Sum-of-products for a function

Question

I am not quite sure if I am understanding this correctly or not. Here is the problem:

"Find the simplest sum-of-products form for the function $f$ using the don't-care condition $d$, where

$f = x_1(x_2 \bar x_3 + x_2 x_3 + \bar x_2 \bar x_3 x_4) + x_2 \bar x_4(\bar x_3 + x_1)$

and

$d = x_1 \bar x_2(x_3 x_4 + \bar x_3 \bar x_4) + \bar x_1 \bar x_3 x_4$"

The problem that I am having is that I'm not sure where to go. I do have a question regarding $d$. In $d$ I see $(x_3 x_4 + \bar x_3 \bar x_4)$. Does that expression simplify to just 1? I am thinking this because the property $(x + \bar x) = 1$ Or does this only apply when there is one input?

I tried this problem using a K-Map and here is my answer, but I am not sure it is correct:

$f = x_2 \bar x_3 + x_1 \bar x_2 \bar x_3 + x_1 x_3 x_4 + x_1 x_2 x_3$

Since my last two terms have an $x_1$ and $x_3$ in common, can I simplify to:

$f = x_2 \bar x_3 + x_1 \bar x_2 \bar x_3 + x_1 x_3( x_4 + x_2)$

Can anyone help point me in the right direction or tell me if I am doing this wrong? Thanks.

Answer 1

Your example translates to the following Karnaugh map by writing the minterms of $f$ and the don't care terms of $d$ in the same map:

               x1 x2
          00  01  11  10
         +---+---+---+---+
      00 | 0 | 1 | 1 | d |
         +---+---+---+---+
      01 | d | d | 1 | 1 |
x3 x4    +---+---+---+---+
      11 | 0 | 0 | 1 | d |
         +---+---+---+---+
      10 | 0 | 0 | 1 | 0 |
         +---+---+---+---+

The terms in the map can be covered by

$$f = x_2 \bar x_3 + \bar x_3 x_4 + x_1 x_2$$

Alternative solution:

$$f = x_2 \bar x_3 + x_1 \bar x_3 + x_1 x_2$$