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Minterms/maxterms; K-map Σm(0,1,2,5,8,14)+d, NOR

Short answer

Minterm: product of all variables, equals 1 for one combination. Maxterm: sum of all variables, equals 0 for one combination. Simplified F = A' C' + B' D' + A C D'. NOR-only: rewrite terms as (A+C)', (B+D)', (A'+C'+D)', feed to a NOR for F', then NOR inverter. Uses 7 NOR gates.

Define minterms and maxterms. Simplify the following using k-map and implement the result using NOR gates only. F(A,B,C,D) = Σm(0,1,2,5,8,14) + d(4,10,13)

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

F(A,B,C,D)=Σm(0,1,2,5,8,14)+d(4,10,13)F(A,B,C,D) = \Sigma m(0,1,2,5,8,14) + d(4,10,13)

        CD
AB      00  01  11  10
      +---+---+---+---+
 00   | 1 | 1 | 0 | 1 |
      +---+---+---+---+
 01   | X | 1 | 0 | 0 |
      +---+---+---+---+
 11   | 0 | X | 0 | 1 |
      +---+---+---+---+
 10   | 1 | 0 | 0 | X |
      +---+---+---+---+
  • Quad m(0,1,4,5) → A‾ C‾\overline{A}\,\overline{C}
  • Quad m(0,2,8,10) → B‾ D‾\overline{B}\,\overline{D}
  • Pair m(10,14) → ACD‾AC\overline{D}
F=A‾ C‾+B‾ D‾+ACD‾F = \overline{A}\,\overline{C} + \overline{B}\,\overline{D} + AC\overline{D}

Check: 0, 1, 5 by A‾ C‾\overline{A}\,\overline{C} (with 4); 2, 8 by B‾ D‾\overline{B}\,\overline{D} (with 10); 14 by ACD‾AC\overline{D} (with 10). Don't-care 13 is not used. (The minimal POS needs 4 terms, so the SOP is simpler.)

NOR-only implementation

For NOR-only realization of an SOP expression, write each product term as a NOR of complemented literals (X‾ Y‾=X+Y‾\overline{X}\,\overline{Y} = \overline{X+Y}), feed these to a NOR gate (which gives F‾\overline{F}), and invert with one more NOR whose inputs are tied.

A‾ C‾=A+C‾B‾ D‾=B+D‾ACD‾=A‾+C‾+D‾F=P1+P2+P3‾‾\begin{aligned} \overline{A}\,\overline{C} &= \overline{A+C} \\ \overline{B}\,\overline{D} &= \overline{B+D} \\ AC\overline{D} &= \overline{\overline{A}+\overline{C}+D} \\ F &= \overline{\overline{P_1+P_2+P_3}} \end{aligned}
A ──[NOR]── A'   (inputs tied)
C ──[NOR]── C'   (inputs tied)

A, C      ──[NOR]── P1  ─┐
B, D      ──[NOR]── P2  ─┤──[NOR]── F'
A', C', D ──[NOR]── P3  ─┘

F' ──[NOR]── F   (inputs tied)

Gates: 2 NOR inverters, two 2-input NORs, one 3-input NOR, a 3-input NOR giving F‾\overline{F} and a NOR inverter: 7 NOR gates.

✓ Answer checked by Claude Opus 5.5 (high effort)

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