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 $$ V_1 = V_3 \qquad \text{ and } \qquad V_2 = V_4. $$

Applying the usual Ohm's law we see that: $$ V_2 - V_1 = r I_1, \\ V_3 - V_2 = V_1 - V_2 = 2r I_2, \\ V_4 - V_3 = V_2 - V_1 = r I_3 $$ 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.

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.