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K-map Σ(0,1,4,8,10,11,12)+d, realize circuit

Short answer

The simplified function is F = B' + C'D'. The circuit uses a NOT gate for B, a NOR gate for C and D to produce (C+D)' = C'D', and an OR gate to combine B' and the NOR output, yielding F.

Simplify the function using K-map F = Σ(0,1,4,8,10,11,12) and D = Σ(2,3,6,9,15). Also realize the simplified logic circuit.

Given F(A,B,C,D)=Σm(0,1,4,8,10,11,12)+d(2,3,6,9,15)F(A,B,C,D) = \Sigma m(0,1,4,8,10,11,12) + d(2,3,6,9,15).

K-map simplification

        CD
AB      00  01  11  10
      +---+---+---+---+
 00   | 1 | 1 | X | X |
      +---+---+---+---+
 01   | 1 | 0 | 0 | X |
      +---+---+---+---+
 11   | 1 | 0 | X | 0 |
      +---+---+---+---+
 10   | 1 | X | 1 | 1 |
      +---+---+---+---+

Groups (don't-cares 2, 3, 9 complete the octet; 6 and 15 are not used):

  • Octet m(0,1,2,3,8,9,10,11) → B‾\overline{B}
  • Quad m(0,4,8,12) → C‾ D‾\overline{C}\,\overline{D}
F=B‾+C‾ D‾F = \overline{B} + \overline{C}\,\overline{D}

Check: the octet covers 0, 1, 8, 10, 11 and the quad covers 0, 4, 8, 12.

Logic circuit

FF needs one OR gate, one AND gate and inverters for B, C, D:

B ──[NOT]── B' ─────────────┐
                            [OR]── F
C ──[NOT]── C' ──┐          │
                 [AND]─ C'D'┘
D ──[NOT]── D' ──┘

Since C‾ D‾=C+D‾\overline{C}\,\overline{D} = \overline{C+D}, the AND gate with two inverters can be replaced by a single NOR gate:

B ──[NOT]── B' ─────────┐
                        [OR]── F
C ──┐                   │
    [NOR]── C'D' ───────┘
D ──┘

The simplified circuit uses only 3 gates (NOT, NOR, OR), compared with 7 four-input AND terms for the unsimplified expression.

A B C D (example)B'C'D'F
0 1 0 0 (m4)011
0 1 0 1 (m5)000
1 0 1 1 (m11)101

✓ Answer checked by Claude Opus 5.5 (high effort)

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