Boolean algebra and Karnaugh maps
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| de morgan's theorem/də ˈmɔːɡənz ˈθɪərəm/ | Breaking a complement bar changes AND to OR and vice versa: (A·B)' = A' + B'. |
| karnaugh map/ˈkɑːnɔː mæp/ | A truth table laid out in Gray-code order so adjacent cells differ in exactly one variable. |
Boolean and Karnaugh: adjacency is simplification
- Absorption
A + AB = Aand De Morgan's theorem(AB)' = A' + B'do the shortening; Karnaugh map groups of $2^k$ cells eliminate k literals; don't-cares only ENLARGE groups. - Every answer checks by exhaustive evaluation — 8 rows is nothing.
$\sum m(1,3,5,7)$ (the odd rows) is just C: all four minterms share only C, and the 8-row check confirms.

{exercise:uni-ee3x-boolean}
The Karnaugh group for Σm(1, 3, 5, 7) of A, B, C reads as —
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Odd-indexed rows are exactly the C = 1 rows: a clean 4-cell group eliminating A and B. Exhaustive check over 8 rows confirms F = C.
The 4-variable majority function (1 when at least 3 inputs are 1) has minterms —
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Count 1s in each 4-bit code: 0111(7), 1011(11), 1101(13), 1110(14), 1111(15) - exactly the codes with three or four 1s. Simplified: AB(C+D) + CD(A+B), verified on all 16 rows.
A K-map group of four cells keeps which variables?
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A variable that changes inside the group is eliminated (both values appear, so it cannot distinguish). The odd-function group keeps only C because A and B both change inside it.