The infinite resistor grid

Every node is an integer lattice point, every edge is a 1 Ω resistor, and the grid runs forever in all directions. Measure across one resistor and you read half an ohm — a number you can derive in three steps, with no algebra.

Why adjacent nodes read ½ Ω

The trick is superposition: split the measurement into two problems that symmetry solves by inspection, then add them back together.

1. Inject 1 A at A, let it drain to infinity. The lattice looks identical in all four directions from A, so each resistor carries ¼ A.

2. Separately, draw 1 A out at B, supplied from infinity. By the same symmetry each of B’s resistors carries ¼ A inward.

3. Add the two. Everywhere else currents partly cancel, but the single A–B resistor carries ¼ + ¼ = ½ A.

current through the A–B resistor ½ A its resistance 1 Ω so the voltage across it ½ V measured resistance R = V / I = ½ V / 1 A = 0.5 Ω

The orange node is where current enters, the blue node where it leaves. Steps 1 and 2 are each unanswerable alone — current vanishing into infinity — but their sum is exactly the measurement you wanted, and in the sum the infinities cancel.

The diagonal is where π appears

Move to the diagonal neighbour and the symmetry argument stops working: there is no single resistor joining them, and the current divides across every path in the lattice. The answer is 2/π, and it needs the lattice Green's function rather than a picture.

That value is checkable without any of the theory. Build a finite L×L grid, solve Ohm's law, grow L, and extrapolate:

Finite grids converging on the infinite-lattice values. Boundary error falls like 1/L², measured at p = 1.993 from the data rather than assumed.
Grid	Adjacent (1,0)	Diagonal (1,1)	Two apart (2,0)
11 × 11	0.504606	0.646193	0.745641
41 × 41	0.500326	0.637273	0.728066
161 × 161	0.500021	0.636662	0.726845
321 × 321	0.500005	0.636630	0.726782
extrapolated	0.50000	0.6366197	0.7267604
exact	1/2	2/π	2 − 4/π

π recovered as 2 / R_diagonal from the 321×321 grid 3.1415402470 after extrapolation 3.1415929618 true π 3.1415926536

Nothing transcendental went in. Every resistor is 1 Ω and every node sits on an integer coordinate; π comes out of the lattice itself.

The same number counts spanning trees

Effective resistance across an edge equals the fraction of all spanning trees of the graph that contain that edge — verified elsewhere to 2 × 10⁻¹⁶ against brute-force enumeration. So these resistances are also combinatorial statements.

Read the ½ that way: in the limit, a given edge of the square lattice appears in exactly half of all spanning trees. Foster's theorem forces it — the effective resistances over all edges sum to n − 1, a large grid has about 2n edges, and every edge is equivalent by symmetry, so each must carry ½. The same conservation law gives 2/z on any edge-transitive lattice: ⅓ for triangular, ⅔ for honeycomb, and ½ again for kagome, which shares the square lattice's coordination number without sharing anything else.