How can you tell the distances by road between the settlements of Ten Towns in Icewind Dale?

I’ve read through Chapter 1 of Icewind Dale: Rime of the Frostmaiden and have not found any guidance on how to judge distances and travel times between the settlements of Ten Towns.

Other than a good old fashion ruler compared to the map scale, is there a quick way to determine the distances between the settlements of Ten Towns?

What benefit (if any) do horses or carriages offer when travelling long distances?

The rules for Travel Pace in the PHB and Basic Rules are good and straightforward, with players able to travel on foot for 8 hours per day without over-exerting themselves, choosing a pace of Slow (2 miles/hour, 18 miles/day; can move stealthily), Normal (3 m/h, 24 m/d) or Fast (4 m/h, 30 m/d, take a penalty to pass Perception).

In this same section, there are two paragraphs about mounts and vehicles; they specify that mounts can gallop at twice the normal fast pace for an hour, and with frequently freshened mounts – typically only available in highly populated areas – a rider can travel fast over long distances. However when it comes to vehicles, only water going vessels or flying mounts get you travelling faster; for commonly available forms of transport:

Characters in wagons, carriages, or other land vehicles choose a pace as normal.

Is this right? Rules as written it seems there is no benefit to travelling by wagon or carriage, and certainly it takes the same amount of time as walking. In my current game of Dragon of Icespire Peak, the players have spent quite a bit of time demanding horses or a carriage in order to travel the 65 miles from Phandalin to Butterskull Ranch, believing it would get them there faster (as the quest there seems urgent). But looking up the travelling rules it really seems there’s no speed benefit to them paying for a ride or hiring horses, except that they might shave an hour off their travel time each day in the latter case. That makes sense for a wagon which is designed only for hauling stuff, not speed, but what about a carriage or a riding horses?

I’d like to know if there are any additional benefits (beyond the purely narrative) that I am missing for travel by carriage, or any ways that mounts or vehicles can decrease their travelling time (beyond the double fast pace for one hour).

I’m also interested in any house rules you may have used to give players a speed boost through riding over walking.

Can two wizards use Leomund’s secret chest to transport items between vast distances? [duplicate]

Related to this question. Assuming that two wizards cast Leomund’s Secret Chest on the same chest and then go in different directions could they use this chest to exchange items over vast distances? To clarify a bit.

1- Wizard A and Wizard B cast Leomund’s secret chest on the same chest.

  1. Wizard A leaves to venture into a dungeon while Wizard B stays at the city.
  2. Wizard A finds a huge hoard of gold but cannot carry it. Using
    sending to reach Wizard B he asks Wizard B to summon their chest
    every five minutes and dump it’s contents.
  3. Wizard A summons the chest and fills the empty chest with gold. Sends it to Ethereal Plane
  4. Wizard B summons the chest and empties the content before sending it to Ethereal Plane Rinse and repeat till the entire hoard is taken
    back to the city within the span of an hour.

CLRS closest-pair $L_m$ distances

I am studying algorithms and datastructures, and in CLRS chapter 33.4, the exercise 33.4-4 states the following:

We can define the distance between two points in ways other than euclidean. In the plan, the $ L_m$ distance between points $ p_1$ and $ p_2$ are given by the expression $ (|x_1-x_2|^m+|y_1-y_2|^m)^{1/m}$ . euclidean distance, there, is $ L_2$ -distance. Modify the closest pair algorithm to use the $ L_1$ -distance, which is also known as the Manhattan distance.

I understand its an easy modification, however when I’ve been comparing my intuition with that of solutions provided online (e.g. https://sites.math.rutgers.edu/~ajl213/CLRS/CLRS.html), all I can find is that they all point to the number of points in the $ \delta$ x $ 2\delta$ is increased to 10 (from 8), adding one point to the middle of both squares formed by the rectangle (here marked with red below):

enter image description here

Can anyone explain to my why this is the case? Intuitively I cannot see why it should be the case that the number of points in the rectange should increase, when switching to another $ L_m$ -distance?

The Manhatten distance $ \geq$ Euclidean distance from what I can see in all instances, but why should this infer, an increase in points found in the $ \delta$ x $ 2\delta$ space.

What am I missing?

Usual distances on DFAs (Deterministic Finite Automata)?

I’ve been searching in the literature for examples of distances defined on the set of the DFAs (or on the set of minimal DFAs) that are defined on a given alphabet sigma.

Since the languages they describe (regular languages) can potentially have an infinite size, defining a distance is not a trivial matter.

Nevertheless, having a distance on these objects can be useful, in order to fit these in metric spaces, which allows for a range of things (in my case to assess the performance of an algorithm).

My only consistent idea so far is to create a distance similar to the edit-distance in labeled graphs on the minimized DFAs.

Does someone have ever heard of other distances ?

In Descent into Avernus, how are distances between locations determined?

I’m currently running Descent Into Avernus and I was wondering:

How are distances between locations determined?

I know the rules say that a normal travel is considered a forced march and the Con saving throw that must be made is 10 + hours spent traveling to overcome a fatigue… but I don’t see anywhere in the book how long it takes to get from one place to another or the calculation it takes to determine it. Is this something that I’m just making up?

How to find the distances between two adjacent maxima in a solution from NDSolve

I try to find a general approach to plot the time evolution of horizontal distances from maximum to maximum in a solution of PDE. The solution u[x,t] normally have multiple maximum and minimum in space x, which move in space x and evolve in time t.

