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shaurya vardhan
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Digital Logic
Oct 9, 2017
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Let there are 12 minterms in a function in which 8 minterms are covered by 2 Essential Prime Implicants. Each of the remaining 4 mintermshave 2 Non- Essential Prime Implicants. Then the total number of minimal expressions is
a)4
b)8
c)16
d)1
can someone pls explain this with an example ? answer is 16.
shaurya vardhan
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Digital Logic
Oct 9, 2017
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shaurya vardhan
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shaurya vardhan
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Oct 10, 2017
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Someone pls answer this ..
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arch
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Oct 24, 2017
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8 min terms are covered by essential prime implicants means essential prime implecants must be choose we dont have any choice now other 4 minters we have 2 choice for each minterms so, ans is 2*2*2*2= 16
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shaurya vardhan
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Oct 24, 2017
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Isn't 16 the number of ways we select the remaining 4 min terms ? What about the 8 mimterms that are covered by two essential prime implicants ?
Since we have to find out total number of minimal expression .. won't those two essential prime implicants be included .. ?
Kindly clear this doubt ..
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arch
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Oct 24, 2017
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essential prime implicants should be included u dont have choice but non essential prime implicants have choice among 2 non prime implicants.
for 8 mintermscoverd by essential prime implicant we have nly 1 choice so
1*1*1*1*1*1*1*1*2*2*2*2
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shaurya vardhan
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Oct 24, 2017
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Ok got it , thanks a lot :)
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