<P> For example: in the first column, with minterm 4, there is only 1 "X". This means that m (4, 12) is essential . So we place a star next to it . Minterm 15 also has only 1 "X", so m (10, 11, 14, 15) is also essential . Now all columns with 1 "X" are covered . </P> <P> 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, one can combine the essential implicants with one of the two non-essential ones to yield one equation: </P> <Dl> <Dd> f A, B, C, D = B C ′ D ′ + A B ′ + A C (\ displaystyle f_ (A, B, C, D) = BC'D' + AB'+ AC \) </Dd> </Dl> <Dd> f A, B, C, D = B C ′ D ′ + A B ′ + A C (\ displaystyle f_ (A, B, C, D) = BC'D' + AB'+ AC \) </Dd>

Explain the advantages and disadvantages of qm algorithm