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K-map Σ(2,3,6,7,9,10,11)+d in SOP and POS

Short answer

SOP form: F = A' C + A B' (using don't-cares 5,12,8). POS form: F = (A+C)(A'+B') (using don't-cares5,12 as 0, 8 as 1). Both expressions describe the same function with minimal two-level gate implementation.

Simplify F(A,B,C,D) = Σ(2,3,6,7,9,10,11) and d = Σ(5,8,12) using K-map. Result in SOP and POS form.

Given F(A,B,C,D)=Σm(2,3,6,7,9,10,11)+d(5,8,12)F(A,B,C,D) = \Sigma m(2,3,6,7,9,10,11) + d(5,8,12).

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

SOP form (grouping the 1s)

  • Quad m(2,3,6,7) → A‾C\overline{A}C
  • Quad m(8,9,10,11) → AB‾A\overline{B}
F=A‾C+AB‾F = \overline{A}C + A\overline{B}

Don't-care 8 completes the AB‾A\overline{B} quad; 5 and 12 are taken as 0.

POS form (grouping the 0s)

The 0-cells are 0, 1, 4, 13, 14, 15. Don't-cares 5 and 12 are used as 0 here (don't-care 8 is not grouped, so it stays 1):

  • Quad of 0s M(0,1,4,5) → (A+C)(A+C)
  • Quad of 0s M(12,13,14,15) → (A‾+B‾)(\overline{A}+\overline{B})
F=(A+C)(A‾+B‾)F = (A+C)(\overline{A}+\overline{B})

Check

Expanding the POS:

(A+C)(A‾+B‾)=AA‾+AB‾+A‾C+B‾C=AB‾+A‾C+B‾C=AB‾+A‾C\begin{aligned} (A+C)(\overline{A}+\overline{B}) &= A\overline{A} + A\overline{B} + \overline{A}C + \overline{B}C \\ &= A\overline{B} + \overline{A}C + \overline{B}C \\ &= A\overline{B} + \overline{A}C \end{aligned}

(B‾C\overline{B}C is the consensus of AB‾A\overline{B} and A‾C\overline{A}C, so it drops out.) Both forms describe exactly the same function: each sets don't-care 8 to 1 and don't-cares 5 and 12 to 0.

Each form needs only two 2-input gates plus one 2-input output gate (and inverters).

✓ Answer checked by Claude Opus 5.5 (high effort)

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