There is an easier and far more elegant way to solve this than the solution given. Consider the circuit as two superimposed elements; one with a current being injected at the first point and flowing outward to a sink at infinity, and the second with current flowing in from a source at infinity and exiting at the second point. (For the sake of argument, say the current is 1 amp).
The current flow patterns in each case are easy to calculate because of the symmetry of each problem.
Now add the two superimposed elements together, and the sources and sinks at infinity cancel out, leaving only the point source and sink.
You now know the current through the overall circuit and the currents through each resistor, and because you know the values of the resistors, you also know the voltages across each resistor. Add up the voltages along any simple path between the two points to get the voltage between the points, and since you also know the overall current, you can now calculate the equivalent resistance.