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So the Prime implicants are those largest square or rectangle made up of bunch of adjacent minterms Essential Prime implicants are those prime implicants which also cover a particular midterm/s. But I can't find much definition about when the function has don't care. This question asks about implicant, but not the essential ones

Let's say we have a function as f(x,y,z)=m(0,1,6,7)

the K-map as:

--------------------------------------------------
| c\ab |      |      |     |     |
--------------------------------------------------
|      |   1  |      |  1  |     |
--------------------------------------------------
|      |   1  |      |  1  |     |
--------------------------------------------------

From the k-map, we can see there are 4 implicants and 2 prime implicants and 2 essential prime implicants, right ?

My questions is: Are don't-cares considered as 1/0 when we counting essential/prime /implicants as well once we mark them as 1/0?

Now we have a similar function as f(x,y,z)=m(0,1,6,7) + d(2,3) so the k map as:

--------------------------------------------------
| c\ab |      |      |     |     |
--------------------------------------------------
|      |   1  | x(1) |  1  |     |
--------------------------------------------------
|      |   1  | x(1) |  1  |     |
--------------------------------------------------

In this k-map, we have 6 implicants for each one: and 2 prime implicants from

1 x(1)
1 x(1)

of the left and

 x(1) 1
 x(1) 1

Of the right.

and 2 essential prime implicants from those prime implicants.

so 6 implicants and 2 prime implicants and 2 essential prime implicants for f(x,y,z)=m(0,1,6,7) + d(2,3).

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  • \$\begingroup\$ You have only one (single) prime implicant for both cases. \$\endgroup\$
    – jay
    Commented Oct 15, 2021 at 1:57
  • \$\begingroup\$ Are the only prime implicants the essential one for each case? \$\endgroup\$ Commented Oct 17, 2021 at 10:42
  • \$\begingroup\$ This explains all about implicants of K-Map. do't-care is conveniently taken as either '0' or '1'. \$\endgroup\$
    – jay
    Commented Oct 17, 2021 at 12:51

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