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The max no. of prime implicants in the minimized expression with n-variable is 2^n-1 . Can someone please explain how ?
in Digital Logic by Junior (533 points)
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How many prime implicant are there :

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Anu007 6 PRIME IMPLICANTS

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yes anu, there are 6 prime implicants in this k-map.
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I got EPI as 2n-1 .Then how PI will be 2n-1

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@Anu007  its not minimized yet thats why

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ok i got it...

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For this we should know the basic definition of prime implicant in K Map..A prime implicant is :

A subcube in K Map which is not a part of  some other larger subcube completely..

Now for hence for maximum no of prime implicants we require that the cells itself are prime implicants and hence not able to be covered by other subcube..

So the arrangement is like this for 3 variable say:

In 1st column we have 1 at 2nd row , then at 2nd column we will have 1 at 1st row and continuing in this manner alternately bottom and up column wise we have 4 minterms and they themselves are prime implicants as well.

Similarly for 4 variable K Map  , we have 8 maximum no of prime implicants possible..

Hence this can be generalised for n variables as = 2n-1 prime implicants..

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