Devilius

@ByteCommander thank you very much

hello, for having flash on firefox, is the recommendation provided in askubuntu.com/questions/562271/…, which directs to instructions on webupd8.org/2014/05/install-fresh-player-plugin-in-ubuntu.html, still up to date?

@NathanOsman thanks again! I'm heading off to bed then, bybye

it worked! good to know for next time. Interesting to note that I had use the same shell the first command was executed in, as the other would give me: "pid 2993 is not a child of this shell"

ok transfer is 95% done....

I've successfully fetched the PID. I'll try it now and see if it works

thank you very much!

that sounds like exactly what I need

ahah

yes

specifically I am currently transferring files using scp. I want it so that when file transfer 1 is done, file transfer 2 starts. Can I request something like that after transfer 1 was requested?

thanks for the link

that makes a lot of sense

ah right

I've been running tests, that's pretty much what made it clear as to why the random pivot choice also had this pathological case

Java

also, thanks for your answer. I've been reading about it and am slowly understanding exactly what you've. Essentially, we do median of three because we want to get the median, but cant do it really efficiently, so settle on an "approximate" median based on three integers. correct?

film 2

from harry potter

it's ginny weasley

Hello, about quicksort median of three: I understand the logic of median of three for an already sorted array: it lets us pick the middle element, and thus avoid the worst case O(n^2). However, why do we do the median of three? What would be the problem with always picking the middle element? what is the advantage in sorting these three values?

proves that sleeping on it works like a charm¸ I've been stuck on this since last night!

I got it

nevermind

specifically, how can one prove that is it not regular, without setting restrictions on |y|? I find myself unable to pick any fitting words that would prove this

hello, I am having some trouble with the pumping lemma for regular languages, specifically for proving that the language $w \in \{a, b\}^∗ : w^R \neq w$ is not regular

thank you very much

I didn't realise you could use the complement to prove the regularity of a language

alright

that is correct terminology

does this work "both" ways, as in if a language is not regular, its complementary is also logically not regular?

other than $|y| \geq 1$

without ever setting conditions on |y|

well, I don't see how to do it then

ok

$|xy| \leq m $

ah, no

well, if I understand correctly, the y would be b, or ab, or aab, etc

so, the language $ \{ w \in \{a,b\}^*: w^R \neq w \}$

reverse

then figured the language a^{m-1}ba^{m} should work: no matter how I pick the y, I will always end up removing the b (correct?)

but couldn't figure out how to do it without setting conditions on |y|

I had to prove that the language \{ w \in {a,b}^*: w^R \neq w \} was not regular

no worries

but sleeping on it must have solved it

I've been stuck on it since last night

thank you though

I actually just solved my problem

hello, I am having some trouble with the pumping lemma for regular languages, is anyone familiar with it?

Hey, is anyone familiar with the pumping lemma for regular languages?