--- title: "Minterms/maxterms; K-map Σ(1,3,5,8,10,11,12)+d" subject: "Digital Logic" chapter: 3 key: "semester_3/digital_logic:minterm-maxterm-kmap-1358" source: https://www.bctengineer.com/q/semester_3/digital_logic/minterm-maxterm-kmap-1358 site: BCT Engineering --- # Minterms/maxterms; K-map Σ(1,3,5,8,10,11,12)+d Digital Logic · Chapter 3: Combinational Logic Circuits - Asked in: 2081 Baisakh, 2+4 marks - Video: [Lec -13: 4 Variable K-Map with examples | Design K-Map | Minimization in Digital Electronics](https://www.youtube.com/watch?v=0vEvpcY8O3o) — Gate Smashers - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/minterm-maxterm-kmap-1358 - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-3 ## Question Define min-terms and max-terms. Simplify the function using K-map, F = Σ(1,3,5,8,10,11,12) and D = Σ(2,9,15) and also realize the simplified logic circuit. ## Short answer A minterm is an AND term with all variables, true for one input; a maxterm is an OR term with all variables, false for one input. Simplified: F = B'C + A'C'D + AC'D'. The circuit uses AND-OR gates with inverters. ## Answer #### Min-terms and max-terms A **minterm** is a product (AND) term that contains every variable of the function exactly once, in true or complemented form. It equals 1 for exactly one input combination. A variable is written true if its value is 1 and complemented if it is 0; e.g. for ABC = 101 the minterm is $m_5 = A\overline{B}C$. A function in **canonical SOP** is a sum of minterms: $F = \Sigma m(\dots)$. A **maxterm** is a sum (OR) term that contains every variable exactly once, in true or complemented form. It equals 0 for exactly one input combination. A variable is written true if its value is 0 and complemented if it is 1; e.g. for ABC = 101 the maxterm is $M_5 = \overline{A}+B+\overline{C}$. A function in **canonical POS** is a product of maxterms: $F = \Pi M(\dots)$. $M_i = \overline{m_i}$, and a function's minterm list and maxterm list are complementary (together they hold all $2^n$ numbers). #### K-map simplification $F(A,B,C,D) = \Sigma m(1,3,5,8,10,11,12) + d(2,9,15)$ ```text CD AB 00 01 11 10 +---+---+---+---+ 00 | 0 | 1 | 1 | X | +---+---+---+---+ 01 | 0 | 1 | 0 | 0 | +---+---+---+---+ 11 | 1 | 0 | X | 0 | +---+---+---+---+ 10 | 1 | X | 1 | 1 | +---+---+---+---+ ``` - Pair m(1,5) → $\overline{A}\,\overline{C}D$ - Pair m(8,12) → $A\overline{C}\,\overline{D}$ - Quad m(2,3,10,11) → $\overline{B}C$ $$ F = \overline{B}C + \overline{A}\,\overline{C}D + A\overline{C}\,\overline{D} $$ Check: 3, 10, 11 by $\overline{B}C$ (with don't-care 2); 1, 5 by $\overline{A}\,\overline{C}D$; 8, 12 by $A\overline{C}\,\overline{D}$. Don't-cares 9 and 15 are not needed. #### Logic circuit (AND-OR) ```text B' ─┐ C ─┴─[AND]── B'C ────┐ A' ─┐ │ C' ─┼─[AND]── A'C'D ──┼─[OR]── F D ─┘ │ A ─┐ │ C' ─┼─[AND]── AC'D' ──┘ D' ─┘ ``` Complements $\overline{A}, \overline{B}, \overline{C}, \overline{D}$ come from NOT gates. The circuit needs one 2-input AND, two 3-input ANDs and one 3-input OR (plus inverters). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the minterm m5 for three variables? A. A'BC' B. AB'C C. A + B' + C D. A'+ B + C' Answer: B. 101 gives A true, B complemented, C true: AB'C. (Key point: Minterm for ABC=101) 2. Which quad gives B'C using don't-care 2? A. m(2,3,10,11) B. m(1,3,5,7) C. m(8,9,12,13) D. m(0,1,2,3) Answer: A. Cells 2,3,10,11 have B=0 and C=1; cell 2 is a don't-care. (Key point: Quad using don't-care) 3. What is the minimal SOP of Σ(1,3,5,8,10,11,12) + d(2,9,15)? A. B'C + A'C'D + AC'D' B. B'C + A'D + AD' C. B'C + C'D D. A'C'D + AC'D' + CD Answer: A. B'C covers 3,10,11; A'C'D covers 1,5; AC'D' covers 8,12. (Key point: Minimal SOP) 4. Which don't-cares are not used in the simplification? A. 2 only B. 9 and 15 C. 2 and 9 D. 15 only Answer: B. Don't-care 2 is used in the quad; 9 and 15 are not needed. 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