In the given network, Fig. the equivalent resistance between A and B is 

In fig. the five resistances form a balanced Wheatstone bridge. Their effective resistance R₁  is given by 
${\mathrm{R}}_1=\frac{2\mathrm{R}\times 2\mathrm{R}}{2\mathrm{R}+2\mathrm{R}}=\mathrm{R}$
This is in parallel with the resistance R between A and B
Therefore, total resistance between A and B
$=\frac{\mathrm{R}\times \mathrm{R}}{\mathrm{R}+\mathrm{R}}=\frac{\mathrm{R}}{2}$