Bounds for the Lonely Runner Problem via Linear Programming

In this note we develop a linear programming framework to produce upper and lower bounds for the lonely runner problem.


The Lonely Runner Problem
Suppose you are competing in race on a circular track of perimeter L with n − 1 other runners. Assume all competitors have distinct constant speeds. The gap of loneliness is the largest length such that at some time t in the future (assuming the race continues forever) the closest runner to you is at distance . The lonely runner conjecture states that L n .
This problem was introduced independently by Wills [8] (1967) and Cusick [2] (1973) in the context of view obstruction problems. The conjecture is known to be true for n 7 runners. Moreover, speeds can be assumed to be distinct integers. For more on the history of partial results see Bohnman et al. [1] (2001), Perarnau & Serra [6] (2016) and Tao [7] (2018). By Galilean relativity your speed can be assumed to be zero and the conjecture takes the following equivalent formulation: Let x = min n∈Z |x−n| denote the distance to the nearest integer. For a vector x ∈ R n−1 let µ(x) = min{ x 1 , ..., x n−1 }.
Then for any vector v = (v 1 , ..., v n−1 ) ∈ Z n−1 of distinct integers show that where T = R/Z. In general, since t → µ(tv) is piece-wise linear with slopes drawn from {v 1 , ..., v n−1 }, the set of local maxima of t → µ(tv) is contained in the intersection of any two line segments. Therefore, if t satisfies µ(tv) = gap(v) then t ∈ 1 q Z, where q is a factor of some (v j ± v i ) with i = j, and therefore gap(v) = a/b where b divides q. (III) f (k) 0 if k / ∈ {0, v 1 , ..., v n−1 }.
We denote by Λ (v) the class of trigonometric polynomials satisfying (I),(II) and (III). We write Theorem 1. Let v = (v 1 , ..., v n−1 ) be a vector of increasing positive integers. Then and Theorem 2. Let v = (v 1 , ..., v n−1 ) be a vector of increasing positive integers. Then equality is attained in (1) if one of the following conditions hold: (i) All v i 's are odd. In this case f (x) = cos(πv i x) 2 is optimal for any i = 1, ..., n − 1; (ii) There exist coprime integers a, m 1 such that a{1, ..., m − 1} ⊂ {v 1 , ..., v n−1 } and all integers in {v 1 , ..., v n−1 } \ a{1, ..., m − 1} are not divisible by m. In this case f (x) = K m (ax) is optimal, where K m is Fejer's kernel (3). (iii) There exists integer a 1 such that v = av and v satisfies condition (i) or (ii). In this case if f (x) is optimal for v then f (ax) is optimal for v.
In the range n 20 and max(v 1 , ..., v n ) 40 we have performed a computer search in v in conjunction with Gurobi's linear programming solver [4] to approximate λ + (v). The sign conditions of f was modelled with sampling. This produced reliable numerical approximations to what we believe is the true value of λ + (v). In this way we check that the only cases where for |v| ∞ 40 and n 20 were the ones contemplated by Theorems 2. This leads to the following conjecture.
Conjecture. Equality is attained in (1) if and only if one of the conditions in Theorem 2 hold.
It is unfortunate that the bounds generated for λ − (v) do not seem to be nearly as good as the bounds generated by λ + (v), and we believe this is because condition (II) seems to be very strong for = −1. We need high degree polynomials and a large number of sampling points for feasibility of the linear program. In Section 3 we propose an improved version of this lower bound.
We note that proving exact bounds is not hard as if some numerical f satisfies f −δ in some region, but f should be nonnegative in that region, then all we have to do is use g = f + δ as this would be admissible for Λ ± (v) and g(0)/g(0) = f (0)/f (0) + O(δ).

Proofs for the main results
We start by recalling that Dirichlet's Approximation Theorem implies the Lonely Runner Conjecture is sharp; a rephrasing of Dirichlet's theorem is max t∈T µ(t(1, 2, ..., n − 1)) = 1 n .
The maxima is attained for t = a/n for a coprime with n. We now observe that inequality (1) is also tight in case v = (1, 2, ..., n − 1), and Fejér's kernel is the unique optimum. Optimality can easily be checked by hand, while uniqueness (modulo scaling) comes from the proof of Theorem 1. Essentially, because f 0, we must have f (k/n) = f (k/n) = 0 for k = 1, ..., n and Fejér's kernel is the only even trigonometric polynomial of degree n − 1 with these properties.
Proof of Theorem 1. Let δ = gap(v) = µ(tv) and h(x) = (δ − |x|) + be a hat function. Since h(x) = (sin(πδx)/(πx)) 2 we have which proves the upper bound. For the lower bound ( = −1), first recall that t = p/q, δ = a/b and b divides q, while q is a factor of some v j ± v i with j > i (both fractions in lowest terms). In particular This proof is inspired by the analytic proof of Dirichlet's approximation theorem due to Montgomery [5] (1994). We observe that equality is attained in (1) or (2) if and only if there is f ∈ Λ s (v) such that: On the other hand, since a is coprime with m we have µ( 1 m au) = 1 m , and since 1 m w has no integer coordinate we have µ( 1 m w) 1 m . Therefore µ( 1 m v) = 1 m and we obtain that gap(v) = 1 m . Now it is easy to check that f (x) = K m (ax) belongs to Λ + (v) and is optimal. Condition (iii) is trivial.

Improved lower bounds
Let V q be the class of vectors v of increasing positive integers such that the global maxima of t → µ(tv) is attained at some point t = p/q ∈ (0, 1 2 ) (in lowest terms). By the proof of Theorem 1, we see that if we let Λ − (v, q) be the class of functions satisfying the above conditions (I), (III) and ( for any v ∈ V q . For instance (1, ..., n − 1) ∈ V n , but some other examples of vectors in V n can be extracted from Goddyn and Wong [3], (2006). They present conditions for v to be tight, that is, gap(v) = 1 n . Some of these tight vectors characterized in [3, Theorem 2.3] belong to V n , for instance: Then equality is attained in (4) if one of the following conditions hold: (i) There exist integer a 1 coprime with q such that a{1, ..., q − 1} ⊂ {v 1 , ..., v n−1 } and all integers in {v 1 , ..., v n−1 } \ a{1, ..., q − 1} are not divisible by q. In this case is optimal, where K q is Fejer's kernel (3). (ii) There exists an integer a 1 such that v = av and v satisfies condition (i). In this case if f (x) is optimal for v then f (ax) is optimal for v.
Proof. Assume condition (i). By the same discussion in the proof of Theorem 2 we have gap(v) = 1/q. It is easy to see that f satisfies conditions (I),(II'),(III). Moreover its mass equals the mass of f (x/a), which in turn, by exact Gaussian quadrature, equals the mass of K q (x), which is 1. Also f (0) = q. This shows optimality. Condition (ii) is trivial.
3.1. Numerics. We used Gurobi [4] to compute λ ± (v) and λ − (q, v) for 2429 different velocity vectors v = (v 1 , ..., v 5 ) selected randomly from 0 < v 1 < ... < v 5 50. In Figure  1 we plot points (x, y) which are numerical approximations to (λ − (v, q), gap(v)) in blue dots, (λ − (v), gap(v)) in yellow triangles and (λ + (v), gap(v)) in green squares. We took q = denominator(t max ) where µ(t max v) = gap(v), and t max is the smallest with this property. The diagonal blue line is x = y and the gray vertical line is x = 1/6. We note that we get much better lower bounds if we know a priori that v ∈ V q , as the yellow triangles are clearly much closer to the line x = y than the blue dots. The plot appears to have some interesting emergent structures: rays of triangles and parabola-like green structures.