Will any charge flow in this circuit? 
I had to find the resistance of this circuit between A and B. My teacher said that since the blank wires have zero resistance, the potential difference across the ends of the blank wire is zero(according to v = ir) and thus 1,4 are equipotent and 2,5 are equipotent and the circuit is thus simplified.The equivalent resistance comes out to be 2r/5.
MY QUESTION:If the points 1,4 and 2,5 are equipotent how can current flow through the horizontal circuit. 
MY REASONING:
The potential difference across each branch of a parallel circuit is same and in this case it is 0 and hence no current can flow through the circuit.
 A: You can explicitly compute the currents to see this.
Let us conventionally assume the currents are positive when they flow from left to right,
let $V_i$ denote the potential at node $i$, and $I_1, I_2, I_3$ the currents respectively between the nodes 1-2, 2-4 and 4-5.
The blank wires shortcircuit some of the nodes, so that we have
$$ V_1 = V_4 \qquad \text{ and } \qquad V_2 = V_5. $$
Applying the usual Ohm's law we see that:
$$
V_2 - V_1 = r I_1, \\
V_4 - V_2 = V_1 - V_2 = 2r I_2, \\
V_5 - V_4 = V_2 - V_1 = r I_3
$$
from which it follows that $ I_1 = I_3 = - 2 I_2. $
As you can see the shortcircuits in the example do not prevent a flow of current in the circuit.
A: It does not matter whether you have the same potential somewhere in your circuit. The main point is that you have a potential difference between A and B. Current will  flow between A and B split over all possible paths, in your case you have three paths: 1-2-5, 1-4-5, 1-4-3-2-5.
The network of resistors can be replaced by an equivalent resistance which you correctly calculated to be 2R/5. 
If you reshape the funny circuit in your question you will see that it is completely equivalent to the following more familiar circuit
A: 
This can surely help. As you can see potential drop across different resistors is V. Therefore current should flow in each resistors.
