Can You Solve This Rather Pedestrian Puzzle?

The Riddler for February 14, 2020. Both halves redesign the sidewalks of Riddler City, a vast grid of unit blocks with City Hall at the centre and employees’ homes scattered uniformly. The Express replaces the grid sidewalks with diagonal ones; the Classic replaces them with a Barcelona-style octagonal pattern. Each asks what fraction of employees then have a shorter walk home.

Riddler Express

Riddler City is a vast grid of unit-side square blocks with City Hall at the centre; employees’ homes are scattered uniformly across it. The traditional sidewalks run along the streets, so reaching the intersection $(x,y)$ blocks from City Hall takes $|x| + |y|$ blocks of walking. A proposal replaces them with sidewalks running diagonally, at $45^\circ$ to the streets. What fraction of employees would then have a shorter walk home?

The Riddler, FiveThirtyEight, February 14, 2020(original post)

Solution

The diagonal sidewalks run in the two $45^\circ$ directions $\tfrac{1}{\sqrt2}(1,1)$ and $\tfrac{1}{\sqrt2}(1,-1)$. To walk to $(x,y)$ you cover $|x+y|/\sqrt2$ along the first diagonal and $|x-y|/\sqrt2$ along the second, a total of $$\frac{|x+y| + |x-y|}{\sqrt2} = \frac{2\max(|x|,|y|)}{\sqrt2} = \sqrt2\,\max(|x|,|y|),$$ using the identity $|a+b| + |a-b| = 2\max(|a|,|b|)$. Compare this with the grid walk $|x| + |y|$.

Work in the octant where $x \ge y \ge 0$ (the other seven are mirror images). There the diagonal walk is $\sqrt2\,x$ and the grid walk is $x + y$, so the diagonal route is shorter exactly when $$\sqrt2\,x < x + y \quad\Longleftrightarrow\quad y > (\sqrt2 - 1)\,x.$$ The boundary line $y = (\sqrt2 - 1)x$ makes angle $\arctan(\sqrt2 - 1) = 22.5^\circ$ with the street, precisely half of the $45^\circ$ octant. So in every octant the diagonal sidewalks win for half the directions, and because the homes are scattered uniformly (so their direction from City Hall is uniform), the city splits evenly: $$\boxed{\tfrac12}.$$ Homes near the diagonals $y = \pm x$ prefer the new sidewalks; homes near the streets prefer the old. Exactly half fall on each side.

The computation

Encode the geometry: scatter homes uniformly over a disk, compute both the grid walk $|x|+|y|$ and the diagonal walk $(|x+y|+|x-y|)/\sqrt2$, and measure how often the diagonal is shorter.

import numpy as np, math
rng = np.random.default_rng(0); N = 2_000_000
a = rng.uniform(0, 2*math.pi, N); r = np.sqrt(rng.uniform(0, 1, N))
x, y = r*np.cos(a), r*np.sin(a)
grid = np.abs(x) + np.abs(y)
diag = (np.abs(x + y) + np.abs(x - y)) / math.sqrt(2)
print(np.mean(diag < grid))                  # ~0.5

The simulated fraction is $0.500$, matching the boxed $\tfrac12$.

Riddler Classic

The petitioners try again, now asking for sidewalks in an octagonal pattern like the chamfered blocks of Barcelona’s Eixample. Under this proposal, what fraction of employees would have a shorter walk home?

The Riddler, FiveThirtyEight, February 14, 2020(original post)

Deferred (figure-dependent geometry)

The answer is the same $\tfrac12$ as the Express, and for the same reason: the octagonal pattern is shorter exactly for homes above the line $y = (\sqrt2 - 1)x$ (and its mirror images), the very boundary that split the Express in half. The official derivation shows that on a block where $x > y > 0$ the diagonal chamfers save a distance $2yd(\sqrt2 - 1)$ while the straight runs add $(x - y)d(2 - \sqrt2)$, where $d$ is the chamfer length; these balance precisely when $y = (\sqrt2 - 1)x$, and the chamfer length $d$ cancels.

We defer the full write-up rather than reconstruct it from text alone. Faithfully encoding the octagonal walk requires the exact geometry of the chamfered-block sidewalk network from the puzzle’s diagram (the lengths and connections of the diagonal corner segments), which the prose does not pin down well enough to model independently without re-deriving the very boundary line we would be checking. The mathematical content is identical to the Express above, which we author in full; the result is $\tfrac12$.

Status

Deferred as figure-dependent. Recorded answer: $\tfrac12$ of employees prefer the octagonal sidewalks, with the same boundary line $y = (\sqrt2 - 1)x$ as the diagonal Express.