Black Hole Singularities

Question: If two black holes collide and then evaporate, do they leave behind two naked sigularities ore? If there are two, can we know how they interact?
Comment: They will not leave anything.
Answer: Since classical gravity is much stronger than quantum gravity, the collision will take place on much faster time scales than the eventual Hawking radiation. So, the two black hole will experience a fairly violent process, and eventually settle down to a static situation which is most likely a new black hole. That black hole will evaporate on much longer time scales, eventually either evaporating completely, or leaving behind some long-lived "remnant". Nobody really knows what the black hole leaves behind after evaporating, but I don't think it will be a naked singularity.
Answer: The answer is neither two nor one. Black holes that collide and subsequently evaporate leave just a bunch of radiation. There is no remaining 'naked singularities'.
In the past there has been a lot of confusion about black hole evaporation. But questions like "what happens with the information describing the micro state of the black hole" have been answered. Key idea is the holographic principle and black hole complementarity. If you want to learn more about this, Lenny Susskind's book 'The Black Hole War' is a good starting point.
Comment: they leave a single black hole plus some radiation, not just radiation.
Comment: @Jerry Schirmer - References?
Comment: http://arxiv.org/abs/0710.1338 , for one.  an arXiv search for "binary black hole coalescence" will give a lot of hits.
Comment: You are wrong. Nowhere in this paper the authors claim that the black hole remaining after coalescence is immune to Hawking radiation.
Comment: The answer is good with the caveat that any question like this that relates to quantum gravity is not settled. However, the most widely held opinion given what we think we know, is that black holes would eventually evaporate in an explosion of elementary particles leaving no other remnant. The holographic principle and its realisation in string theory has provided new understanding of how this can work without violating the laws of quantum mechanics and thermodynamics.
Answer: you can't produce any naked singularity in any process in 3+1 dimensions. This statement was originally coined by Roger Penrose under the slogan "Cosmic Censorship Conjecture" (CCC). A naked singularity would be a singularity that could affect the world around it because it would have no horizon; I will explain these words later.
Because dynamics seems problematic near a singularity - it's hard to calculate it by classical general relativity or even the descriptions of quantum gravity that we know - it looked like the Universe would be unpredictable if there could be "naked singularities", i.e. singularities that are unprotected by event horizons. That was viewed as a problem by Penrose (and others), a potential inconsistency that the Universe should avoid.
This thinking due to Penrose has been showed partly incorrect, at least in higher dimensions. A proper theory of quantum gravity simply can (and has to) produce predictions - at least in principle - what happens to an observer who can observe any singularity. And indeed, there have been counterexamples - thought experiments in higher dimensions - in which one can produce a naked singularity at the end.
As far as I know, it is still true that the CCC holds in 3+1 dimensions. One can't create naked singularities by any process - whether it involves 0 black holes, 1 black hole, or 2 black holes. Black holes usually have a singularity inside them but they are not naked exactly because they're black holes. A defining feature of a black hole is that they're holes that are black - which means that light can't escape from them. It can't escape from them because it is confined by the event horizon. The event horizon are the "clothes" in which the singularity is "dressed" - so it is not naked.
So despite some popular misconceptions, the defining feature of the black holes is the event horizon, and not a singularity. A singularity is just a typical addition that comes with the black hole "package". In Schwarzschild black holes, the singularity is spacelike. Nevertheless, once the black hole evaporates, nothing is left. At the very final stages of a black hole decay, the tiny black hole looks pretty much indistinguishable from a heavy particle species in ordinary particle physics (or string theory), and it decays to a few particles that we know from the colliders.
Two stabilized black holes, when they collide, emit some gravitational waves, and when they're sufficiently close (so that they don't just continue as two black holes), the rest of their mass=energy that hasn't been radiated away collapses into a single black hole that quickly gets stabilized as well. This black hole will proceed with the Hawking radiation and after a very long time, it disappears as well. Nothing is left. The research in the last 20 years or so has made it pretty much clear that there can be no remnants left etc.
All the best
Lubos
Comment: An extreme fine-tuning of initial conditions can produce naked singularities, as shown in the critical collapse simulations of the '90s.  The modern rephrasing of the Cosmic Censor Conjecture states that evolve to horizon-ed solutions form a dense set in the phase space of physicially reasonable initial conditions, so it would take an implausibly precise fine-tuning to produce a naked singularity.
Comment: As I read the question, the implicit assumption is that black hole evaporation produces the naked singularity, not the collision of the black holes...I like the part about the singularity/horizon distinction, as you know this causes infinite amount of confusion.
Comment: I believe the suitably refined but still general statement is still as of today a conjecture.  You can prove it in certain scenarios with extra symmetry.  But the black ring solutions in higher D definitely leaves the door open in 3+1D for exotic possibilities
Answer: The rules of general relativity give us a very specific set of rules for the interaction of the two black holes.  We can integrate thee equations to know what would happen if, for example, we had two black holes orbiting each other, or on a head-on collision course.
The result that we get is that the two colliding black holes will not evaporate or form a naked singularity.  Their horizons will coalesce into a single distorted horizon, which will then smooth itself out by emitting gravitational radiation.  At late times, you will have a single black hole with a mass less than the combined masses of the two initial black holes (the difference in mass will be the energy that is radiated away as gravitational radiation).  According to numerical simulations of collisions, no naked singularity should be formed for a generic collisions.
Comment: @Ted Bunn: Yes.  I didn't quite understand what colliding black holes had to do with Hawking radiation, so I answered the question from the classical GR side.  The only way that I could see a naked singularity involved is from the fact that the critical solution in critical collapse phenomena is a zero mass naked singularity.  But in any regard, we'd need a full theory of quantum gravity to be able to talk about the late stages of Hawking radiation.
Comment: I assume that the "evaporation" in the question is Hawking radiation. The end stage of Hawking radiation is not well-understood, but nobody expects any naked singularities to be involved.
Comment: it is ok to say that we can't have absolute certainty about the late stages of the Hawking radiation process without a full QG theory, but to deny Hawking radiation (see your comment on my answer), is obviously incorrect.
Comment: @Johannes: sure, but we can't be absolutely sure what the final state is.  It's completely plausible that the radiation will eventually reach equilibrium with the black hole, for one example.  If the horizon evaporates quickly enough, a naked singularity is certainly possible for some length of time.  I also hesitate to say how sure we are about an effect that has never been observed, however much it is supported by simple generalizations of existing theory.  Perhaps Hawking radiation ends up suppressed, like how QM ultimately prevents electrons from spiraling into protons.
Answer: The answer is zero. When black holes evaporate due to Hawking radiation they will not leave any singularity. The singularities in black holes are inside the event horizon.
However, when two black holes collide their singularities do merge to form one singularity inside the event-horizon of the merged black hole. Perhaps that is what you really wanted to know.

Black Hole Singularities


Running both VMWare and VirtualBox on Ubuntu

Question: Currently I use VMWare Player on Ubuntu 10.04 (64bit). Since I am not content with the performance of a virtual Windows XP machine, I'd like to test VirtualBox.
Is it possible to install VirtualBox on Ubuntu 10.04, when VMWare Player is already installed? I remember that the kernel was newly compiled when installing VMWare Player, and thus hesitate to install VirtualBox: the VMWare Player must stay functional.
Do you have experience with such setups? Are there some special steps to follow with such a scenario?
Answer: Yes i've installed in my debian squeeze machine both VMWare Player and VirtualBox OSE, there is not incompatibility problem between them .
I only never run a virtual machine with VirtualBox and another with VMWare Player at the same time, for performance related issue, but yes they can be installed together in the same system and no special steps to follow.


