The lonely runner conjecture is the following open problem:

Conjecture 1Suppose one has runners on the unit circle , all starting at the origin and moving at different speeds. Then for each runner, there is at least one time for which that runner is “lonely” in the sense that it is separated by a distance at least from all other runners.

One can normalise the speed of the lonely runner to be zero, at which point the conjecture can be reformulated (after replacing by ) as follows:

Conjecture 2Let be non-zero real numbers for some . Then there exists a real number such that the numbers are all a distance at least from the integers, thus where denotes the distance of to the nearest integer.

This conjecture has been proven for , but remains open for larger . The bound is optimal, as can be seen by looking at the case and applying the Dirichlet approximation theorem. Note that for each non-zero , the set has (Banach) density for any , and from this and the union bound we can easily find for which

for any , but it has proven to be quite challenging to remove the factor of to increase to . (As far as I know, even improving to for some absolute constant and sufficiently large remains open.)

The speeds in the above conjecture are arbitrary non-zero reals, but it has been known for some time that one can reduce without loss of generality to the case when the are rationals, or equivalently (by scaling) to the case where they are integers; see e.g. Section 4 of this paper of Bohman, Holzman, and Kleitman.

In this post I would like to remark on a slight refinement of this reduction, in which the speeds are integers of *bounded size*, where the bound depends on . More precisely:

Proposition 3In order to prove the lonely runner conjecture, it suffices to do so under the additional assumption that the are integers of size at most , where is an (explicitly computable) absolute constant. (More precisely: if this restricted version of the lonely runner conjecture is true for all , then the original version of the conjecture is also true for all .)

In principle, this proposition allows one to verify the lonely runner conjecture for a given in finite time; however the number of cases to check with this proposition grows faster than exponentially in , and so this is unfortunately not a feasible approach to verifying the lonely runner conjecture for more values of than currently known.

One of the key tools needed to prove this proposition is the following additive combinatorics result. Recall that a *generalised arithmetic progression* (or ) in the reals is a set of the form

for some and ; the quantity is called the *rank* of the progression. If , the progression is said to be *-proper* if the sums with for are all distinct. We have

Lemma 4 (Progressions lie inside proper progressions)Let be a GAP of rank in the reals, and let . Then is contained in a -proper GAP of rank at most , with

*Proof:* See Theorem 2.1 of this paper of Bilu. (Very similar results can also be found in Theorem 3.40 of my book with Van Vu, or Theorem 1.10 of this paper of mine with Van Vu.)

Now let , and assume inductively that the lonely runner conjecture has been proven for all smaller values of , as well as for the current value of in the case that are integers of size at most for some sufficiently large . We will show that the lonely runner conjecture holds in general for this choice of .

let be non-zero real numbers. Let be a large absolute constant to be chosen later. From the above lemma applied to the GAP , one can find a -proper GAP of rank at most containing such that

in particular if is large enough depending on .

We write

for some , , and . We thus have for , where is the linear map and are non-zero and lie in the box .

We now need an elementary lemma that allows us to create a “collision” between two of the via a linear projection, without making any of the collide with the origin:

Lemma 5Let be non-zero vectors that are not all collinear with the origin. Then, after replacing one or more of the with their negatives if necessary, there exists a pair such that , and such that none of the is a scalar multiple of .

*Proof:* We may assume that , since the case is vacuous. Applying a generic linear projection to (which does not affect collinearity, or the property that a given is a scalar multiple of ), we may then reduce to the case .

By a rotation and relabeling, we may assume that lies on the negative -axis; by flipping signs as necessary we may then assume that all of the lie in the closed right half-plane. As the are not all collinear with the origin, one of the lies off of the -axis, by relabeling, we may assume that lies off of the axis and makes a minimal angle with the -axis. Then the angle of with the -axis is non-zero but smaller than any non-zero angle that any of the make with this axis, and so none of the are a scalar multiple of , and the claim follows.

We now return to the proof of the proposition. If the are all collinear with the origin, then lie in a one-dimensional arithmetic progression , and then by rescaling we may take the to be integers of magnitude at most , at which point we are done by hypothesis. Thus, we may assume that the are not all collinear with the origin, and so by the above lemma and relabeling we may assume that is non-zero, and that none of the are scalar multiples of .

with for ; by relabeling we may assume without loss of generality that is non-zero, and furthermore that

where is a natural number and have no common factor.

