Is there a scenario where a gnoll flesh gnawer can move at least 45 feet during its Rampage bonus action?

The base movement speed of a gnoll flesh gnawer is 30 feet, and if it activates rampage by bringing a creature to 0 hit points, it can move an additional 15 feet for a total of 45 feet during “normal” combat.

But is there a way that rampage could be triggered for the flesh gnawer after it gains a movement speed of 90 feet from its Sudden Rush action? This should allow the flesh gnawer to traverse 135 feet in one turn.

Ideally this would require only actions/abilities possessed by the various gnoll types, but I am open to other first-party/non-homebrew solutions for Dungeons and Dragons 5e that may make this possible.

Potentially relevant text from some of the gnoll actions are listed below:

Sudden Rush

Until the end of the turn, the gnoll’s speed increases by 60 feet and doesn’t provoke opportunity attacks.

Incite Rampage (Possessed by the Gnoll Pack Lord)

One creature the gnoll can see within 30 feet of it can use its reaction to make a melee attack if it can hear the gnoll and has the Rampage trait.

Rampage

When the gnoll reduces a creature to 0 hit points with a melee attack on its turn, the gnoll can take a bonus action to move up to half its speed and make a bite attack.

How to create a user with the least privileges/permissions but enough to do ssh tunneling?

I want to use SSH Tunneling on my local machines to bypass government restrictions. I’m talking about creating a socks proxy server using a ssh connection.

ssh -f -N -D 1080 admin@server1.example.com 

This works perfectly right now. But I want to pass this to few other people (friends and family members). The thing is right now I’m using an admin user to do this. I thought I should probably just create a non-admin/guest user for each person.

I’m a bit worried if they decide to ssh normally and mess up with the server, or if they lose the login credentials or it gets into a hacker’s hand.

That’s I wanna take it one step further and just limit them to the point where they literally can’t do anything or harmful, but with just enough functionality to run the ssh tunneling.

So what do I do after adduser <username>?

Are there rules (or at least relevant statements) in 5e about conservation of mass with magic?

Since some spells (such as enlarge/reduce) seem to reduce or increase the amount of mass that is present in game-play with no explanation as to how this happens, some interpret that to mean that conservation of mass is “thrown out the window” in 5e.

However, another view is that the RAW is silent on how those spells work, which leaves it up to the DM to either enforce or discard conservation of mass. The detailed mechanics of enlarge, for example, might be that it draws molecular matter from deep within the earth and/or from the environment or atmosphere, or even from another plane, such that conservation of mass holds true. Or then again, it might not. The spell description doesn’t say.

In a couple of answers about 3.5e (such as this one and this one), it has been softly asserted that conservation of mass is not really respected in D&D, and the topic was touched upon in comments to an answer to a 5e question, but the issue has not been definitively put to rest, that I can see.

Is this indeed ambiguous according to RAW, perhaps intentionally (perhaps the rules authors are avoiding over-managing the campaign settings of various tables)? To put it differently: Is the RAW silent on conservation of mass? Or is there a definitive RAW answer to this somewhere that I have not come across?

Regular expression with exactly 2 a’s and at least 2 b’s [duplicate]

This question already has an answer here:

  • How do I find a regular expression for a particular language? 2 answers

Find the regular expression with exactly 2 a’s and at least 2 b’s.

I’m kind of stuck on this one. I know that the shortest strings that I can get are

aabb, abba, bbaa, baab, baba, abab but how can I build a regular expression from this?

Least Significant Bit (LSB) vs Little Endian – Are they equivalent in anyway!

For a multiple choice question: What do we call the LSB? (i)Little Endian (ii)Upper bit (iii)Big Endian (iv)Lower Bit

I feel ideally none of them is a true correct choice, but my best bet was (iv) Lower Bit.

However, our professor claims that the correct choice should be (i)Little Endian. He even said “LSB stands for Least Significant Bit which is right most bit of a binary equivalent and it is Little endian format.

I don’t agree with the last part i.e. “..it is Little endian format.”

