How Do You Identify XOR in Karnaugh Maps?

[STEMucator]

Homework Statement

Using XK-map techniques, obtain the minimum Boolean expression for the following map:

Homework Equations

The Attempt at a Solution

So I've been doing some reading and I found this has to do with Reed Muller logic. This was unfortunately not covered in the course I'm taking, so I'm a tad bit confused.

How exactly am I supposed to circle this map? I know I have to circle groups of zeroes and ones such that the parity of the ones is odd and the parity of the zeroes is even.

For reference, the answer is listed as:

$$Z = \bar a \bigoplus b \bigoplus c \bigoplus d$$

I have experimented with a few circling combinations, but cannot find one that gives the answer above.

 

[mfb]

You could "circle" the four "0" for not A i guess. They all have odd parity for B,C,D. Similar for A.
That does not give nice square blocks - it cannot, you are looking for XOR.

 

[STEMucator]

After staring at this for a few more minutes, I came up with this:

So the black box would be $\bar a$. The red box would be $b$. The blue box would be $c$. The green box would be $d$.

So $z = \bar a \oplus b \oplus c \oplus d$.

 

[Momason]

I am in high school and my Computer Science teacher teach us this and I figure it out by kinda like a Math way.
First I list out all the 1s.
A'B'C'D'+A'B'CD+A'BC'D+A'BCD'+ABC'D'+ABCD+AB'C'D+AB'CD'
=(A'B'C'D'+A'BCD'+ABC'D'+AB'CD')+(A'B'CD+A'BC'D+ABCD+AB'C'D)
=D'(A'B'C'+A'BC+ABC'+AB'C)+D(A'B'C+A'BC'+ABC+AB'C')
=D'(C'[A'B'+AB]+C[A'B+AB'])+D(C[A'B'+AB]+C'[A'B+AB'])
RECALL THAT
A'B + AB' = A XOR B
A'B' + AB= A XNOR B
=D'(C'[A XNOR B]+C[A XOR B]) + D(C[A XNOR B]+C'[A XOR B])
WHILE XNOR = (XOR)'
=D'(C'[A XOR B]'+C[A XOR B]) + D(C[A XOR B]'+C'[A XOR B])
=D'([A XOR B] XNOR C)+ D([A XOR B] XOR C)
=D'(A XOR B XOR C)' + D(A XOR B XOR C)
=A XOR B XOR C XNOR D
=A ⊕ B ⊕ C ⊕ D'
OR IF U PULL OUT 'A' AT FIRST IT WILL BE THE SAME