Here is a simple example, in which the maxima and minima are periodic. But in my real problem they are not periodic and the distances between different pairs of adjacent max are different at a given t, also the distances between two adjacent max can change with t.

sol = NDSolve[{D[u[x, t], t] + u[x, t] D[u[x, t], x] + D[u[x, t], x, x] +  0.4*D[u[x, t], {x, 3}] + D[u[x, t], {x, 4}] == 0, u[-4 \[Pi], t] == u[4 \[Pi], t], u[x, 0] == 0.1*Sin[x]}, u, {t, 0, 20}, {x, -4 \[Pi], 4 \[Pi]}]  Plot3D[Evaluate[u[x, t] /. First[sol]], {t, 0, 10}, {x, -4 Pi, 4 Pi}, PlotRange -> All, PlotPoints -> 100] 

enter image description here

I have tried to use Table[FindMaximum[Evaluate[u[x, t] /. First[sol]], {x, x0}][[2, 1, 2]], {t,0,tend,0.01}] with an initial position x0 to find a local maximum. But I don’t know how to find two adjacent maxima simultaneously in order to plot the time evolution of their distance.

inequality of distances in a graph

Certainly it’s obvious but I can’t catch the reason behind it.

Why do we have :

Let $ D= (V,A)$ be a directed graph, $ w:A \to \mathbb R$ be arc weights and $ s \in V$ . Denote with $ d(s,v)$ the length of the shortest path from $ s$ to $ v$ in $ D$ , subject to $ w$ .

If there are no negative cycles in $ D$ , then we have $ $ \forall (u,v) \in A : d(s,v) \leq d(s,u) + w(u,v) \iff d(s,v)- d(s,u) \leq w(u,v).$ $

Why?

How can I send long messages over long distances?

For communicating short messages (of 25 words or less) over long distances, we have the Sending spell. However, if I wanted to communicate, say, the contents of a 1-page letter (i.e. long enough that simply casting Sending multiple times isn’t good enough) to someone far away (on the same plane, probably on the same continent, but many weeks away by land travel), is there a spell or other magic that can accomplish this? A solution might involve physically teleporting the letter itself, or simply conveying the message via some other means. Although sending the letter as a physical object would probably be preferable, unless you’re expecting the recipient to be constantly prepared to take dictation at any time.

It can be assumed that the intended recipient is living in a city and not traveling around much, so for example a spell that sent the message to their place of residence rather than directly to them would be fine, as long as the message isn’t so ephemeral that it appears and then disappears before they get home from their grocery shopping. It would also be acceptable if some reasonable amount of setup was required to create a “receiving station”. I, on the other hand, am a busy adventurer constantly on the move, so an elaborate or non-portable setup on the sending end is probably not viable.

One obvious method of accomplishing this task is to use a 7th level spell slot to teleport to the recipient and deliver the message in person. So that puts a sort of upper limit on what should be required: a better method should be more efficient than teleporting there yourself.

Since this is pretty open-ended, you can impose reasonable additional constraints in your answer, such as requiring the recipient to also be a spellcaster. Just be clear about what your solution requires.

Relationship between distances on homogeneous spaces and their Lie groups

Consider the (round) sphere $ M=\mathbb{S}^{n-1}$ as a homogeneous $ O(n)$ -space. Then for $ x,y\in\mathbb{S}^{n-1}$ there is $ g\in O(n)$ such that $ y=g\cdot x$ . Denote the Riemannian distance on $ \mathbb{S}^{n-1}$ by $ d_{\mathbb{S}^{n-1}}$ . Intuitively, if $ y$ and $ x$ are not far apart then $ g$ should be almost the identity (because the $ O(n)$ -action is smooth). I am able to explicitly construct such a rotation $ g$ so that \begin{align} \|g-\operatorname{Id_{\mathbb{R}^n}}\| \leq d_{\mathbb{S}^{n-1}}(y,x) \end{align} where $ \|\cdot\|$ is the operator norm for matrices.

Analoguously, if $ M=\mathrm{Gr}_m(\mathbb{R}^n)$ is the Grassmannian, then by a similar construction using principle angles I can find a rotation $ g$ such that $ F=g\cdot E$ for $ m$ -planes $ E,F$ and \begin{align} \|g-\operatorname{Id_{\mathbb{R}^n}}\| \leq 2m\; d_{\mathrm{Gr}_m(\mathbb{R}^n)}(F,E) \end{align} where $ d_{\mathrm{Gr}_m(\mathbb{R}^n)}$ is the angle metric on $ \mathrm{Gr}_m(\mathbb{R}^n)$ (e.g. here).

However, I find these constructions rather unsatisfying and would like to understand if there is a more abstract underlying principle at play.

Here is my question: Given a homogeneous $ G$ -space $ M$ , are there always metrics on $ M$ and $ G$ such that there is a quantitative estimate \begin{align} d_G(g,e) \leq C \; d_M(g\cdot y, x) \end{align} for all $ x,y\in M$ and $ g\in G$ ?

Feel free to add any hypotheses (such as compactness etc) that apply to $ \mathbb{S}^{n-1}$ , $ \mathrm{Gr}_m(\mathbb{R}^n)$ and $ O(n)$ .