Alone on the track

Picture a circular track of length one, with a number of runners starting from the same line. Each keeps running forever at a constant speed, every speed being different. Runner by runner, ask a very simple question. Is there some moment at which this runner is lonely, in that everybody else is at least a certain distance away? With $k + 1$ runners in all, the lonely runner conjecture says that a distance of $1/(k + 1)$ can always be achieved, whatever the speeds are.

The statement has the feel of a schoolroom exercise, and it first appeared in number theory, with Jörg Wills in 1967, as a question about approximating several numbers at once and keeping them away from whole numbers rather than near them. Thomas Cusick independently found it in 1973, in a question about whether a ray from the origin can avoid a lattice of cubes, and Luis Goddyn gave it its name. It is a natural companion to the essays on more things than boxes, since the first good bound on it is a pigeonhole in time. But unlike those it is unfinished, in spite of being proved case by case at a speed of about one new runner per decade, until a sudden run of new results in 2025.

Reducing it to one still runner #

Fix one runner and subtract that runner’s speed from everybody. The chosen runner now stands still, and the others keep their constant relative speeds, some possibly negative. Reversing direction does not change distance, so only sizes matter. So the conjecture becomes: for any $k$ distinct positive whole numbers $v_1, \ldots, v_k$, there is a time $t$ for which

$$\| t v_i \| \ge \frac{1}{k + 1} \quad \text{for every } i,$$

where $\|x\|$ is the distance from $x$ to the nearest whole number. Speeds can be taken to be whole numbers because anything else is either a rational speed that can be scaled up, or it reduces to an easier case by a theorem about independent directions of the kind used in two numbers approximated at once.

The first figure shows why the constant could not be any better. Take speeds $1, 2, \ldots, k$ and time $t = 1/(k + 1)$. Then the runners are at positions $1/(k+1), 2/(k+1), \ldots, k/(k+1)$, spaced evenly round the track with the still runner in the remaining place. That is the most even arrangement possible, and the nearest other runner is at $1/(k+1)$. The search below shows it is the best those speeds can manage. So if the conjecture holds at all, this set of speeds is a worst case, and the number $1/(k + 1)$ is not an arbitrary choice but the exact threshold.

A curve for each set of speeds #

For a given set of speeds, how lonely can the still runner become? Plot the distance to the nearest other runner against time. It is the smallest of $k$ sawtooth-like tents $\|t v_i\|$, and that makes it piecewise straight.

That shape gives an exact computation. A highest point of a minimum of tents has to occur at the peak of one tent, where $t = (2j + 1)/(2 v_i)$, or at a crossing of two, where $t(v_i \pm v_j)$ is a whole number. That is a finite list of rational times, and evaluating the curve at each of them gives the exact best distance. For speeds 1, 2, 3 the best is exactly $1/4$, at $t = 1/4$. For 1, 3, 4 it is $2/7$, at $t = 3/7$. For 2, 3, 7 the best is $2/5$. So the set 1, 2, 3 sits on the dashed line, and the others rise above it by various amounts. Loneliness is not a smooth function of the speeds. It depends heavily on their arithmetic, in the same way as the gaps in the three-gap theorem depend on an angle’s continued fraction, not on its size.

A pigeonhole on time #

A short proof gets exactly half of the way. The set of times when runner $i$ is within $\delta$ of the still one takes up a share $2\delta$ of all time, whatever the speed: a fast runner passes more often, and more briefly, and the share is the same. So the times when any of $k$ runners is within $\delta$ add up to at most $2k\delta$. If that is less than 1, some time is left over, and at that time every runner is at least $\delta$ away. So a loneliness of

$$\delta = \frac{1}{2k}$$

is always possible. This is the pigeonhole principle in its continuous form. It is the same reasoning as the measure argument that forces an orbit to come back, and it uses nothing about the speeds except that they are not zero.

The figure shows why there is a gap between this and the conjecture. At $\delta = 1/8$, the marked times add up to 1.000, and if they did not overlap at all they would just cover everything. In reality they overlap heavily, and 23% of time remains free. They always overlap, since all tents vanish at time nought, and that is why the bound $1/(2k)$ has room to spare. To reach $1/(k + 1)$ the argument needs to show that the marked sets overlap by a definite amount, even though their total length is nearly twice the time available. That requires knowing something about how the times of different runners correlate, which is a question about arithmetic relations among speeds. No argument of a simple kind reaches it. Yong-Gao Chen and Cusick improved the $1/(2k)$ by a small amount in 1999, for prime $2k - 3$, and Terence Tao in 2018 obtained an extra term, of size about $\log k/(k^2 \log \log k)$. At large $k$, it remains the case that all known general bounds are very close to half of the conjecture.