SVM - Predict the response of one of the observations

Question: I apply k fold cross validation using svms ( classification or regression, it doesn't matter) and I optimize my parameters based on the mean testing error. If I want to predict the value of one of my samples what do I do? I choose randomly one of the (perhaps best) k fold realizations where this sample happened to be on the testing set and predict its value?Or is it better if I apply the leave one out methodology in order to have one "testing phase" prediction for each sample?
Answer: Neither: you typically train a new model with the optimal hyperparameters on the entire training set and use that for prediction. Usually, it will perform marginally better than your cross-validation estimate because you are using more data for the final model.
An alternative that is sometimes used is to create an ensemble of the models you trained on each fold and then use the ensemble for predictions.
Comment: Yeah you are right. Thanx!
Comment: One more question. If I do the ensemble thing, is it better to find the optimal parameters for each fold? Or should I stick with the optimal parameters extracted from all folds?
Comment: It is best to stick with the optimal parameters you found from all folds.
Comment: Done. Thanx again


Wireless problem in ubuntu 12.04

Question: i purchased a dell inspiron 15R a few days back.
it had ubuntu 12.04 preinstalled. the guys at the shop installed win8 on it 4 me.
but later i decided to shift 2 ubuntu.
& I'm kinda a new user.
i installed it back using the recovery disk.
now everything is working fine except the wi-fi.
can any1 please help me out??
Answer: Trying to figure out what's going on with wireless can be quite tricky if all we have to go by is "the wireless isn't working". One of the biggest problems can be hardware compatibility issues. I would hope since your computer shipped with 12.04 lts, that the wireless card is supported in that version, but it is a 2 year old version of ubuntu at heart. I suggest you try a live usb drive of a newer version and see if the wireless works then, this will help eliminate hardware issues and with a new machine you will generally have more stuff work out of the box.
You can download ubuntu 14.04 here:
http://www.ubuntu.com/download/desktop
Make a live usb stick in ubuntu:
https://help.ubuntu.com/community/Installation/FromUSBStick
or
Make a live usb stick in windows:
http://www.ubuntu.com/download/desktop/create-a-usb-stick-on-windows
Answer: I have had the same problem if it is connect in the network panel, and if it is not connect in the network panel you have to delete the old one, and add a new hidden wifi.  and you have to in the terminal type:
<code>    rfkill list
    lspci
    lsmod wlan0
    lsusb -D wlan0
</code>
and then scan the port, and in the terminal type:
<code>    nmap -v -iR 10000 -Pn -p port
</code>
and this might work fine this way.

Wireless problem in ubuntu 12.04


MAF and MAP location

Question: I have a 2000 Pontiac sunfire se 2.2 I'm having trouble locating the MAP and MAF sensors. The mechanic I took it to said he can't work on my car cause it didn't stall enough while driving my car.. told me catalytic converter, plugs and wires, o2 sensors need replaced due to it misfiring
Comment: Welcome to Motor Vehicle Maintenance & Repair!
Comment: please don't go changing all these things as a troubleshooting procedure.  Also, O2 sensors are in the exhaust system
Answer: Try GM forums for help because the 2.2L 4cyl engine is universal to most GM models. No maf sensor as its always mounted between the air filter assembly and throttle body, in the plastic air tubing between the two. The map sensor is mounted on the intake manifold to measure vacuum while the engine's running, as a load sensor. A simple catalytic converter test, if its internally damaged and blocking exhaust flow) is to remove the exhaust manifold mounted O2 sesnor (upstream of the catcon) and start the engine. While LOUD, if the engine suddenly has renewed power and can be driven around the block better than before removing the O2 sensor, the catcon is blocking exhaust flow; effectively a constipated engine that can't exhale exhaust gases. No improvement, it's not a catcon problem. Stalling can be a worn out fuel pump (a fuel pressure gauge is needed to measure pressure against specs), clogged fuel filter not replaced every 100k miles, spark plugs not replaced every 100k miles, one or both ignition coils, ignition control module, crank positioin sensor. An expert in EFI systems can diagnose and troubleshoot your problem without throwing parts at the problem the average mechanic would do without experience.
Answer: I believe the following diagram should look like your engine:

MAP is at location #2. The MAP will have three wires in the connector. There may be a plastic cover over this area of the engine.
As @FDryer stated, no MAF.
Comment: Awesome thanks so much..

MAF and MAP location


How can I open a terminal window on OSX?

Question: I am doing some testing on a computer running Mac OSX Yosemite.
How can I open a terminal? - note the command key doesn't work either.
Ive googled, but no luck.
Answer: If you open a finder window and go to the Applications folder and into the Utilities subfolder, you should find terminal there. Double click and Terminal should start.
Another option is to use spotlight search by either clicking the search (magnifying glass) icon in the top right corner of the mac's top bar or use the default CMD+Space to bring it (spotlight search). Then start typing "Terminal" and you should get it as a suggestion.
Comment: Alright - got there! Thanks for your your prompt response - really saved me alot of unecessary frustration!
Answer: There is a System shortcut available [but I don't think activated by default] in System Prefs > Keyboard > Shortcuts > Services to open a New Terminal [or tab] at the current folder.

You can set this to work just in the right click menu, under Services, or add a hot key.


How much Weight can a drywall ceiling hold

Question: I want to hang something and I want to make sure it doesn't fall
Comment: I wouldn't do anything more than a few oz. Your adhesive should fail before the drywall does.
Comment: Do not recommend hanging anything from drywall ceiling. What are you wanting to hang?
Answer: It is recommended to not hang anything just from the ceiling drywall by itself. Particularly anything you have concern that it could fall due to its weight or safety of people or things under the hanging object.
Instead you need to locate the structure that holds the drywall in place and screw the hooks / brackets / retainers into that structure. The type of things you screw in would be dependent upon whether the ceiling structure was wood, metal rails or concrete. But since we have no guidance from you yet as to what type of structure you have suggestions are limited.
Comment: I have pinned Acuity lighting sensors into drywall ceilings, that being said, if they fell on someone, it is unlikely a serious injury would occur as they are plastic and lightweight.   It would not surprise me if the OP intended to hang something heavy enough (ceiling fan etc.) that could seriously injure someone.


Why are there teardrops on PCB pads?

Question: What's the use of the teardrop shapes around some PCB pads?
Comment: Possible to show a picture on what you mean?
Comment: @Dean - added a picture (I think you need a minimum rep to do so)
Comment: Going to guess to improve connectivity between track and pad.
Comment: Good question. I've never used them, and I see that teardrops are sometimes optional in EDA suites.
Answer: There are primarily two reasons to use teardrops:

It avoids a pocket (where the trace meets the pad) that could collect acid from the PCB etching process which would later do bad things.
It reduces mechanical & thermal stress resulting in less hairline cracks in the trace.

That being said, in professionally made PCB's teardrops are rarely needed. It's almost more of an aesthetic thing than a solution to a real problem. I've done many boards with and without teardrops and I have yet to notice a difference. In my opinion, they are more trouble than they are worth.
Comment: It's worth noting that teardrop-pads *are* necessary in some specialist situations, namely *flex-pcb*. Teardrops on flex-pcb layouts serve to reduce the stress bottleneck where the trace joins the pad.
Comment: I see how they could be little use; but how are they trouble? (as in "more trouble than they are worth.")
Comment: @Javier  It really depends on how your CAD software handles them.  What you don't want is to have to draw the teardrops manually for every pad/via.  Years ago I used Protel that would make good looking teardrops, but had some bugs that caused problems in the gerber output when teardrops were used.  Now I use Cadence Allegro, which has lots of issues.  At the very least, the teardrops simply take longer to draw and makes editing the PCB more cumbersome (assuming that you have a super complex PCB to begin with).
Comment: I've seen a number of PCBs with broken pads that cracked at the point where the teardrop touches the trace.
Comment: also there could be issues with electromigration.

Electromigration is the transport of material caused by the gradual movement of the ions in a conductor due to the momentum transfer between conducting electrons and diffusing metal atoms.
Answer: The teardrop is to help the drilling process. It prevents drill breakout where the trace joins the via or through-hole. Sometimes it's not necessary to do it because the manufacturer can do it for you.
Where I work, our repair department recommends we do it because it increases the strengthening of connections between pads and tracks.
Comment: It might help the drilling process, but I've seen plenty of them on SMT pads.
Comment: We have never had any issues that I know of with pads getting pulled up when drilling after etching.
Comment: @Kellenjb: it's not that the pads pull up, it's that the drill can be misaligned by a larger distance and still not create an open-circuit by drilling through the narrow track.
Answer: If you look at the picture the Ben Jackson posted in answer to a question about solder mask expansion, you'll see that the drilled holes can be significantly off-center from the pad centers.
In extreme cases, you could actually end up with holes that leaves no or very little annulus to make the connection to outgoing traces.  The tear drop ensures that there's enough pad material left over to have a solid connection to the pad.
Comment: I understand that the hole isn't always dead in the center, but if it's so far off that it goes off the pad I think I would go and find another manufacturer.
Comment: Drilled holes might be a little off-center, but not necessarily in the direction of the teardrops. Teardrops are oriented to the direction of the track and this may be different for each pad. But if there is a misalignment between drill holes and pads, the direction of misalignment may be the same for all or most pads. If you want to be sure to get a minimum annulus in every case, you just have to use larger pads.
Comment: It might be a cost-quality tradeoff.

