A fairly well-known puzzle asks for the resistance between two adjacent nodes of an infinite square resistor grid, like the points A and B in the drawing. If the resistors are all $1\Omega$, the answer is $\frac12\Omega$, which can be seen by superposition of the two situations where you inject a current in node A, or an opposite current in node B, and use symmetry to argue that in each situation $1/4$th of the current flows through each of the directly adjacent resistors.
Question: Is there also a simple way to see the impedance between diagonally adjacent nodes A and D? (And maybe even for other cases, like A to C, or the knight-jump A to E, etc.?)
Those resistance values can be calculated with higher mathematics but that method is not convincing to many people (for instance some will insist that points A and D should give a resistance of $\frac12\sqrt2$). For completeness, nevertheless, here we'll derive the values in that way. We denote nodes on the grid by vectors ${\bf x}=(n,m)$ with integer coordinates. From Kirchhoff's and Ohm's laws we then have for the voltage $V({\bf x})$ on each node: $$ I({\bf x}) = 4V({\bf x})-V({\bf x}-{\bf e}_1)-V({\bf x}+{\bf e}_1) -V({\bf x}-{\bf e}_2)-V({\bf x}+{\bf e}_2) \tag{Eq1} $$ with ${\bf e}_1$ and ${\bf e}_2$ the basis vectors $(1,0)$ and $(0,1)$, and $I({\bf x})$ an external current injected in node ${\bf x}$. Since this is essentially the discrete Laplace operator, we can transform to the Fourier domain and write:
$$ V({\bf x}) = \int_{-\pi}^{\pi} \frac{dk}{2\pi} \ \int_{-\pi}^{\pi} \frac{dq}{2\pi} \ e^{i{\bf p}\cdot{\bf x}}\ \hat V({\bf p}) $$ where the transformed $\hat V$ in Fourier space now depends on ${\bf p}=(k,q)$. Doing the same for $I({\bf x})$ we can invert the discrete Laplace, and write [Eq1] as: $$ \hat V({\bf p}) = \frac1{4-2\cos k-2\cos q} \ \hat I({\bf p}) $$ $$ \Rightarrow \quad V({\bf x}) = \int_{-\pi}^{\pi} \frac{dk}{2\pi} \ \int_{-\pi}^{\pi} \frac{dq}{2\pi} \ \frac{e^{i{\bf p}\cdot{\bf x}}}{4-2\cos k-2\cos q} \ \hat I({\bf p}) \tag{Eq2} $$ If we now inject two opposite unit currents at points A and B, a trivial Fourier transform from ${\bf x}$ space to ${\bf p}$ space will give us: $$ \hat I({\bf p}) = e^{-i {\bf p}\cdot{\bf x_B}} - e^{-i {\bf p}\cdot{\bf x_A}} $$ Using [Eq2] we then transform back and find the potential difference between any two other points C and D, as: $$ V({\bf x}_D)-V({\bf x}_C) = \int_{-\pi}^{\pi} \frac{dk}{2\pi} \ \int_{-\pi}^{\pi} \frac{dq}{2\pi} \ \frac{e^{i\mathbf{p}\cdot\mathbf{x}_D} -e^{i\mathbf{p}\cdot\mathbf{x}_C} } {4-2\cos k-2\cos q} \ (e^{-i {\bf p}\cdot{\bf x}_B} - e^{-i {\bf p}\cdot{\bf x}_A}). $$
For the resistor puzzle, points C and D are the same as A and B, and we can choose one of them to be the origin, which leaves us with one vector ${\bf x} = {\bf x}_B $. The numerator in the integral then becomes:
$$ (1-e^{-i {\bf p}\cdot{\bf x}}) \ (1-e^{i {\bf p}\cdot{\bf x}}) = 2-2\cos({\bf p}\cdot{\bf x}) = 2-2\cos(n k + m q) = $$ $$ = 2-2\cos(n k)\cos(m q)+2\sin(n k)\sin(m q). $$ Since the denominator is totally symmetric in $q$ and $k$ we can keep only the symmetric part of this last expression. Also we'll subsequently use four times the integral in only one quadrant. If we then denote $\vec B-\vec A={\bf x}=(n,m)$ we get:
$$ R(n,m)= 8 \int_0^\pi \frac{dk}{2\pi} \ \int_0^\pi \frac{dq}{2\pi} \ \ \frac{1-\cos(n k)\cos(m q) } {4-2\cos k-2\cos q}. $$
For the case $m=0$, doing the $q$ integral leaves us with:
$$ R(n,0)= \int_0^\pi \frac{dk}{\pi} \ \frac{1 - \cos(k n)} {\sqrt{3-4\cos k +\cos^2 k}}. $$
Doing also the $k$ integral then shows us $R(1,0)=1/2$, which we already knew, just like $R(0,0)=0$, but we also see that for instance $R(2,0)= 2-4/\pi$, which is new! We can of course redo the $q$ integral for $m=1$. This gives:
$$ R(n,1)= \int_0^\pi \frac{dk}{\pi} \Big(\cos(k n) + \frac{(\cos k - 2) \cos(k n) + 1} {\sqrt{3-4\cos k +\cos^2 k}} \Big). $$ and this finally gives us the "diagonal" resistance: $R(1,1)=2/\pi$, which is clearly 10% smaller than the $\frac12\sqrt2\ $ some might have expected, and the knight jump: $R(2,1)=4/\pi-1/2$. In summary:
$$ R(0,1) = \frac12 $$ $$ \qquad R(0,2) = 2-\frac4\pi $$ $$ \quad\qquad R(0,3) = \frac{17}2-\frac{24}\pi $$ $$ R(1,1) = \frac2\pi $$ $$ \qquad R(1,2) = \frac4\pi-\frac12 $$ $$ \qquad R(1,3) = \frac{46}{3\pi}-4 $$
Of course the procedure can nicely be generalized, e.g. for the anisotropic case where the horizontal resistors have a different value than the vertical ones, or for more than $2$ dimensions. And the approach is of course standard practice for particle physicists, but it is still an interesting question whether at least some of these values, like the diagonal $R(1,1)$, have a simpler derivation...
