Identifying self dual function

Question

Given answer: D
I am not getting how to approach this question.

Answer 1

A Self dual function is a function such that$f=f^D$

option D). $f=\Sigma_m(1,2,4,7)$

$F= A'B'C+A'BC'+AB'C'+ABC = (A'B'+AB).C+(A'B+AB').C'$

$= \overline{(A\oplus B)}.C+(A\oplus B).\overline{C}= (A\oplus B)\oplus C$

Now its dual (Dual is obtained by complementing the input variables and then by complementing the function)

$f^D= ((AB+A'B')C'+(AB'+A'B)C)'$
$= ((A\oplus B)'C'+(A\oplus B)C)' =\overline{((A\oplus B)\odot C)}$

$=(A\oplus B)\oplus C$

Comments

I have read somewhere that dual is obtained by replace OR -> AND, AND -> OR, 0 -> 1 and 1 -> 0. Is your definition of dual same as this definition?
and for solving such questions do I need to check this for each option?

In the explanation of this question this was given:
"Every cyclic K-map given cyclic function and every cyclic function can be the self-dual function."
I couldn't understand it, but I think directly by looking at the K-map we can identify whether the function is self-dual or not. Please explain if anyone has an idea about this.

Answer 2

There are two conditions for Self Dual fn->

1.must be neutral fn(number of maxterm=number of minterm)

2.no two mutually exclusive term should be present(for three variable(0,7)both should not be present)

now here 1st condition is satisfied ..so checking 2nd condition

A-> minterms 3 and 4 both are present and are ME(sum is 7) so not a self dual function

B->minterms 0 and 7 are present so no SD fn

C->minterms 2 and 5 are present so again no SD fn

D->no ME terms are present

so option D is correct..

Answer 3

Here is the simple answer " A function cannot be self-dual if dual pair exist in it"

What are dual pairs

000 (0)->111(7)   0 and 7 are dual pair

001(1)->110(6)    1 and 6 are a dual pair

010(2)->101(5)    2 and 5 are a dual pair

011(3)->100(4)   3 and 4 are a dual pair

so option A has (3,4) as a pair so it cannot be self-dual

B has (0,7) as a pair so it cannot be self-dual

C has (2,5) as a pair so it cannot be self-dual

D has no such dual pair so as a pair so it is self-dual

 

To generalize it we can say that in an 'n' variable function there should not be a pair whose sum is = 2$^{n}$-1

eg 0+7= 7, 1+6=7, 2+5=7, 3+4=7

Answer 4

A function can be recognized that weather it is self dual or not by following two conditions:-

1>   IT should have 2^(n-1) min terms ..................... where n = no. of terms in the fn

              eg.    f(x,y,z)= summation (1,2,4,7) .........................here n = 3................as  no. of terms are x y z so

                       2^(3-1)=4 which is our no. minterms in fn

2>  If  mi is the corresponding mean term in the fn

     then    m((2^n-1)-i) should not belong to our fn

          eg.      in above example f(x,y,z)=summation(1,2,4,7)  

                                                m1 is present and our i=1 and from above n =3

                                 so by rule 2

                        m((2^3-1)-1)=m6 which do not belong to our f ......... similarly check for 2,4,7

if any os the above rule fails then it is not self dual

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hope u got this