Condition for the use of perpendicular symmetry in solving resistance problems 
It is possible to use parallel symmetry even when there is more than one battery, but what about perpendicular symmetry?

I tried to solve some examples using KVL but it didn't work out as I expected. If it is not possible to use it in all the cases then in which cases is it applicable? Also I didn't quite understand how we just separate the wires from a junction if we have established perpendicular symmetry; kindly provide an intuitive proof. 
Finally, this all came in my mind because of the following problem:

(The question was too long to solve by using KVL as there would be 7 loops)
NOTE: The resistance of the middle resistance is not specified; it is R (not 2 ohm).
 A: This is a homework problem, so I'm not going to completely solve it for you but I'll give some hints that might apply to other problems.
First imagine you did use all Kirchoff's laws, and solved for $I$ the current in the middle branch. Kirchoff's laws are linear, so in terms of $V_{bottom}=100V$ and $V_{top}=50V$, your solution will have the form
$$I=A V_{bottom}+B V_{top}$$
for some coefficients $A,B$ which we don't know until we solve the problem.
So what this means is that if we solve a new circuit where we set the bottom voltage to zero (i.e. replace the battery with a plain wire) the new current in the middle branch $I_{top}$ is $$I_{top}=B V_{top}.$$ And similarly if we replace the top battery and solve for the middle branch $$I_{bottom}=A V_{bottom}.$$
But if we take out the top battery, it is easy to see by symmetry or by thinking about potential drops that there will be no current in the middle branch.
$$I_{bottom}=A V_{bottom}=0.$$
So $I=I_{top}$. We can pretend the bottom battery is not there at all as far as the middle branch is concerned!
Then by symmetry we can not only replace the bottom battery by a wire, we can completely take out that branch since the potentials it is connecting would be equal anyway. This then lets you use series and parallel resistor rules to simplify the circuit even more, and by this point the circuit has been simplified enough to solve in a couple lines with Kirchoff's laws.
