Related Concepts: 3.01 Boolean Algebra Foundations & Duality Principle | 3.04 Mathematical Conversions & Expansion of SOP and POS | 3.05 Karnaugh Map (K-Map) Optimization & Don’t Care States | 3.06 Tabular Method (Quine-McCluskey) & Prime Implicants

3.07 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 (PYQ 2017 — 11 marks; 2020 — 10 marks)

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

This is the only K-map question in the course that runs backwards — you are handed the answer and asked what extra freedom made it legal. The logic is simple once seen: a simplification can only cover more ground than the original if the extra cells were don’t cares.

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 (PYQ 2022 — 09 marks)

Question (verbatim): Give three possible ways to express the following Boolean function with eight or fewer literals: $F=B^{\prime }{C}^{\prime }{D}^{\prime }+A{B}^{\prime }C{D}^{\prime }+B{C}^{\prime }D+A^{\prime }BCD$

Why this is a factoring question, not a minimisation question

Expanding gives $F=\sum m(0,5,7,8,10,13)$. Plotted on a K-map, the only adjacencies are four isolated pairs — $\{0,8\}$, $\{8,10\}$, $\{5,7\}$, $\{5,13\}$ — so the minimal two-level SOP is

$F=B^{\prime }{C}^{\prime }{D}^{\prime }+A{B}^{\prime }{D}^{\prime }+A^{\prime }BD+B{C}^{\prime }D\text{(12 literals — over budget)}$

The minimal SOP cannot meet the budget. Any question that sets a literal or gate ceiling is asking you to move to a multi-level, factored form. See 3.04 Mathematical Conversions & Expansion of SOP and POS §5 for the K-map partition that produces the factorisation below.

Solution:

Way 1: Factor on $B$ and $D$ — the target answer

Group the original terms in pairs and factor: $F=B^{\prime }{D}^{\prime }(C^{\prime }+AC)+BD(C^{\prime }+A^{\prime }C)$ Apply the absorption identity $X^{\prime }+XY=X^{\prime }+Y$: $\boxed{\mathbf{F}={\mathbf{B}}^{\prime }{\mathbf{D}}^{\prime }(\mathbf{A}+{\mathbf{C}}^{\prime })+\mathbf{BD}({\mathbf{A}}^{\prime }+{\mathbf{C}}^{\prime })}\text{— 8 literals}$

Way 2: Factor on $C$

$\mathbf{F}={\mathbf{C}}^{\prime }({\mathbf{B}}^{\prime }{\mathbf{D}}^{\prime }+\mathbf{BD})+\mathbf{C}(\mathbf{A}{\mathbf{B}}^{\prime }{\mathbf{D}}^{\prime }+{\mathbf{A}}^{\prime }\mathbf{BD})\text{— 12 literals}$

Way 3: Factor on $D$

$\mathbf{F}={\mathbf{D}}^{\prime }(\mathbf{A}{\mathbf{B}}^{\prime }+{\mathbf{B}}^{\prime }{\mathbf{C}}^{\prime })+\mathbf{D}({\mathbf{A}}^{\prime }\mathbf{B}+\mathbf{B}{\mathbf{C}}^{\prime })\text{— 10 literals}■$

Only Way 1 actually meets the eight-literal budget

Ways 2 and 3 are valid alternative expressions of the same function and are worth showing as your “three ways”, but count their literals honestly — 12 and 10 respectively. If the examiner marks strictly on the ceiling, Way 1 is the one that scores, so present it first and derive it properly from the $BD$ partition.

Literal counting rule: every appearance of a variable counts, complemented or not. $B^{\prime }{D}^{\prime }(A+C^{\prime })$ contains $B^{\prime }$, $D^{\prime }$, $A$, $C^{\prime }$ = 4 literals; the whole of Way 1 is $4+4=8$.

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3. 4-Variable XNOR Expansion Proof

The printed question is internally inconsistent — say so in your answer

The 2017 paper reads: “Show that $A\oplus B\oplus C\oplus D=\sum (0,3,5,6,9,10,12,15)$.” As printed this is false. Chained XOR outputs 1 when an odd number of inputs are 1, so $m_0$ (0000) gives $0\oplus 0\oplus 0\oplus 0=0$ and cannot be in the list. The listed set is the even-parity set, which belongs to chained XNOR ($\odot$). The underlying textbook problem (Mano) uses $\odot$.

How to answer safely and score full marks:

1. State the identity that is true: $A\odot B\odot C\odot D=\sum (0,3,5,6,9,10,12,15)$, and prove it (below).
2. Add one line noting that $A\oplus B\oplus C\oplus D=\sum (1,2,4,7,8,11,13,14)$, the complement set.
3. Note that the two are related by $A\odot B\odot C\odot D=\overline{(A\oplus B\oplus C\oplus D)}$.

Answering both readings costs three extra lines and protects you whichever version the examiner intended. Silently “correcting” the question without flagging it is the risky option.

Major PYQ Proof (PYQ 2017 — 10 marks)

Question (as it should read): 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)

Question (as asked)	Years	Marks	Solved in
$BE+B^{\prime }D{E}^{\prime }$ 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?	2017, 2020	10–11	§1
Give three possible ways to express $F=B^{\prime }{C}^{\prime }{D}^{\prime }+A{B}^{\prime }C{D}^{\prime }+B{C}^{\prime }D+A^{\prime }BCD$ with eight or fewer literals	2022	9	§2
Show that $A\oplus B\oplus C\oplus D=\sum (0,3,5,6,9,10,12,15)$ (as printed)	2017	10	§3

Pattern to notice: these three are the chapter’s “thinking” questions rather than its drill questions. Each is worth 9–11 marks and each rewards recognising what kind of problem it is:

• §1 — a set-difference problem disguised as a K-map problem
• §2 — a factoring problem disguised as a minimisation problem
• §3 — an identity proof where the printed operator and the printed minterm list disagree

All three reward writing down the method explicitly before computing. Since the reverse don’t-care puzzle has now appeared twice, it is the one worth rehearsing.