At $\mathrm{A}$current is distributed and at$\mathrm{B}$currents are collected. Between $\mathrm{A}$ and$\mathrm{B}$, the distribution is symmetrical, as shown in the figure. It appears that current in $\mathrm{AO}$ and $\mathrm{OB}$ remains the same. At $\mathrm{O}$, current $i_4$ returns back without any change. If we detach $\mathrm{O}$ from $\mathrm{AB}$  there will not be any change in the distribution.
Now, $\mathrm{CO}$ & $\mathrm{OD}$will be in series hence it's total resistance = 2 $\mathrm{\Omega }$
It is in parallel with CD, so, equivalent resistance = $\frac{2\times 1}{2+1}=\frac{2}{3} \mathrm{\Omega }$
This equivalent resistance is in series with AC & DB, so, the total resistance $=\frac{2}{3}+1+1=\frac{8}{3} \mathrm{\Omega }$

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Now,  $\frac{8}{3} \mathrm{\Omega }$is parallel to $\mathrm{AB}$, that is, $2 \mathrm{\Omega , }$so total resistance

$=\frac{8/3\times 2}{8/3+2}=\frac{16/3}{14/3}=\frac{16}{14}=\frac{8}{7} \mathrm{\Omega }$