Equivalent resistance of the following network is ________ Ω .

In the above diagram, $6Ω$ is short circuit. Therefore, we can remove it.

In the left part, $2Ω$ and $2Ω$ are in parallel combination, which will give us $1Ω$ which again in turn is in series combination with the $2Ω$(adjacent to $3Ω$ resistor), therefore net contribution of this part will be $1Ω + 2Ω = 3Ω$. 

Now, all three $3Ω$ are in parallel across point A & B, therefore ${R}_{eq} = 3 \times \dfrac{1}{3} = 1Ω$