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Minterms/maxterms; K-map Σm(0,1,2,4,5,9,12,15)+d, NAND

Short answer

A minterm is an AND term with every variable (e.g., A'B'C'D'), and a maxterm is an OR term with every variable (e.g., A+B+C+D'), each corresponding to a single input combination. K-map simplification yields Y = A' B' D' + B C' + C' D + B D. NAND implementation uses NAND inverters for complements, four first-level NAND gates for the product terms, and a final NAND to produce Y.

What are the minterms and the maxterms? Simplify the following logic function using K-map and implement the result using NAND gates only. Y(A,B,C,D) = Σm(0,1,2,4,5,9,12,15) + d(7,10,13)

Minterms and maxterms

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 m5=AB‾Cm_5 = A\overline{B}C. A function in canonical SOP is a sum of minterms: F=Σm(… )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 M5=A‾+B+C‾M_5 = \overline{A}+B+\overline{C}. A function in canonical POS is a product of maxterms: F=ΠM(… )F = \Pi M(\dots).

Mi=mi‾M_i = \overline{m_i}, and a function's minterm list and maxterm list are complementary (together they hold all 2n2^n numbers).

K-map simplification

Y(A,B,C,D)=Σm(0,1,2,4,5,9,12,15)+d(7,10,13)Y(A,B,C,D) = \Sigma m(0,1,2,4,5,9,12,15) + d(7,10,13)

        CD
AB      00  01  11  10
      +---+---+---+---+
 00   | 1 | 1 | 0 | 1 |
      +---+---+---+---+
 01   | 1 | 1 | X | 0 |
      +---+---+---+---+
 11   | 1 | X | 1 | 0 |
      +---+---+---+---+
 10   | 0 | 1 | 0 | X |
      +---+---+---+---+
  • Pair m(0,2) → A‾ B‾ D‾\overline{A}\,\overline{B}\,\overline{D}
  • Quad m(4,5,12,13) → BC‾B\overline{C}
  • Quad m(1,5,9,13) → C‾D\overline{C}D
  • Quad m(5,7,13,15) → BDBD
Y=A‾ B‾ D‾+BC‾+C‾D+BDY = \overline{A}\,\overline{B}\,\overline{D} + B\overline{C} + \overline{C}D + BD

Check: 0, 2 by A‾ B‾ D‾\overline{A}\,\overline{B}\,\overline{D}; 4, 5, 12 by BC‾B\overline{C}; 1, 9 by C‾D\overline{C}D; 15 by BDBD (with don't-cares 7, 13). Don't-care 10 is not used.

NAND-only implementation

An SOP expression maps directly to a two-level NAND-NAND circuit (De Morgan: X+Y=X‾⋅Y‾‾X + Y = \overline{\overline{X}\cdot\overline{Y}}). Each product term becomes a first-level NAND and the output NAND replaces the OR gate. Complemented inputs are made with NAND gates whose inputs are tied.

Y=A‾ B‾ D‾‾⋅BC‾‾⋅C‾D‾⋅BD‾‾Y = \overline{\overline{\overline{A}\,\overline{B}\,\overline{D}}\cdot\overline{B\overline{C}}\cdot\overline{\overline{C}D}\cdot\overline{BD}}
A ──[NAND]── A'   (inputs tied)
B ──[NAND]── B'   (inputs tied)
C ──[NAND]── C'   (inputs tied)
D ──[NAND]── D'   (inputs tied)

A', B', D' ──[NAND]── P1  ─┐
B, C'      ──[NAND]── P2  ─┤
                           ├──[NAND]── Y
C', D      ──[NAND]── P3  ─┤
B, D       ──[NAND]── P4  ─┘

Gates: 4 NAND inverters, 4 first-level NANDs and one 4-input output NAND.

✓ Answer checked by Claude Opus 5.5 (high effort)

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