Semi streaming algorithm for 2 vertices connectivity

Let $ G=(V,E)$ be an undirected graph. given a pair of vertices $ s,t \in V$ , how can we construct a semi-streaming algorithm which determines is $ s$ and $ v$ are connected? Is there any way to construct such an algorithm which scans the input stream only once?

Note that a semi-streaming algorithm is presented an arbitrary order of the edges of $ G$ as a stream, and the algorithm can only access this input sequentially in the order it is given; it might process the input stream several times. The algorithm has a working memory consisting of $ O(n⋅polylogn)$ .

What class this graph connectivity problem belong to

We are given a Boolean circuit $ C(x_1, \dots, x_n,y_1, \dots, y_n)$ with $ 2n$ inputs and we have to determine if the graph $ G_C = (V_C, E_C)$ (specified below) is connected: $ $ V_C = \{0,1\}^n, \ E_C = \{((a_1, \dots, a_n), (b_1, \dots, b_n)) : C(a_1, \dots, a_n,b_1, \dots,b_n) = 1\}.$ $

To which class this problem belongs? I know that graph connectivity can be done so fast (as fast as to be in P) and using a small amount of space. Does this is different in this particular graph? Is this problem in P, EXP, PSPACE?

Fastest algorithm for connectivity problem

Let $ G = (V,E)$ be any undirected graph. Let $ k$ be some number and $ C = |u \longrightarrow v|$ where $ u \longrightarrow v$ means there is a path from $ u$ to $ v$ . We want to add $ k \subseteq V \times V\ E$ edges into $ G$ such that $ C$ is maximised in the new graph.

Question : What is the fastest algorithm for this problem?

I can solve the above problem in $ n^{O(k)}$ time by brute force method.

Business Data Connectivity Service is not accessible

I am working on sharing service applications across farms ( SP2013 Publishing Farm, SP 2010 Consuming Farm). Services like Search, MMS and Secure Store are working perfectly fine after sharing. However, Business Data Connectivity Service is not working. Following error is thrown when trying to create an External content type from SP 2010 Designer:

Event viewer shows following error: The BDC Service application Connection to: SP 2013 Business Data Connectivity Service is not accessible. The full exception text is: The formatter threw an exception while trying to deserialize the message: There was an error while trying to deserialize parameter The InnerException message was ‘Invalid enum value ‘OData’ cannot be deserialized into type ‘Microsoft.BusinessData.SystemSpecific.ThrottleScope’. Ensure that the necessary enum values are present and are marked with EnumMemberAttribute attribute if the type has DataContractAttribute attribute.’. Please see InnerException for more details.

One of the entries in the ULS logs apart from the above is “No ThrottlingRules found”. So far I have tried following fixes without any success:

  1. Made sure all authentication methods except for Anonymous and Windows are Disabled for security token service in both SP 2010 and SP 2013 farm.
  2. Specified administrator for the service and given full control to it.
  3. Started “Request Management” service in SP 2013 farm. Not sure if it is needed. However, Request Management and Throttling rules seem to have a relationship with each other.
  4. Ran BCS throttle config Power shell commands in SP 2013 which also looks OK.

Hamming connectivity of regular languages

Call a language $ L$ Hamming connected iff, for every pair of strings $ x, y \in L^2$ , where $ |x|=|y|$ , $ x$ may be transformed into $ y$ by a sequence of single symbol in-place replacements, so that after every replacement, the resulting string is in $ L$ .

Is Hamming connectivity a decidable property for regular languages? Note that the brute-force algorithm of checking all string lengths fails, as $ L$ may contain infinitely many strings.


Define $ L$ on the binary alphabet to be the set of strings that do not contain $ 101$ or $ 010$ as a substring. $ L$ is regular. We now show that $ L$ is Hamming connected.

Informal Proof: We can grow internal “runs” out to the ends to transform any string in $ L$ into the all 1’s or all 0’s string without creating a forbidden substring, e.g., 001100 -> 011100 -> 111100 -> 111110 -> 111111. This transformation can also be done in reverse to reach any string. Thus, $ L$ is Hamming connected.

Help! zlib-1.2.9 – Killed AppArmor and all network connectivity, cannot fix

So I was installing Flexihub ( and I had an error noting that the package “zlib-1.2.9” was unavailable, so I tracked it down from proper sources, installed it, and after a reboot, lost all network connectivity.

When I try to connect to a WiFi network, it will start to try and connect, then drop; the same will happen with any wired LAN connection. Most of my applications start fine, but I suddenly could not run any of my KVM virtual machines, I get the error “Error starting domain: internal error: cannot load AppArmor profile” and then it lists the profile name.

I’ve tried to uninstall zlib-1.2.9 but it cannot find the package. Any ideas on what to do? I cannot download any fixes to the system thanks to no internet connections D:

It is Ubuntu 19.04 with KDE on a laptop, any ideas on what I can do? I would strongly prefer against reinstalling the OS as it took me ages to set up my current KDE Rice, days even.

Wired Network Connectivity Issue

I am facing some strange issue with my wired internet connection. Whenever the Default Route is : and DNS is : I am not able to access the internet. But if I keep on connecting the same network again and again the Default Gateway changes to something else like : and DNS too changes and the internet starts working. Can someone please help me with what is happening here? If any info is needed please comment I will add the info in question. I am on Ubuntu 18.04.3.

Network density vs connectivity

I was reading this paper where they mention about undirected networks:

“The total connectivity of a network is defined as $ C=\frac{E}{N(N-1)}$ where E is the number of edges and N the total number of nodes.”

However, this is just another way of talking about network density is it not? You could just change the denominator and refer to network density for directed and undirected networks.

Or is this “total connectivity” anything different than density?

Verifying connectivity of a graph in O(n^2)

I solved the following problem:

We have vertices which represents cities and a textfile with edges between vertices representing roads. How many roads do we need to build to make the graph connected – you can travel by road to each city?

Using the BFS algorithm. Basically, I’m calling the BFS algorithm after adding each edge to the graph until the graph is connected and counting the number of BFS calls. However, since the BFS itself has a time complexity of O(n^2) and I’m calling it n-times, I have a worstcase time complexity near O(n^3).

How would I solve this issue in O(n^2)?