GATE CSE
First time here? Checkout the FAQ!
x
+2 votes
100 views

Given $x,y,z$ are Boolean variables and $f(x,y,z)=f(y',x',z)$. How many such functions are possible with $x, y, z$?

  1. $0$
  2. $2^2$
  3. $2^4$
  4. $2^6$ 
asked in Set Theory & Algebra by Junior (945 points)   | 100 views
yes $2^6$
i want explanation i didn't get the explanation which was given in source.
unable to understand the que?? wht is que asking??

plz explain

1 Answer

+5 votes
Best answer

In a three variable boolean function if all of the truth table rows are independent w.r.t the O/P, then total no of boolean function possible is $2^{2^3} = 256$

  • because in the O/P of a truth table we have $2^n= 2^3$ places to fill with 0 or 1. 
  • This is possible because we have not specified any constraint on the boolean function.

 The question here provide us with a constraint equation $f(x,y,z) = f(y',x',z)$

because of this constraint, all 8 rows of the truth table are not independent now. They are in 6 groups.

  1. $f(0,0,0) \ \ \ \text{ and } \ \ f(1,1,0)$ are having same O/P.
  2. $f(0,0,1) \text{ and } f(1,1,0)$ are having same O/P.
  3. $f(0,1,0) ,f(0,1,1) ,f(1,0,0) ,f(1,0,1) ,$ are independent.

So in these 6 rows of the truth table, we can fill up O/P with 0 or 1. 

$\Rightarrow$ total $2^6$ functions possible.

note : we can also solve by calculating no of minterms and taking combinations of them. 

answered by Veteran (36.2k points)  
selected by
highly appreciated !

question has same logic as Self dual function

Top Users Jan 2017
  1. Debashish Deka

    8322 Points

  2. sudsho

    5166 Points

  3. Habibkhan

    4718 Points

  4. Vijay Thakur

    4468 Points

  5. Bikram

    4420 Points

  6. saurabh rai

    4212 Points

  7. Arjun

    4072 Points

  8. santhoshdevulapally

    3732 Points

  9. Sushant Gokhale

    3514 Points

  10. GateSet

    3336 Points

Monthly Topper: Rs. 500 gift card

19,155 questions
24,065 answers
52,872 comments
20,288 users