
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 Req=3×31=1Ω
