So re-computing the ranks in the fused together tree, we see that again the rank

at the route at s1 is 2., same thing as it was before it didn't go up.

So, what's the general principle here?

Well, in general, suppose you have these two trees and

suppose the first ones route s1 has a bigger rank, say rank r.

What does that mean?

So, that means that there exists a path from a leaf to the root

s1 that requires a full r hops.

And in the second tree by virtue of its rank being less than r most r minus 1.

It says that every single path from a leaf to the root s2 of that

tree uses only r minus 1 hops.

So when you install s2 as a child under s1 the length of every leaf to root

path now the root is S1, it's only gone up by 1.

You just have to get to s2 and then 1 more hop you make it all the way to s1.

So if all those paths took almost r minus 1 before,

all the paths all the way to s minus 1 to s1 now, taken most are hops now.

So the worst case path link has not gone up.

That's exactly what happens in this example so it's true that when we install

s2 under s1 we have yet another path that requires two hops that goes through s2.

But we had one previously so the rank doesn't go up.

The worst case path length is exactly the same as before.