A Problem from my book-8

Relevant wiki: Transformation of Resistances (Star to Delta and Delta to Star)

THE SOLUTION TO THIS PROBLEM IS A LONG ONE; SORRY FOR THE INCONVENIECE

Although it is possible to use Kirchhoff´s laws to solve for the equivalent resistance, it is much easier to look for resistance conected in series and parallel and progressively reduce the circuit.

Since the the original circuit is symetrical with respect to the diagonal that passes through points A and B, a node from a side of the diagonal is at the same potential as the node that is its reflexion with respect to the diagonal. This could also be thought as follows: a source of potential (like a battery) is connected to points A and B, the current "I" flowing through the resistors immediately connected to point A should be the same, because both resistors have the same resistance and the circuit is symetrical, as shown in the upper-left image. Hence, there is an equal potential difference through these resistors. A similar analysis can be made to the rest of the resistors.

Knowing this, we could look for nodes with the same potential and, hence, resistors with the same potential difference (resistors in parallel). In the upper-centered image, nodes with the same potential have the same color.

The original circuit can be reduced by "combining" the resistors that are in parallel, the resultant circuit is shown in the upper-right image (the resistors in that image are 1 $\Omega$ each).

The circuit can be reduced further by "adding" resistors in series and parallel and it can be rearranged into an unbalanced, Wheatstone bridge with a resistor of $\frac{5}{3} \Omega$ and the other four (including the bridge) of 1 $\Omega$ as it is shown by the lower-left and lower-centred images.

The Wheatstone bridge is unbalanced so a current flows through the bridge and the circuit can´t be further reduced unless an equivalent network is used. Such equivalent circuit can be easily found be using an Delta-Star transformation on the three lower resistors (which are arranged in a "delta" shape) of the circuit shown in the lower-centred image.

After the transformation, the circuit shown in the lower-right is obtained, where the three lower resistors are arranged in a "star" shape and are $\frac{1}{3} \Omega$ each (due to the transformation). This circuit is a "simple" one and the equivalent resistance can be obtained by "combining" the resistors in series and parallel.

From the image: $R=(( \frac{1}{5/3+1/3}+\frac{1}{1+1/3} )^{-1} + \frac{1}{3} ) \Omega = \frac{17}{15} \Omega$ Where "R" stands for equivalent resistance.

Hence the answer is 17+15=32

Use symmtery method and find the equipotential junctions and draws its equivalent circuit . it will form a unbalanced wheatstone setup with one resistance of 5/3 ohm and other 4 resistance of 1 ohm each. apply kirchoffs law and solve for it.

The 4 Deltas of the first Fig. are converted to Stars with StarPoint P,Q,R,S added gives Fig 2. With pairs of series resistances replaced with equivalent resistances as shown in auxiliary Figs. But Q and R are at same potentials due to symmetry, so PQ is removed.
Solving the series parallel net as shown in auxiliary Figs. the equivalent resistance between A and B is 17/15.
We could have used symmetry in Fig 1 itself by splitting D into two disconnected points, but that does increase our calculations.
17+15=32.