Theoretical and Numerical Results Regarding the General Case of the Lonely Runner

Conjecture

Rahul Bhargava^{1} Mathematics Department

Princeton University Princeton, NJ 08544

February 4, 2003

1E-mail: rbhargav@princeton.edu

Abstract

For x real, let x be the fractional part of x (i.e. x = x − ⌊x⌋). In this paper, theoretical and numerical results are presented regarding the following conjecture (The Lonely Runner Conjecture). Given any k positive reals v_{1}, …, v_{k} there exists t ∈ ℝ such that 1/(k + 1) ≤ v_{i}t ≤ k/(k + 1) for i = 1, …, k.

Contents

1 Introduction 2

2 Known Results 3 2.1 Preliminary Results . . . . . . . . . . . . . . . . . . . . . . . . . 3 2.2 Proof of k=2,3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.3 Proof k=4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3 Theoretical Findings 7 3.1 Interval Overlap Approach . . . . . . . . . . . . . . . . . . . . . 8 3.2 Locked Steering Wheel Approach . . . . . . . . . . . . . . . . . 9

4 Computational Results 11

Chapter 1

Introduction

The Lonely Runner Conjecture can be stated informally: at the same time, k runners depart from a common starting point and run repeated laps around a circle of circumference one. Each runner maintains a constant speed, and all the speeds are distinct. Does there exist a time such that all the runners are at a distance of 1/(k + 1) from the starting point. The “Lonely Runner” reference is due to another statement of the problem in which there are k + 1 runners. Does there exist a time such that one runner is ’lonely’ - the shortest distance from the runner to any other runner is at least 1/(k + 1).

The conjecture was introduced by J. Wills [WI] in the context of diophantine approximations, and once again independently by T. Cusick [Cu] in application to View Obstruction Problems. A formal statement of the conjecture is given:

Conjecture 1.0.1. For any collection of v_{1}, v_{2}, …, v_{k} ∈ ℝ^{+} there exists t ∈ ℝ^{+} such that the following holds:

\{v_{i}t\} ∈ [\frac{1}{k + 1}, \frac{k}{k + 1}] for i = 1, …, k  (1.1)

The conjecture has been proven for the case k ≤ 4 by Cusick and Pomerance [CP], and once again by Bienia, Goddyn, Gvozdjak, Sebo, and Tarsi [BGGST]. Bohman, Holzman, and Kleitman [BHK] have shown the conjecture to be true for k = 5. [BGGST] used their result to prove a theorem on nowhere zero flows in graphs. It has been noted that a proof of the Lonely Runner Conjecture for higher cases would have similar applications to flows in regular matroids.

Known Results

2.1 Preliminary Results

It has been shown by Wills [WI] that the case of irrational speeds can be reduced to the rational case. Thus it can be assumed that v_{1}, …, v_{k} ∈ ℕ, where v_{i} is the number of laps that the runner completes in one unit of time. As a result, if the conjecture does not hold for a time t ∈ [0, 1), then the conjecture will not hold for any time t ∈ ℝ^{+}.

We introduce some notation. Denote the number of runners by k. Let v := (v_{1}, …, v_{k}) ∈ ℕ^{k} be the speed vector, v_{i} ≠ v_{j} if i ≠ j. The position of runner i at time t is given by {v_{i}t}. The position of all k runners at time t is given by the vector {vt} = ({v_{1}t}, …, {v_{k}t}). A vector x = (x_{1}, …, x_{k}) is a position if there exists t ∈ [0, 1) with x = {vt}. Denote the set of all such positions as X. Lastly, if there exists t ∈ [0, 1) such that (v_{1}, …, v_{k}) satisfies condition 1.1, we say (v_{1}, …, v_{k}) is distant.

We now state two propositions given by [BGGST] that will be helpful in approaching this conjecture.

Proposition 2.1.1. If x_{1}, x_{2} ∈ X and α ∈ ℤ, then the vector x = {x_{1} + αx_{2}} ∈ [0, 1)^{k} is also in X. If, x_{1} = {vt_{1}}, x_{2} = {vt_{2}}, and t ≡ t_{1} + αt_{2} (mod 1), then x = {vt}.

