Related Concepts: 3.03 SOP, POS, Canonical & Standard Forms | 3.05 Karnaugh Map (K-Map) Optimization & Don’t Care States | 3.06 Tabular Method (Quine-McCluskey) & Prime Implicants

3.04 Mathematical Conversions & Expansion of SOP and POS

Concept Overview

Converting standard Boolean expressions into canonical Sum of Products (SOP) or Product of Sums (POS) form is accomplished algebraically via Missing-Variable Expansion.

graph TD
    Expression[Non-Canonical Expression] --> Type{Expansion Target}
    
    Type -->|Expand to SOP| SOPRule[Multiply term by x + x' = 1]
    SOPRule --> ExpandSOP[Apply Distributive Law: AB x+x' = ABx + ABx']
    ExpandSOP --> ListMinterms[List Sum of Minterms Σm]
    
    Type -->|Expand to POS| POSRule[Add xx' = 0 to sum term]
    POSRule --> ExpandPOS[Apply Distributive Law: X + YZ = X+Y X+Z]
    ExpandPOS --> ListMaxterms[List Product of Maxterms ΠM]

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1. Algebraic Expansion Rules

• SOP Expansion Rule: Multiply each non-canonical AND product term by $(x+x^{\prime })=1$ for every missing variable $x$.
• POS Expansion Rule: Add $(x\cdot x^{\prime })=0$ to each non-canonical OR sum term for every missing variable $x$, then expand via distribution: $A+BC=(A+B)(A+C)$.

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2. Canonical Expansion (Major PYQ)

Method Demonstration

Question: Express $F(A,B,C,D)=(A+B^{\prime }+C)(B+D^{\prime })$ in Product of Maxterms ($\Pi M$) and Sum of Minterms ($\Sigma m$) notation.

Solution:

Part 1: Product of Maxterms ($\Pi M$)

• Term 1 $(A+B^{\prime }+C)$ is missing $D$: $(A+B^{\prime }+C+D{D}^{\prime })=(A+B^{\prime }+C+D)(A+B^{\prime }+C+D^{\prime })$ $\text{Binary: }0100_2=M_4,0101_2=M_5$

• Term 2 $(B+D^{\prime })$ is missing $A$ and $C$: $B+D^{\prime }+A{A}^{\prime }+C{C}^{\prime }$ Expanding via distribution yields 4 maxterms:

  • $(A+B+C+D^{\prime })\to 0001_2=M_1$
  • $(A+B+C^{\prime }+D^{\prime })\to 0011_2=M_3$
  • $(A^{\prime }+B+C+D^{\prime })\to 1001_2=M_9$
  • $(A^{\prime }+B+C^{\prime }+D^{\prime })\to 1011_2=M_{11}$

$\text{Product of Maxterms: }\mathbf{F}(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})=\prod \mathbf{M}(1,3,4,5,9,11)$

Part 2: Sum of Minterms ($\Sigma m$)

The minterm indices are the remaining numbers out of $0\dots 15$: $\text{Sum of Minterms: }\mathbf{F}(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})=\sum \mathbf{m}(0,2,6,7,8,10,12,13,14,15)■$

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3. The Actual Exam Expressions

The examiner has used a different expression almost every year, but only two shapes ever appear: a product of sum-terms (expand each bracket) or a sum of product-terms (expand each product). Both are solved below.

3.1 Three-Variable POS Input

Worked Exam Problem (PYQ 2017 — 12 marks; 2021 — 10 marks)

Question (verbatim): Express the following functions in a sum of Minterms and a product of Maxterms: (i) $F(A,B,C)=(A^{\prime }+B)(B^{\prime }+C)$ (ii) $F(x,y,z)=(xy+z)(y+xz)$

Part (i) — the input is already POS, so expand each sum term over the missing variable:

