Grid of resistors with removal Suppose I have an infinite square grid of ideal 1 ohm resistors and want to measure the resistance of the grid between two points, but one or more resistors have been removed. How does this affect the outcome?
Please retag. I'm spitballing with linear algebra.
 A: A special case of this problem has been posed in brain-teaser books:  The network of $1\Omega$  resistors is an infinite Cartesian grid, but the resistor between $(0,0)$ and $(0,1)$ has been removed.  What is the net resistance between $(0,0)$ and $(0,1)$?
This problem can be solved by a combination of symmetry, superposition, and the parallel resistance formula.  By symmetry, in a complete grid of $1\Omega$ a 1 amp current sourced at $(0,0)$ will split into $\frac14$ amp along each of the four resistors emanating from that point, and a 1 amp current sinked at $(0,1)$ will split into $\frac14$ amp going into that point. Superposing the two, we find that the net resistance between $(0,0)$ and $(0,1)$ in the complete network is $\frac12\Omega$.
Now let our actual grid, with the missing resistor, have resistance $R$.  Then we can create the complete grid by adding our grid to a $1\Omega$ resistor going from  $(0,0)$ two $(0,1)$. The complete grid is a combination of that resistor and our desired grid, in parallel.  Thus
$$
\frac11 + \frac1R = \frac1{1/2} \implies R = 1
$$