We now define a variant of by the map

where the are real numbers that are linearly independent over , whose precise value will not be of importance in our argument. This is a linear map with the property that , so that consists of at most distinct real numbers, which are non-zero since none of the are scalar multiples of , and the are linearly independent over . As we are assuming inductively that the lonely runner conjecture holds for , we conclude (after deleting duplicates) that there exists at least one real number such that

We would like to “approximate” by to then conclude that there is at least one real number such that

It turns out that we can do this by a Fourier-analytic argument taking advantage of the -proper nature of . Firstly, we see from the Dirichlet approximation theorem that one has

for a set of reals of (Banach) density . Thus, by the triangle inequality, we have

for a set of reals of density .

Applying a smooth Fourier multiplier of Littlewood-Paley type, one can find a trigonometric polynomial

which takes values in , is for , and is no larger than for . We then have

where denotes the mean value of a quasiperiodic function on the reals . We expand the left-hand side out as

From the genericity of , we see that the constraint

occurs if and only if is a scalar multiple of , or equivalently (by (1), (2)) an integer multiple of . Thus

By Fourier expansion and writing , we may write (4) as

The support of the implies that . Because of the -properness of , we see (for large enough) that the equation

and conversely that (7) implies that (6) holds for some with . From (3) we thus have

In particular, there exists a such that

Since is bounded in magnitude by , and is bounded by , we thus have

for each , which by the size properties of implies that for all , giving the lonely runner conjecture for .

## 51 comments

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13 May, 2015 at 10:27 pm

domotorpI’m in a hurry, but we have also investigated results similar to Lemma 4: http://arxiv.org/abs/0912.0424. One can show that a single exponential bound in d is not sufficient using this result of Alon and Vu: http://www.math.tau.ac.il/~nogaa/PDFS/av1.pdf.

14 May, 2015 at 8:54 pm

domotorpI’m sorry, now after taking a proper look, our results seem unrelated, they just contain similar words and bounds.

14 May, 2015 at 1:24 am

AnonymousIn conjecture 2, it seems that “the origin” should be “the integers”.

[Corrected, thanks – T.]14 May, 2015 at 4:42 am

BogdanYou proved that if the conjecture is false for a given n, then it false for SOME v_1, …, v_n of size n^(Cn^2). Can similar methods to be extended to prove a stronger result, that if the conjecture is false for a given n, then it false for MANY v_1, …, v_n of size n^(Cn^2), say, for a positive proportion (independent of n) of all tuples v_1, …, v_n of this size. Then, for the first unproved case n=8, we could check the conjecture for some RANDOM tuples bouded by 8^64C (if C is not too lagre), and, if true, confirm the original conjecture for n=8, with probability, arbitrary closed to 1.

14 May, 2015 at 10:28 am

Terence TaoOne can certainly establish the conjecture for “most” choices of ; for instance, if the are linearly independent over (which occurs for almost every choice of reals ), then the behave as independent random variables and the claim is quite easy to show in this case. However, all of the difficulty of the problem is concentrated in “special” choices of , and knowing that the conjecture is true for “most” choices doesn’t make much impact on this most difficult case. (An analogy is with Fermat’s last theorem: it is easy to show that the equation is false for “most” choices of $a,b,c,n$, for just about any notion of “most”, but this does not really make a dent in solving the problem.)

15 May, 2015 at 2:22 am

BogdanYou completely misundersood my comment :) I wil try again.

Your theorem is: If there is at least one counterexample for ANY v_1, …, v_n, then there is at least ONE counterexample for bounded v_1, …, v_n.

Now, assume we have a stronger result, namely:

Statement A: If there is at least one counterexample for any v_1, …, v_n, then positive proportion of ALL bounded v_1, …, v_n are all counterexamples.

Then, trying at random, we could rule out a possibility that positive proportion of ALL bounded v_1, …, v_n are all counterexamples, and

use statement A to prove the result in full.

Example how this may work.

Statement B: If there is at least one counterexample N to odd Goldbach’s conjecture, then at least N/log N even numbers in the region from 2 to N are ALL counterexamples to even Goldbach’s conjecture.

Proof is trivial. If N is a counterexample to odd Goldbach’s conjecture, then N-p is a counterexample to the even Goldbach’s conjecture for any odd p less than N.

Now, we can check even Goldbach’s conjecture for sufficiently many random numbers below exp(3100), and the odd Goldbach’s conjecture for numbers below exp(3100) follows from statement B. See https://rjlipton.wordpress.com/2014/06/06/is-this-a-proof-2/ for a variant of this argument.

