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.