While mmc's answer is correct, the result of applying Kirchhoff's laws is that you get the Graph Laplacian problem.
Given a graph $G$ with each edge $E_i$ given a resistance $R_i$, the weighted graph Laplacian is given by considering the operation on functions which takes $\phi(V)$ to
$$ \nabla^2 \phi = \sum_{<W,V>} {1\over R_i} (\phi(W)-\phi(V)) $$
Where the sum extends over all $W$ which are graph neighbors of $V$. This graph Laplacian then can be solved with discrete delta-function (kronecker delta) sources. The interpretation of the Laplacian is that $\phi$ is the voltage at each point, the Laplacian is summing the currents entering each node (with sign) and setting it to zero.
If you force a certain current into one node and extract the same current at another node, you are solving the equation
$$ \nabla^2 \phi = \delta_{V,V_1} - \delta_{V,V_2} $$
The solution gives the electrostatic potential configuration for all the nerd-sniping problems. For finite graphs, this is a matrix inversion. For infinite homogenous graphs with a translation symmetry, you can reduce the translation invariant directions by Fourier transforms, which allows you to solve any homognenous grid. For general graphs, you study the graph Laplacian, which is related to the random walk on the same graph, which has many bounds on the eigenvalues, which give you the asymptotic behavior of the potential in nearly all the cases of interest.
My answer is r. 2 resistances on the vertical line are redundant as no current flows via that route due to the symmetry of the problem.
Consider this circuit joined to the source with A joined to the positive terminal, then current will equally split along two possible routes from A. Due to symmetry of the problem, the electric potential energy above the vertical line is same as the point below it. So no current flows via that part of the circuit. So that section of wire can be removed, yet giving same equivalent resistance. In our simplified circuit, 2 resistors of resistance 2r are in parallel giving equivalent resistance r.
Best Answer
There are a number of techniques that can be employed to analyse in an efficient way complex passive networks by means of paper-and-pencil calculations. Of course, for a computer, matrix methods are the way to go.
Here are a few:
[1] R. D. Middlebrook, "Null Double Injection and the Extra Element Theorem", IEEE Trans. Edu., 32, 167-180, 1989 online copy
[2] V. Vorperian, Fast Analytical Techniques for Electrical and Electronic Circuits, Cambridge University Press, 2011.
[3] R. D. Middlebrook, V. Vorperian, J. Lindal, "The N Extra Element Theorem", IEEE Transactions on Circuits and Systems I: Fundamental Theory and Applications, 45, 919-935, 1998.
[4] R. W. Erickson, "The $n$ Extra Element Theorem", online.