Perhaps the main goal is to slap a product together cheaply - they might be willing to go with a lower-precision fab, and design the board to accomodate that choice.

It might also be that there is only a small handful of parts that need extra densely placed PTH's that doesn't justify paying for higher precision fab for the other 99% of the board.   The teardrop might be the "compromise" solution for those cases.
Comment: @stevenvh, most fabs I've used give a minimum annular ring spec that allows for something like 20% breakout of the hole (which they derive from their copper and hole registration capabilities). If you really want to never have a hole go off the pad, you just need to use a larger pad size than they gave you as the minimum.
Comment: A fast-turn manufacturer I use says the only way to safety use an 18mil ring with an 8mil hole, to avoid breakout, is to use teardrops. In this situation they are a real benefit on tight boards.
Comment: The purpose of the annulus is to connect the via barrel to the trace with imprecision in the drill center.  There is both offset for all holes in a given board as well as wander, variation from hole to hole.  IPC specs allow some portion of the via barrel to 'break out" from the pad.  This "allowed" amount can be enough to cut the trace from the annulus when in the right position (not allowed).  Tear drops prevent this for smaller annulus.  They also mitigate "acid trapping" which may no longer be a problem. I've never seen evidence this is a design problem and not a quality control issue.
Answer: They are mainly used on single-sided boards, to increase reliability. I often use them on my home-made PCBs.
Comment: Is that because it's single-sided, or (with all respect) because it's non-professional (DIY)? In other words, are they also useful if you have your PCBs produced in a shop?
Comment: It's mainly because single-sided boards don't have PTH.
Comment: that was going to be my answer... The bit about it not having plated through holes which results in the pad being able to be pulled up a lot easier or move around with shock resulting in breaks over time.
Comment: In my dictionary PTH = Pin Through Hole, but I guess you mean Plated Through Hole. :-)
Answer: The simple answer is for Strain Relief.
Like a plug with a thin cable, it needs a graduated support to prevent the shear forces of thermal expansion when desoldering a component or when a human interface force is involved ( eg mic jack or a heavy part on a board.)
Where you do not have strain or any thermal or mechanical stress in the application, it is unnecessary.  But I have seen dozens of times when this would have prevented a track failure that was hard to see with the naked eye.  One example was a $2K Mac vertical monitor with the flyback transformer on the main board...  It created a micro-sheer crack impossible to see, but whenever the CRT went blank a slap to the side of the case fixed it. Which lasted for a few days, until the secretary cried.... help.. so I came to the rescue with a soldering iron. and said to myself... I wish the board designer knew about tear drop pads.
Comment: If it's for strain relief can't you simply use a larger pad? A lot easier and more effective, as it goes all the way round the pin.
Comment: Absolutely not. It is the track that needs to be enlarged , or rather the track pad interface... When you put teardrops into the standard library for interface devices its not hard. It is not needed for every via.
Answer: I thought it had to do with IPS-6012C for rigid board.  Specifically related to section 3.4.2 Annular Ring Breakout.  For example, Class 2 allow for an annular ring to break out (be missing) for 90 degrees.  This is due to drill misalignment. On large vias or plated through holes, this is not likely to happen in modern fab facilities.  When dealing with 10 mil finished drill holes and 20 mil pads this is much more probable.  If the breakout happens on the side of the trace exiting the pad, Class 2 states that it can cut the trace down to 50um and still be acceptable.  If there was a teardrop (or fillet), the misdrill would not cut into the trace.
Answer: A teardrop via make it easy to remove embedded resistors when using ECL logic.  This was done by Harris Computer Systems about 1987 to make it easier to repair memory PCA's that cost upwards of $15,000 (or more) in their h1000 and h1200 systems.  To remove the embedded resistor that was connected to the via all one needed was a pin vice and then drill a hole at the point of the teardrop.  Then add a wire to a spare embedded resistor or add an external resistor. I remember building the light box to see the teardrop via.

Why are there teardrops on PCB pads?


Disable dragging on eGalax touchscreen on Ubuntu 14.04

Question: I have tried almost every imaginable option I could find online to put in the 10-evdev.conf file in the xorg.conf.d directory so as to disable the dragging action of my touchscreen (eGalax Inc. Touch) but I have not been able to.
The touchscreen is recognized as a tablet device but I can live with that.
If I give some Options such as Option "Ignore" "on" under the following section in 10-evdev.conf then it works (meaning, it disables the touchscreen). If I do the same under the touchscreen catchall, it does not work! Also, in the xorg log, I can see that it is recognized as a tablet (I believe this is a bug in the kernel but it really doesn't matter as long as I can configure the touchscreen and have it working properly)
Section "InputClass"
  Identifier "evdev tablet catchall
  MatchIsTablet "on"
  MatchDevicePath "/dev/input/event*"
  Driver "evdev"
EndSection

Now, my problem is that no matter what Option I've tried, I am not able to disable dragging. If I touch the screen and drag my finger across the screen, it draws a rectangle or actually drags the icons on the desktop.
All I need from the touchscreen is click (tap) functionality since I'm going to be using is as a kiosk.
Can anyone give me the right configuration to disable dragging?
Thanks
Nahom
Answer: OK, found the solution:
You need the Linux eGalax driver from here ....
http://www.eeti.com.tw/drivers_Linux.html
After you unzip this download, look in the Programming Guide in the /Guide directory. It contains lengthy instructions. I did not follow them all, but note that my PC only has a single point touchscreen. What I did, which worked:
Run sudo sh setup.sh as indicated at section 3.1.
then reboot
Hope you succeed as well.
Comment: The eGalax driver from the manufacturer has issues. I was hoping to be able to disable dragging while still using the evdev driver that came with Ubuntu.


Is a cylinder a Lipschitz domain?

Question: I'm wondering if the domain $(0,T)\times \Omega$ is a Lipschitz domain ($T$ is a positive real number), provided that $\Omega$ is an open bounded subset of $\mathbb{R}^n$ with Lipschitz boundary, and how to prove or disprove this fact.
Thank you
Answer: Yes, it is. There are three parts of the boundary to check.

$(0,T)\times \partial \Omega$.  Locally, and in a right coordinate system, $\Omega$ is the set above the graph of some Lipschitz function. Multiplying it by a line segment maintains this property: the new Lipschitz function can be taken to be independent of $t$.
$\{0,T\}\times \partial \Omega$. Begin as above, and then slightly tilt the coordinate system so that its axes are not parallel to the hyperplane $t=0$. Then the plane $t=0$ is also the graph of a Lipschitz function. The product domain lies above the graph of the maximum  of these two Lipschitz  functions: one linear, one coming from $\partial \Omega$.
$\{0,T\}\times \Omega$ is flat.
Comment: And the corners?
Comment: Yes what about the corners?!

Is a cylinder a Lipschitz domain?


How long does it take for Apple to bill me for purchases on iTunes?

Question: I've noticed that I'll often get an email about 24 hours after my purchase took place on iTunes. Does Apple have some sort of billing cycle, or does the withdrawal frm my account actually happen immediately?
Comment: They withdraw funds immediately. The invoice they provide can sometimes take days and doesn't appear to be on any kind of special cycle. I imagine it's issued depending on the load (number of purchases they need to process).
Answer: Per the iTunes Store — Terms and Conditions

If a credit card or your PayPal account is being used for a transaction, Apple may obtain preapproval for an amount up to the amount of the order. Billing occurs at the time of or shortly after your transaction. If you are using 1-Click purchasing or your PayPal account, your order may be authorized and billed in increments during one purchasing session, so it may appear as multiple orders on your statement. If an iTunes Card, iTunes Store Gift Certificate, or Allowance Account is used for a transaction, the amount is deducted at the time of your transaction. When making purchases, content credits are used first, followed by Gift Certificate, iTunes Card, or Allowance Account credits; your credit card or PayPal account is then charged for any remaining balance.
Answer: From my experience and from checking the last couple of months of receipts my average is 24-72 hours with the longest being just under 2 weeks.