Proposition 2.1.2. Let k ≥ 3. If gcd(v_{1}, …, v_{k−1}) does not divide v_{k}, then there exists a time when (v_{1}, …, v_{k}) is distant if and only if there exists a time when (v_{1}, …, v_{k−1}) is distant

Proof: Let d ≥ 2 be the gcd defined in the statement, and suppose without loss of generality that gcd(d, v_{k}) = 1. Hence at time t = j/d where j = 0, 1, …, d − 1, {v_{i}t} = 0 for i = 0, 1, …, k − 1. Since d does not divide v_{k},

\{\{tv_{k}\} : t = \frac{j}{d}, j = 0, 1, …, d − 1\} = \{\frac{0}{d}, \frac{1}{d}, …, \frac{d − 1}{d}\}

Let x = {tv} be a position where (v_{1}, …, v_{k−1}) is distant. This implies that (v_{1}, …, v_{k−1}) will be distant at the d positions given by \{x + \frac{j}{d}v\}, j = 0, …, d− 1. As k ≥ 3 and d ≥ 2, \frac{1}{d} is at most the length of the interval [\frac{1}{k+1}, \frac{k}{k+1}], the interval of distant positions. Hence, v_{k} is distant in one of the d positions.

2.2 Proof of k=2,3

Theorem 2.2.1. The Lonely Runner Conjecture is true for k ≤ 3

Proof: The case k = 1 is trivial. We present the proof given by [BGGST] for the two remaining cases individually.

For k = 2 a stronger statement of the theorem is proven:

Suppose v_{1}, v_{2} ∈ ℕ are relatively prime speeds. At any time t, the nearer runner has distance at most ⌊\frac{v_{1}+v_{2}}{2} ⌋/(v_{1} + v_{2}). Moreover, the bound is achieved at time t = \frac{τ}{v_{1}+v_{2}} for some τ ∈ N.

When the distance from the starting point to the nearest runner is maximum, {tv_{1}} = 1 − {tv_{2}}. This will occur if and only if t is an integer multiple of 1/(v_{1} + v_{2}). At such t, the runners are at a distance a/(v_{1} + v_{2}) for some integer a ≤ ⌊\frac{v_{1}+v_{2}}{2} ⌋. Since gcd(v_{1}, v_{1} + v_{2}) = 1 the congruence v_{1}τ ≡ ⌊(v_{1} + v_{2})/2⌋ mod (v_{1} + v_{2}) can be solved to obtain a time when the bound is obtained.

For k = 3

Assume v_{1}, v_{2}, v_{3} have no common factor. If all three speeds are odd, then at t = 1/2, (v_{1}, v_{2}, v_{3}) will be distant. Hence we may assume without loss of generality that v_{2} is even. By Proposition 2.1.2, we may also assume v_{1} and v_{2} are odd. At time t_{1} = 1/2 therefore our position vector is x_{1} = (\frac{1}{2}, 0, \frac{1}{2}). Define T to be the time interval for which v_{2} is for the first time in the region [\frac{1}{4}, \frac{3}{4}]. For r = 1, 3, define T_{r} = \{t ∈ [0, 1) : \{tv_{r}\} ∈ [\frac{1}{4}, \frac{3}{4}]\}. If T \(T_{1} ∪ T_{3}) ≠ ∅, pick t_{2} ∈ T \(T_{1} ∪ T_{3}). By Proposition 2.1.1, since only v_{2} is distant at t_{2}, (v_{1}, v_{2}, v_{3}) is distant at time t_{1} + t_{2}. Thus we may asuume, T ⊆ T_{1} ∪ T_{3}. Suppose T ⊆ T_{i}, i ∈ {1, 3}. This implies T is contained in one of the closed intervals of T_{i} which implies v_{2} ≥ v_{i}. However, v_{i} first becomes distant no later than v_{2}, so v_{2} ≤ v_{i}, and as a result v_{2} = v_{i}, contradiction. The remaining case is T ⊆ T_{1} ∪ T_{3}, T ∩ T_{i} ≠ ∅ (i = 1,3). Both T ∩ T_{1} and T ∩ T_{3} consist of disjoint closed intervals, and their union is T. Thus, ∅ ≠ (T ∩ T_{1}) ∩ (T ∩ T_{3}) = T ∩ T_{1} ∩ T_{3}, and we have obtained the desired result.