The figure uses speeds 1, 3, 4 and 7 because they are remarkable. With $\delta = 1/5$, those marks cover all of time exactly. So this set, like $1, 2, 3, 4$, is at the limit.

Every set of speeds up to a bound #

The exact method runs quickly, and that allows a complete search over all sets of speeds below a bound. Scaling does not change anything, so only sets with no common factor are needed.

No set breaks the conjecture, as expected, since all these values of $k$ are already theorems. The interest lies in the sets that are exactly tight. For two moving runners there is just $(1, 2)$, and for three there is just $(1, 2, 3)$, out of 8,410 sets. At four, a second one appears: $(1, 3, 4, 7)$. At five there is $(1, 3, 4, 5, 9)$. At six nothing else appears, in 74,144 sets up to speed 22. And at seven there are two extra: $(1, 2, 3, 4, 5, 7, 12)$ and $(1, 4, 5, 6, 7, 11, 13)$.

These sets look patternless, and in fact there is a construction. Luis Goddyn and Erick Wong studied such sets in 2006. They showed how some tight sets can be changed into others by replacing certain speeds, under arithmetic conditions on the multiples of $k + 1$. It means that the extreme cases are not a single family, and any proof must allow for those too. Any argument that uses the specific structure of $1, \ldots, k$ cannot be enough. The existence of a list like this also makes an argument by induction less likely, since a proof must be exactly tight at all these places at once.

How rare the hard cases are #

Among 10,061 sets of four speeds up to 24, exactly two are tight. Only 9.4% have a best gap below 0.30, and the most common gaps cluster round $1/3$ and above. The histogram has marked spikes, since a best time tends to come at simple fractions, and that is a consequence of the piecewise-linear structure. For a random set of $k$ speeds, Sebastian Czerwiński showed that the gap tends to $1/2$ with high probability. Random sets are nowhere near the threshold.

This is the same phenomenon as in other problems at the edge of the pigeonhole principle, and it is why all the proofs have needed so much arithmetic. For almost every set the conjecture has room to spare, and its entire difficulty is concentrated in the sets that have many short linear relations among their speeds. In a set with many relations the runners’ tents are highly correlated, and they can be arranged to cover nearly all of time. So the work amounts to classifying how much relation is possible. This has a formal version. Matthias Beck, Serkan Hoşten and Matthias Schymura translated the conjecture into a question about zonotopes, the polytopes built from a set of line segments, and the tight instances appear there as a covering problem for lattice points.

The record, and how it moved #

The conjecture has been proved for each small number of runners, and it has been a slow story. With $k = 2$ moving runners, the proof is elementary. Betke and Wills did $k = 3$ in 1972. Cusick and Carl Pomerance did $k = 4$ in 1984, with computer assistance. Tom Bohman, Ron Holzman and Daniel Kleitman did $k = 5$ in 2001, and Javier Barajas and Oriol Serra did $k = 6$ in 2008. Then nothing happened for a long time.

Tao’s paper of 2018 changed the logic. He proved that, for each $k$, it would be enough to check speeds up to a bound of the shape $k^{O(k^2)}$, and that turned each case into a finite problem in principle. In 2024, Romanos Malikiosis, Francisco Santos and Matthias Schymura improved that to a bound on the product of the speeds. Then in September 2025, Matthieu Rosenfeld and the undergraduate Tanupat Trakulthongchai each gave a computer-assisted proof for eight runners, independently and within about a day of each other. Trakulthongchai went on to nine and ten runners, using a sieve to dispose of most cases, and Rosenfeld gave his own proof for nine. So what stood at seven runners from 2008 to 2025 moved to ten in a few months, by a combination of the finite bound with new sieving.

That is a striking case of how a finite reduction changes what a problem is. These proofs are not the ones anybody expected: they do not explain why the conjecture is true, and each new case costs far more than the one before. For the search shown here, a bound of 22 or 24 is small compared with what these proofs must rule out; the proofs instead argue that a counterexample must have a product of speeds divisible by many primes. A finite check for each $k$ would still leave the conjecture for all $k$ untouched, and that is a different problem.

