Find the equivalent resistance across the two ends $A$ and $B$ of this circuit.

(1) From the diagram,we observe that the following pairs of resistors are in parallel:
$R_1$ and $R_2$,$R_3$ and $R_4$,$R_5$ and $R_6$,and $R_7$ and $R_8$.
Since each resistor has a value of $2 \, \Omega$,the equivalent resistance for each parallel pair is:
$\frac{1}{R_{12}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2} + \frac{1}{2} = 1 \, \Omega \implies R_{12} = 1 \, \Omega$.
Similarly,$R_{34} = 1 \, \Omega$,$R_{56} = 1 \, \Omega$,and $R_{78} = 1 \, \Omega$.
Now,$R_{12}$ and $R_{34}$ are in series,so their equivalent resistance is:
$R_{1234} = R_{12} + R_{34} = 1 + 1 = 2 \, \Omega$.
Also,$R_{56}$ and $R_{78}$ are in series,so their equivalent resistance is:
$R_{5678} = R_{56} + R_{78} = 1 + 1 = 2 \, \Omega$.
Finally,$R_{1234}$ and $R_{5678}$ are in parallel between points $A$ and $B$. The net resistance $R_{eq}$ is:
$\frac{1}{R_{eq}} = \frac{1}{R_{1234}} + \frac{1}{R_{5678}} = \frac{1}{2} + \frac{1}{2} = 1 \, \Omega^{-1}$.
Therefore,the equivalent resistance is $1 \, \Omega$.