Seg Faults and Instantiation Errors while running PyQGIS on OS X

Question: I'm having issues getting off the ground with running PyQGIS on my OS X. I'm frankly very confused as some errors occur sporadically while executing the same code.
When I try the following:
QgsApplication.setPrefixPath('/Applications/QGIS.app/Contents/MacOS', True)
QgsApplication.initQgis()
QgsProviderRegistry.instance().providerList()

I either get one of three scenarios:

Segmentation fault: 11 when returning the list
A blank list returned
Or a this error after initQgis()...
QCoreApplication::applicationDirPath: Please instantiate the QApplication object first
QWidget: Must construct a QApplication before a QPaintDevice

A workaround I found to work is to instantiate a QgsApplication object first as the error suggests with something like
app = QgsApplication([], True)

However, the problem here is that this creates a GUI which I don't need - I'm just trying to do some scripting. If I change the instantiation parameter to False, then I get an error with initQgis() of:
QWidget: Cannot create a QWidget when no GUI is being used

Also, in a lot of examples I've found online you don't need the instantiation step... is there something funky with my setup?
I'm using a venv if that makes any difference... all commands I want to run work normally in the QGIS Python console, it's just the external script that's giving me issues.
Any thoughts?
Python: 2.7.10,
QGIS: 2.10.1-Pisa,
OS X: 10.10.3
Answer: I had also some troubles with this, here is my solution for OSX 10.9.5:
import os, sys
sys.path.append('/Applications/QGis.app/Contents/Resources/python')
from qgis.core import *

app = QgsApplication([],True)
QgsApplication.setPrefixPath(r"/Applications/QGIS.app/Contents/MacOS", True)
QgsApplication.initQgis()

layer = QgsVectorLayer('/tmp/myshape.shp', 'test', 'ogr')
print layer.isValid()

Answer: You need to add first Applications/Qgis.app/Contents/Resources/python to the PYTHONPATH
First solution outside Python
export PYTHONPATH=/Applications/Qgis.app/Contents/Resources/python
# then
python
....
#Import QGIS
from qgis.core import *

Second solution in Python
import os, sys
sys.path.append('/Applications/QGis.app/Contents/Resources/python')
#Import QGIS
from qgis.core import *

In both cases you need to initialize the app (otherwise crash)
app = QgsApplication([],True)
QgsApplication.setPrefixPath(r"/Applications/QGIS.app/Contents/Plugins", True)
QgsApplication.initQgis()

You can now use PyQGIS
layer = QgsVectorLayer('/Users/Shared/test.shp', 'test', 'ogr')
layer.isValid()
True

In the same way you can add the path of the plugins to sys, use the processing algorithms and...
Comment: _export PYTHONPATH=/Applications/Qgis.app/Contents/Resources/Python_ should be **python** with lowercase p in Python


Is there a pause shorter than a comma?

Question: I'm writing a poem, and like how it's flowing. But I find that a tiny pause shorter than a comma would do tremendous. Is there such a mark or multitude of symbols to make a short pause?
Comment: In a word, no . . .
Comment: The movement from one line to the next can imply a short pause, even if there's no punctuation. Or write "(a tiny pause)" in the line? :-)
Comment: A demisemiquaver rest? https://en.m.wikipedia.org/wiki/Thirty-second_note
Answer: I really believe
You should use
Line breaks
To give people
A chance
To think.
Maybe  you could use multiple spaces?
Comment: Unfortunately this isn't an option in places where line-breaks aren't permitted, especially in inline technical documentation.
Answer: Depending on your poetic style, you can add a syllable to a word to increase the length of the sentence or line. However, it's poetry; that probably won't work. But because it's poetry, there's nothing against using two or more spaces after a word. The reader will automatically slow down if for nothing else because it's unusual.
Answer: Semicolons and colons have been relegated to the fringes more and more in the last hundred years. A comma seemed relatively short in the old days, but now it can seem oppressive. An unintended consequence of the continuing distaste for periodic writing.

Is there a pause shorter than a comma?


What's the difference between "$\exists F_\epsilon :m^*(E\setminus F_\epsilon)<\epsilon$ for each $\epsilon$," and "$\exists F:m^*(E\setminus F)=0$"?

Question: Let $E\subseteq \mathbb{R}$ be given, and let $m^*$ denote the outer measure. For each $\epsilon$, there exists a closed set $F_\epsilon\subseteq E$ such that $m^*(E\setminus F_\epsilon)<\epsilon$.  Does this imply that there exists a closed set $F\subseteq E$ such that $m^*(E\setminus F)=0$?
Comment: What is $E/F$?  Do you mean $E\setminus F$?
Comment: I think $m^*$ is a more usual notation for outer measure.
Answer: There exists a $F_\sigma$ set such that this is true, take $\cup_{n=1}^\infty F_{\frac{1}{n}}$.
However, the statement is not true in general. Take $E = (0,1)$. Then taking $F_\epsilon = [\frac{\epsilon}{3},1-\frac{\epsilon}{3}]$ results in $m^* (E \setminus F_\epsilon) < \epsilon$.
Suppose $F \subset E$ is closed. Let $I=[\inf F, \sup F]$. Clearly, $F \subset I$, and $0< \inf F, \sup F < 1$, hence $m^*(E\setminus F) > \frac{\inf F}{2}> 0$.
Answer: No.  If $E$ is closed then the statement clearly holds, so the only possible counterexample must be non-closed.  What's the simplest possible example of that?  A bounded open interval.  No closed subset $F$ of a bounded open interval $E$ can be such that $m^*(E\setminus F)=0$.  This is because $E\setminus F$ is open, thus is either empty or contains an open interval.  It can't be empty since that would imply $E=F$ which would make $E$ clopen, which is impossible (the only clopen subsets of $\mathbb{R}$ are the empty set and the whole space).  So it contains an open interval, and hence has positive outer measure.
Answer: It might be useful to see how these statements can be rewritten to better show the difference in their logical form.
The 1st statement becomes: $\;\; \left(\forall \, \epsilon > 0 \right) \left( \exists F \subseteq {\mathbb R} \right ):$ $\;\;\;F$ is closed and $m^{*}\left(E-F\right) < \epsilon $
The 2nd statement becomes: $\;\; \left( \exists F \subseteq {\mathbb R} \right )\left(\forall \, \epsilon > 0 \right): $ $\;\;\;F$ is closed and $m^{*}\left(E-F\right) < \epsilon $
Of course, the fact that the 2nd statement is a logically stronger $\exists \; \forall$ uniform statement doesn't mean that, in this specific context, we get a mathematically stronger statement. However, we do in fact get a mathematically stronger statement, as the other answers show.


Delta formula for FX vanilla option

Question: What value do you use for annual dividend yield?
It does not apply in case of FX.
Answer: When pricing FX options, the underlying is the spot or forward exchange rate.
The foreign currency is analogous to a stock where the owner of the foreign currency receives a "dividend yield" equal to the risk-free rate in the foreign currency.

Delta formula for FX vanilla option


Connect to Salesforce using wsdl in Drupal 8

Question: I am trying to connect to Salesforce (SF) with wsdl to pull data.
I can't use SF suite module or Salesforce rest client service.
In the D7 file, I have:
require_once ('soapclient/SforceEnterpriseClient.php');
require_once('soapclient/SforceHeaderOptions.php');
ini_set("soap.wsdl_cache_enabled", "0");

$sfdc = new SforceEnterpriseClient();
$SoapClient = $sfdc->createConnection($curPath.'/enterprise.wsdl');

It create the connections without any problems.
D7 References from here: https://developer.salesforce.com/page/PHP_Toolkit_11.0_Getting_Started and learned to how to use it from here: https://github.com/developerforce/Force.com-Toolkit-for-PHP/blob/master/soapclient/SforceEnterpriseClient.php
In the D8, I am trying to figure out how to connect to SF in the same way:
namespace Drupal\mymodule\Controller\Soapclient;
$sfdc = Soapclient::SforceEnterpriseClient();
$SoapClient = $sfdc->createConnection('../../includes/soapclient/enterprise.wsdl');

But got the SforceEnterpriseClient class not found error.
Stuck on it for a while, any help will be so much appreciated.
Anybody knows how to connect to SF in the same way like this:https://github.com/developerforce/Force.com-Toolkit-for-PHP/blob/master/soapclient/SforceEnterpriseClient.php
Answer: It seems the problem you are facing is due to the way you are referring the library; please refer the following question; they have explained how you should be using external libraries in Drupal 8 custom modules.
Comment: I don't think they are connected. What i did in my D7 was download the php files from: https://developer.salesforce.com/page/PHP_Toolkit and use it but right now they don't have latest php version available yet so should i write my own soapclient class or D8 have some other way to handle it? That's my questions. Thanks in advance.