• $(A^{\prime }+B)$ is missing $C$: $(A^{\prime }+B+C{C}^{\prime })=(A^{\prime }+B+C)(A^{\prime }+B+C^{\prime })$
  • $(A^{\prime }+B+C)\to$ zero when $A=1,B=0,C=0\to 100_2=M_4$
  • $(A^{\prime }+B+C^{\prime })\to$ zero when $A=1,B=0,C=1\to 101_2=M_5$
• $(B^{\prime }+C)$ is missing $A$: $(A{A}^{\prime }+B^{\prime }+C)=(A+B^{\prime }+C)(A^{\prime }+B^{\prime }+C)$
  • $(A+B^{\prime }+C)\to 010_2=M_2$
  • $(A^{\prime }+B^{\prime }+C)\to 110_2=M_6$

$\mathbf{F}=\prod \mathbf{M}(2,4,5,6)=\sum \mathbf{m}(0,1,3,7)$

Part (ii) — despite the brackets this is not POS; multiply it out first:

$F=(xy+z)(y+xz)=xy\cdot y+xy\cdot xz+zy+z\cdot xz=xy+xyz+yz+xz$

$xyz$ is absorbed by $xy$, leaving the majority function:

$F=xy+xz+yz$

Expanding each 2-literal term over its missing variable: $xy\to m_6,m_7$; $xz\to m_5,m_7$; $yz\to m_3,m_7$.

$\mathbf{F}=\sum \mathbf{m}(3,5,6,7)=\prod \mathbf{M}(0,1,2,4)$

The bracket trap

Brackets do not mean POS. $(xy+z)(y+xz)$ is a product of two non-canonical sum terms containing products — you must multiply out before you can talk about maxterms at all. Attempting to expand it directly as if each bracket were a maxterm produces nonsense. Check every bracket: a true sum term contains only single literals joined by $+$.

3.2 Single-Direction Conversion

Worked Exam Problem (PYQ 2024 — 09 marks)

Question (verbatim): What is meant by canonical and standard form? Express the Boolean function $F=xy+xz$ in a product of maxterm form.

(The definitions half is answered in 3.03 SOP, POS, Canonical & Standard Forms.)

Expand each product term over its missing variable:

• $xy$ is missing $z$: $xy(z+z^{\prime })\to m_7(111),m_6(110)$
• $xz$ is missing $y$: $xz(y+y^{\prime })\to m_7(111),m_5(101)$

$F=\sum m(5,6,7)$

The maxterms are every index not in that list:

$\mathbf{F}=\prod \mathbf{M}(0,1,2,3,4)$

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4. The $F=fg$ Problem

Worked Exam Problem (PYQ 2018 — 10 marks; 2022 — 07 marks)

Question (verbatim, 2018): With the use of K-map, find the simplest sum-of-products form of the function $F=fg$, where $f=ab{c}^{\prime }+c^{\prime }d+a^{\prime }c{d}^{\prime }+b^{\prime }c{d}^{\prime }$ $g=(a+b+c+d^{\prime })(b^{\prime }+c^{\prime }+d)(a^{\prime }+c+d^{\prime })$

This looks intimidating but is purely mechanical. $F=f\cdot g$ means the minterms of $F$ are the minterms present in $f$ AND in $g$ — an intersection, nothing more.

Step 1: Minterms of $f$ (an SOP — expand each product term)

Term	Missing	Minterms
$ab{c}^{\prime }$	$d$	$m_{12},m_{13}$
$c^{\prime }d$	$a,b$	$m_1,m_5,m_9,m_{13}$
$a^{\prime }c{d}^{\prime }$	$b$	$m_2,m_6$
$b^{\prime }c{d}^{\prime }$	$a$	$m_2,m_{10}$

$f=\sum m(1,2,5,6,9,10,12,13)$

Step 2: Maxterms of $g$ (a POS — each bracket is zero for one input pattern)

Sum term	Zero when	Maxterm(s)
$(a+b+c+d^{\prime })$	$a=0,b=0,c=0,d=1$	$M_1$
$(b^{\prime }+c^{\prime }+d)$	$b=1,c=1,d=0$, $a$ free	$M_6,M_{14}$
$(a^{\prime }+c+d^{\prime })$	$a=1,c=0,d=1$, $b$ free	$M_9,M_{13}$

$g=\prod M(1,6,9,13,14)$

So $g$ is 0 at exactly minterms $\{1,6,9,13,14\}$ and 1 everywhere else.

