homework and exercises - How does symmetry allow a rapid determination of the current between $A$ and $B$?

The following was originally given to me as a homework question at my physics 2 course:

Consider the following circuit

The difference of potentials between the point $V_{1}$and the point $V_{2}$ is $4.4$ volts, the resistance of all the resistors is the same $R=1\Omega$.

Find the current between point $A$ and point $B$.

The answer given is simply $0$ and the argument was just the pair of words ``using symmetry''.

I don't really understand the answer:

First, it is not completely symmetric: There is a difference of potentials so the potential at the point $V_{1}$ is not the same as the potential at the point $V_{2}$.

Secondly: How can I see that the symmetric structure will give me that the current between $A$ and $B$ is $0$ ?

Also, I would appreciate to see a calculation of this current to get a better feel for whats going on, I know the rule $V=IR$ (which seems the most useful here, but I also know other rules that can be used), but I don't understand how to use this rule to find the current.

Answer

This is your circuit:

The current that comes from the source, when reaches the point that must choose it's way, sees no difference between the two paths (symmetry) , so half of it flows through one way and the other part flows in the second way. It means that, $I_1=I_2$ , So the potential difference across yellow resistors is the same. It means that the potential of point $\mathbf{A}$ is equal to potential of point $\mathbf{B}$ :

$$I_1=I_2\to V_A=4.4-I_1R \text{ , }V_B=4.4-I_2R\to V_A=V_B$$

So there isn't any potential difference across the blue resistor, and the current through it is 0, and it can be omitted from the circuit without any change in the behavior of the circuit.