Can You Hop in a Spiral?

Frankie the frog hops on a vast hexagonal lattice of lily pads, each pad of diameter $1$, so neighbouring centres sit $1$ apart. Starting from pad $A$, she hops to neighbouring pads in a counterclockwise spiral, subject to two rules: each pad must be in a more counterclockwise direction from $A$ than the previous one, and each pad must be farther from $A$ than the previous one. After spiralling around, she finds herself once again due east of $A$. How close to $A$ can she be?

The Fiddler, Zach Wissner-Gross, January 30, 2026(original post)

Solution

She can be no closer than $$\boxed{64}.$$

The two rules say her bearing from $A$ strictly increases and her distance from $A$ strictly increases at every hop. To come back due east she must wind her bearing all the way from $0^\circ$ round to $360^\circ$. The lattice has six-fold symmetry, with pads lined up along the six rays at $0^\circ, 60^\circ, \dots, 300^\circ$. Crossing from one ray to the next, the tightest legal corner Frankie can turn, still gaining both bearing and distance, is a $30$–$60$–$90$ triangle, which doubles her distance from $A$. Six such sectors carry her once around, multiplying her distance by $2^6 = 64$: starting one pad out, she returns due east no nearer than $64$.

The computation

Encode the two rules on the lattice: process pads in order of distance from $A$ (a valid order, since each legal hop strictly increases distance) and mark a pad reachable if some marked neighbour hops to it while gaining both distance and counterclockwise bearing. The nearest reachable due-east pad past the start is $64$.

import numpy as np
from math import atan2, pi, hypot
S3 = np.sqrt(3)/2
NB = [(1,0),(-1,0),(0,1),(0,-1),(1,-1),(-1,1)]   # 6 lattice neighbours
def rad(a, b): return hypot(a + b/2, b*S3)
def prin(a, b):
    t = atan2(b*S3, a + b/2); return t if t >= 0 else t + 2*pi
def ccw(P, Q):                                   # bearing strictly increases
    return ((prin(*Q) - prin(*P) + pi) % (2*pi) - pi) > 1e-9
RMAX = 80.0; M = int(RMAX) + 2
pts = sorted([(a, b) for a in range(-M, M+1) for b in range(-M, M+1)
              if 1e-9 < rad(a, b) <= RMAX], key=lambda p: rad(*p))
reach = {(1, 0)}
for P in pts:                                    # radius order = valid topological order
    if P not in reach: continue
    for da, db in NB:
        Q = (P[0]+da, P[1]+db)
        if rad(*Q) > rad(*P)+1e-9 and rad(*Q) <= RMAX and ccw(P, Q): reach.add(Q)
east = sorted(p[0] for p in reach if p[1] == 0 and p[0] > 1)
print(east[0])                                   # 64

Extra Credit

Frankie stores food on pad $A$. Every second, the food on each pad splits into six equal portions that jump to its six neighbours. After how many seconds $N>2$ does pad $A$ hold less than $1$ percent of the original amount?

Solution

The fraction left on $A$ after $N$ seconds is exactly the probability that a random walk on the lattice, stepping to a uniformly chosen neighbour each second, sits back at its start after $N$ steps. Propagating the distribution second by second, this return fraction falls below $1$ percent for the first time at $$N = \boxed{28}.$$ At $N=27$ it is still $1.00\%$, at $N=28$ it is $0.97\%$. (The source’s value is behind its paywall; this is my own.)

The computation

Diffuse the food directly: hold the amount on every pad and, each second, split each pad’s amount evenly to its six neighbours. Watch the amount back on $A$ until it drops below $1$ percent.

from collections import defaultdict
cur = {(0, 0): 1.0}
for N in range(1, 30):
    nxt = defaultdict(float)
    for (a, b), m in cur.items():
        for da, db in NB: nxt[(a+da, b+db)] += m/6
    cur = nxt
    if N > 2 and cur[(0, 0)] < 0.01:
        print(N, round(cur[(0, 0)], 5)); break    # 28 0.00967