How many shortest distances change when adding an edge to a graph? Let $G=(V,E)$ be some complete, weighted, undirected graph. We construct a second graph $G&#39;=(V, E&#39;)$ by adding edges one by one from $E$ to $E&#39;$. We add $\Theta(|V|)$ edges to $G&#39;$ in total.

Every time we add one edge $(u,v)$ to $E&#39;$, we consider the shortest distances between all pairs in $(V, E&#39;)$ and $(V, E&#39; \cup \{ (u,v) \})$. We count how many of these shortest distances have changed as a consequence of adding $(u,v)$. Let $C_i$ be the number of shortest distances that change when we add the $i$th edge, and let $n$ be the number of edges we add in total.


  How big is $C = \frac{\sum_i C_i}{n}$?


As $C_i = O(|V|^2)=O(n^2)$, $C=O(n^2)$ as well. Can this bound be improved? Note that I define $C$ to be the average over all edges that were added, so a single round in which a lot of distances change is not that interesting, though it proves that $C = \Omega(n)$.

I have an algorithm for computing a geometric t-spanner greedily that works in $O(C n \log n)$ time, so if $C$ is $o(n^2)$, my algorithm is faster than the original greedy algorithm, and if $C$ is really small, potentially faster than the best known algorithm (though I doubt that).

Some problem-specific properties that might help with a good bound: the edge $(u,v)$ that is added always has larger weight than any edge already in the graph (not necessarily strictly larger). Furthermore, its weight is shorter than the shortest path between $u$ and $v$.

You may assume that the vertices correspond to points in a 2d plane and the distances between vertices are the Euclidian distances between these points. That is, every vertex $v$ corresponds to some point $(x,y)$ in the plane, and for an edge $(u,v)=((x_1,y_1),(x_2,y_2))$ its weight is equal to $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2.}$
Consider the following linear chain with $n+1$ nodes, $n$ edges and viciously chosen weights:


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Clearly, the edges could have been added in order of their weights and there are $n \in \mathcal{O}(|V|)$ of them. Adding the dashed edge (which is legal) creates shorter paths for all pairs $(u_i,b_j)$ with $i,j = 1,\dots,k$. As $k \approx \frac{n}{4}$ and assuming that $n \in \Theta(|V|)$, both first and last row contain $\Theta(|V|)$ many nodes each and the addition causes $\Theta(|V|^2)$ many shortest path changes.

We can now move "outwards" now, i.e. add the next edge with weight $n+2$ between $u_{k-1}$ and $b_{k-1}$ and so on; if we continue this to $(u_1,b_1)$, we cause in total $\Theta(|V|^3)$ shortest path changes.

If this does not convince you, note that you can actually start this "process" with $(c_1,c_2)$ and work outwards from there; this way you add $\approx n$ edges which cause in total $\approx \sum_{i=1}^{n}i^2 \in \Theta(n^3) = \Theta(|V|^3)$ many shortest path changes---this is just impossible to draw to fit on one screen.