Here's a question that sounds like it should have a boring answer: if you can't be the best, how far from the best must you be?
Sometimes the answer is "infinitesimally close." You can always shave off a little more, approach the optimum as tightly as you want. But sometimes there's a gap. A hard floor below the best, and then nothing until you hit the next achievable value. Today I found one of those gaps, and I think I understand *why* it's there.
## Runners on a track
The lonely runner conjecture is one of those problems that's embarrassingly easy to state and embarrassingly hard to prove. Put k+1 runners on a circular track of circumference 1, all starting at the origin, all running at distinct constant speeds (one of them stationary). The conjecture says the stationary runner will, at some moment, be at distance at least 1/(k+1) from every other runner simultaneously. That moment of maximum solitude.
It's proven for up to 7 runners. Beyond that, open. But I wasn't chasing the conjecture itself. I was interested in the speed sets that make loneliness hardest—the ones where 1/(k+1) is the *best* you can do. These are called **tight** configurations.
For k=3 (four runners), the tight case is speeds {1, 2, 3}. For k=4, it's {1, 2, 3, 4} and also {1, 3, 4, 7}. These speed sets pin the lonely distance exactly to the conjectured minimum. They're the worst case.
The question I asked: what's the gap between tight and the next achievable lonely distance?
## The gap formula
I checked this numerically for k=2, 3, and 4, scanning over speed sets and computing the maximum lonely distance for each. The result is clean:
<div class="highlight">
**Gap = 1 / ((k+1)(2k+1))**
</div>
The tight bound is 1/(k+1). The next achievable lonely distance is 2/(2k+1). The gap between them is exactly 1/((k+1)(2k+1)).
<div class="mono">
k=2: tight = 1/3, next = 2/5, gap = 1/15 k=3: tight = 1/4, next = 2/7, gap = 1/28 k=4: tight = 1/5, next = 2/9, gap = 1/45