# Digital - Boolean Algebra - 9

+1 vote

How many implicants exist in the following KMap?

(A) 5
(B) 6
(C) 7
(D) 8

reshown May 14

There's a finer point that I missed while framing this question:

An implicant need not be a minterm but a minterm is always an implicant.

If the question asked for no. of minterms, 8 would've been the correct answer. But no. of implicants can be more than the no. of minterms. After all, prime implicants are also implicants.

The number of implicants = 8 + # of sets of two adjacent 1's + the central box of 4 1's.

An implicant, i, of a function, f, is essentially a term that implies : IF i is true THEN f is true.
answered May 15 by (660 points)
selected May 16
Sir, you are correct that every minterm is an implicant but as per the rules for framing implicant we always prefer as big as possible subcube. By your statement we always have to include even implicant of size 1 which is minterm which is not as per rules of  kmap.
Every subcube is considered as an implicant .Hence even subcube  consisting two adjacent 1's present inside the central box of 4 1's should be considered
hence
total no. of implicants=minterms+#of sets of two adjacent 1's+# of sets of 4 adjacent 1's
=8+8+1=17
5 is the right answer according to me.

1)4 1's in the middle which form a square.

each vertex 1 of square forms an implicant with adjacent 1 to the vertex not belonging in that square.

so total=1+4=5
answered May 14 by (1,950 points)
I think the answer should be 8 every single term represents an implicant. If they would have asked prime implicants, then the answer would be 5 as done by you
@tssharmaayush361 I agree with this .While calculating implicants we have to consider every single term .As in question it is clearly mention number of implicants answer should be 8.
–1 vote

Every Sub-Cube in K-Map is an IMPLICANT(Size may be 1,2,4,8,16, ..... ie powers of 2)

But Since No option Match with Answer which is actually Reqd by the Question(No of IMPLICANTS)

So May be Question Wants to ask No. of Prime Implicants

Then It would be 5 (4 pairs and 1 QUAD in Middle

Also No. of EPI=4 (NO OPTION)

Therefore BEST ANS IS 5 !

answered May 14 by (730 points)

4 Essential Prime Implicants(EPI) which are A'C'D , ABC' , A'BC , ACD .

Remaining 5 Implicants are A'BD , BCD , BC'D , ABD & BD .

So, totally, there are 4 + 5 = 9 implicants. But 9 is not given in the options.

Actually, I marked option (A) 5 because there are 4 EPI and 1 implicant BD .

BD covers A'BD , BCD , BC'D , &  ABD .

Ans.:

answered May 14 by (6,910 points)
edited May 14
There are 8 implicants only in this question. 4 are EPI,PI are 5,1 is RPI. Check example 1 from this link which is exact question like the above one-
@getgatebook

Why marks given to all in this question?

Ans should be 8.

And if not then what is the answer and explanation. Please guide us.
answered May 15 by (11,660 points)
+1 vote

P is called an implicant of F if F also takes the value 1 whenever P equals 1.

where

For instance, the function

is implied by , by , ,  and many others; these are the implicants of .
In the karnaugh map it is easy to see. All the singletons, pairs, octets....are implicants.
There are 17 implicants.

answered May 15 by (31,410 points)

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no option correct but here if we find out

Total implicabt =17

Total pI= 5

Total EPI= 4

answered Jun 23 by (24,690 points)