How to format a SP Online List Name ( Header ) using JSON via Format View?

Question: I have used JSON to format most of the columns on a list and then some using "Format Current View" and "Format This Column" however I've scoured the web looking for a solution to formatting the List Title on a SharePoint Online List and can't find a definitive way to accomplish this. Is there a way to do this using JSON?
Answer: Yes, we can use JSON to format the column in the list. But we cannot use JSON to format the list title.
We can add the custom code in the modern list and format the list title.
More references:
Use an SPFx Application Customizer to add JavaScript (e.g. header) to every page in a site
SPFx Applications Customiser CSS Injection
Comment: Please forgive me Amy. I missed you answer all this time. My sincere apologies and gracious thanks for your assistance.


Is the marginal distribution of a maximum entropy distribution also a maximum entropy distribution?

Question: I am considering under what circumstances maximum entropy distributions are closed under marginalization. The main case I am interested is the following setting:

a finite set of discrete random variables $X_{1},\ldots,X_{n}$. To simplify notation, in the examples below I just use two random variables $X,Y$. The random variables are not independent.
a set of sufficient statistics $T_{1}, \ldots, T_{m}$ where $T_{i}(x_{1},\ldots,x_{n})$ returns a real number
a set of mean values $\mu_{1},\ldots,\mu_{m}$

I would like to start with the maximum entropy distribution $P^{\ast}(X,Y)$ that satisfies the moment matching constraints
$$E_{P}[T_{i}(X,Y)] = \mu_{i}.$$
It is well-known that this distribution can be represented in exponential form. Now consider the marginal sufficient statistics
$$T'_{i}(x) = \sum_{y} T_{i}(x,y).$$ with associated moment matching constraints $E_{Q}[T'_{i}(X)] = \mu_{i}$ and maximum entropy solution $Q^{\ast}$. Letting $P^{\ast}(x) = \sum_{y} P^{\ast}(x,y)$ be the marginal distribution for $P^{\ast}$, do we have that
$$P^{\ast}(x) = Q^{\ast}(x)?
$$
If not true in general, are there conditions on the sufficient statistics that guarantee this condition?
Answer: Since $Q^*$ is of the form
$$Q^*(x)\propto \exp\left\{\sum_j \lambda_j \mathbb{E}[T_j(x,Y)|X=x]\right\}$$ and $P^*$ is of the form
$$P^*(x)\propto\sum_{y\in\mathcal Y} \exp\left\{\sum_j \zeta_j T_j(x,y)\right\}$$
and
there is no reason this marginalisation property holds.
Comment: Thank you for the reply. Actually $Q^{*}$ is the solution for the *marginal* sufficient statistics. So it is of the form

$$Q^*(x)\propto \exp\left\{\sum_j \zeta_j T'_j(x)\right\}$$
Comment: OK, I switched the notations then.


Who else can reply to an approval request beside the designated approver?

Question: For the standard Salesforce approval processes, who else can approve a request besides the designated approver?
Answer: Roles
If you've set the approval to "anybody from the Queue" make sure you're OK with who's in the Queue exactly. Easiest is to click "view all members". Roughly speaking expect that all managers of your queue members (all the way to CEO or whatever is your top role) will have access too.
I honestly don't know if it also works when an User is set as the actual approver (can his Manager do it? can somebody higher in the hierarchy do it?)
Edit: Ralph has tested this thoroughly and while higher Roles do become indirect queue members - this fact alone doesn't mean they can approve. Explicit access must be given or "modify all" permission for the object.
Profiles / Permission sets
As pointed out by Jesse - anybody with "modify all data".
Last but not least
User's own delegated approver, if anybody was set. That's for "what to do when I'm on holidays". Delegated approvers don't get notifications (unless they wish to do so, they can set it on their user record) and don't see the requests on home page. As far as I know if hey don't want spam - easiest for them is to periodically check reports.
Sysadmins (actually "modify all data" again) can view Administrative -> All pending approval requests
Delegates can view Activity -> My delegated approval requests.
Comment: Regarding modify all data, ran some tests and you can approve anything as long as you have modify all for the particular object
Comment: So i tried out the queue approach and only the members, not their managers are users higher in the role hierarchy, could approve.  Are you sure about that one?
Comment: It was at my previous client's place so I don't have access anymore but I recall that if Queue was approving - people higher could too. They weren't getting email notifications (that was sent only to "regular" members) but if they stumbled upon record in approval they were able to. I'm almost 100% sure they didn't have "modify all" on the object. And the approver was just a queue, nothing like "My Manager" or "let me choose manually". I will try to contact old colleagues.
Comment: pretty sure on this one, i setup a test approval process, assigned it to a queue with someone lower in the hierarchy, and then logged in as their manager (couldn't approve), and a user higher in the role hierarchy (couldn't approve)
Comment: Thx @Ralph, updated! You're 100% right... The person with rogue approval I had to inspect some time ago was higher in the food chain but also had "Modify all" on the object...
Answer: System Administrators can approve the records as well. I believe it is based on the Modify All permission on the object with the approval process.
Comment: anyone else besides administrators?
Answer: https://login.salesforce.com/help/doc/en/approvals_change_approval_user_pref.htm
A delegated user can also approve or reject .he cannot reassign but he can approve or reject


How can I encourage members of a small organization to contribute to a wiki?

Question: I am the web site administrator for a local ham radio club. Our old site was in a blog format, and only one active member of the club did anything with it. Almost two years ago, I set up a new wiki-based site as a test, and we later decided to make it our main site. It is set up to require an account to edit; we create accounts for any member who wants one. The goal is for as many club members as possible to add content; that way, the responsibility is shared among several people.
The problem is that nobody edits except me. I'm not sure what the exact ratio is, but the vast majority of edits on the site are mine. How can I encourage more people to edit the site?
I've already tried asking people if they want accounts. Most people say yes, so I create their accounts, then they never use them. I also tried writing out note cards with ideas for articles to write, setting those cards out at a meeting, and asking people to take any they might be interested in; one person took one card, but I don't think he ever wrote anything about it on the site.
I don't think the reason is that people are afraid of making mistakes. I've explained at length that it is very easy to fix bad edits, so I doubt that's the issue.
Most of our active members were at the meeting where we passed a resolution to "create a wiki account with full editing privileges for any member who requests one" and "encourage all members to make use of these privileges by contributing to the site, even if these contributions may need further improvements." Almost everyone, if not everyone, at that meeting voted in favor of that resolution, so I believe they do know that they have these privileges.
How can I encourage club members to start editing the web site?
Answer: I have a similar problem.  I've found that introducing new members to the site in person helps enormously.
Sit beside them and talk them through their first edit.
They soon become confident and start making contributions.  Also, they start to help others.
It was a slow process to get the ball rolling and I felt it might never happen, but in time it worked.


Is there a recommended logging path/folder for ASP.NEt applications?

Question: Currently we use log4net and create a new folder (usually below C:) with write and create access rights for the worker process user. Is there perhaps a standard windows folder we should or could be using such as LOCALAPPDATA? I'm guessing that LOCALAPPDATA is a bad choice in the general case because IIS users tend to be non-interactive users and thus I don't think they have the usual user folder structrues available.
Answer: Put your logs in the App_Data folder below your web site root.
Microsoft guidance:
To improve the security of the data used by your ASP.NET application, a new subfolder named App_Data has been added for ASP.NET applications. Files stored in the App_Data folder are not returned in response to direct HTTP requests, which makes the App_Data folder the recommended location for data stored with your application, including .mdf (SQL Server Express Edition), .mdb (Microsoft Access), or XML files. Note that when using the App_Data folder to store your application data, the identity of your application has read and write permissions to the App_Data folder.
Comment: The web application does not have write access to that directory by default! But here's now to add that: http://stackoverflow.com/a/7334485/143684
Answer: I would not use c:\ at all. You never know how big your logfiles may grow and one day your System Partition is full.
Other then that I don't know of a standard logging folder, but why not use the Database?
Comment: Logging to the database is not a good idea. In addition to adding a lot of extra database overhead, you wouldn't be able to log database connection failures. Writing files is the most robust way to log information.
Answer: Perhaps you might consider logging to a directory inside the app root. There the worker process certainly has permissions and the log data will be easily discoverable, perhaps even servable.