Step 3: Intersect

Delete from $f$‘s list every minterm where $g=0$:

$\{1,2,5,6,9,10,12,13\}-\{1,6,9,13\}=\{2,5,10,12\}$

$F=\sum m(2,5,10,12)$

Step 4: K-Map

       cd
  ab \ 00   01   11   10
    +----+----+----+----+
 00 |  0 |  0 |  0 |  1 |   m2
    +----+----+----+----+
 01 |  0 |  1 |  0 |  0 |   m5
    +----+----+----+----+
 11 |  1 |  0 |  0 |  0 |   m12
    +----+----+----+----+
 10 |  0 |  0 |  0 |  1 |   m10
    +----+----+----+----+

Only $m_2$ and $m_{10}$ are adjacent (top and bottom rows wrap): $ab=00,10⟹b^{\prime }$; $cd=10⟹c{d}^{\prime }$. Both $m_5$ and $m_{12}$ are completely isolated.

$\mathbf{F}={\mathbf{b}}^{\prime }\mathbf{c}{\mathbf{d}}^{\prime }+{\mathbf{a}}^{\prime }\mathbf{b}{\mathbf{c}}^{\prime }\mathbf{d}+\mathbf{ab}{\mathbf{c}}^{\prime }{\mathbf{d}}^{\prime }$

Do not multiply $f\times g$ algebraically

Expanding $(4\text{ terms})\times (3\text{ brackets})$ by distribution generates twelve products and takes twenty minutes. Converting both sides to minterm index lists and intersecting takes three. The 2022 version of this question is worth only 7 marks — it is not designed to take long.

Careful with the 2022 wording: its first bracket is $(a+b+c^{\prime }+d^{\prime })$, not $(a+b+c+d^{\prime })$. One character changes $M_1$ into $M_3$ and therefore changes the whole answer. Re-derive rather than reciting.

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5. The Literal-Budget Problem

Worked Exam Problem (PYQ 2022 — 09 marks)

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

Step 1: Expand to minterms

$B^{\prime }{C}^{\prime }{D}^{\prime }\to m_0,m_8$; $A{B}^{\prime }C{D}^{\prime }\to m_{10}$; $B{C}^{\prime }D\to m_5,m_{13}$; $A^{\prime }BCD\to m_7$.

$F=\sum m(0,5,7,8,10,13)$

Step 2: Notice that minimal SOP will not meet the budget

       CD
  AB \ 00   01   11   10
    +----+----+----+----+
 00 |  1 |  0 |  0 |  0 |
    +----+----+----+----+
 01 |  0 |  1 |  1 |  0 |
    +----+----+----+----+
 11 |  0 |  1 |  0 |  0 |
    +----+----+----+----+
 10 |  1 |  0 |  0 |  1 |
    +----+----+----+----+

The only adjacencies are $\{m_0,m_8\}$, $\{m_8,m_{10}\}$, $\{m_5,m_7\}$ and $\{m_5,m_{13}\}$ — four pairs, no quads. The minimal two-level SOP is therefore

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

This is the whole point of the question. It is not asking for minimal SOP; it is asking you to factor, i.e. to move to a multi-level expression.

Step 3: Factor by partitioning on $B$ and $D$

Split the map into its four $BD$ regions and read what is left in each:

Region	Cells	$F$ in that region
$B^{\prime }{D}^{\prime }$	$m_0,m_2,m_8,m_{10}$	1 except at $m_2$ $⟹A+C^{\prime }$
$BD$	$m_5,m_7,m_{13},m_{15}$	1 except at $m_{15}$ $⟹A^{\prime }+C^{\prime }$
$B^{\prime }D$	$m_1,m_3,m_9,m_{11}$	all 0
$B{D}^{\prime }$	$m_4,m_6,m_{12},m_{14}$	all 0

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

Step 4: The other two “ways”

The question asks for three expressions. Form 1 above is the one that meets the eight-literal budget; two further valid factorisations (on $C$ and on $D$, at 12 and 10 literals) are worked out in 3.07 Boolean Algebra Puzzles & Exam Proofs §2 — present Form 1 first, since it is the one that actually satisfies the constraint.

