According to the article, the algorithm for deleting edges runs in amortized O(log n) time, which relies on the fact that if you delete an edge which breaks the

According to the article, the algorithm for deleting edges runs in amortized O(log n) time, which relies on the fact that if you delete an edge which breaks the graph into two halves (which is the worst-case scenario), the next time you’ll only have half as many vertices in each component. However, the OP is not actually deleting the edges, just checking if deleting them would disconnect the graph. Therefore, unless I’m missing something, the amortization would not work.