Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.
I can reduce the problem to 3 linear equations in 3 unknowns, which is easily solvable. First, label the resistors R1...R3 across the top, left to right, then R4...R7 across the middle, left to right, then R8, R9 across the bottom, left to right. Now imagine applying 1 volt at A while B is grounded. If we knew the total current from A to B we would know the effective resistance. Let V1 be the voltage at the junction of R1 and R2, V2 be the voltage at the junction of R2 and R3, and V3 be the voltage at the junction of R8 and R9. V1, V2, and V3 are the three unknowns I mentioned at the beginning. Furthermore they comprise all the remaining nodes besides A and B. So if we know V1, V2, and V3 then we know all the currents through every resistor, so we can compute the current from A to B just by knowing the currents through R1 and R7. So all the remains is to find three linear equations in V1, V2, and V3. Those three equations are simply the equations that express the fact that the current into any one of those junctions is equal to the current out of that junction. For example, at V1 we have (1 - V1)/R amps in and (V1 - V2)/R + (V1 - V3)/R + (V1 - 0)/R amps out. That's one equation. Make similar equations for current equality at node V2 and V3. Solve as described above.