How to spot a "factor it" question

Any question that sets a literal budget (“with eight or fewer literals”, “using no more than N gates”) is asking for a factored, multi-level form, not a minimal SOP. The tell is that the minimal SOP visibly exceeds the budget. Look for a variable pair whose regions each collapse to a short expression — here $B$ and $D$ — and factor on those.

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6. Class Test 01 Master Problem: SOP vs POS Gate Optimization

Class Test 01: SOP vs POS Gate Optimization

Question: Given $F=\Pi M(0,2,5,7,8,10,13,15)$: a) Convert into canonical SOP. b) Which implementation (SOP or POS) requires fewer logic gates?

Part A: Canonical SOP The minterm list is the exact inverse of the given maxterm list for a 4-variable system (0-15):

• Minterms = $1,3,4,6,9,11,12,14$
• Canonical SOP: $\mathbf{F}=\mathbf{\Sigma m}(1,3,4,6,9,11,12,14)$

Part B: Gate Optimization Justification To determine which implementation requires fewer gates, simplify both using K-maps:

• SOP Simplification: Plotting 1s yields a quad $(1,3,9,11)\to B^{\prime }D$ and a quad $(4,6,12,14)\to B{D}^{\prime }$. $F_{\text{SOP}}=B^{\prime }D+B{D}^{\prime }=\mathbf{B}\oplus \mathbf{D}$
• POS Simplification: Plotting 0s yields $F^{\prime }=B^{\prime }{D}^{\prime }+BD=(B\oplus D{)}^{\prime }$. Complementing gives: $F_{\text{POS}}=(B+D)(B^{\prime }+D^{\prime })=\mathbf{B}\oplus \mathbf{D}$
• Conclusion: Both simplified expressions mathematically reduce to the exact same logic (an XOR gate, or 2 ANDs + 1 OR). Therefore, both SOP and POS require the exact same number of logic gates. $■$

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Past Year Questions (PYQs)

Question (as asked)	Years	Marks	Solved in
Express $F(A,B,C,D)=(A+B^{\prime }+C)(A+B^{\prime })(A+C^{\prime }+D^{\prime })(A^{\prime }+B+C+D^{\prime })(B+C^{\prime }+D^{\prime })$ in sum of minterms and product of maxterms	2015	12	§2 method
Given $f(A,B,C)=(A+B+C^{\prime })(A+B^{\prime }+C^{\prime })(A^{\prime }+B+C^{\prime })(A^{\prime }+B^{\prime }+C^{\prime })$, construct the truth table and express in maxterm and minterm forms	2016	12	§2 method
Express in sum of minterms and product of maxterms: (i) $F(A,B,C)=(A^{\prime }+B)(B^{\prime }+C)$, (ii) $F(x,y,z)=(xy+z)(y+xz)$	2017, 2021	10–12	§3.1
Express $F(w,x,y,z)=y^{\prime }z+wx{y}^{\prime }+wx{z}^{\prime }+w^{\prime }{x}^{\prime }z$ in a sum of minterms and a product of maxterms	2019	9	§2 method
Convert into SOP and POS: (i) $(AB+C)(B+C^{\prime }D)$, (ii) $x^{\prime }+x(x+y^{\prime })(y+z^{\prime })$	2022	10	§3.1 method
What is meant by canonical and standard form? Express $F=xy+xz$ in product of maxterm form	2024	9	§3.2
Find the simplest SOP of $F=fg$ (full expressions in §4)	2018, 2022	7–10	§4
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	§5
Class Test 01: $F=\prod M(0,2,5,7,8,10,13,15)$ — canonical SOP, and which of SOP/POS needs fewer gates	CT	10	§6

Pattern to notice: minterm/maxterm expansion appears in every single paper, worth 9–12 marks, and the expression changes every year — so there is nothing to memorise beyond the two expansion rules in §1. Three sub-skills cover all of it:

1. Multiply out anything that is not already a clean SOP or POS (the bracket trap, §3.1)
2. Expand each term over its missing variables
3. Read the complementary index list for the other form

Highest-risk item: the $F=fg$ question (§4) — two appearances, and students routinely burn twenty minutes multiplying it out algebraically instead of intersecting index lists.