2.3 Proof k=4

We shall present the proof given by [BGGST] for the case k = 4. First we introduce notation and present two more preliminary results. For k ≥ 2, partition the circular track C = [0, 1) as 0 ∪ C_{1} ∪ C_{2} where

C_{1} := (0, \frac{1}{k + 1}) ∪ (\frac{k}{k + 1}, 1) ∪ \{\frac{1}{k + 1}, \frac{2}{k + 1}, …, \frac{k}{k + 1}\} and
C_{2} := (\frac{1}{k + 1}, \frac{2}{k + 1}) ∪ (\frac{2}{k + 1}, \frac{3}{k + 1}) ∪ ⋯ ∪ (\frac{k − 1}{k + 1}, \frac{k}{k + 1})

Denote the set of runners R = {1, …, k}. Define D := {r ∈ R : (k + 1)|v_{r}}, and let x ∈ X. We can partition the runners R as R_{0}(x) ∪ R_{1}(x) ∪ R_{2}(x) where

R_{0}(x) := D ∪ {r ∈ R : x_{r} = 0}
R_{1}(x) := {r ∈ R\D : x_{r} ∈ C_{1}}
R_{2}(x) := {r ∈ R\D : x_{r} ∈ C_{2}}

Proposition 2.3.1. Let k + 1 be prime, and suppose there exists x ∈ X in which D is distant, and |R_{2}(x)| < |R_{0}(x)|. Then there exists a time when R is distant.

Proposition 2.3.2. Suppose that k+1 is prime, and the only speed which it divides is v_{2}. If there exists d ∈ ℕ dividing at least k/2 different speeds, but not dividing v_{2}, then there exists a time when R is distant.

Theorem 2.3.3. The Lonely Runner Conjecture is true for k = 4.

Proof: If |D| = 0, then time \frac{1}{5} will suffice. Suppose 2 ≤ |D| ≤ 3. Then by induction on k there is a position y where |D| is distant. Either y is a position for which R is distant, or there is some runner in R\D that is not distant. This implies |R_{0}(y)| ≥ |R_{2}(y)| and we are done by 2.3.1. Hence we may asuume D = {2}, 2 ∈ R_{0}(x) for every position x.

If 2 is the fastest runner, we are once again done by 2.3.1. We may assume v_{1} > v_{2}, v_{3}, v_{4}. At least one of v_{3}, v_{4}, without loss of generality, v_{3}, is not equal to v_{1} − v_{2}. If the gcd(v_{1}, v_{3}) > 1 and divides v_{2}, we are done by 2.1.2; if not then we are done by 2.3.2.

As a result, we may assume gcd(v_{1}, v_{3}) = 1 ⟹ ∃ α ∈ ℕ, αv_{3} ≡ 1 mod v_{1}. Let x be the position at time \frac{α}{v_{1}} . If 2 is distant in x, then we are done by 2.3.1. So we may assume 2 is not distant in x.

There are two consequences to this. First, the distance of x_{2} from 0 is at least twice that of x_{3}. Second, if a runner has distance δ ≤ 1/4 from 0 in position z ∈ X, then it has position 2δ in position {2z}. Let x′ be the first position in the sequence {2x}, {4x}, … in which 2 is distant. As before, 1,2 ∈ R_{0}(x′), and in addition x′_{3} ∈ (0, 1/5) ⟹ 3 ∈ R_{1}(x′). Thus we are done by 2.3.1.

Theoretical Findings

This chapter provides theoretical results for the general case of the Lonely Runner Conjecture, as well as two approaches to the solution of the general case. We first introduce some notation. Denote by I_{k}(v) the times at which the runner with speed v is in the interval [\frac{1}{k+1}, \frac{k}{k+1}].

I_{k}(v) = ⋃_{n=0}^{+∞} [\frac{n(k + 1) + 1}{v(k + 1)}, \frac{n(k + 1) + k}{v(k + 1)} ]

Since we are only concerned with t ∈ [0, 1)

I_{k}(v) = ⋃_{n=0}^{v−1} [\frac{n}{v} + \frac{1}{v(k + 1)}, \frac{n}{v} + \frac{k}{v(k + 1)}]

Hence, the Lonely Runner Conjecture can be restated as:

⋂_{i=1}^{k} I_{k}(v_{i}) ≠ ∅, ∀ (v_{1}, …, v_{k}) ∈ ℕ^{k}

Note that the intervals in the definition of I_{k}(v) are disjoint as \frac{k−1}{v(k+1)} < \frac{1}{v} implies \frac{n}{v} + \frac{k}{v(k+1)} < \frac{n+1}{v} + \frac{1}{v(k+1)}. Furthermore, since \frac{n}{v} + \frac{1}{v(k+1)} < \frac{n}{v} + \frac{1}{vk} < \frac{n}{v} + \frac{k−1}{kv} < \frac{n}{v} + \frac{k}{v(k+1)} it is true that I_{k}(v) ⊂ I_{k+1}(v).

