Development and Production on the same RDS instance, but under different databases?

I was told to create a single RDS instance (amazon web services RDS) onto which I should create two users, development and production, and two databases, development and production. The idea is that the development user only has access to the development database and production user only has access to the production, which would make it secure.

In other projects we’ve had separate RDS instances for development and production, but they want to save costs. I’m not too sure about it.

Is it a bad idea?

Is it practical to have several subnets under a ethernet port for a dhcp configuration?

Does it make any sense to set several subnets on a ethernet port of DHCP server. The configuration on dhcpd.conf is listed below and the server has no complain, but it is doubtful:

subnet netmask { #ethernet port enp5s0; range; range; option domain-name-servers,; option routers; default-lease-time 86400; max-lease-time 259200; } subnet netmask { #ethernet port enp5s0; range; range; option domain-name-servers,; option routers; default-lease-time 86400; max-lease-time 259200; } subnet netmask { #ethernet port enp7s0; range; range; option domain-name-servers,; option routers; default-lease-time 86400; max-lease-time 259200; }

For route command output, what does an asterisk under interface column mean?

For my Kubernetes nodes, I see following entry with a star under the interface column. I do not see this mentioned in “route” command’s documentation. The only star mentioned there is for gateway. This entry has been created by Calico for the gateway on current VM.

Destination Gateway Genmask Flags Metric Ref Use Iface — — — U — 0 — 0 — 0 — *

Under Poincaré-Bendixson hyphotesis can a $\omega(p)$-limit set be like this picture?

Last class we were proving Poincaré-Bendixson theorem in $ \mathbb R^2$ which states that:

Assume that the positive orbit $ \mathcal O^+(p)$ is contained in a compact subset $ K$ of the planar domain $ D$ of the differential equation $ x’=X(x)$ . Assume further that $ X$ has only finitely maney fixed points in $ K$ . Then one of the following is satisfied:

a) $ \omega(p)$ is a periodic orbit;

b) $ \omega(p)$ is a single fixed point

c) $ \omega(p)$ consistis of a finite number of fixed points, together with a finite set of orbits such that for each orbit its $ \alpha$ -limit set is a single fixed point and its $ \omega$ -limit set is also a single fixed point.

During the proof, he stopped the class and drawed the following picture in the black board: enter image description here

and asked, can $ \omega(p)$ be like that?

So I assume that he asked that in the context given by the hyphotesis of Poincaré-Bendixson theorem. Also he intented to picture with those dots the singularities of the field. I want to say that the answer is no, because I think that every singularity must be connected by an orbit, which does not happen in this picture. But the problem is: I can’t justify that. I’ve read the poincaré-bendixson demonstration quite a few times, but I can’t find this justificative there. Any insight would be very helpful. Thank you

RHEL KVM stuck when trying to create VM under VMware Fusion

I am trying to use KVM on RHEL 7 VM but after I try to create a new VM it goes well until I click Finish button, at that moment the window stuck and does not respond anymore, so I can’t create a VM!

That’s what I used: – VMware Fusion Professional Version 11.0.0 (10120384) – RHEL 7 + KVM – Mac book pro with Mojave – I enabled the VTx feature on VMware Fusion.

Here you can see a screenshot of the problem:

enter image description here

Any help will be appreciated, thanks.

Closure properties of a non-regular language under complement?

  • Assume I have L1 which is a regular language, so we know since regular language is closed under complement, the complement of L1 is also a regular language.
  • But let’s say if the complement of L1 is a non-regular language, is it safe to conclude that L1 is a non-regular language as well?

Since I’m trying to prove a language L1 is not a regular language, and the pumping lemma doesn’t work well with this case. But I can easily prove the complement of L1 is not regular, I’m wonder if that option is possible.

Visiting the US after applying for a visa under another name

I live in Europe (can’t say where) about 7 or 8 years ago I applied for a American visa and I got given one but didn’t travel due to family issues. I did my fingerprints at the embassy in Europe. Two years later I went to Africa and went to the US embassy for a visa again with another name. (It was a stupid mistake, made by a child at the time, family pressure.) All my information came up from the earlier application. The application was refused.

Years later I am still living in Europe and have a European passport. I am planning on going for Holiday with my son to see my family in the US. My only concern is, if I do fingerprints at the airport will that same situation come up again and if so what are they going to do about it? I am travelling with a young child and I am very stressed out about this.

Conditions that imply closure under intersection of Context-Free languages

Context Free languages are not closed under intersection.

Suppose $ L_1, L_2 \in CF \setminus REG$ ($ L_1,L_2$ are context fere but not regular)

Are there well-known theorems (and/or whole papers/research topics) that try to shape the sufficient/necessary conditions for $ L_1,L_2$ to make $ L_1 \cap L_2$ context free?

Is fleeing from battle considered “acting under attack”?

A short movement is’nt enough to flee from battle, since the opponent can do a short movement AND an attack: “A character can also try to make a short move and take another (relatively simple) physical action, like make an attack” (p. 208)

So to completely flee from combat a long movement is required.

Page 227 of Cypher-system states that anything but moving grant an immediate extra attack from the opponent. But it is also specified that the character “is assumed to be moving slowly and carefully out of the fight”.

While this is surely true for a short movement, this is also valid for a long movement?