So I want to now split the proof into an easy case and a tricky case.

To explain why the easy case is easy lets, lets observe a property that this

greedy clustering algorithm has. Now the algorithm's philosophy is that

the squeaky wheel should get the grease. That is, the separated pair of points

that are closest to each other are the ones that should get merged.

So for this reason, because it's always the closest separated pair that get

merged, if you look at the sequence of point pairs that get merged together,

that determine the spacing in each subsequent iteration, the distances

between these sort of worst separated points is only going up over time.

At the beginning of the algorithm, the closest pair of points in the entire

point set are the ones that get directly merged.

Then those are out of the picture, and now that some further away pair of

points are separated, it determines the spacing,

then they get merged. Once they've been coalesced, then there

is still some further away pair of points, which is now the smallest

separated. They get merged, and so on.

So if you look at the sequence of distances between the pairs of points

that are directly merged by the greedy algorithm, that is only going up over

time. And this sequence culminates with the

final spacing S of the greedy algorithm. At some sense, the spacing of the output

of the greedy algorithm is the distance between the point period that would get

merged if we ran the greedy algorithm one more in moderation but unfortunately

we're not allowed to do that. Okay?

So the point is, for every pair of points directly merged by the greedy algorithm,

they're always a distance at most S away from each other.

So the easy case, then, is when this pair of points, pq,

which, on the one hand, lie in a common greedy structure,

but on the other hand, in different clusters with c hats.

If they were, at some point, not merely in the same cluster, but actually

directly merged by the greedy algorithm. If, at some iteration, they determined

the spacing, and were picked by the greedy algorithm to have their, clusters

merged. Then we just argued that the distance

between p and q is no more than the space in capital s of the greedy clustering.

And since p and q lie in different clusters of the c hats.

It's separated by the C hats and therefore they upper bound the spacing of

the C hats. Maybe there's some even closer separated

pair by the C hats. But the very least P and Q are separated

so they upper bound the spacing of the C hat clustering.

So that's what we wanted to prove. We wanted to show that this alternative

spacing didn't have better spacing than our greedy spacing.

It had to be at most as big. It had to be at most capital S.

So in this easy case, when P and Q are directly merged by the greedy algorithm,

we're done. So the tricky case is when P and Q are

only indirectly merged, and you may be wondering at the moment, what does that

mean? How did two people wind up in the same

cluster if they weren't, at some point, directly merged?

So let's draw a picture and see how that can happen.