The equivalent resistance between $$A$$ and $$B$$ is ______ $$\Omega$$.

Let the center node be O.

From A to O:

• $$8Ω\ in\ parallel\ with\ 2Ω$$

$$R_{AO}=\frac{8\cdot2}{8+2}=\frac{8}{5}$$

From O to B:

• $$4Ω\ in\ parallel\ with\ 6Ω$$

$$R_{OB}=\frac{4\cdot6}{4+6}=\frac{24}{10}=\frac{12}{5}$$

These are in series:

$$R_{middle}=\frac{8}{5}+\frac{12}{5}=4Ω$$

So middle branch is $$4Ω$$

Now all branches between A and B are parallel:

Top outer branch:

$$1.5+0.5=2\Omega$$

Top horizontal:

$$12Ω$$

Middle branch:

$$4Ω$$

Bottom horizontal:

$$6Ω$$

Bottom outer:

$$1+1=2\Omega$$

So

$$\frac{1}{R_{eq}}=\frac{1}{2}+\frac{1}{12}+\frac{1}{4}+\frac{1}{6}+\frac{1}{2}$$

LCM 12:

$$\frac{6+1+3+2+6}{12}=\frac{18}{12}=\frac{3}{2}$$

Therefore

$$R_{eq}=\frac{2}{3}$$