15 May, 2015 at 7:25 am

Terence TaoAh, I see what you are saying now. I doubt that this sort of “counterexample dispersion” effect is present in this problem; roughly speaking, the only symmetry we have available in this problem to convert one counterexample into many counterexamples are Freiman homomorphisms. (There is a technicality here in that there is some loss in the lower bound on runner separation every time one applies such a homomorphism, based on the quality of that homomorphism, but let me ignore this technicality for sake of current discussions.) There are enough homomorphisms present to “rectify” any finite set of speeds into a set of bounded integers, but not enough to obtain a positive proportion of such sets. As mentioned in my previous comment, for randomly selected sets of speeds (which all lie, more or less, in a single Freiman isomorphism class) the claim is trivial to prove, but this does not have much bearing on all the other Freiman isomorphism classes (and if the conjecture is to fail, it is likely to come from a very special, carefully constructed, Freiman isomorphism class, which will not be detectable by random search).

14 May, 2015 at 6:39 am

AnonymousIt seems that the (closely related) Littlewood conjecture complements this conjecture for (where is constrained to be an integer.)

15 May, 2015 at 12:21 am

AnonymousCorrection: In conjecture 1 (to be exactly equivalent to conjecture 2), it seems that “different speeds” should be interpreted as “there is at least one runner whose speed is different from the speeds of the other runners”.

[The formulations are equivalent, as can be seen by replacing multiple runners with the same speed with a single runner of that speed. -T.]15 May, 2015 at 8:05 am

AnonymousThe simple identity implies that conjecture 2 has the following analytic formulation

It is easy to verify that the (outer) maximum is attained at an intersection point of two sine curves (with corresponding velocities ), i.e.

for some

Implying that is rational (a multiple of or ).

Therefore (without loss of generality) may be assumed to be rational (with denominator being the sum or difference of two integer velocities).

15 May, 2015 at 3:01 pm

arch1Is Conjecture 2 equivalent to asking whether one can see infinitely far from the origin in an n-space occupied by opaque n-cubes of side length (n-1)/(n+1) centered at half-integer coordinates?

..and no fair peeking down a coordinate axis:-)

17 May, 2015 at 12:15 am

Terence TaoYes; indeed, the original literature on the lonely runner conjecture was motivated by such “view-obstruction” formulations of the problem.

17 May, 2015 at 9:06 am

arch1Thanks. As a layperson this motivates me to try understand more of this posting. At first blush, it’s very surprising to me that such a basic view obstruction question, in such an apparently simple, symmetrical, static setting, is so hard.

16 May, 2015 at 9:18 am

Drive by shootingWould you buy this argument?

We can generalize the velocities to be arbitrary real numbers, for example by choosing a frame of reference traveling with an arbitrary runner. The system then satisfies the constraints of the Poincare recurrence theorem, for the positions, and the velocities. But any group of $n-1$ runners must then also fulfill the constraints of the Poincare recurrence theorem. Thus, fix a reference frame to an arbitrary runner, and consider a volume of the position space of the remaining runners, of length for each of the runners, and exactly opposite to the origin of the chosen runner. By the Poincare recurrence theorem the runners must revisit this volume infinitely often. We then have for any each runner is separated from all the other runners by that amount infinitely often.

17 May, 2015 at 12:15 am

Terence TaoIn order to apply the Poincare recurrence theorem, the region of space considered must be visited

at least oncebefore one can conclude that it is visited infinitely often. (With stronger hypotheses on the dynamical system, such as minimality, one does not need this additional hypothesis, however the system will only be minimal in the easy case when the relative velocities are linearly independent over the rationals.)17 May, 2015 at 1:42 am

James SmithA small thing that confused me: you say in your definition of the problem that all the runners start at the origin. To say they start at the same place might be a little clearer. And I wonder now I come to think of it if the circle were indeed replaced with a disc how the problem would unfold.

[ denotes the additive unit circle, rather than the multiplicative (or geometric) unit circle – one can think of it as the geometric unit circle transformed by a logarithmic change of variables. In particular, the identity element 1 of the multiplicative unit circle becomes the identity element 0 of the additive unit circle. -T.]17 May, 2015 at 1:18 pm

James SmithI never made the distinction before. Thanks for clearing it up!

17 May, 2015 at 7:33 pm

victorWhat if the velocities are random, ie, the velocity of the $i$th runner is $v_0 \exp( \mu t + \sigma B^i_t)$ where $B^i_t$ is standard Brownian Motion and is independent across the runners? The deterministic problem is the limiting case where $\mu = 0$ and $\sigma= \infty$. Would it make sense to solve the stochastic problem first?

18 May, 2015 at 9:14 am

Terence TaoThis is an irreducible Markov process (where the state space tracks both position and velocity) and so the stochastic runners will almost surely traverse a dense set of states, which amongst other things establishes the lonely runner property for such a set of runners. The difficulty in this conjecture lies not with such “generic” situations, but rather with very special “arithmetically structured” situations in which the runner velocities all have a rigid relationship with each other (e.g. they are all rational multiples of each other with small height).