Looking for aux cable that splits surround into standard left/right/ground cables

Question: I have an old Dell Dimension 8300, and it comes with a 5.1 surround + subwoofer system. The speakers have proprietary connections that connect to the subwoofer, where the subwoofer has 3 audio jack inputs. I can only assume one is left, one is right, and one is ground. I want to know if there is an audio splitter for one aux port (or USB) that sustains the quality of the sound and the surround-ness of the sound.
Comment: 5.1 "Suround-ness" is dependent on **5** speakers FL, FR, CENTER, RL, RR, Plus a Sub woofer (**.1**)  to play sound that was mixed  to have different sounds in that mix sent to the the 5 speakers and sub woofer.  "*I can only assume one is left, one is right, and one is ground*"  Do NOT assume, get the manual or find it online. This will tell you what the connections are and possibly options for stereo sound. You can not really have surround sound out of 2.1  speaker set up.
Comment: If they are RCA ports then each wire/port has a  **+ and -**, (*positive and a minus*). they would not share a ground (-). Some older stereo's did that but it is unlikely that your system does. We do not know who made your system, or model #.   **Get the manual, do some research, join an Audio forum**.  https://www.avsforum.com/forums/
Answer: Welcome to Sound Design.
The three audio input jacks on the subwoofer are each two channel 3.5 mm TRS (tip-ring-sleeve) connections colour-coded according to the PC 99 standard:

Lime Green - Front left and right
Black - Rear left and right
Orange - centre and subwoofer (both mono)

The subwoofer channel is known as LFE (low frequency effects) in the Dell documentation. The outer braid in each cable carries the ground return.
These are all analogue audio connections, not USB. If you don't have the original cables 3 x stereo 3.5 mm to 3.5 mm TRS plug cables will be fine. Connect each cable between each one of the input jacks on the subwoofer and the output jack with the same colour surround on the back of the PC.
This will maintain the 5.1 surround sound.
If you want a 2.0 left/right stereo connection (without surround sound) just use a single cable to the lime green jack on your PC and tell your operating system that you are using 2.0 or stereo speakers. You must do this so that the rear left/right and subwoofer signals are sent to your stereo speakers. (In Windows use Control Panel -> Sound to configure your speaker connection.)
You can't have a single left/right stereo connection and maintain the surround sound effect.
I don't understand your mention of an audio splitter for an aux port connection. If I haven't addressed this above perhaps you could clarify your meaning. What is it that you are trying to do?


Should the bias value be added after convolution operation in CNNs?

Question: Should we add bias to each entry of the convolution then sum, or add bias once at end of calculating the convolution in CNNs?
Comment: what do you mean by "sum"? Can you express your question mathematically, its a bit vague from words.
Answer: Short answer: the bias is added once after the convolution has been calculated.
Long answer: discrete convolution that you see in CNNs is a linear function applied to pixel values in a small region of an image. The output of this linear function is then jammed through some nonlinearity (like ReLU). For a region $\mathbf{x}$ of size $i \times j$ of an image and a convolutional filter $\mathbf{k}$, and no bias term, this linear function $f$ would be defined as:
$$
f(\mathbf{x}, \mathbf{k}) = \mathbf{x}*\mathbf{k} = \sum_{i,j} k_{i,j} x_{i,j}
$$
Without a bias term, this linear function $f$ must go through the origin. In other words, if $\mathbf{x}$ or $\mathbf{k}$ is all zeroes, the output of $f$ will be zero as well. This may not be desirable, so we add a bias term $b$. This gives the model more flexibility by providing a value that is always added to the output of the convolution, regardless of the values of $\mathbf{x}$ and $\mathbf{k}$ -- in other words, it's the intercept value.
$$
f(\mathbf{x}, \mathbf{k}, b) = b + (\mathbf{x}*\mathbf{k}) = b + \sum_{i,j} k_{i,j} x_{i,j}
$$
If this value was added to each entry of the convolution, it would not achieve its purpose as $f$ would still necessarily go through the origin.
Answer: Based on the answer here and the blog post here there are two variants for using bias in convolutional layers. Tied biases if you use one bias per convolutional filter/kernel  and untied biases if you use one bias per kernel and output location.


Subsets of $\{1,2 \dots n\}$ with no consecutive integers

Question: How many subsets with cardinality k of $\{1, 2, \dots n\}$ contain no consecutive integers? I know that there are $F_{n+2}$ subsets of $\{1, 2, \dots n\}$ with no consecutive integers, but I do not know how to go about finding the number for a given $k$.
Answer: This is equivalent to chooosing a sequence of $k$ ones and $n-k$ zeroes with no adjacent ones. An example with $n=8$ and $k=3$ is
$$
00101001, \text{ corresponding to the set }\{3,5,8\}
$$
To choose such a sequence, start with a string of $n-k$ zeroes, with $n-k-1$ spaces between the zeroes, plus two extra spaces before and after, for $n-k+1$ spaces total:
$$
\;\_\; 0\;\_\;0\;\_\;0\;\_\;0\;\_\;0\;\_\qquad,\text{ with $8-3+1=6$ gaps}.
$$
Each of the $k$ $1$'s goes into exactly one gap. We need to choose $k$ of these gaps to put a $1$ in. This can be done in
$$
\binom{n-k+1}{k}\text{ ways.}
$$
Answer: Without loss of generality, let your $k$ elements from a selected subset be $x_1<x_2<x_3<\dots<x_k$
Given such a $k$-tuple $(x_1,x_2,\dots,x_k)$, construct the related $(k+1)$-tuple $(x_1-1, x_2-x_1, x_3-x_2,\dots, x_k-x_{k-1}, n-x_k)$ describing the distance between each number and/or the boundaries in the case of the first and last numbers.
Renaming those values in the $(k+1)$-tuple $(y_1,y_2,\dots,y_{k+1})$ we can recognize that $y_1+y_2+\dots+y_{k+1} = n-1$ as the sum telescopes.
Now, consider the related problem of finding the number of integer solutions to the system:
$$\begin{cases}y_1+y_2+\dots+y_{k+1} = n-1\\ y_1\geq 0\\ y_{k+1}\geq 0\\ y_i\geq 2~~~\text{for all other }i\end{cases}$$
The inequalities here coming from that the elements may not be consecutive.  This should now be in a known problem format for you or can be slightly modified further with another change of variable to be in a known format and the problem can be completed using stars-and-bars.
Answer: We select the first value of the set:
$$\frac{z}{1-z}$$
followed by $k-1$ differences that are at least two:
$$\frac{z}{1-z} \left(\frac{z^2}{1-z}\right)^{k-1}$$
and we conclude by collecting the count for all subsets with maximum
element $\le n:$
$$[z^n] \frac{1}{1-z} \times \frac{z}{1-z}
\left(\frac{z^2}{1-z}\right)^{k-1}.$$
This is
$$[z^n] \frac{z^{2k-1}}{(1-z)^{k+1}}
= [z^{n+1-2k}] \frac{1}{(1-z)^{k+1}}
= {n+1-2k+k\choose k}$$
or equivalently
$$\bbox[5px,border:2px solid #00A000]{
{n+1-k\choose k}.}$$
We get for the total
$$\sum_{k=0}^{\lfloor (n+1)/2 \rfloor} {n+1-k\choose k}
= \sum_{k=0}^{\lfloor (n+1)/2 \rfloor} [z^{n+1-2k}] \frac{1}{(1-z)^{k+1}}
\\ = [z^{n+1}] \frac{1}{1-z}
\sum_{k=0}^{\lfloor (n+1)/2 \rfloor} z^{2k} \frac{1}{(1-z)^{k}}.$$
Here the coefficient extractor enforces the range and we may continue
with
$$[z^{n+1}] \frac{1}{1-z}
\sum_{k\ge 0} z^{2k} \frac{1}{(1-z)^{k}}
\\ = [z^{n+1}] \frac{1}{1-z} \frac{1}{1-z^2/(1-z)}
\\ = [z^{n+1}] \frac{1}{1-z-z^2}
= [z^{n+2}] \frac{z}{1-z-z^2} = F_{n+2}.$$
The above construction works for $k\ge 1.$ For $k=0$ we get the empty set, for a total count of one. Note however that ${n+1\choose 0} = 1$ so the formula holds there as well.
Answer: Hint: Choose $k$ pairs of consecutive numbers from $\{1, 2, \ldots, n, n+1\}$, then choose the lowest number in each pair.


