Exercises — Sum of products (SOP) form

SOP = the outermost operator is OR, and each piece it joins is a pure AND of literals (a single literal counts as a one-literal product). Use the precedence rules above to find the outermost operator: OR binds loosest, so it is applied last and therefore sits on the "outside".

• (a) Yes. Outer is $+$; pieces $A\bar{B}$ and $\bar{A}C$ are ANDs. ✓
• (b) No. Outer operator is AND (two bracketed sums multiplied). That is POS shape, not SOP.
• (c) Yes. A single product term is a valid (one-term) SOP.
• (d) Yes. Outer is $+$; each piece is a single literal (a product of one literal). ✓

A minterm equals 1 only when every literal is 1 simultaneously.

• $A=1$ (plain $A$), $\bar{B}=1\Rightarrow B=0$, $C=1$. So the row is $A=1,B=0,C=1$ — i.e. binary $101$ = row 5. Every other row makes at least one literal 0, so the AND is 0.

Step 1 — pick the 1-rows. $F=1$ at rows $01$ and $10$. Step 2 — write each minterm using the rule "0 gets a hat" — the overbar $\stackrel{ˉ}{}$ (variable is 0 → draw the hat/bar; 1 → leave bare):

• Row $01$: $A=0,B=1\Rightarrow \bar{A}B$
• Row $10$: $A=1,B=0\Rightarrow A\bar{B}$ Step 3 — OR the winners: $F=\bar{A}B+A\bar{B}$ This is the XOR function (see logic gates). In the $\sum m$ shorthand (OR the minterms of these row numbers), the 1-rows are $01$ = row 1 and $10$ = row 2, so canonical form $=\sum m(1,2)$.

Note the brackets are load-bearing: by the precedence rules $AB+\bar{C}$ would mean $(AB)+\bar{C}$, a different function — the brackets force the OR to happen first. The shape "variable AND a bracketed OR" is a product-of-sums seed. We use the distributive law ($X(Y+Z)=XY+XZ$) because it is the one law that turns an AND-over-OR into an OR-of-ANDs — exactly the SOP direction. $F=A\cdot B+A\cdot \bar{C}=AB+A\bar{C}$ Outer operator is now OR, each piece an AND → valid SOP with two terms.

Group and factor using Boolean Algebra Laws. First group the two terms sharing $B$: $\bar{A}B+AB=B(\bar{A}+A)=B\cdot 1=B$ (used complement law $\bar{A}+A=1$ — the green stripe on the left of Figure s02). Now: $F=B+A\bar{B}$ Apply absorption in the form $X+\bar{X}Y=X+Y$ (here $X=B$, $Y=A$, and $A\bar{B}=\bar{B}A$): $F=B+A$ Gate count: original = three 2-input ANDs + one 3-input OR. Final = a single 2-input OR gate. Same truth table, far cheaper circuit.

Method — simplify $F_1$ and compare. Group first two (share $\bar{A}$): $\bar{A}\bar{B}+\bar{A}B=\bar{A}(\bar{B}+B)=\bar{A}$. $F_1=\bar{A}+AB$ Absorption $\bar{A}+AB=\bar{A}+B$ (since $X+\bar{X}Y=X+Y$ with $X=\bar{A},Y=B$). $F_1=\bar{A}+B=F_2✓$ Cross-check by truth table (both give $1$ except when $A=1,B=0$): rows $00,01,11\to 1$, row $10\to 0$. Identical.

Step 1 — list the 1-rows (two or three switches on). Reading each row as the input triple $(ABC)$, the 1-rows are $011,101,110,111$. Step 2 — minterms (0→bar, 1→plain):

• $(ABC)=011\to \bar{A}BC$
• $(ABC)=101\to A\bar{B}C$
• $(ABC)=110\to AB\bar{C}$
• $(ABC)=111\to ABC$ Step 3 — canonical SOP: $F=\bar{A}BC+A\bar{B}C+AB\bar{C}+ABC=\sum m(3,5,6,7)$ Step 4 — simplify by pairing $ABC$ with each of the other three (it shares two literals with each): $ABC+\bar{A}BC=BC(A+\bar{A})=BC$ $ABC+A\bar{B}C=AC(B+\bar{B})=AC$ $ABC+AB\bar{C}=AB(C+\bar{C})=AB$ Why is OR-ing these three factored results the same as the original four minterms? We use the idempotent law $X+X=X$ (right side of Figure s02): OR-ing a term with itself changes nothing, so we may duplicate $ABC$ for free: $ABC=ABC+ABC+ABC$. Rewrite the four-minterm sum as $F=\underset{=BC}{\underbrace{(ABC+\bar{A}BC)}}+\underset{=AC}{\underbrace{(ABC+A\bar{B}C)}}+\underset{=AB}{\underbrace{(ABC+AB\bar{C})}}.$ Every one of the four original minterms still appears at least once inside those brackets ($\bar{A}BC$, $A\bar{B}C$, $AB\bar{C}$ each once, and $ABC$ three times), and no new row is introduced — each bracket only pairs $ABC$ with an existing minterm. So the OR of the three factored terms lights up on exactly the same four rows. Therefore: $\boxed{F=AB+BC+AC}$ Read it: "any two switches together." The K-map in Figure s03 gives the same three pairs visually — each looped pair of adjacent 1-cells collapses to one two-literal term.

