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I had got a following question

let $f(A,B,C,D)=Π(2,3,5,9,11,12,13)$. The total number of prime implicants and essential prime implicants are denoted by P and Q respectively. What is the value Q%P where ′%′ denotes the modulo operator?

I solved it using $POS$ $k-map$ and got $3$ as answer. But the answer given is $2$. I am not sure whether I am correct or not.

Can it be solved using $k-map$ of $POS$ alone or I should use $SOP$ and then find essential PI and PI of $(1,4,6,78,10,14,15)$?

Mr. Sigma.
  • 245

1 Answers1

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The function is defined in POS form (product of sums). To answer the question about implicants, it has to be converted to SOP form (sum of products):

disjunction | binary | inverted 
------------+--------+----------- 
     2      | 0010   | 1101 
     3      | 0011   | 1100 
     5      | 0101   | 1010 
     9      | 1001   | 0110 
    11      | 1011   | 0100 
    12      | 1100   | 0011 
    13      | 1101   | 0010

From the bit-wise inverses of the defined disjunctions, we know the seven maxterms of the function: 2, 3, 4, 6, 10, 12, 13 The minterms of the functions are the remaining nine terms: 0, 1, 5, 7, 8, 9, 11, 14, 15

The is depicted in the following Karnaugh map:

             cd
       00  01  11  10
      +---+---+---+---+
   00 | 1 | 1 | 0 | 0 |
      +---+---+---+---+
   01 | 0 | 1 | 1 | 0 |
ab    +---+---+---+---+
   11 | 0 | 0 | 1 | 1 |
      +---+---+---+---+
   10 | 1 | 1 | 1 | 0 |
      +---+---+---+---+

The list of prime implicants:

!B !C         essential
!A !C D
!A B D
B C D
A C D         essential
A B C         essential
A !B D

$P\%Q = 7\%3 = 1$

There might be a different result, if the bits $ABCD$ are counted differently. Apart from remaining flaws, the solution process should be ok.

Axel Kemper
  • 4,943