A Cubic Resistance.

Actually this question answers only when there is another resistor between B and E ..... only then does Vaibhav's solution goes right .

The points F,D & B are at same potential hence they can be added together such that resistances AB,AD,AF ,becomes parallel . Similarly the points E,C& G .too are at same potential and they too can be joined .Hence we get a structure with a 6 resistances parallel and a pair of three resistances parallel.hence equivalent resistance is $R_{eq}=\frac{r}{3}+\frac{r}{6}+\frac{r}{3}$ .Hence solving this we get $R_{eq}= \frac{5r}{6}$ hence putting the value of $r$ we get $\boxed{R_{eq}=0.833...}$

Assume that 3 i current enters the circuit .Now we know that the resistance being the same through out, the current will divide equally as i at first and then as i/2 (at B,F and D).Now these will combine to give current i (at G,E and C) and as 3 i at H. Now we get ,1.V(A) - V(D)=i r (Applying V=I R). 2.V(D) - V(G)=i/2 r. 3.V(G) - V(H)=i r. Hence we get ,

V(A) - V(H)=i r+i/2 r+i r=5/2 i*r.

IF Equivalent resistance is assumed to be R,

Then ,we get V(A) - V(H) = 3 i R.

or ,3 i R=5/2 i r.

R = 5/6*r.(Here r=1).So R=5/6=0.83.

Here V(A) - V(D) means Potential difference across A and D.

Divide the current and apply kirchoff's law to find the eq resistance.