-4
$\begingroup$

I can only write down 16 and cannot go further.

8 of

$f(x_{1},x_{2},x_{3})=0$

000 = 0

001 = 0

010 = 0

110 = 0

111 = 0

110 = 0

101 = 0

011 = 0

and

8 of

$f(x_{1},x_{2},x_{3})=1$

000 = 1

001 = 1

010 = 1

110 = 1

111 = 1

110 = 1

101 = 1

011 = 1

That's the maximum I can go. How do I proceed?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ We have no idea how you proceed because we have no idea what you've done so far. However, since there are 256 Boolean functions of 3 outputs, it really shouldn't be difficult to find 16 of them. $\endgroup$
    – David Richerby
    Oct 11 '15 at 7:06
  • $\begingroup$ See here. Duplicate? $\endgroup$
    – Raphael ♦
    Oct 11 '15 at 7:08
  • $\begingroup$ I don't understand what you mean by "8 of $f(x_1,x_2,x_3)=0$". There is only one function that maps all possible inputs to 0. I think the problem is your understanding of the basic concepts, such as what a function is. I don't think we can really help you with that here, since such basic concepts can really only be taught interactively and trying to do that in the comments here would be a nightmare. Assuming you're studying this at a school or university, I recommend you have a chat with your course instructor or one of your fellow students. $\endgroup$
    – David Richerby
    Oct 11 '15 at 11:06
2
$\begingroup$

There are $2^{2^3}=256$ different boolean functions of 3 inputs.

Here is the truth table for one of the functions.

Every different permutation of column $X$ is a different function. There are a total of 256 permutaions.

Improve this answer
$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.