Approximation concerning Asymmetric TSP, Symmetric TSP, and Metric TSP

I always considered Symmetric TSP to be inapproximable in general, and thus by extension Asymmetric TSP as well. Once you add the condition of the triangle inequality however, you obtain Metric TSP (which can be Symmetric or Asymmetric), which is approximable (e.g. Christofides algorithm).

However, I’m having doubts after finding the following paper :

An improved approximation algorithm for ATSP
Vera Traub, Jens Vygen (https://arxiv.org/pdf/1912.00670.pdf)

In their paper, there is no mention of Metric TSP, or the triangle inequality. Does this mean that I’m misunderstanding, i.e. Asymmetric TSP is in fact approximable, even without the triangle inequality?

How are metric TSP and Eulerian cycle equivalent?

In paper : https://arxiv.org/pdf/1908.00227.pdf it is stated that

In the metric TSP problem, which we study here, the distances satisfy the triangle inequality. Therefore the problem is equivalent to finding a closed Eulerian connected walk of minimum cost.

They don’t seem to be equivalent at all, since for a complete graph of size 5, with all edge costs 1, a metric TSP solution would be of cost (and length) 5, whereas an Eulerian cycle would be of cost (and length) 10. Am I misunderstanding?

Geodeics in 2+1 Schwarzschild metric

Would appreciate if someone could explain the query that, I am trying to understand the code for timelike geodesic in 2+1 Schwarzschild $ (\theta=\frac{\pi}{2})$ where we have taken $ M=1$ . From this how to get orbit plots? Is this possible to draw orbit without involving effective potential and angular momentum? Here I have taken $ \dot{t}, \dot{r}, \dot{\phi}$ such way initial condition that satisfy the timelike condition

$ (-\dot{t}^{2}+ \dot{r}^{2}+\dot{\phi}^{2}=-1)$ . We are currently using NDSolve. $ $ ds^{2}=-(1-\frac{2M}{r})dt^{2}+(1-\frac{2M}{r})^{-1}dr^{2}+r^{2}d \phi^{2}$ $

code image

Reference: Introduction to Black Hole physics, Valeri P. Frolov and Andrei Zelnikov, Chapter-7, section-7.2.3 (pg.no:192). We are not exactly looking at this picture, similar to this but more concessions on trajectories.

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Is there a metric or distance of two languages?

Given a language $ L$ , I am finding a method to evaluate the advantage of an automaton to decide $ L$ .

My goal is to decide a language $ L$ (and maybe it is not decidable for automata). If one constructs an automaton $ A_{1}$ whose laguage is $ L_{1}$ , I want to know the advantage of $ A_{1}$ for decisidng $ L$ . If there is a distance or metric of two languages $ d(\cdot, \cdot)$ , I can define $ \mathrm{Adv}_{L}(A_{1}) = d(L_{1}, L)$ . Thus, we can say $ A_{1}$ is better than $ A_{2}$ respect of $ L$ if $ \mathrm{Adv}_{L}(A_{1}) > \mathrm{Adv}_{L}(A_{2})$ .

In my opinion, the follow conditions are satisfied at least.

1 It is non-negativity and identity of indiscernibles, $ d(L_{1}, L_{2}) \geq 0 ~\text{iff}~L_{1} = L_{2}$ .

2 It is symmetry, $ d(L_{1}, L_{2}) = d(L_{2}, L_{1})$ .

3 If $ L_{1} \cap L \subseteq L_{2} \cap L$ and $ L_{1} \setminus L = L_{2} \setminus L$ , then $ d(L,L_{1}) \leq d(L,L_{2})$ .

4 If $ L_{1} \cap L = L_{2} \cap L$ and $ L_{1} \setminus L \subseteq L_{2} \setminus L$ , then $ d(L,L_{1}) \geq d(L,L_{2})$ .

Set metric for wlan0 network in XUBUNTU

I need to set the metric 0 for the wlan network

$   ifconfig   wlp2s0 .... 

if I use the command:

sudo ifmetric wlp2s0 1 

Everything works until the system reboots.

already tried to configure the file: /etc/network/interfaces

auto wlp2s0 iface wlp2s0 inet dhcp   metric 0 

failed, when the system starts the network wlp2s0 is disabled.

How can I make this permanent?

I need to leave the wireless (wlp2s0) network with priority 0.

Homeomorphism between LP-Spaces on Metric Spaces and Lp-Spaces on Euclidean Space

Let $ (X,d_X,\mu)$ and $ (Y,d_Y)$ are complete metric spaces and $ \mu$ be doubling; moreover suppose that $ (X,d_X)$ is homeomorphic to a Euclidean space $ E^D$ and $ (Y,d_Y)$ is homemorphic to $ E^n$ .

Fix an arbitrary $ x_0 \in X$ and define the $ L^p$ -space $ L^p(X,Y;\mu)$ are the equivalence class of all measurable functions $ f$ for which $ $ \int_{x \in X} d^p(f(x),f(x_0)) \mu(dx)<\infty. $ $ I know that this is a complete metric space (standard result).

Since $ X$ is homeomorphic to $ E^D$ , then is $ L^p(X,Y;\mu)$ homeomorphic to $ L^p(E^d,E^n;\nu)$ for some measure $ \nu$ ?(probably $ \nu$ should be the push-forward of $ \mu$ along the homeomorphism relating $ (X,d_X,\mu)$ to $ E^D$ ).

In short, is $ L^p(X,Y;\mu)$ homeomorphic to a separable Banach space?

PDE System problem to find the metric

I have a big problem to solve this system $ \Delta f-hf^2=0$

$ p|\nabla f|^2+hf^3=0$

where $ h$ and $ p$ are constants, $ f$ is a scalar function defined on a 3-manifold ($ f:M \rightarrow \mathbb{R}^3$ , where $ M$ is a 3-manifold), $ \Delta f$ is Laplacian of $ f$ and $ \nabla f$ is the gradient of $ f$ , for the metric $ g$ (where $ g$ is the metric of $ M$ ).

Should I find $ f$ and $ g$ ..is it possible? is there a solution?

Professor Robert Bryant has found the solution here (where $ M$ is a 2-manifold): Pde system problem

Pseudo metric on the orthogonal group

Let $ O(n)$ be the set of all $ n\times n$ orthogonal matrices. Define an equivalence relation $ \sim$ on $ O(n)$ as follows: $ U\sim V$ iff there exists a permutation matrix $ \Pi$ and a diagonal matrix $ \Lambda$ where all diagonal entries are either 1 or -1, such that $ U=V\Pi\Lambda$ . Let $ \tilde{O}(n)$ be the quotient set $ O(n)/\sim$ .

The question is that: does there exist a psedo metric $ \rho$ on $ O(n)$ such that there is a well-defined metric $ \tilde{\rho}$ on $ \tilde{O}(n)$ such that $ \tilde{\rho}([U],[V])=\rho(U,V)$ , where $ [U],[V]\in\tilde{O}(n)$ are the equivalent classes with representative elements $ U,V\in O(n)$ respectively.