3.1 Interval Overlap Approach

We shall arrange the speeds in ascending order, i.e. v_{1} < … < v_{k−1} < v_{k}. By induction assumption,

⋂_{i=1}^{k−1} I_{k−1}(v_{i}) ≠ ∅

Fix a point

p ∈ ⋂_{i=1}^{k−1} I_{k−1}(v_{i})

At time p, the k − 1 runners are in the interval BC as shown in Fig. 1. The maximum distance a runner can travel and still be assured of remaining in the interval AD is \frac{1}{k+1} (i.e. the length of interval CD). The time it takes the fastest runner, with speed v_{k−1}, to cover that distance is \frac{1}{k(k+1)v_{k−1}} . Since v_{k−1} > v_{i}, 1 ≤ i < k − 1, the distance travelled by the other k − 2 runners is less than \frac{1}{k+1}. Hence we have established the following result:

Proposition 3.1.1. Suppose p ∈ ⋂_{i=1}^{k−1} I_{k−1}(v_{i}). Then

[p − \frac{1}{k(k + 1)v_{k−1}} , p + \frac{1}{k(k + 1)v_{k−1}} ] ⊆ ⋂_{i=1}^{k−1} I_{k}(v_{i})

Let D_{i} = {v ∈ (v_{1}, …, v_{k}) : i|v} We have the following proposition:

Proposition 3.1.2. If ∃ l, 2 ≤ l ≤ (k + 1) such that |D_{l}| = 0, then (v_{1}, …, v_{k}) is distant at time \frac{1}{l}.

Hence given (v_{1}, …, v_{k}) such that |D_{k}| = 0, the Lonely Runner Conjecture holds.

Conjecture 3.1.3. If there exists a time when (v_{1}, …, v_{k}) is distant, then there exists a time when (v_{1}, …, kv_{i}, …, v_{k}) is distant.

Given any set of speeds (v_{1}, …, v_{k}), there is a corresponding set of speeds (v′_{1}, …, v′_{k}), where v_{i} = v′_{i}k^{j_{i}}, j_{i} the highest power of k in v_{i}. Since |D_{k}| = 0 for (v′_{1}, …, v′_{k}), by Proposition 3.1.2 there is a time when (v′_{1}, …, v′_{k}) is distant. By multiplying the speeds of v′_{1}, …, v′_{k}) by k, we can obtain (v_{1}, …, v_{k}), which by Conjecture 3.1.3 is distant. Hence, a direct consequence of the proof of Conjecture 3.1.3 is the proof of the Lonely Runner Conjecture.

Given a speed v, the gaps in the intervals of I(kv) are of length \frac{2}{k(k+1)v}.

By Proposition 3.1.1 the interval [p − \frac{1}{k(k+1)v_{k−1}} , p + \frac{1}{k(k+1)v_{k−1}} ] is contained in

⋂_{i=1}^{k−1} I_{k}(v_{i}). The length of this interval is \frac{2}{k(k+1)v_{k−1}} . If \frac{2}{k(k+1)v} < \frac{2}{k(k+1)v_{k−1}} , then there will be an overlap and ⋂_{i=1}^{k−1} I_{k}(v_{i}) ∩ I_{k}(kv) ≠ ∅. This is the case when v > v_{k−1}. We thus have the following result:

Proposition 3.1.4. Conjecture 3.1.3 holds when v_{i} = v_{k}, i.e. when we multiply the fastest speed by k.

Corollary 3.1.5. If (v_{1}, …, v_{k−1}) is distant and v_{k} > kv_{k−1}, then (v_{1}, …, v_{k}) is also distant.

Proposition 3.1.4 guarantees an overlap when we multiply the largest speed by k. However, the proposition does not ensure that

⋂_{i=1}^{k−1} I_{k}(v_{i}) ∩ I_{k}(v) ∩ I_{k}(kv) ≠ ∅  (3.1)

This is a stronger form of Conjecture 3.1.3; in addition to stating that there is a time when (v_{1}, …, v_{i}, …, v_{k}) is distant, the times that the two sets of speeds are distant overlap. Examining Fig. 2, we note that the endpoints of the intervals of I_{k}(v) are in I_{k}(kv). As a result, there are two immediate conditions that will satisfy equation 3.1.

