A path in the unit square that "doubles back" on itself in a nice way

Question

Given a path $P$ in the unit square, and two points $p_{1},p_{2}$ located on $P$, let $d_{P}(p_{1},p_{2})$ denote the distance from $p_{1}$ to $p_{2}$ traversed along $P$. Given $a>1$, I am looking for the shortest $P$ that satisfies the following property:

For any point $x$ in the unit square, there exist $p_{1},p_{2}$ on $P$ such that $d_{P}(p_{1},p_{2})\geq a(\|p_{1}-x\|+\|p_{2}-x\|)$.

This is clearly a very difficult optimization problem, so I'm just looking for some practical suggestions, not necessarily provably optimal (Spirals? Sawtooths?). The motivation for this problem is as follows: imagine a car driving along $P$ at unit velocity containing several people. We want it to be possible for a person to leave the car at a certain point $p_{1}$, walk at speed $1/a$ to $x$, and then rendezvous back with the car at $p_{2}$ without requiring the car to stop.

Answer 1

This is just about the asymptotic behavior for $a\to+\infty$. I claim that the minimal length $\ell$ is about $\sqrt{2a}$ for large $a$.

The upper bound

Consider $2n$ horizontal lines splitting the square into strips of width $\frac 1{2n-1}$ Now travel along the odd-numbered lines (in the natural enumeration from the top) in the natural way: left to right on line 1, down to line 3, right to left on line 3, doun to line 5, left to right on line 5, and so on. Then return to line 2 in constant time and repeat the process going over even lines in the same fashion. Notice that every strip is bounded by one odd line and one even line and the time between going over those lines is about $n$ (with constant error). So, if $x$ is in the strip, we can drop off at the nearest point on the odd line and get picked up at the nearest point on the even line walking the total distance $\frac 1{2n-1}$. Thus if $\frac 1{2n-1}a<n-C$, we have the desired property, i.e., we can take $n\approx \sqrt{a/2}$ for large $a$ giving the length $\approx 2n\approx \sqrt{2a}$.

The lower bound

Assume we have an admissible curve $\gamma$ of length $\ell$. Parameterize it by the arc length $t\in[0,\ell]$. Suppose we have any (reasonably decent) function $f(t)$ with the property $f(t')+f(t'')\ge |t'-t''|/a$. Then every point $x$ must lie in the union of disks $D_t$ centered at $\gamma(t)$ of radius $f(t)$ (if we can reach $x$ by getting out at $t'$ and getting back on at $t''$, then, since we walked at most $|t''-t'|/a$, either $x$ is $f(t')$ close to $\gamma(t')$, or $x$ is $f(t'')$ close to $\gamma(t'')$). Thus the area of the union of those disks should be at least $1$. On the other hand, under some mild smallness and regularity assumptions on $f$, we can bound it by $2\int_0^\ell f(t)\,dt+o(1)$. Now just use $f(t)=a^{-1}|t-\frac\ell 2|$ to get the bound $\ell^2/2\ge a(1+o(1))$, so $\sqrt{2a}$ cannot be improved much for large $a$.

Answer 2

This is not an answer and adds little, but ... It maybe easier to consider a surrounding disk rather than a square. I like the OP's idea of a spiral. Concentric circles allow $a>1$ shortcuts:

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          ^{$2 \sin \theta / 2 < \theta$.}

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A tight spiral approximates concentric circles.

Answer 3

If $n=\lceil a/\sqrt{2}\rceil$, then there is a path of length at most $n+3$: the car can take $\lceil n/2 \rceil +1$ trips around the small square of side length $1/2$ with lower left corner at $(1/4,1/4)$. For an injective path, we can perturb this by $\epsilon$.

From any point $x$, the closest point $p$ on the small square has a distance below $1/2\sqrt{2}$. So the round-trip distance is below $1/\sqrt{2}$, and the round-trip time is below $a/\sqrt{2}$, which is below $n$.

Let $p_1$ be the first time that the path visits $p$, and let $p_2$ be $\lceil n/2 \rceil$ visits later. Since each trip around the small square takes time $2$, there is enough time to walk from $p$ to $x$ and back before the car makes its last visit.