A simple circuit and its modified version

The $2\times 2$ grid circuit is symmetrical between points $A$ and $B$ . Due to this symmetry, the voltages at $C$ and $C'$ are the same, those of $D$ , $D'$ , and $D"$ , those of $E$ and $E'$ are also the same. Therefore, they can be consider as single points of $C$ , $D$ , and $E$ and the equivalent circuit as is on the right. (For convenience, I have use thick red lines to denote resistance and thin black lines as conductors.). Then we have:

$\begin{aligned} R_1 & = 1||1 + 1||1||1||1+1||1||1||1 + 1||1 = \frac 12 + \frac 14 + \frac 14 + \frac 12 = \frac 32 \ \Omega \end{aligned}$

With resistance $R_{AC'}$ remove, the circuit can be simplified as the circuit in the middle with $R_{CE} = 1+1 = 2 \ \Omega$ and $R_{DE'} = 1||3 = \dfrac {1\times 3}{1+3} =\dfrac 34 \ \Omega$ . Converting the delta circuit $\triangle CDE$ into star circuit $R_{C1} = R_{C2} = \dfrac {2\times 1}{2+1+1} = \dfrac 12 \ \Omega$ and $R_{C3} = \dfrac {1\times 1}{2+1+1} = \dfrac 14 \ \Omega$ and we get the equivalent circuit on the right.

Then we have

$\begin{aligned} R_2 & = 1 + \frac 12 + \left(\frac 12 + 1\right) \bigg|\bigg| \left(\frac 14 + \frac 34 + 1\right) = \frac 32 + \frac 32 \bigg|\bigg| 2 = \frac 32 + \frac {\frac 32 \times 2}{\frac 32 + 2} = \frac 32 + \frac 67 = \frac {33}{14} \end{aligned}$

Therefore $R_2 - R_1 = \dfrac {33}{14} - \dfrac 32 = \dfrac 67$ $\implies a+b = 6+7 = \boxed{13}$ .