Understanding Karnaugh Maps: Two Ways of Representation

[DiamondV]

Well, I've recently been studying karnaugh maps and I've noticed there's two sorts of ways to represent them. In my learning materials, sometimes they are expressed in one way,sometimes in another way.
For example:

http://puu.sh/luJBp/4b2c878122.png

Now, what I don't understand is with this method is that do I assume the variables to be initially 1? Otherwise the contents/minterms of the map itself don't really make sense. Also for this example no boolean expression was given.
I understand the other way of expressing karnaugh maps

In this method, the numbers above the cells indicate the value of the variables whereas in the first method they don't? So do I assume the variables to initially be at 1? I've read several websites and watched videos and I can't seem to figure it out.

 

[mfb]

Now, what I don't understand is with this method is that do I assume the variables to be initially 1?

There is no initial state where such a question would make sense.
All three tables are just different notations for the same thing (well, the third table corresponds to a different logic with different input variables).
"1" or "0", and "$Z$" or"$\overline Z$", represent the same thing.

 

[DiamondV]

There is no initial state where such a question would make sense.
All three tables are just different notations for the same thing (well, the third table corresponds to a different logic with different input variables).
"1" or "0", and "$Z$" or"$\overline Z$", represent the same thing.

So z =1, and z bar = 0. Is that what I assume for the second table

 

[mfb]

The label "$Z$" means "if $Z$ is 1".
The label "$\overline Z$" means "if $\overline Z$ is 1".
Perfectly symmetric.

 

[DiamondV]

The label "$Z$" means "if $Z$ is 1".
The label "$\overline Z$" means "if $\overline Z$ is 1".
Perfectly symmetric.

if Z bar is 1 doesn't that mean z=0

 

[mfb]

Sure.

And Z is 1 means Zbar is 0.