{B}}^{\prime }{\mathbf{D}}^{\prime }+\mathbf{BD})+\mathbf{C}(\mathbf{A}{\mathbf{B}}^{\prime }{\mathbf{D}}^{\prime }+{\mathbf{A}}^{\prime }\mathbf{BD})\text{(8 literals)}$

Way 3: Group by $D^{\prime }$ and $D$

$\mathbf{F}={\mathbf{D}}^{\prime }(\mathbf{A}{\mathbf{B}}^{\prime }+{\mathbf{B}}^{\prime }{\mathbf{C}}^{\prime })+\mathbf{D}({\mathbf{A}}^{\prime }\mathbf{B}+\mathbf{B}{\mathbf{C}}^{\prime })\text{(8 literals)}■$

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3. 4-Variable XNOR Expansion Proof

OCR Exam Scanner Typo Correction

PYQ 2017 text writes $A\oplus B\oplus C\oplus D=\Sigma m(0,3,5,6,9,10,12,15)$. This is an OCR typo. Minterm 0 ($0000$) evaluates to 0 in XOR. The true textbook question (Morris Mano 4-23) uses XNOR ($\odot$).

Major PYQ Proof (2017 - 10 Marks)

Question: Show that $A\odot B\odot C\odot D=\sum m(0,3,5,6,9,10,12,15)$.

Proof Solution:

Let $X=A\odot B=AB+A^{\prime }{B}^{\prime }$ and $Y=C\odot D=CD+C^{\prime }{D}^{\prime }$. $F=X\odot Y=XY+X^{\prime }{Y}^{\prime }$

• $X^{\prime }=(A\odot B{)}^{\prime }=A\oplus B=A^{\prime }B+A{B}^{\prime }$
• $Y^{\prime }=(C\odot D{)}^{\prime }=C\oplus D=C^{\prime }D+C{D}^{\prime }$

Substitute into $F$: $F=(AB+A^{\prime }{B}^{\prime })(CD+C^{\prime }{D}^{\prime })+(A^{\prime }B+A{B}^{\prime })(C^{\prime }D+C{D}^{\prime })$ $=ABCD+AB{C}^{\prime }{D}^{\prime }+A^{\prime }{B}^{\prime }CD+A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }+A^{\prime }B{C}^{\prime }D+A^{\prime }BC{D}^{\prime }+A{B}^{\prime }{C}^{\prime }D+A{B}^{\prime }C{D}^{\prime }$ Evaluating these 8 product terms: $\mathbf{F}=\sum \mathbf{m}(0,3,5,6,9,10,12,15)■$

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Past Year Questions (PYQs)

• [PYQ 2017, 2020]: 5-variable Reverse Don’t Care condition puzzle. (11 Marks)
• [PYQ 2022]: Express Boolean function with 8 or fewer literals. (09 Marks)
• [PYQ 2017]: Show that $A\odot B\odot C\odot D=\Sigma m(0,3,5,6,9,10,12,15)$. (10 Marks)