## Lonely Runner Conjecture. General case.

We consider (LRC) .

Conjecture Suppose $n>1$ runners having distinct constant speeds $v_k > 0$ start at a common point (origin) and run laps on a circular track with circumference 1. Then, there is a time when no runner is closer than $\frac{1}{n+1}$ from the origin.

W.l.o.g. we can assume $v_k > v_m \forall k>m$.

One can formulate LRC as follows. Suppose $l_k \in \mathbb{N}_0$ is the number of whole laps (including 0) runner k passed on a track, than $\exists t \in \mathbb{R}_+$ and $\exists l_k | l_k+\frac{ 1 }{n+1} \leq v_k t \leq l_k+ \frac{n}{n+1}$.

Case n=1, two runners is trivial. At time $t=\frac{1}{2 v_1}$ runner 1 is exactly distance $\frac{1}{2}$ ($l_1= 0$).

Case n=2, three runners is a special case.

$\left\{ \begin{array}{ll} \frac{l_1+1/3}{v_1} \leq t, & t \leq \frac{l_1+2/3}{v_1} \\ \frac{l_2+1/3}{v_2} \leq t, & t \leq \frac{l_2+2/3}{v_2} \end{array} \right.$

Since $t \in \mathbb{R}_+$ we can eliminate it if all combination of left parts on the left columns are smaller than any right parts in the right column of the table. e.g.
$\left\{ \begin{array}{l} \frac{l_1+1/3}{v_1} \leq \frac{l_1+2/3}{v_1} \\ \frac{l_1+1/3}{v_1} \leq \frac{l_2+2/3}{v_2} \\ \frac{l_2+1/3}{v_2} \leq \frac{l_1+2/3}{v_1}\\ \frac{l_2+1/3}{v_2} \leq \frac{l_2+2/3}{v_2} \end{array} \right.$

The first and the last inequality are trivially correct. From the second and third inequality we would like to express $l_2$
$\frac{v_2}{v_1} \left( l_1 + \frac{1}{3} \right) -\frac{2}{3} \leq l_2 \leq \frac{v_2}{v_1} \left( l_1 + \frac{2}{3} \right) -\frac{1}{3}.$
In other words,
$\frac{v_2}{v_1} \left( l_1 + \frac{1}{3} \right) -\frac{2}{3} \leq l_2 \leq \frac{v_2}{v_1} \left( l_1 + \frac{1}{3} \right) -\frac{2}{3} + \frac{1}{3}\frac{v_2}{v_1} +\frac{1}{3}.$

There are 2 sub-cases.

1. Sub-case $\frac{v_2}{v_1} \geq 2.$
$\frac{1}{3}\frac{v_2}{v_1} +\frac{1}{3} \geq 1$ and $\exists l_2 \in \mathbb{N}_0$ satisfying inequality
2. Sub-case $\frac{v_2}{v_1} < 2 .$
In this case $l_1=l_2=0$ lead to (remeber that $\frac{v_2}{v_1} > 1$ )
$\frac{1}{3} \frac{ v_2}{v_1} - \frac{2}{3} \leq 0 \leq \frac{2}{3} \frac{v_2}{v_1} - \frac{1}{3}.$

Case n=3, four runners is an illustration for a general case.

$\left\{ \begin{array}{ll} \frac{l_1+1/4}{v_1} \leq t, & t \leq \frac{l_1+3/4}{v_1} \\ \frac{l_2+1/4}{v_2} \leq t, & t \leq \frac{l_2+3/4}{v_2} \\ \frac{l_3+1/3}{v_3} \leq t, & t \leq \frac{l_3+3/4}{v_3} \end{array} \right.$

Let’s express $l_2$ in terms of $l_1$
$\frac{v_2}{v_1} \left( l_1 + \frac{1}{4} \right) -\frac{3}{4} \leq l_2 \leq \frac{v_2}{v_1} \left( l_1 + \frac{3}{4} \right) -\frac{1}{4}.$
Rewriting it we obtain
$\frac{v_2}{v_1} \left( l_1 + \frac{1}{4} \right) -\frac{3}{4} \leq l_2 \leq \frac{v_2}{v_1} \left( l_1 + \frac{1}{4} \right) -\frac{3}{4} + \frac{1}{2}\frac{v_2}{v_1} +\frac{1}{2}.$
$\frac{1}{2}\frac{v_2}{v_1} +\frac{1}{2} >1,$ since $\frac{v_2}{v_1} > 1.$ In other words, $\forall l_1 \exists l_2 \in \mathbb{N}_0$ satisfying inequality.

Now, let express $l_3$ in terms of $l_1$ and $l_2$
$\left\{ \begin{array}{l} \frac{v_3}{v_1} \left( l_1 + \frac{1}{4} \right) -\frac{3}{4} \leq l_3 \leq \frac{v_3}{v_1} \left( l_1 + \frac{3}{4} \right) -\frac{1}{4} \\ \frac{v_3}{v_2} \left( l_2 + \frac{1}{4} \right) -\frac{3}{4} \leq l_3 \leq \frac{v_3}{v_2} \left( l_2 + \frac{3}{4} \right) -\frac{1}{4}. \end{array}\right.$