Each term is missing a variable. Multiply the missing variable in as $(X+\bar{X})=1$ — this changes nothing (that is the complement law, the green stripe in Figure s02: $X+\bar{X}$ is always 1, and multiplying by 1 is harmless) but it reveals the hidden minterms. $A=A(B+\bar{B})=AB+A\bar{B}$ $\bar{B}=(A+\bar{A})\bar{B}=A\bar{B}+\bar{A}\bar{B}$ Combine and drop the duplicate $A\bar{B}$ (by the idempotent law $X+X=X$): $F=AB+A\bar{B}+\bar{A}\bar{B}=\sum m(0,2,3)$ Rows: $00\to \bar{A}\bar{B}$ (m0), $10\to A\bar{B}$ (m2), $11\to AB$ (m3). ✓

$F=1$: every row outputs 1, so OR all minterms: $F=\bar{A}\bar{B}+\bar{A}B+A\bar{B}+AB=\sum m(0,1,2,3)$ (Simplifies right back to $1$: $\bar{A}(\bar{B}+B)+A(\bar{B}+B)=\bar{A}+A=1$.) $F=0$: there are no 1-rows, so there are no minterms to OR. The canonical SOP is the empty sum, which by convention equals $0$. Why must an empty OR equal 0? An OR asks "is at least one input 1?". Its neutral/identity value is 0, because OR-ing $0$ onto anything leaves it unchanged ($X+0=X$). So starting an OR "accumulator" at 0 and then adding no terms leaves it at 0 — nothing ever turned it on. (Set-algebra analogy: the union of no sets is the empty set; nobody was ever added, so nothing is inside.) You cannot list winners when nobody wins — the degenerate case.

Step 1 — find the forbidden row. $AB=10$ means $A=1,B=0$ → row $10$ = m2. So $F=0$ only at m2; $F=1$ everywhere else. Step 2 — canonical SOP over the remaining rows $00,01,11$: $F=\bar{A}\bar{B}+\bar{A}B+AB=\sum m(0,1,3)$ Step 3 — simplify. Group first two (share $\bar{A}$): $\bar{A}(\bar{B}+B)=\bar{A}$. $F=\bar{A}+AB\stackrel{X+\bar{X}Y=X+Y}{\to }\bar{A}+B$ Step 4 — gate count: just one NOT (for $\bar{A}$) and one 2-input OR. Compare: the complement view says the single 0-row $10$ gives POS $F=(\bar{A}+B)$ directly — the same answer, confirming both methods agree.

Expand each term to full minterms (insert the missing variable via $(X+\bar{X})$):

• $AB=AB(C+\bar{C})=ABC+AB\bar{C}\to$ rows $111$ (m7), $110$ (m6).
• $\bar{A}C=\bar{A}C(B+\bar{B})=\bar{A}BC+\bar{A}\bar{B}C\to$ rows $011$ (m3), $001$ (m1). Collect: $\{m1,m3,m6,m7\}$. $F=\sum m(1,3,6,7)✓$ Truth-table check — build the full table for $F=AB+\bar{A}C$ and read off where it is 1:

The 1-rows are exactly $\{1,3,6,7\}$ — a one-to-one match with the four minterms. ✓

1. Outer operator that identifies SOP? → OR.
2. Which bit gets the bar (the "hat") in a minterm? → The 0 bit.
3. SOP of $F=0$? → Empty sum $=0$ (no minterms; 0 is OR's identity).
4. Law used to inject a missing variable? → Multiply by $(X+\bar{X})=1$ (complement law).
5. Law $X+\bar{X}Y=?$ → $X+Y$ (absorption).
6. Law that lets you duplicate a term for free? → Idempotent $X+X=X$.
7. Precedence order strongest→weakest? → NOT, then AND, then OR.