This is an interesting question posed at http://www.sgforums.com/forums/2297/topics/329946
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Answer:
First, we need to divide the circuit. The top white coloured resistor has no current passing through it, because the potential is the same between two ends. There's no potential difference across the resistor, and hence no colour.
The bottom resistor in pink is only to be considered in the end; we will touch it later.
The next diagram shows the top resistor removed.
Follow the circuit along the route of the green resistors. Notice that the centre is also at the same potential as the blue resistors. Hence, the wires can be separated as shown below.
If you trace the path of the current flowing through the green resistors, you will realise that the centre is at the same potential as the centre of the blue resistors. This only works because the ratio of the resistance from between P¢re and between Q¢re is the same for both the blue and the green path.
Since they are of same potential, joining together with wires or not (or even if we put a resistor in between, like what happened to the top white resistor) doesn't matter. We can safely separate them without any effect.
The resistors in the red box are in parallel, and thus can be simplified as follows:
The rest of the resistors are 1 ohms. Hence, total resistance is
(1/2 + 1/2 + 3/8)-1 = 8/11 ohms = 0.73 ohms