18 May, 2015 at 9:22 am

victorThanks for clarifying Prof Tao.

27 October, 2015 at 6:23 am

AntoineHello, with all my respect, Dr. Tao, your point of the optimal and relation to the Dirichlet approximation theorem. well i haven’t seen that, from the stated theorem at wiki, is what precisely in the theorem? thanks.

[The would correspond to the “” in the Wikipedia formulation of the theorem (and the time would correspond to , and to ). -T]</I<20 November, 2015 at 3:37 am

AntoineThanks sorry i didn’t expect a reply!, anyway the $q$ depends on $\alpha $ in the theorem no, so for a fixed $q$ maybe there exist $\alpha$ such that the less or equal is bigger or equal, did i miss something? t.y.

20 November, 2015 at 3:39 am

AntoineNo Tex sorry.

20 November, 2015 at 8:36 am

Terence TaoThe Dirichlet approximation theorem (with the substitutions indicated) tells us that for any real , there exists a between and (depending on ) such that . Setting for , we conclude that Conjecture 2 fails if the quantity is replaced by any smaller quantity.

20 November, 2015 at 8:47 am

AnonymousThank you.

19 December, 2015 at 8:20 am

AntoineThanks

12 January, 2016 at 12:08 pm

ElliottProf. Tao, can’t the bound 1/(n+1) be seen to be optimal without referring to the Dirichlet approximation theorem? For speeds 1,…,n let t = 1/(n+1). Then ||1/(n+1)|| and ||n/(n+1)|| both equal 1/(n+1).

12 January, 2016 at 4:16 pm

Terence TaoThis doesn’t immediately preclude Conjecture 2 holding with 1/(n+1) replaced by a larger bound, because there could be a different value of t (other than t = 1/(n+1)) in which the numbers t, 2t, …, nt are further away from the origin than in your example. (But this scenario is ruled out by the Dirichlet approximation theorem.)

25 April, 2019 at 2:52 am

Stephen Mc GuireDirichlet’s theorem tells us that not that . So we can, at best, conclude that all runners are within of the lonely runner i.e. the runner stationary at 0. Perhaps I’m misunderstanding something?

26 April, 2019 at 1:32 pm

Terence TaoThe Dirichlet approximation theorem does indeed show that for all and all natural numbers . (For some reason the Wikipedia entry only gives the weaker estimate , but the stronger bound is what actually comes out of carefully optimising the proof, and in particular observing that some pair amongst must be at a distance at most from each other.)

20 November, 2015 at 12:14 am

koza3301Could you explain how to prove that “ for a set of reals of (Banach) density ” from the Dirichlet approximation theorem?

20 November, 2015 at 8:35 am

Terence TaoCover the n-torus by boxes of sidelength . By the Dirichlet box principle (pigeonhole principle), one of these boxes will contain for a set of of positive density. Now subtract to get the claim. (This argument is identical to the one used to prove the usual Dirichlet approximation theorem. There are some technical issues regarding the nature of density used here, but by using some equidistribution theory one can show that all the usual notions of density (lower Banach, upper Banach, etc.) agree for the sets in question.)

20 November, 2015 at 8:43 am

koza3301Thanks :)

19 December, 2015 at 8:32 am

AntoineHi, I’ve got another question Prof. Tao if no bother; a result such as: their exist infinite times when say the $k$ runners are all distant from each other by at least $a\dfrac{1}{k}$ with $a$ less than $1,$ can be bigger than $\dfrac{1}{2}$ and it depends on the speed of the $k$ runners, Where it stands on a crowded scale for this problem, thanks all respect.

23 March, 2016 at 9:53 pm

David RWhat do you think of this argument Prof. Tao? Suppose we use the following three lemmas to prove the slowest runner becomes lonely:

Lemma 1: Let for . Then becomes lonely if and only if there exists such that

Proof:: Straightforward in both directions.

Lemma 2: Let and . Then if and only if for all ..

Proof: (). Proof by contradiction.

(). Proof by contrapositive.

Lemma 3: Let for .Then there exists such that

for , .

Proof: Try a proof by induction. Proving the base case follows easily from the fact that the LRC has been proven for . Proving that looks to be quite difficult though.

We can then make the following argument:

Theorem 1: Let . Then becomes lonely.

Proof: By Lemma 3, there exists such that

for all , . Hence, there exists such that

for all , . By setting

and because for we have from Lemma 2,

.

