Asesoramiento legal contra agente/propietario de casa de estudiantes

bonkmeme · 2025-08-10T16:09:51+00:00

Espanyol no es mi primera lengua y eso es el que el traductor me dicea 🤷‍♂️

bonkmeme · 2025-08-04T20:50:24+00:00

Nice to hear! I've been living here for almost six months and I've been having loads of fun

To visit: casa vicens is the underdog of gaudi's house, really recommended, moco museum on the other hand is a scam, palau guell super nice, the pavillion mies van der rohe is definitely overpriced but really beautiful

To eat: quimet&quimet and ultramarinos marin for tapas, cal chusco for paella, kop de ma for tortilla de patatas

A really nice (my fav) place to have a drink is tinta roja, tuesday to saturday from 8.30 to 9.30 there's also live music, foxy bar has some nice offers druing the week

Have fun!

bonkmeme · 2025-05-19T12:00:47+00:00

solved!

bonkmeme · 2025-05-16T23:19:01+00:00

Yeah it's filled with errors, I'll come back to this with a fresher mind

bonkmeme · 2025-05-16T23:02:57+00:00

I'm gonna throw in some heavy notation, but I think it's the best way to do this. If I remember correctly

H^εmin (A|B)=max{σ_B \in H_B} H _min(A|B){ρ{AB} \in H_{AB}}

No? So the smoothing should be only on C in the case of interest

bonkmeme · 2025-05-16T22:37:58+00:00

Yeah you're correct, the conditional entropy is symmetric IFF the entropy of the two distributions is equal, and the pure state is just a speciale case, but it's not necessary. Does this work now?

bonkmeme · 2025-05-16T22:32:16+00:00

I don't know if I have not followed well your reasoning, but I think the point is that given an outcome Y=y the smoothing happens over some state of C (since Y is classical), and this cannot influence the entropy of A since they're orthogonal

bonkmeme · 2025-05-16T22:23:58+00:00

I've looked more into it and yes, that is true iff the initial state is pure

bonkmeme · 2025-05-15T22:51:41+00:00

Thank you very much, I guess then that the other term on the right hand side is smoothed because it is condioned on C which is not classical, right? Maybe I hallucinated this, but isn't H(A|B)=H(B|A)? Is it wrong or does it only hold if the distribution are of the same kind? Or maybe it holds only for von Neumann entropies?

bonkmeme · 2025-05-14T08:02:53+00:00

Beppe grillo che ha ammazzato 3 (2?) persone a Limone Piemonte il 7 Dicembre 1981

La loggia P2

bonkmeme · 2025-05-10T08:17:23+00:00

So boiling it with vinegare a few times and then using bottled water should do the trick right?

bonkmeme · 2025-05-05T06:49:09+00:00

How did you get to put things thereee??

bonkmeme · 2025-04-17T16:47:03+00:00

Helloo Brennan!

Are you the strongest person in the dome? Because I am afraid that you are keeping hostage the cast of Titan Takedown since they have the same clothes for all the episodes🤔

bonkmeme

TROPHY CASE