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Max how many no. Of function is possible if

f(x,y,z)= f(x',y,z')

 

Options

a)8

b)16

c)4

d)32
asked in Digital Logic by (369 points)   | 50 views

2 Answers

+2 votes
Best answer

Here according to the equation we need to see how the constraint works here actually..

Given ,

f(x,y,z)  =  f(x',y,z')

So f(0,0,0)  =  f(1,0,1)

     f(0,0,1)  =  f(1,0,0)

     f(0,1,0)  =  f(1,1,1)

     f(0,1,1)  =  f(1,1,0)

   

So we can see that each minterm(possible combination of x,y,z) is covered exactly once..

Now say f(0,0,0) can be 0 or 1..But 

             f(0,0,0)  = f(1,0,1)

So having selected f(0,0,0) we have one choice for f(1,0,1)..

Likewise for other 3 pairs also..

Hence number of functions possible  = 2 * 2 * 2 * 2 [ 2 choices for each pair ]

                                                      = 16 functions

Hence B) should be the correct answer..

answered by Veteran (68.8k points)  
selected ago by
0 votes
answered by Active (1.2k points)  
The answer given there is wrong ..Plz see that there is a keyword called "constrained"

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