By Lemma 1, becomes lonely for all T in the intersection.

The thing that’s interesting about this argument is that if we can prove Lemma 3, then we can repeat essentially the same argument to prove that runners with some intermediate speed as well as runners with the fastest speed become lonely. The argument for proving that the fastest runner becomes lonely would be almost identical to the argument for the slowest runner.

7 June, 2016 at 5:48 am

Klaus85Dear Terry, I have just found this arxiv-paper (http://arxiv.org/abs/1606.01783) claiming to proof the lonely-runner conjecture. It is the highest-rated paper of today on scirate, thus i thought it might be serios; but i would love to hear more details by an expert. Does it investigate the same question as mentioned here, or is it a weaker variation? Thanks so much!

7 June, 2016 at 4:33 pm

Terence TaoThe paper has been withdrawn due to an error on the last line.

7 June, 2016 at 4:58 pm

anthonyquasCan you say in more detail where the error was?

7 June, 2016 at 5:38 pm

Terence TaoThe final inequality is not true in the case.

7 June, 2016 at 8:50 pm

David Luido you think this is an easily fixable problem, or would it require significant effort to fix, or is the proof outright wrong?

8 June, 2016 at 8:22 am

Terence TaoProbably the author is the best person to answer this. But one test can come from the observation that the lonely runner conjecture is only barely true in the case (as pointed out in above in the blog post) or permutations or affine transformations thereof. As such, any proof of the lonely runner conjecture should also only barely work in this case (in particular, any key inequality in the proof should hold with equality in this case). This did not seem to be the case in this most recent attempt to prove the conjecture, which suggests at minimum that a significant modification of the argument would be required.

7 June, 2016 at 4:31 pm

anthonyquasI have read through the Beck paper once without finding any problems in it. It does indeed address the full Lonely Runner Conjecture as described in this blog post. I am about to re-read.

7 June, 2016 at 8:41 pm

Curious PartyWhat time was the mistake found? A wager was placed on this result being false.

7 June, 2016 at 11:07 pm

Mike MillerI find this to be in extremely poor taste.

8 June, 2016 at 7:42 am

David Luiis it asking when it was found, or the wager part that you find to be in poor taste?

8 June, 2016 at 2:38 pm

Classy MofoPerhaps it is your own taste that is poor?

29 August, 2016 at 2:52 am

giorgiHello terence tao. (1,2…m-1 and i are indexes ) i just proof what conjecture is equivalent to this conjecture (i mean if we proof “”this” then lonely runner conjecture is true too): q1, q2, …qm-1 is any positive numbers , if q1, q2, …qm-1

we can Multiplicate some number k (k>0) and get this type of similar proportion numbers: C1+N1, C2+N2 , …. Cm-1+Nm-1 , were Ci are numbers between 1/m<=Ci<=(m-1)/m and Ni are natural numbers then "lonely runner" is true too. what you think about it?

p.s. im just lover of mathematics and sorry for my english and "latex"

7 September, 2016 at 8:44 am

একেলা দৌড়বিদের কনজেকচার (The Lonely Runner Conjecture) - বিজ্ঞান স্কুল[…] আরও পড়ার জন্য দেখতে পারেন এখানে https://terrytao.wordpress.com/2015/05/13/a-remark-on-the-lonely-runner-conjecture/ […]

10 January, 2017 at 1:40 pm

Some remarks on the lonely runner conjecture | What's new[…] to Contributions to discrete mathematics. I had blogged about the lonely runner conjecture in this previous blog post, and I returned to the problem recently to see if I could obtain anything further. The results […]

22 October, 2018 at 11:30 am

dspThis will probably turn out to be a very stupid question, but anyway: I have been thinking about what happens when you replace the Euclidean distance in the lonely runner conjecture with the p-adic one, i. e., what can one say about the set of vectors of distances , where are fixed rational numbers, ranges over and denotes the p-adic absolute value of ? I’ve come up with something I can’t understand. I guess my (as I said, probably stupid) question boils down to this:

1) For an algebraic number , the product of all the p-adic absolute values of equals .

2) For any ,

So if is a sufficiently large positive integer, since the Euclidean distance between, say, 1 and is smaller than 1, for at least one , we must have that the p-adic distance between the two numbers is by 1). But by 2), this distance is at most , and since it is a root of unity.

I hope this isn’t too far off topic.

24 October, 2018 at 9:40 am

Terence TaoI’m not sure one can make sense p-adically of if is real, since will then in general just be a transcendental complex number. However, there is some work on studying function field analogues of the lonely runner conjecture which you may be interested in: https://arxiv.org/abs/1711.01207