A ray in a lattice of cubes #

Cusick’s form of the question is the source of the surprising connection. Put a cube of side $1 - 2\delta$ at the centre of every cell of the unit lattice in $k$-dimensional space, offset by half a cell. A ray from the origin in a direction with all positive coordinates runs through the lattice, in the way a line spells a word in a cube. The question is how large the cubes can be before they obstruct every ray. Reducing coordinates modulo 1, a ray in direction $(v_1, \ldots, v_k)$ at time $t$ is at $(t v_1, \ldots, t v_k)$. That point lies in a cube at the centre exactly when every $\|t v_i\|$ is at least $\delta$. So the lonely runner conjecture says that a ray always hits a cube of side $(k - 1)/(k + 1)$, and this is called view obstruction. A second reformulation by Bienia, Goddyn, Gvozdjak, Sebő and Tarsi, in 1998, ties it to nowhere-zero flows on graphs, the circulations that the flow essays route through networks. In that setting the bound is the conjectured shape of a circular flow number, and no connection could look less likely from the original picture of running on a track.

The limit of these pictures #

The exhaustive runs are limited in two ways. Every speed tested is small, and it is exactly among large speeds with intricate arithmetic that a counterexample would be expected, if one exists. The tight sets found here are only those below the bounds shown, so the list is not complete. In addition, a search cannot say anything at all about whether the tight sets stop appearing, or how many there are at larger $k$. The figure’s seven-speed row is the largest that was tried. The histograms and curves are all finite, and the gap’s limit at $1/2$ for random sets is not visible in them.

The times tested for each set come from exact formulas, though they were evaluated in double precision with a tolerance of $10^{-9}$. For speeds this small the rational values involved have very small denominators, so that causes no difficulty.

What was computed and what was quoted #

The best gaps are computed from the complete list of candidate times, at peaks and crossings, for each set, and the speeds $1, \ldots, k$ were checked for $k$ up to 8. The census was exhaustive: 489, 8,410, 10,061, 15,246, 74,144 and 31,788 sets, for $k = 2$ to 7 and bounds as shown. The covering figure’s free shares come from a grid of 4,000 times. The tight sets are as found. Their relation to Goddyn and Wong’s construction is cited and was not derived here.

All the theorems are quoted: Betke and Wills, Cusick and Pomerance, Bohman, Holzman and Kleitman, Barajas and Serra, Tao, Malikiosis, Santos and Schymura, Rosenfeld, Trakulthongchai, Czerwiński, Chen and Cusick. The recent computer-assisted proofs are very new. Their dates and the priority between them are as stated in their preprints, and they have not been checked here in any independent way.

Still open: every number of runners #

Is the conjecture true for eleven runners, and for all numbers? Each new case is now a finite problem, in the sense that a large enough computer could do it. The computation grows quickly enough that this strategy cannot simply be pushed much further, and it has no prospect of reaching all $k$. A proof for every $k$ needs a new idea, and it must be exactly tight on $1, 2, \ldots, k$ and on the sets such as $(1, 3, 4, 7)$, which limits what it could look like.

For large $k$ the distance between $1/(2k)$ and $1/(k + 1)$ is a factor of about two, and the question of how much of that factor can be gained is a separate one: is there any constant $c > 1/2$ such that a loneliness of $c/k$ is always achievable? Tao’s gain is too small for that, and no such constant is known. A complete classification of the tight sets is also open. They might be a very thin family. If that were so, they might offer a way into the general statement, as any proof would have to pass through every one of them.

Half of a distance #

A runner is lonely at distance $1/(2k)$ by a pigeonhole on time, and the conjecture asks for $1/(k + 1)$, nearly twice as much, which the speeds $1, 2, \ldots, k$ show is the best possible. Testing every set of speeds up to the bounds here finds the conjectured value exactly, with no exceptions, and it finds that the extremal sets are few and curious.

The structure of the problem is a good example of how the pigeonhole principle runs out. It is a powerful argument because it uses so little, and here it uses only the total measure, with nothing about how the runners’ times overlap. Doubling the result means knowing how they overlap, and that is an arithmetic question about speeds. Almost every set has room to spare, and a handful like $(1, 3, 4, 7)$ have none. A proof has to distinguish them.