∃p ∈ ⋂_{i=1}^{k−1} I_{k−1}(v_{j}) s.t.

p ∉ I_{k}(v) but [p − \frac{1}{k(k + 1)v_{k−1}} , p + \frac{1}{k(k + 1)v_{k−1}} ] ∩ I_{k}(v) ≠ ∅ (3.2)

and

p ∈ I_{k}(v) but [p − \frac{1}{k(k + 1)v_{k−1}} , p + \frac{1}{k(k + 1)v_{k−1}} ] ⊈ I_{k}(v)  (3.3)

3.2 Locked Steering Wheel Approach

We now shift to a different approach. Consider the position of the k runners with speeds (v_{1}, …, v_{k}) at time t_{0} = \frac{1}{3}. We will assume that |D_{3}| = 1. Denote the speed that is a multiple of three by v_{i}. Then at time t_{0}, the runners whose speeds are congruent to one modulo three will be at position \frac{1}{3} and the runners whose speeds are congruent to two modulo three will be at position \frac{2}{3}. The runner with speed v_{i} will have returned to the starting point. Assume that v_{i} is in fact the fastest runner, i.e. i = k. Without loss of generality we may assume that the runner with speed v_{k−1} is at position \frac{2}{3}. The distance from \frac{2}{3} to \frac{k}{k+1} is \frac{k−2}{3(k+1)}. For k ≥ 5, \frac{k−2}{3(k+1)} ≥ \frac{1}{k+1}. Hence at time t_{0} + \frac{1}{(k+1)v_{k}}, the k − 1 runners will be in the interval (\frac{1}{3}, \frac{k}{k+1}) and the runner with speed v_{k} will be at position \frac{1}{k}. So we may assume that i ≠ k.

Suppose that v_{k} is at position \frac{2}{3}. The time it will take for the runner to reach \frac{k}{k+1} is \frac{k−2}{3(k+1)v_{k}}. The other k − 2 runners will also be in the interval (\frac{1}{3}, \frac{k}{k+1}] since v_{k} is the fastest runner. Thus if the runner with speed v_{i} can travel a distance of \frac{1}{k+1} in that time, (v_{1}, …, v_{k}) will be distant at time t_{0} + \frac{k−2}{3(k+1)v_{k}}. For this to be

the case, \frac{(k−2)v_{i}}{3(k+1)v_{k}} ≥ \frac{1}{k+1} ⟹ v_{i} ≥ \frac{3v_{k}}{k−2}, k ≥ 5.

To obtain a better lower bound on v_{i}, we can go in the opposite direction. Once again we will assume v_{k} is at position \frac{2}{3}. Denote the speed of the fastest runner at position \frac{1}{3} by v_{m}. The distance from position \frac{2}{3} to position \frac{1}{k+1} in the clockwise direction is \frac{2k−1}{3(k+1)}. The time it will take the runner with speed v_{k} to cover that distance is \frac{2k−1}{3(k+1)v_{k}}. The distance from position \frac{1}{3} to position \frac{1}{k+1} in the clockwise direction is \frac{k−2}{3(k+1)}, and the time it will take the runner with speed v_{m} to cover it

is \frac{k−2}{3(k+1)v_{m}}. If \frac{k−2}{3(k+1)v_{m}} < \frac{2k−1}{3(k+1)v_{k}}, then if \frac{(k−2)v_{i}}{3(k+1)v_{m}} ≥ \frac{1}{k+1}, (v_{1}, …, v_{k}) will be distant at time t_{0} + \frac{k−2}{3(k+1)v_{m}}. This is the case when v_{i} ≥ \frac{3v_{m}}{k−2}. Otherwise, our

requirement is \frac{v_{i}(2k−1)}{3(k+1)v_{k}} ≥ \frac{1}{k+1} ⟹ v_{i} ≥ \frac{3v_{k}}{2k−1}.

A further decrease in the lower bound on v_{i} would require some of the k − 1 runners crossing the starting point. However we must ensure that the other runners remain in the interval [\frac{1}{k+1}, \frac{k}{k+1}]. No theoretical result as of yet has been found that meets this requirement.

