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K-map Σ(0,2,3,4,7,8,10,13)+d, logic circuit

Short answer

Simplified Y = A' C + B' D' + B C'. Logic circuit: three 2-input AND gates for terms A'C, B'D', BC' and one 3-input OR gate; inverters generate A', B', C', D'. (Alternatively, the B'D' AND can be replaced by a NOR if desired.)

Simplify the following expressions using K-map and also draw the logical circuit. Y(A,B,C,D) = Σ(0,2,3,4,7,8,10,13) and d = Σ(5,6,12)

Given Y(A,B,C,D)=Σm(0,2,3,4,7,8,10,13)+d(5,6,12)Y(A,B,C,D) = \Sigma m(0,2,3,4,7,8,10,13) + d(5,6,12).

K-map simplification

        CD
AB      00  01  11  10
      +---+---+---+---+
 00   | 1 | 0 | 1 | 1 |
      +---+---+---+---+
 01   | 1 | X | 1 | X |
      +---+---+---+---+
 11   | X | 1 | 0 | 0 |
      +---+---+---+---+
 10   | 1 | 0 | 0 | 1 |
      +---+---+---+---+
  • Quad m(2,3,6,7) → A‾C\overline{A}C
  • Quad m(0,2,8,10) → B‾ D‾\overline{B}\,\overline{D}
  • Quad m(4,5,12,13) → BC‾B\overline{C}
Y=A‾C+B‾ D‾+BC‾Y = \overline{A}C + \overline{B}\,\overline{D} + B\overline{C}

Check: 2, 3, 7 by A‾C\overline{A}C (with 6); 0, 2, 8, 10 by B‾ D‾\overline{B}\,\overline{D}; 4, 13 by BC‾B\overline{C} (with 5, 12). All 8 minterms are covered with three quads.

Logic circuit

A ─[NOT]─ A' ─┐
C ────────────┴─[AND]── A'C ──┐
B ─[NOT]─ B' ─┐               │
D ─[NOT]─ D' ─┴─[AND]── B'D' ─┼─[OR]── Y
B ────────────┐               │
C ─[NOT]─ C' ─┴─[AND]── BC' ──┘

Gates: 4 NOT gates, three 2-input AND gates and one 3-input OR gate. (As B‾ D‾=B+D‾\overline{B}\,\overline{D} = \overline{B+D}, that AND with two inverters can be replaced by one NOR gate.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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