We can express $l_2$ with inequalities obtained earlier
$\left\{ \begin{array}{l} \frac{v_3}{v_1} \left( l_1 + \frac{1}{4} \right) -\frac{3}{4} \leq l_3 \leq \frac{v_3}{v_1} \left( l_1 + \frac{3}{4} \right) -\frac{1}{4} \\ \frac{v_3}{v_2} \left( \frac{v_2}{v_1} \left( l_1 + \frac{1}{4} \right) -\frac{3}{4} + \frac{1}{4} \right) -\frac{3}{4} \leq l_3 \leq \frac{v_3}{v_2} \left( \frac{v_2}{v_1} \left( l_1 + \frac{3}{4} \right) -\frac{1}{4} + \frac{3}{4} \right) -\frac{1}{4}. \end{array}\right.$
Collecting terms we obtain
$\left\{ \begin{array}{l} \frac{v_3}{v_1} \left( l_1 + \frac{1}{4} \right) -\frac{3}{4} \leq l_3 \leq \frac{v_3}{v_1} \left( l_1 + \frac{3}{4} \right) -\frac{1}{4} \\ \frac{v_3}{v_1} \left( l_1 + \frac{1}{4} \right) -\frac{3}{4} -\frac{1}{2} \frac{v_3}{v_2} \leq l_3 \leq \frac{v_3}{v_1} \left( l_1 + \frac{3}{4} \right)-\frac{1}{4}+ \frac{1}{2} \frac{v_3}{v_2}. \end{array}\right.$

We can see that first inequality is always stronger that the second one (meaning that if $l_3$ satisfies first inequalities it will satisfy second inequalities). But the first inequalities are the same as previous inequalities for $l_2$ with relabelling. Therefore, $\forall l_1 \exists l_2, l_3 \in \mathbb{N}_0$ satisfying initial inequalities and LRC holds.

General case

We start with $2 n$ inequalities

$\left\{ \begin{array}{lll} \frac{l_k+\frac{1}{n+1}}{v_k} \leq t, & t \leq \frac{l_k+\frac{n}{n+1}}{v_k} & k=1..n \end{array} \right.$

Now we expressing $l_{k+m_1}, m_1 \geq 1$ in terms of $l_k$ and $l_{k+m_2}, m_2 > m_1$ in terms of $l_{k+m_1}$ and $l_{k}.$
$\frac{v_{k+m_1}}{v_k} \left( l_k + \frac{1}{n+1} \right) -\frac{n}{n+1} \leq l_{k+m_1} \leq \frac{v_{k+m_1}}{v_k} \left( l_k + \frac{n}{n+1} \right) -\frac{1}{n+1}.$
$\frac{v_{k+m_2}}{v_{k+m_1}} \left( l_{k+m_1} + \frac{1}{n+1} \right) -\frac{n}{n+1} \leq l_{k+m_2} \leq \frac{v_{k+m_2}}{v_{k+m_1}} \left( l_{k+m_1} + \frac{n}{n+1} \right) -\frac{1}{n+1}.$
$\frac{v_{k+m_2}}{v_k} \left( l_k + \frac{1}{n+1} \right) -\frac{n}{n+1} \leq l_{k+m_2} \leq \frac{v_{k+m_2}}{v_k} \left( l_k + \frac{n}{n+1} \right) -\frac{1}{n+1}.$

Now we can substitute first inequalities into the second ones.
$\frac{v_{k+m_2}}{v_{k+m_1}} \left( \frac{v_{k+m_1}}{v_k} \left( l_k + \frac{1}{n+1} \right) -\frac{n}{n+1} + \frac{1}{n+1} \right) -\frac{n}{n+1} \leq l_{k+m_2} \leq \frac{v_{k+m_2}}{v_{k+m_1}} \left(\frac{v_{k+m_1}}{v_k} \left( l_k + \frac{n}{n+1} \right) -\frac{1}{n+1} + \frac{n}{n+1} \right) -\frac{1}{n+1}.$
Rearranging terms we obtain
$\frac{v_{k+m_2}}{v_k} \left( l_k + \frac{1}{n+1} \right) -\frac{n}{n+1} - \frac{n-1}{n+1} \frac{v_{k+m_2}}{v_{k+m_1}} \leq l_{k+m_2} \leq \frac{v_{k+m_2}}{v_k} \left( l_k + \frac{n}{n+1} \right) -\frac{1}{n+1} +\frac{n-1}{n+1} \frac{v_{k+m_2}}{v_{k+m_1}}.$
Compare it to
$\frac{v_{k+m_2}}{v_k} \left( l_k + \frac{1}{n+1} \right) -\frac{n}{n+1} \leq l_{k+m_2} \leq \frac{v_{k+m_2}}{v_k} \left( l_k + \frac{n}{n+1} \right) -\frac{1}{n+1}.$
The second inequality is always stronger.

Therefore, we left with the set of inequalities (when they are satisfied LRC holds when LRC holds they are satisfied!!! )
$\frac{v_{k}}{v_1} \left( l_1 + \frac{1}{n+1} \right) -\frac{n}{n+1} \leq l_{k} \leq \frac{v_{k}}{v_1} \left( l_1 + \frac{n}{n+1} \right) -\frac{1}{n+1},\ \forall k=2..n.$

Lets rewrite it again
$\frac{v_{k}}{v_1} \left( l_1 + \frac{1}{n+1} \right) -\frac{n}{n+1} \leq l_{k} \leq \frac{v_{k}}{v_1} \left( l_1 + \frac{1}{n+1} \right) -\frac{n}{n+1} + \frac{n-1}{n+1} \left( \frac{v_k}{v_1} +1 \right)$

$\frac{n-1}{n+1} \left( \frac{v_k}{v_1} +1 \right) > 2 \frac{n-1}{n+1} = 1+ \frac{n-3}{n+1} \geq 1$ if $n \geq 3 .$

It is not clear from this why $\frac{1}{n+1}$ is tight distance.

11.2.2016
It is clear why it is not clear. The above just tell that with speed ratio more the one we can always find a lap for faster runner. The tightness comes from from ratios less than one. Most probably from the interactions or the ratios coming from the same pairs.