Computational Results

In this section we present conjectures regarding the Interval Overlap Approach based upon computational results. The computational results were simulations of the problem for various values of k and different sets of speeds. No theoretical proofs for these conjectures have of yet been developed.

It should be noted that the Lonely Runner Conjecture is sharp. If in the statement of the problem, the endpoints of the interval, \{\frac{1}{k+1}, \frac{k}{k+1}\}, were not included than the conjecture would be false. The simplest example of this is the set of runners with the speeds (1, …, k). There exist only two times when the runners are distant: \{\frac{1}{k+1}, \frac{k}{k+1}\}. At these times, runners with the speeds 1 and k are located at the endpoints of the interval. Thus if the endpoints are not included in the statement of the conjecture, there would not exist a time when the runners are distant. An instance of the problem where the runners are distant only at times when two of the runners are at the endpoints of the interval (one runner at each endpoint) is referred to as an extremal example. [BHK] showed that there are other sporadic extremal examples for particular values of k. Let us consider however, only the extremal example (1, …, k − 1).

Suppose that we have k runners, k −1 of which have the speeds (1, …, k −1). Let the k^{th} runner have speed v. Clearly if v ≤ k, then (1, …, k − 1, v) is distant. If v > k and k does not divide v, then the runners are distant at time \frac{1}{k}. Hence we may assume v = k^{j}m, where j is the highest multiple of k in v. If m > k then the set of speeds (1, …, k − 1, m) is distant at time \frac{1}{k}. By Proposition 3.1.4, the set of speeds of the form (1, …, k − 1, k^{i}m) are distant. Now if m < k, then v does not have a factor of k+1 and as a result the runners will be distant at time \frac{1}{k+1}.

Denoting the speeds of k − 1 runners as (v_{1}, …, v_{k−1}), we may assume that (v_{1}, …, v_{k−1}) ≠ (1, …, k − 1). Furthermore, let us assume that (v_{1}, …, v_{k−1}) is not a sporadic extremal example. By induction, ⋂_{i=1}^{k−1} I_{k−1}(v_{i}) ≠ ∅, and ∃ [a, b] ∈ ⋂_{i=1}^{k−1} I_{(}k − 1)(v_{i}). Let the k^{th} runner have speed v < v_{k−1}. We return to Conjecture 3.1.3.

Conjecture 4.0.1. If ⋂_{i=1}^{k−1} I_{k}(v_{i}) ∩ I_{k}(v) ≠ ∅ and

I_{k−1}(v) ∩ ⋂_{i=1}^{k−1} I_{k−1}(v_{i}) = ∅ ⟹ I_{k−1}(kv) ∩ ⋂_{i=1}^{k−1} I_{k−1}(v_{i}) ≠ ∅

Taking Conjecture 4.0.1 as given, we may assume

I_{k−1}(v) ∩ ⋂_{i=1}^{k−1} I_{k−1}(v_{i}) ≠ ∅

Furthermore, we may assume Conditions 3.2 and 3.3 are false. Choose [a, b] ∈ ⋂_{i=1}^{k−1} I_{k−1}(v_{i}) such that (b − a) is maximum. For

[a − \frac{1}{k(k + 1)v_{k−1}} , b + \frac{1}{k(k + 1)v_{k−1}} ] ∩ I_{k}(kv) = ∅ ⟹

(b − a + \frac{2}{k(k + 1)v_{k−1}} ) < \frac{2}{k(k + 1)v}

While this will produce a better lower bound on the size of v, no definite bound has as of yet been found. This would require finding a lower bound on the value of (b − a). The numerial results suggest that the value of (b − a) is large enough to ensure an overlap. No characterization of the intervals of ⋂_{i=1}^{k−1} I_{k−1}(v_{i}) has yet been developed that would allow a theoretical lower bound to be found.

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[BHK] T. Bohman, R. Holzman, D. Kleitman, Six Lonely Runners, The Electronic Journal of Combinatorics 8(2) (2001), R3.

[Cu] T.W. Cusick, View-obstruction problems, Aequationes Math. 9 (1973), 165 − 170.

[CP] T.W. Cusick, C. Pomerance, View-obstruction problems. III, J. Number Theory 19 (1984) 131 − 139.

[WI] J.M. Wills, Zwei Satze uber inhomogene diophantische Approximation von Irrationalzahlen, Monatsch. Math. 71 (1967) 263 − 269.