GATE CSE 2015 Set 3 | Question: 43

Question

The total number of prime implicants of the function $f(w, x, y, z) = \sum (0, 2, 4, 5, 6, 10)$ is __________

Comments

Make use of Quine-McCluskey Method :

 

 

 

 

 

 

 

Prime Implicants are the crossed one $(\times)$ : $m_{2}+m_{10}|m_{4}+m_{5}|m_{0}+m_{2}+m_{4}+m_{6}$

Total of 3 Prime implicants.

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answer is 3

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Quine-McCluskey Method :

not required for GATE

Answer 1

As you can see that there is  one $4$ -set and two $2$ -set that are covering the star marked $1's$ (i.e. the ones that are not covered by any other combinations).

So, the answer is $3$.

Comments

@shivangi EPI are term which are not covered and they are 4 marked by *

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EPI should be 3 not 4.

As we are able to cover all 1's in 3 PI.

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I agree that no of EPI = 3

Answer 2

(Can be solved using K-Map also. )

Place all minterms that evaluate to one in a minterm table.

Input (first column for no. of 1's)

0	m0	0000
1	m2 m4	0010 0100
2	m5 m6 m10	0101 0110 1010

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Combine minterms with other minterms. If two terms vary by only a single digit changing, that digit can be replaced with a dash indicating that the digit doesn't matter. Terms that can't be combined any more are marked with a "*". When going from Size 2 to Size 4, treat '-' as a third bit value. For instance, -110 and -100 or -11- can be combined, but -110 and 011- cannot. (Trick: Match up the '-' first.)

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First Comparison

0	(2, 0) (4, 0)	00-0 0-00
1	(6, 2) (10, 2) (5, 4) (6, 4)	0-10 -010 010- 01-0

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Second Comparison

0

(6, 4, 2, 0)

0--0

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Prime Implicants

(6, 4, 2, 0)	0--0
(10, 2) (5, 4)	-010 010-

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Answer: Total number of prime implicants 

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Source: Finding prime implicants - Quine-McCluskey algorithm - Wikipedia

Comments

is Quine McCluckey algo in gate syllabus

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na ,not in syllabus

Answer 3

all the min terms can only be covered by three diffrernt color so there is only three prime imlecants

Answer 4

I think this is bit easier to understand.
The hint here is that we try to find the minimum number of groups(octa(eight 1's) ,quad (four 1's),dual(two 1's) that can be formed and that represents the minimum number of implicants that covers F which is Prime Implicant.

Here, we can form 1 quad and 2 duets.3 prime implicants.

 

Comments

The blue combination is wrong. There is two bits of difference between the blue combination.

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@Harsh Kumar-Yes, this is wrong. Thanks :)

Corrected Now. Please verify.