Lonely Runner Redux

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The other day I attempted to prove under rather lax conditions that

There exists a time t s.t. all the moving runners are exactly at a distance 1/2 from the still runner.

Turns out, I can't (and I think my claim is provably false). The trouble is with any runner who's speed is a factor of 2. But I think I can show that the strategy works absent such a runner. So let me update the claim.

If the runners' speeds are odd relative primes then there exists a time t s.t. all the moving runners are exactly at a distance 1/2 from the still runner.

August 31, 2019

(The words "relative primes" in the claim above were thus crossed-out.)

Proof. At time t = 1/2 any runner with odd speed is directly across the starting point on the circle of unit circumference.

February 26, 2020

Note to self: The strategy mentioned at the bottom of this post might still work for cases with an even-speed runner in the mix. So that I don't forget (and maybe encourage myself to give it another go when I find the time), here's how I think you can approach it:

We might be able to approximate an even-speed runner as an odd-speed runner on a different time scale. For example, if we multiply all the runners' speeds by 99 (an odd number), then the timescale becomes 1/99th of before, and the odd-speed runners remain odd. To make the even-speed runners odd we add one to the result of multiplying their speeds by by 99.

(You can skip the rest below.)

In fact, let us describe an algorithm for calculating this time. As before, we consider time in discrete steps. Consider pmax, the maximum runner speed in the set {p1, p2, .., pn}.

Scale the track to be of length 2pmax. while keeping the runners' speeds constant. (You can think of this as our changing the units of time.) Consider the track to consist of 2pmax bins of length 1.

As before, label the bins from 0 to 2pmax -1. The position (bin) bi of any runner with speed pi at time t is

bi = t pi  mod 2pmax

Observe at time t = 1, and every odd time thereafter, the fastest runner (with speed pmax) is directly across the starting point (at bin pmax).

Similarly, for each runner i, find the smallest positive solution to

 pmax = ti pi  mod 2pmax

  
Because the pi's are odd coprimes there always exist solutions for ti . And these ti 's are always odd. Note as with the fastest runner, any odd multiple of ti is also a solution.

The time t1/2 when all runners are directly across the starting point then is

t1/2 = ∏ ti             ; ti ≠ ti-1, ti ≠ ti-2, ..

where the product is taken over distinct ti's. And after renormalizing our track length to 1

T1/2 = (1/ 2pmax) ∏ ti

Examples:

1. {3, 5, 7, 11} gives T1/2 = 11/22 =  1/2

(Here, the distinct ti's are 1 and 11.)


2. {7, 11, 13, 15} gives T1/2 = (13 . 15) / (2 . 15) =  13/2


~

So the runner with speed 2 is special.

Now one thing to note is that if this argument is correct, then we can arrive at approximate solutions for the case where there is an even speed runner, by scaling the speeds to a high odd factor and then tweaking these values to be odd coprimes.

Addenda:

August 15, 2019

The claim can be relaxed so that the only requirement is that the runners' speeds be odd. This is because adding a runner to the race whose speed divides the speed of any the original runners does not change the method of calculating T1/2 . So the updated claim is

If the runners' speeds are odd relative primes then there exists a time t s.t. all the moving runners are exactly at a distance 1/2 from the still runner.

August 29, 2019

Not so sure about this last claim that they just be odd. The argument hinges on the fastest runner's speed pmax not being divisible by any of the other runners' speeds the pi 's.

PS I risk making a fool of myself posting thoughts on line. (I've decided it doesn't matter.) Anyway, I must be getting this whole problem wrong, because for any runner with odd speed at time t=1/2  they're directly across the starting point on the circle of unit circumference. So all this calculation, much a do about nothing.