When Will The Fall Colors Peak?

Riddler Express

A solar year is about $365.24217$ days. The Gregorian calendar uses $97$ leap years per $400$, giving $365.2425$. Find $L$ and $N$ (with $N < 400$) so that $L$ leap years every $N$ years makes the average day-count as close as possible to $365.24217$.

Solution

We want the leap fraction $L/N$ as close as possible to the fractional part $0.24217$, since the average year length is $365 + L/N$. Scanning every denominator $N$ below $400$ and choosing the best numerator $L$, the closest is $$\boxed{N = 351,\quad L = 85},$$ giving $\tfrac{85}{351} = 0.2421652\ldots$, off the target by under $2 \times 10^{-5}$ of a day, a good deal sharper than the Gregorian $0.2425$.

The computation

For each $N < 400$ pick the $L$ minimising $|0.24217\,N - L|$ and keep the best pair.

best = (1, None, None)
for n in range(1, 400):
    l = round(0.24217 * n)
    err = abs(0.24217 * n - l)
    if err < best[0]: best = (err, n, l)
print(best[1], best[2])                         # 351 85

Riddler Classic

Each tree starts changing colour at a uniformly random time between the equinox and the solstice, then drops its leaves at a uniformly random later time before the solstice. Across all trees, what is the largest fraction simultaneously in mid-change?

Solution

Scale the season to $[0, 1]$. A tree starts at $S \sim \mathcal{U}[0,1]$ and ends at $E \sim \mathcal{U}[S, 1]$, and it is mid-change at time $t$ when $S < t < E$. The fraction of trees changing at $t$ is exactly $\Pr(S < t < E)$. Conditioning on the start $s < t$, the end exceeds $t$ with probability $\tfrac{1 - t}{1 - s}$, so $$\Pr(S < t < E) = \int_0^t \frac{1 - t}{1 - s}\,ds = -(1 - t)\ln(1 - t).$$ Maximise over $t$: with $u = 1 - t$, this is $-u\ln u$, whose derivative $-\ln u - 1$ vanishes at $u = 1/e$. The peak fraction is therefore $$-u\ln u\Big|_{u = 1/e} = \frac{1}{e} \approx \boxed{0.368}.$$ The colours peak when a fraction $1 - 1/e$ of the season has passed, with just over a third of the trees changing at once.

The computation

Evaluate the changing-fraction $-(1 - t)\ln(1 - t)$ across the season and take its maximum; a direct tree-by-tree simulation lands on the same value.

import numpy as np
t = np.linspace(1e-9, 1 - 1e-9, 1_000_000)
print((-(1 - t) * np.log(1 - t)).max(), 1 / np.e)   # 0.3679, 0.3679