Count bridging edges in a family of two component forests

I am given a (simple, undirected, connected) graph $ G = (V, E)$ and a fixed spanning tree $ T$ in this graph. Removing an edge $ e\in E(T)$ from $ T$ splits it into a spanning forest $ F^e$ with two components $ F_1^e$ and $ F_2^e$ . I am interested in the number $ c(e)$ of edges in $ G$ that connect $ F^e$ i.e. with one vertex in $ F_1^e$ and the other in $ F_2^e$ .

I would like to compute the number $ c(e)$ for all $ e$ in $ T$ simultaneously. This can be done naively in $ O(|V||E|)$ . Is there a faster way to achieve this?

The graphs I am working with are small-world graphs. In particular the average distance between nodes in $ G$ can be assumed to be small.

generalizing ball-bin problem to $k-$universal family

I am trying to solve a question in the book on Probability and computing by Michael Mitzenmacher, Eli Upfal. The question asks to generalize ball-bin problem for 2-universal hashing to k-universal hashing. In the standard bin-ball problem setup there are n bins and n balls. It is shown that for 2-universal hash family, max load is $ l_{max}=1+2\sqrt n$ with probability greater than $ \frac{1}{2}$ . It follows with usual application of Chebyshev’s inequality. Now we try to generalize this to $ k$ -universal family and find $ l_{max}$ so that load is less than that with probability $ \frac{1}{2}$ .

I tried to do following: First set up indicator variable as in previous case for 2-universal problem

$ X_{ij}=1$ if ball i goes in bin $ j$ . Now load for bin $ i$ is $ X=\sum_{j=1}^{n}X_{ij}$ .

Since hash is now k-universal we try to go for Markov on $ k^{th}$ moment instead of Chebyshev (which comes from 2-nd moment of Markov) to get probabilty bound of $ \frac{1}{2n}$ . . In the end we apply union bound. But I am unable to simplify expression for $ k$ -th moment in this context to get an answer.

Definition of $ k$ -universal hash function

$ Pr(h({i_{1}})=h({i_{2}})=\cdots=h({i_{k}}))= \frac{1}{2^{k-1}}$ for any $ k$ distinct elements $ i_1,\ldots,i_k$ .

[ Family ] Open Question : Im so sad?

its hard for me to explain how sad i am my current situation just made it even worse. i wanted to go to las vegas to see mariah careys show, i asked my dad and he said yes. he also wanted to go to see the grand canyon. he told me we would stay at caesars palace, the hotel where the show is. i was so excited because theres a lot of other stuff there too. we’re leaving tomorrow and now he suddenly is acting like im crazy for thinking we would stay there saying its too expensive when he’s the one who kept telling me we would. he keeps trying to make me feel bad saying im making him spend so much money to do this and stuff like that when it was entirely his choice to agree to this or not.  it makes me think of all the things he’s done to me and how much i hate him and my mom. i have so many memories of them doing horrible things to me. telling me they hope i die as they drop me off for school. one time my dad did something really bad to me in a hotel room and the police came and he never even apologized, my parents both blamed it on me. im so sad in life. i feel like nothing will ever work out for me and theres something wrong with me. i told my bestfriend about what my dad did to me in the hotel and she thought it was a joke. i have no one. part of me hopes i dont wake up in the morning. i feel stupid for ever thinking this would work out. 

Should I quit an rpg table because my family disapproves?

Every friday night I go to my best friend house, me and my friends play all-night long Vampire the Dark Ages. But lately,my mom has been interested in issues of my friend’s family, trying to use me as a Nosferatu for her. Since this started, I have the idea of ​​”killing” my vampire, Asking the GM that in the next narrative that there is a serious confrontation against the Tremere, my Cappadocian will die in a way acceptable to history, because I feel wrong being forced to discover problems in others just to satisfy my parents’ curiosity and be able to have fun on Fridays. What should I do? Do I remain active in the story and confront my parents or speak to the GM? We are all under age except our GM, so I always avoided confronting my parents. Has anyone ever experienced this? Could you help me with that?

Plot a family of solutions of ODE with singularity

In the post Use Mathematica to plot the flow of an ODE with discontinuity, the following ODE with discontinuous coefficient was solved

T = 1; Y = ParametricNDSolveValue[{X'[t] == Boole[X[t] > 0], X[0] == x}, X, {t, 0, T}, {x}]; Show[  Table[   ParametricPlot[{Y[x][t], t}, {t, 0, T}],   {x, -1, 1, 0.1}   ],  PlotRange -> All,  AxesLabel -> {"x", "t"}  ] 

I also wish to plot an additional family of solutions displayed in green in the picture below.

  • How can a plot just like the one below (but possibly with the t axis being the vertical one) be done starting from the code above?

  • Also, is it possible to have the “first” (x=t) and the “last” (x=0) of the green lines displayed in a different color (for example, blue instead of green)?

enter image description here

For what family of graphs does Dijkstra’s algorithm achieve the run-time upper bound

The question is in the title of the post. I am hoping to get some validation regarding my solution.

After some trial and error, my idea is as follows.

  1. Worst case complexity of $ \mathcal{O}(E\lg_{}{V})$ is achieved when nodes relax all of their neighbors when dequed.
  2. For the above to occur, longer paths to every neighbor of $ v$ must relax their respective edges into the said neighbors before $ v$ is dequed.

The two points above give rise to the following formalism.

For all paths $ p$ , $ q$ between any two nodes $ s$ and $ t$ , $ l(p) > l(q) \implies l(p \setminus \{t\}) < l(q \setminus \{t\})$ where $ l(x)$ is the length of path $ x$ .

enter image description here

For example, in the graph above, longer path edge $ A \rightarrow C$ is relaxed before the edge $ B \rightarrow C$ of the shorter path $ A \rightarrow B \rightarrow C.$ This property is true for every node in the graph.

There is a difference between malware detection using automata and family behavior graph?

I have a question,

There is a difference between dynamic malware detection using automata and family behavior – graph?

I think that they are both relying on API function calls but I don’t understand if there is any major difference between them.

Please help me,

if you’re not sure what I’m talking about:

automata – https://www.researchgate.net/publication/309710040_Detecting_Malicious_Behaviors_of_Software_through_Analysis_of_API_Sequence_k-grams

family behavior – graph – https://drive.google.com/open?id=1dOZ80FcaBiDHRDW4kusdxXGZw2C9aXfK

of course, they are free

first one – just click on Request full-text and it will download the pdf files. the second one is google drive link.

Thank you.

What is the difference between a collection of Turing Machines and a family of Circuits?

Given a Collection of Turing Machines $ T_1, T_2, T_3,…T_n$ where $ T_1$ denotes that the Turing machine can only take in an input of size 1. Is there any difference in computational power to a family of Circuits $ C_1, C_2, C_3,…C_n$ ?

What if we assumed that each Turing machine, encoded special information for that specific input size to make each instance efficient?

If there is no difference in computational power, then maybe we could use this to define non-uniform algorithms instead of circuits?