My understanding (primarily from the book “Computer Organization and Design” by Patterson and Hennessy,) is that LSB (and MSB) concepts were introduced to clearly identify bits at both extremes while avoiding any confusion arising due to writing styles like vertically or horizontally which can be confusing when read/written from different directions. Further it (LSB) has nothing to do with byte ordering such as Little endianness. For example any number will have both LSB and MSB, and can be stored in a little endian hardware as much as it can be stored in a big-endian hardware.

Pls let me know if I’m misunderstanding the professor’s statement in any way.

Can a character take on additional backgrounds, or at least their benefits?

Can a character take on additional backgrounds, or at least their benefits?

For example, there is a sorcerer character that has the Charlatan background. This sorcerer happens to be Drow and in an adventure he comes across the Bregan D’aerthe and they recruit him. This would essentially make him a faction agent. So would the character then benefit from the Charlatan background as well as any features the Faction Agent background offers, but without the Faction Agent backstory?

I am asking as the DM.

How do I design an efficient and easy layout for a settings page which has at least 200+ options spanning 18+ categories

The web app I am now working on as a pilot project for my learning has a very complex settings page. It has 18-20 main categories and tweaks about 200 configurations within these categories. There are 2 access levels, one is a manager and another is owner. Owner can see all changes, but doesn’t necessarily have to configure all of them. Manager can see only select tweaking options and has to configure things by himself.

Right now I am using a vertical tabbed approach. How can I improve it further?

Here’s the screenshot for reference

https://pasteboard.co/IroZI2x.png

enter image description here

Regular Expression: Writing an expression with at least two characters in length?

A past exam question: (1) Consider the language, L, of strings over the alphabet {x, y} of length at least 2 with the second symbol being x. For example, yx, xxyy, and yxy are members of L while xy, yyy,and yyxx are not in L.

(a) Write a regular expression to describe the language L.

I thought perhaps EXE* would be the correct answer, but then E could be either an empty string or X or Y, and the empty string would make it no longer valid because X wouldn’t be the second symbol. Is there some kind of statement that could (x + y)XE*?

Proof that $G=(V,E)$ is connected, if every node has at least one adjacent edge, $|E|\ge n-1$ and $|V|=n$

Let $ G=(V,E)$ be an undirected graph without self-loops or parallel edges.

Does the statement:
If $ |V|=n, |E|\ge n-1$ and every node has at least one adjacent edge , then $ G$ is connected;
hold?

I’ve proofed it for $ |E|=n-1$ :

Per induction:
Start:
For $ \left|V\right|=1$ the graph is trivially connected.

Induction step:
Let the statement be shown for all graphs $ G=\left(V,E\right)$ where $ \left|V\right|=n-1$ and $ |E| = n-2$ .

Let further $ G=\left(V,E\right)$ with $ \left|V\right|=n$ and $ |E| = n-1$ be given.

We’re now looking for an induced sub graph $ G|_{V^\prime}$ where $ V^\prime\subset V, \left|V^\prime\right|=n-1$ , so that $ G|_{V^\prime}$ has at least $ n-2$ edges.

(Any such sub graph can have at most $ n-2 $ edges, as there’ll always be at least one edge that originally lead to the removed node)

Let’s now assume that every sub graph $ G|_{V^\prime}$ has less than $ n-2$ edges.
Then, the removed node in any sub graph would have at least $ 2$ edges.

Thus, every node must have at least $ 2$ edges, and therefore there’d have to exist at least $ n$ edges in the graph.

Therefore, there’s at least one sub graph $ G|_{V^\prime}$ with $ n-2$ edges, for which our induction assumption holds. And because there is one edge from $ G|_{V^\prime}$ to the erased edge, we get that $ G$ is connected.

Therefore, the induction is proven.

However, if I try to generalize the above proof, the same style leads to an inequality that only holds if $ |E|>|V|$ .

Therefore, if the above proof can be generalized, how would it look? If not, what’s an example where it fails?