GPT 6 Astra Ultra attempt on Math-1187

Rendezvous on a line

Definitions and mathematical statement

Two players start distance two apart on a line, at global coordinates $0$ and $2$, at time zero. Neither knows the direction of the other or has a shared orientation. A strategy is a probability distribution $\mu$ on continuous paths $f:[0,\infty)\to\mathbb R$ with $f(0)=0$ and $|f(t)-f(s)|\le|t-s|$; thus speed is at most one. Paths describe displacement in a player's local coordinates before any meeting.

In the symmetric model both players independently sample $f,g$ from the same distribution $\mu$. Their local orientations $\epsilon_1,\epsilon_2\in\{-1,1\}$ are independent fair signs, independent of the path samples. Define \[ T=\inf\{t\ge0:\epsilon_1 f(t)=2+\epsilon_2g(t)\}, \qquad R=\inf_\mu\mathbb E[T], \] with $\inf\varnothing=+\infty$ for a missed meeting. The problem is to determine $R$ exactly. The infimum is over measurable strategy distributions for which the hitting-time expectation is defined, allowing the value infinity. No shared random bits, preassigned distinct roles, or communication before meeting is allowed. Both players know the separation and start simultaneously. The source's additional proposed value is $R=17/4$ in this normalization. Rescaling initial separation would rescale time, so the specified distance two is essential.

Short English statement

What is the smallest expected meeting time for two indistinguishable players distance two apart on an unoriented line, using the same independently randomized strategy?

Research attempt

Scope and method.

This research attempt by GPT-6 Astra Ultra is dated 2 October 2026. It preserves the catalogue statement and distinguishes elementary proofs below from cited prior work. No novelty claim or formal proof-assistant verification is made. The final paragraph states the precise remaining scope. The finite checks described below are reproducible in scripts/check_attempt_batch03_root_2026_10_02.py.

Keeping the symmetric model explicit.

Both players must independently sample from the same distribution. Assigning one player to wait and the other to search would change the problem. The source [S2] reports bounds much sharper than the elementary ones proved below and proposes $17/4$. Our purpose is to construct a fully admissible baseline strategy and prove a model-specific lower bound, then identify why this strategy cannot attain the proposed value. All times use initial separation two.

Independent excursions give a finite bound.

Each player samples an infinite sequence of independent fair signs. In round $j$, lasting two time units, the player walks one unit in the locally chosen direction and then returns to its initial position. This describes a continuous speed-one path sampled in advance, hence a permitted distribution $\mu$. Multiplying its fair signs by a fixed random local orientation still gives independent fair global directions in each round. The two players both move towards one another with probability $1/4$, meeting exactly one unit into that round. If both move in the same global direction, their separation stays two. If both move away, their separation increases and then returns to two; they do not meet. Every failed round resets both positions. If $J$ is the number of failures before the first success, then $\Pr(J=j)=(3/4)^j/4$, $T=2J+1$, and \[ \mathbb E T=2\frac{3/4}{1/4}+1=7. \] In particular the probability of never meeting is zero. This proves $R\le7$, without using distinct roles, a shared compass or shared random bits.

A lower bound from four orientations.

Fix any two sampled 1-Lipschitz paths $f,g$ starting at zero and condition on them. The four orientation pairs are equally likely. In a pair $(a,b)$, a meeting at time $s$ means $af(s)-bg(s)=2$. No meeting is possible before time one, by the speed bound on relative displacement. Two orientation pairs differing only in the second sign cannot both meet before time two: if their meeting times are $s\le t$, adding their equations and using the Lipschitz bounds gives \[ 4=a(f(s)+f(t))+b(g(t)-g(s))\le s+t+(t-s)=2t. \] The same conclusion for pairs differing only in the first sign follows by adding equations to bound $f(s)-f(t)$ and $g(s)+g(t)$ instead. If both signs differ, the same 2-Lipschitz relative path must take values $2$ and $-2$; their hitting times differ by at least two. Thus at most one orientation pair can have met by any time $t<2$. For $1\le t<2$, conditional meeting probability is therefore at most $1/4$. Averaging over the sampled paths and integrating survival probabilities proves, also for strategies of infinite expectation, \[ \mathbb E T\ge1+\frac34=\frac74. \] This lower bound is intentionally weaker than the published bounds; it is a self-contained consequence of the stated orientation model.

Optimizing this restricted excursion family.

Replace the unit excursion by one of fixed length $a$. For $a\ge1$, a successful round still meets at time one, while every failure lasts $2a$. The expectation is $1+6a$, minimized at $a=1$. For $a<1$, even simultaneous inward motion can close a distance of at most $2a<2$; every round returns to the starting configuration, so meeting never occurs. No member of this fixed-amplitude independent-round family attains $17/4$. Improving it requires changing the path distribution, rather than optimizing that single amplitude.

Verification and final status.

The root script solves the piecewise-linear meeting equations for all four direction pairs and several rational amplitudes. It checks the finite geometric sum with its exact remaining tail. These calculations verify the strategy's expectation, not an optimal strategy theorem. The proved interval is $7/4\le R\le7$; determining the exact infimum remains unresolved by this attempt.

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3447 (OPG-581), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, Rendezvous on a line. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/rendezvous_on_a_line.