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S Nov 16 at 11:01 history bounty ended CommunityBot
S Nov 16 at 11:01 history notice removed CommunityBot
Nov 12 at 13:28 comment added badboul You can also list the 2-vertex cuts of the graph and if X contains two nodes that are not in this list than you know that X does not disconnect the graph (as described here: en.wikipedia.org/wiki/SPQR_tree). You can do the same for more than 2 vertices (as described here: en.wikipedia.org/wiki/K-vertex-connected_graph).
Nov 12 at 13:09 comment added Optimizer @badboul - yes, the case where X has just one node to remove is straight forward constant time operation, but as soon as there are more than 1 holes, it becomes tricky and it seems like relying on floodfill after apply X and Y operations is the only way to go.
Nov 11 at 9:43 comment added badboul Since your graph, X, and Y are very small, it is very unlikely that a data structure with better worst case asymptotic runtime will be faster than breadth first search in practice. I have the following suggestion: Compute the k-connected components of your graph and the associated vertex cuts. Then, by counting how many vertices of each type are contained in X you may be able to determine that the graph remains connected. For example, if X contains only one vertex that is not an articulation point of your graph, then the graph remains connected after removing X.
S Nov 8 at 9:56 history bounty started Optimizer
S Nov 8 at 9:56 history notice added Optimizer Draw attention
Nov 7 at 13:36 comment added Optimizer @ChandraChekuri looks like the paper you linked wont be helpful.. It suggests a theorem to do MST calculation in log^4 (N) but log^4 (N) > N for N < 5000 or so.. My graph would never have V+E > 5000 ..
Nov 7 at 12:07 comment added Optimizer @badboul In my practical use case, i am mostly dealing with single digit X and Ys while V+E can reach 100+ but the main reason why I want to optimize over N floodfills is because N can reach to billions.
Nov 7 at 10:57 comment added badboul Even if the size of X and Y is bounded by a constant, the algorithm of thorup et al mentioned above may not be helpful because there, the insertion/deletion of edges has a good amortized time which does not mean a good worst case time.
Nov 7 at 10:37 comment added badboul Do you have bounds on the size of X and Y? Because if not, deleting X and adding Y can create a completely new graph so there is no use in the information that the original graph was connected.
Nov 7 at 4:06 comment added Optimizer @ChandraChekuri Thanks! I will take a look at this one.
Nov 6 at 21:51 comment added Chandra Chekuri See the paper below. It will give the MST cost when you add/delete edges. MST cost is infinity iff G is disconnected. @article{holm2001poly, title={Poly-logarithmic deterministic fully-dynamic algorithms for connectivity, minimum spanning tree, 2-edge, and biconnectivity}, author={Holm, Jacob and De Lichtenberg, Kristian and Thorup, Mikkel}, journal={Journal of the ACM (JACM)}, volume={48}, number={4}, pages={723--760}, year={2001}, publisher={ACM New York, NY, USA} }
Nov 6 at 20:43 comment added Optimizer @ChandraChekuri None of the literature that I've come across solves for what i am asking in the question. If you've seen something similar, mind sharing the paper titles? Most of the dynamic data structures worry about queries around biconnectivity or connectivity between two vertices. I am asking for something much more simpler and probably that can be done in much more efficient manner. or if you know how I can reuse the literature and retrofit them in my use-case, please add an answer.
Nov 6 at 20:35 comment added Chandra Chekuri There is a lot of literature on dynamic connectivity maintenance for edge removals/additions. One can do this poly-log update time. You can simulate vertex addition/deletion by removing edges incident to the vertices. You can search for dynamic data structure for connectivity.
Nov 6 at 20:01 history edited Optimizer
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S Nov 5 at 19:35 review First questions
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S Nov 5 at 19:35 history asked Optimizer CC BY-SA 4.0