Find req about AB... Also pls tell which symm aur law u are using

Solution
Look at the symmetry of the circuit. The battery (suppose) is connected between points A and B.
From the symmetry current in AO will be same as current in OB
Similarly current CO equal to current in OD
The given circuit can be redrawn as

$$\begin{gathered} R =4+4+\frac{1}{\left(\frac{1}{4}+\frac{1}{8}\right)} =8 +\frac{1}{\left(\frac{2+1}{8}\right)} =8 +\frac{8}{3} =\frac{32}{3} \\ \frac{1}{R_{AB}} =\frac{1}{4} +\frac{1}{8} +\frac{1}{\frac{32}{3}} =\frac{1}{4} +\frac{1}{8} +\frac{3}{32} =\frac{8+4+3}{32}=\frac{15}{32} \\ {R}_{AB} =\frac{32}{15} Ohm \end{gathered}$$