Safety flying in Dubai air space

Question: How safe is flying through Dubai air space considering the current unrest between the USA and Iran?
We are booked through to UK in July.
Answer: Exactly as safe as it ever was. If there were an unacceptably higher risk, the airline would take measures to mitigate the risk, or stop flying in the risky area(s). They absolutely do not want another MH17.
Answer: Currently it looks like Emirates flights are overflying Iraq, avoiding both Iran and Syria. For example, this flight track from EK164 taken a few minutes ago.
AFAIK there are no particular safety concerns flying within the UAE, so I presume you meant flights to/from UAE.
Comment: A few Emirates flights towards Europe do fly over Iran instead of Iraq, with no clear pattern to which. Right now on Flightradar24, I see EK91 (A380 to Milan) EK85 (A380 to Zurich), EK163 (B777 to Dublin), EK181 (B777 to Brussels) over Iran. [At the same time there's a column of nine Emirates A380s in progress north over Iraq, so it looks like _most_ go that way].

Safety flying in Dubai air space


By byte size and number of inputs/outputs, what are the largest transactions in the block chain so far?

Question: Disregarding value, but instead going by size in the protocol, what are the largest transactions so far in the history of Bitcoin, in terms of:

byte size
number of inputs ("vin")
number of outputs ("vout")
total number of inputs and outputs
Answer:
The largest single transaction to date is 465,554 bytes.  It's hash is 659135664894e50040830edb516a76f704fd2be409ecd8d1ea9916c002ab28a2.
The same transaction has the largest number of inputs: 2,585.
Two transactions tie for the largest number of outputs: 623463a2a8a949e0590ffe6b2fd3e4e1028b2b99c747e82e899da4485eb0b6be and 5143cf232576ae53e8991ca389334563f14ea7a7c507a3e081fbef2538c84f6e both have 3,075 outputs
The same two transactions have the largest number of inputs plus outputs, although they only have a single input so the total is 3,076 inputs plus outputs.

Note, however, that the blockchain has been used to store information by manipulating output addresses so it is more than likely that the transactions listed in 3) are not "real", in that they do not contain spendable outputs.
Comment: Limits on transaction sizes suggest that it would be hard to create a transaction this big nowadays. There is a 100K limit on "standard" transactions [See here](https://github.com/bitcoin/bitcoin/blob/master/src/main.h#L57)
Answer: The largest transaction by size to date was the Mega Transaction mined by f2pool on 7 July 2015:
It was 999,657 bytes, had 5,569 inputs and 1 output.
The transaction filled a complete block by itself and that block was widely cited as taking 25 seconds to validate. The transaction caused a lot of discussion back then, and prompted e.g. this blog post about the Mega Transaction's validation time by Rusty Russell.
Answer: Scripts to find it out yourself + preprocessed data
I have uploaded scripts that extract most of the data you ask at: https://github.com/cirosantilli/bitcoin-strings-with-txids including the top 10k transactions for each category.
Those scripts rely on https://github.com/alecalve/python-bitcoin-blockchain-parser which parses the blockchain data offline and without the need for a running local server.
The scripts could quickly update the data at any time efficiently from the last snapshot, and this is what I've found when I last updated on February 2021:

byte size: list at data/tx_size_bytes:  bb41a757f405890fb0f5856228e23b715702d714d59bf2b1feb70d8b2b4e3e08 (Jul 2015) with 999,657 bytes, it regroups a bunch of smaller outputs into one big one. The values are small, so it's likely made for the data and not value.
This transaction is sometimes dubbed "The Megatransaction", and is mentioned e.g. at:

https://en.bitcoin.it/wiki/July_2015_flood_attack:

The July 2015 flood attack was a large "stress test" of the Bitcoin network. The possibly distributed attack has provoked hundreds of thousands of transactions, leaving over 80,000 in the mempool at one time. The attack is subsequent to stress tests executed in June.

Longest-time-verifying block: unsurprisingly it also took the longest to verify

number of inputs: data/tx_nins: f2e197a6d8d088b13afd0f99d4027da36a9413b9f3d7730ba5278132ebc950a7 (Aug 2015) + 8 other tied with 20,000

number of outputs: data/tx_nouts: dd9f6bbf80ab36b722ca95d93268667a3ea6938288e0d4cf0e7d2e28a7a91ab3 (Jan 2016) with 13,107 outputs. Some of them have been spent, so it is not all just data embedding
Answer: i found transactions with 20,000 inputs
cf1032c2213e6faea04f1813aa6890e7f588bb378cb98e7425aec83c11d4457c
30b3b19b4d14fae79b5d55516e93f7399e7eccd87403b8dc048ea4f49130595a
f2e197a6d8d088b13afd0f99d4027da36a9413b9f3d7730ba5278132ebc950a7
8dabbf51f78c1e7286866af1de403118c5ddbe57ca93b54859245916d2bf1063
740ac533882221099e7202bbdafbb99ec589c6e74fd2fe7ca1274b46ea4f0a96
dd6067e71c04cb62f8e5aa52ecc99b01ffcd551a52727d046a2fabb14eb39b4d


Is there such a thing as "Action at a distance"?

Question: What ever happened to "action at a distance" in entangled quantum states, i.e. the Einstein-Rosen-Podolsky (EPR) paradox? I thought they argued that in principle one could communicate faster than speed of light with entangled, separated states when one wave function gets collapsed. I imagine this paradox has been resolved, but I don't know the reference.
Comment: I believe most of these correlations are related to symmetries, or per Emmy Noether, conservation. Is this correct? Also, we are assuming 3 dimensions in assessing our distances. I'm sorry if I only added more questions.
Answer: It's not possible to communicate faster than light using entangled states. All you get out of entanglement is a correlation between the values of two measurements.; the entanglement doesn't allow you to influence the value measured at another location in a non-causal way. In other words, the correlation only becomes evident after combining the results from the measurements afterwards, for which you need classical information transfer.
For example, consider the thought experiment described on the Wikipedia page for the EPR paradox: a neutral pion decays into an electron and a positron, emitting them in opposite directions and with opposite spins. However, the actual value of the spin is undetermined, so with respect to a spin measurement along a chosen axis, the electron and positron are in the state
$$\frac{1}{\sqrt{2}}\left(|e^+ \uparrow\rangle|e^- \downarrow\rangle + |e^+ \downarrow\rangle|e^- \uparrow\rangle\right)$$
Suppose you measure the spin of the positron along this chosen axis. If you measure $\uparrow$, then the state will collapse to $|e^+ \uparrow\rangle|e^- \downarrow\rangle$, which determines that the spin of the electron must be $\downarrow$; and vice versa. So if you and the other person (who is measuring the electron spin) get together and compare measurements afterwards, you'll always find that you've made opposite measurements for the spins. But there is no way to control which value you measure for the spin of the positron, which is what you'd need to do to send information. As long as the other person doesn't know what the result of your measurement is, he can't attach any informational value to either result for his measurement.
Comment: Excellent answer, but I think there is something still missing. You can still naively argue that the results of all the measurements were predetermined when the electron and the positron were emitted, and therefor there is nothing "special" about the correlation. To answer this claim you need Bell's theorem, as explained by [this great answer](http://physics.stackexchange.com/questions/3158/why-quantum-entanglement-is-considered-to-be-active-link-between-particles/3433#3433).
Comment: Are you saying that, after you measure one particle (and get a truly random, no-hidden-variables result), when you later measure the other particle it will give a result that is both truly-random *and* correlated to the first one? Isn't lack of correlations precisely what random means?
Comment: @JuanPerez Absolutely not. A process can be random and still give correlations between different measurements. (The word used in statistics for a lack of correlations is _independent_, not _random_.)
Answer: Well, the problem in that paradox is that yes, one of the parties will measure the entangled particle to get the wave function collapsed and yes it will collapse for the other party. However, the other party will still have to measure the thing to learn what it is or has to wait for the initial party to send them a message telling what the wave function has collapsed to. The first method will result in a 50% +x and 50% -x(if it is spin you are measuring), as the wave function that collapsed can collapse to either one of these states. So the fact that the wave function collapsed does not really transfer any usable information to the other side. The second method is capped with the speed of light anyway.
Comment: Of course they can't transfer information faster then the speed of light because of the no-signalling theorem. However, we should be careful of how we state the first scenario! In some games, for instance CHSH, you can win the game %75 of the times classically, whereas you can do ~%80 using quantum theory!
Answer: Let's be more rigorous. No-signalling has been proven safely and shouldn't be worried about. Nevertheless, you'd notice that the point of EPR paper was to show that if quantum mechanics is considered to be a description of "reality", then it is "incomplete". There is an approach, such as in operationalism, to say quantum mechanics isn't meant to be a description of reality. It's a description of our knowledge of reality, due to Asher Peres. Another approach is to say we can give an ontological model of quantum theory using contextual hidden variables, such as the one in de Broglie-Bohm model.
So conclusion: EPR argument hasn't been resolved if you mean it's gone! Because in fact orthodox quantum mechanics isn't a complete description of reality. However, it doesn't mean one can signal faster than light!
Some interesting extra information: There is an interesting paper which kind of analyses Einstein's argument. It bring historical facts that show Einstein didn't like the EPR and wrote another paper with the same title in correspondence with Schrödinger, and the one with Rosen and Podolsky was never reviewed by him.
This quotation is from a letter of Einstein to Schrödinger, dated June 19, 1935:

