8

It is very easy to say something about the expected length before you get stuck in a loop: if there are $n$ videos, it will (starting from a random video) take in expectation $\Theta(\sqrt{n})$ videos before you loop around (the actual value is around $1.25\sqrt{n}$). This is effectively the birthday problem, since each time you draw a video at random. ...


4

This is hierarchical clustering method. As you remove edges, at some point the graph will become disconnected: the connected components are your top-level communities. As you keep removing edges, the top-level communities become disconnected too, and you can think of the new smaller connected components as sub-communities. So you can think of the algorithm ...


4

You're (basically) computing the square of the graph, in which two vertices are adjacent if there is a path of length 2 (or at most 2, depending on the definition) connecting them. The square will contain at least two connected components, corresponding to the two bipartitions.


4

this may be a bit unexpected but yes, this has been studied in at least one particular context: PRNGs. a PRNG can be visualized as a directed graph, specifically a functional graph (all vertices, single outdegree) of "current value, next value". however most PRNGs are designed to have a single very long cycle. there is some analysis of PRNGs with multiple ...


3

Suppose we want to quantify the extent to which $v$ is between $s$ and $t$. There could be a few ways. One way to describe that extent is the probability of passing through $v$ if we want to reach from $s$ to $t$ by a randomly-selected shortest path. Assuming each shortest path is selected with equal probability, we will get $\frac{\sigma_{st}(v)}{\sigma_{...


3

However it doesn't seem to me that the formula calculates what is defined. The formula is right. The betweenness centrality is a value in an interval $[0, \ldots, 1]$. Thus, if the betweenness centrality of node $v$ is equal to $1$, then all shortest paths between two nodes of this graph pass through $v$. I will explain the correctness of this summation ...


3

This is NP-hard by reduction from Set Cover. The problem remains NP-hard even if edge weights are restricted to $\{1, 3\}$. In the Set Cover problem, we are given a ground set $U$ containing $m$ elements $x_1, \dots, x_m$, a collection $\mathbf S$ of $n$ subsets $S_1, \dots, S_n$ of $U$, and an integer $k$, and our task is to determine whether there exists ...


2

There's an efficient streaming algorithm for computing the average degree. Note that if you have the sum of the degrees and the number of vertices, you can compute the average -- so we'll try to keep track of those two values. Also note that if you delete or insert an edge, it is easy to update the sum of the degrees. If you delete or insert a vertex, it ...


2

Your analysis so far is correct. Let's consider the probability that it will take $k$ attempts to transmit a package. This means that there must be $k-1$ failures followed by a final success. There are two ways a transmission can fail: Transmit succeeds, ACK fails. Call this a type-1 failure. Its probability will be $(1-p)q$. Transmission fails (so ACK is ...


2

Sure, of course. You can define a matrix to contain whatever numbers you want it to contain. There's nothing that prevents you. The real question is whether the result has the properties you want it to have, but since you haven't listed any properties, there's nothing to answer here.


1

You're asking how to compute the shortest path between two vertices in a graph. Solution: use an algorithm for computing shortest paths. In your case, BFS would be a good choice. There's no need for A*.


1

I don't know if there is a standard method to identify which activity matches which arrow, but I was able to complete your PERT chart by examining a few possible cases, while filling the chart from left to right. Each case was accepted or rejected after a while. I found that the higher arrow matches B and the lower C. Note that if you try to match C with the ...


1

Certainly it is possible. For example, in the following study the Indian railway network was analyzed. Small-world properties of the Indian railway network. Parongama Sen, Subinay Dasgupta, Arnab Chatterjee, P. A. Sreeram, G. Mukherjee, and S. S. Manna. Phys. Rev. E 67, 036106 – 2003 In another study, the Chinese railway network was considered. W. Li, ...


1

Assuming for the moment that the two types of person are distinct, your graph is (directed) bipartite, so it makes more sense to store it as a matrix whose rows correspond to people with fishing rights, and whose columns correspond to fishermen. If a person can function in both roles, you can think of their two personas as distinct, i.e., have a row and a ...


1

Sure. Yes, you can capture the packets and view them. That'll work. It's a useful, informative exercise. You can use Wireshark (or some other packet capture tool) to capture the packets involved in the TLS handshake and view the captured packets. It'll show you the raw bytes, and also decode the handshake messages for you. Wireshark has a nice GUI ...


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