Resistor Diamonds Are Forever

Notice the Wheatstone Bridges in the circuit.

Now we just need to add the resistances in parallel.

$\frac{1}{R} = \frac{1}{2} +\frac{1}{2} +\frac{1}{2} + \frac{1}{2} +\frac{1}{2}$

$\frac{1}{R} = \frac{5}{2}$

$R = \frac{2}{5} = 0.4 \Omega$

Energy will split in 5 paths of 2Ω so 0,4 Ω.

Note, you could remove or ignore the 2 vertical resistor as potential are the same on each side.

Due to symmetry of the circuit, there is voltage across the two vertical resistors in the diagram is $0$ and there is no current through them. The two vertical resistors can hence be replaced by a short-circuit (same voltage) or an open-circuit (no current),

Case 1 -- Short-Circuit: Then the resistance across AB is:

$\begin{aligned} R_{AB} & = [( 1 || 1 ) + (1 || 1)] || [1+1] || [( 1 || 1 ) + (1 || 1)] \\ & = [ \frac{1}{2} + \frac{1}{2} ] || 2 || [ \frac{1}{2} + \frac{1}{2}] = 1 || 2 || 1 \\ & = \frac{1}{1+\frac{1}{2}+1} = \frac{1}{\frac{5}{2}} = \frac{2}{5} = \boxed{0.4} \Omega \end{aligned}$

Case 2 -- Open-Circuit: Then the resistance across AB is:

$\begin{aligned} R_{AB} & = (1+1) || (1+1) || (1+1) || (1+1) || (1+1) \\ & = 2 || 2 || 2 || 2 || 2 = \frac{1}{\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}} \\ & = \frac{1}{\frac{5}{2}} = \frac{2}{5} = \boxed{0.4} \Omega \end{aligned}$