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m Make it clear which minterms are essential |
→Step 2: prime implicant chart: better use a different mark for the new meaning; if the rows are removed, they cols they cover should be removed, too |
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Line 169:
! || 4 || 8 || 10 || 11 || 12 || 15 || ⇒ || A || B || C || D
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| style="text-align:left;" | m(4,12) {{color|
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| style="text-align:left;" | m(8,9,10,11) || || {{Ya|✓}} || {{Ya|✓}} || {{Ya|✓}} || || || ⇒ || 1 || 0 || {{sdash}} || {{sdash}}
Line 175:
| style="text-align:left;" | m(8,10,12,14) || || {{Ya|✓}} || {{Ya|✓}} || || {{Ya|✓}} || || ⇒ || 1 || {{sdash}} || {{sdash}} || 0
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| style="text-align:left;" | m(10,11,14,15) {{color|
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To find the essential prime implicants, we look for columns with only one "✓". If a column has only one "✓", this means that the minterm can only be covered by one prime implicant. This prime implicant is ''essential''.
For example: in the first column, with minterm 4, there is only one "✓". This means that m(4,12) is essential (hence marked by {{color|blue|<sup>#</sup>}}). Minterm 15 also has only one "✓", so m(10,11,14,15) is also essential. Now all columns with one "✓" are covered. The rows with minterm m(4,12) and m(10,11,14,15) can now be removed, together with all the columns they cover.
The second prime implicant can be 'covered' by the third and fourth, and the third prime implicant can be 'covered' by the second and first, and neither is thus essential. If a prime implicant is essential then, as would be expected, it is necessary to include it in the minimized boolean equation. In some cases, the essential prime implicants do not cover all minterms, in which case additional procedures for chart reduction can be employed. The simplest "additional procedure" is trial and error, but a more systematic way is [[Petrick's method]]. In the current example, the essential prime implicants do not handle all of the minterms, so, in this case, the essential implicants can be combined with one of the two non-essential ones to yield one equation:
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