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Combinatorics: Take an m x n piece of grid paper, and in each box pick two opposite corners at random to connect. What can be said about the resulting pattern?

Let’s start by applying Euler’s formula [math]V - E + F - C = 0[/math] to this graph with [math](m + 1)(n + 1)[/math] vertices, [math]mn[/math] diagonal edges, [math]2m + 2n[/math] perimeter edges.We can get a simple lower bound on [math]C[/math], the number of edge-connected components of the graph, by noting that each internal vertex connected to no edges adds another component. By linearity of expectation, we expect that [math]\tfrac{1}{16}[/math] of the [math](m - 1)(n - 1)[/math] internal vertices are connected to no edges, so [math]\mathrm E[C] \ge 1 + \tfrac{(m - 1)(n - 1)}{16}[/math].This lets us bound the expected number of faces,[math]\mathrm E[F] = -V + E + \mathrm E[C][/math][math]\ge -(m + 1)(n + 1) + mn + 2m + 2n[/math][math]{} + 1 + \tfrac{(m - 1)(n - 1)}{16}[/math][math]= m + n + \tfrac{(m - 1)(n - 1)}{16}[/math].That means the average area of a face is[math]\lim_{m, n \to \infty} \frac{mn}{F} \le 16[/math].(We can ignore the edge effects of the faces that touch the sides of the rectangle, since there are only [math]m + n[/math] of those.)So an average path has length at most [math]32\tfrac{1}{\sqrt2}[/math] (where each “step” has length [math]\tfrac{1}{\sqrt2}[/math] and contributes area [math]\tfrac12[/math]).By considering not just components of one disconnected vertex, but also two or three, we can reduce the [math]32[/math] to [math]\tfrac{8192}{353} \approx 23.2068[/math] steps. Simulations on a [math]10000[/math][math] \times [/math][math]10000[/math][math][/math] grid suggest that the real answer is about [math]20.3924 \pm 0.0004[/math] steps.There’s clearly a lot more work to do here! I’ll continue investigating this later.

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