"For reasons of language this [paper] was written by Podolsky after many discussions. But still it has not come out as well as I really wanted; on the contrary, the main point was, so to speak, buried by the erudition."

Update: A source of confusion in my answer has been pointed out by Marek. I'll try to clarify here:
Scientific realism assumes there is an underlying objective reality which has attributes regardless of them being measured by an observer.
One can suggest a model which ignores such reality and say "...there is no logical necessity for a realistic worldview to always be obtainable"(Fuchs and Peres, Physics Today 53 (3), 70-71.).
On the other hand, one can offer an ontological model which in this case it can be located in 3 different category as in figure below:

I believe Einstein had in mind to show that quantum mechanics can't give a picture of type (a). Which was successful. Because, even if there is an underlying reality, quantum states can't sharply specify them by any means.
Comment: @Sine: nice overview but one thing has to be made more precise: "quantum mechanics is not complete description of reality" has to be specified as "quantum mechanics doesn't allow us to know everything we want to know". But that's because (as far as we know) reality is inherently quantum in nature (in particular, there can be no local hidden parameter theory) and so quantum theory certainly *is* complete description of reality in this sense.
Comment: that's a great piece of historical trivia. Einstein's most highly cited paper (by a huge margin!) wasn't even written by him.
Comment: @Marek: The update is my attempt to clarify what I meant by completeness.
Comment: @Sina: uh, I have to say I am confused now. I think the picture a) is correct as long as one works in good interpretation of quantum theory (in particular, one has to dispose of the notion of wave-function collapse). Complete state of system indeed is specified by a ray in Hilbert space and there is no more information to be found anywhere. On the other hand, picture b) is surely incorrect if it means to be a hidden parameter theory. As for picture c), I am uncertain about what it means precisely. But again it feels like hidden parameter theory and so is ruled out.
Comment: Hmmm... I understand it's not entirely clear from the figure only! It's derived from the paper I linked in the body of my answer. I was intending to point to where one should look at, in that paper to find an adequate answer. It is explained in page 5. Nevertheless, I guess all this is just complimentary information. I think I made the point by just saying No-signalling theorem shows that we can't signal faster than light using non-local features of quantum mechanics. :-)
Comment: @Sine: thanks, I didn't notice this figure was to be considered part of that paper; I'll look at it. And yes, this is just complementary discussion and I find your answer satisfactory. Still, I always find it useful to discuss QM interpretations and such because it often (re-)convinces me I don't understand quantum theory at all :-)
Answer: One of the fundamental tenets of relativity is that all observers are correct. Two people pass each other near the speed of light and they will each think that the other person's clock is running more slowly than their own. Who's right? They both are. It doesn't matter if you have to bend and warp and crumple the universe itself in order to make them both correct, and in fact that is exactly what Einstein did with general relativity. Now if you have a photon moving at the speed of light, it observes infinite longitudinal length contraction, meaning it sees the universe as an infinitely thin sheet of paper that it is passing through. From its own perspective, it is created and absorbed at a single point in space, all during a single moment of time. What we perceive as spatial extent and temporal duration of the photon's path are both, and only, consequences of relativistic dilation from our own perspective. Now consider three photons having mutually orthogonal trajectories. Each thinks the universe is infinitely thin in its direction of motion, and each traverses that universe in a single instant. They must all be right. This happens only if the universe is itself only a single point which lasts only for a single moment. (If this last conclusion is troubling, it might be helpful to presume that the mutually orthogonal entangled emissions arise together from a single event.)
Now comes the Einstein-Podolsky-Rosen paradox: how can two particles, which themselves have spins in absolutely as-yet undefined directions, after traveling light-years apart, "decide" that the instant one of them is measured, thus collapsing itself into a totally random distinct spin state, then the other one, no matter how far away, and no matter how much time later, will always, once measured, collapse into the opposite spin state in order to preserve total angular momentum? Einstein called this "spooky action at a distance" and believed that it couldn't happen -- that the particles, which cannot communicate with each other (else that information would have to impossibly travel faster than light), must have some "hidden variable" inside them which determines which spin state will arise, once measured. However, the Bell Inequality, which I won't describe here, proves inescapably that there cannot be an Einstein "hidden variable" making the determination. What then are we to do? Even if we have to bend and warp and crumple the universe itself, Einstein and Bell must both be right. And that is exactly what we have to do. The particles do not engage in "spooky action at a distance" because they are both at the same point in space, and at the same moment in time, when they are measured. That's because the universe in which they reside is itself a single point in space and a single moment in time. This is a little different from the "non-locality" of the universe asserted in some literature. Here, the universe is entirely local, but all at a single point of spacetime. This then has been just one person's personal perspective on the EPR paradox, but something similar has been said before: a thousand years is like unto a day, and the heavens can be measured in the palm of the hand.
Michael C. Kleder, 7 Feb 2021
Answer: Everybody misses the point Einstein was trying to make, which makes it all the more remarkable that it's been 80 years since he was working on spooky action at a distance. The no signalling theorems mean nothing, and it's a shame that most answers simply site: no signalling, nothing spooky about it. Bell emphasized that his theorem could be quickly summarized as: there is non-locality. Guess what? Bell was very well aware of the no signalling theorems. The point is not that we can send signals, the point is that there is a signal sent by the photons themselves -- they have to be communicating. How else could they always coordinate their spins? Einstein's whole critique of quantum mechanics was that it needed to be like Bertleman's socks -- e.g. that the spins were already determined before the experiment, or else there would have to be a non-local communication to coordinate the spins. Einstein called it telepathy, and it's been proven by Bell.
If you don't think there's spooky action, then how do you explain that the spins are always coordinated? If you gave Alice and Bob each a quarter and separated them by a large distance and tasked them with choosing heads or tells, and they always came back to you with one choosing heads, the other choosing tails, what might you think? Maybe they talked to each other on a phone and coordinated their results?
Answer: Being silly, in the vein of the season, if you agreed before hand that if in David's example your colleague could open their christmas present on the other side of the world if their electron was spin up, and the same time they measured their electron you measured your positron as spin down, you now know they have opened their present, faster than the speed of light
Instantaneous data transfer!
Comment: @Soulman: you don't need quantum present for this, classical present would do just fine. If you put blue teddy bear in one box and red teddy in the other and then ask your friend to take one of them with them then as soon as you open your present and find blue teddy, BAM you *instanteously* know that your friend has the red one. I think people often don't realize, that classical physics can sometimes contain same paradoxes quantum physics can. And also that problem with entanglement and collapse lies in something completely different (cont).
Comment: This is Berkelmann's always-mismatched socks, used by Bell to demonstrate nonlocal correlation. When you see one is red, you instantly know the other isn't. This isn't how quantum entanglement works, because this type of thing is determined by local hidden variables (the color of the socks) and so can't violate Bell's inequality.
Comment: -1  I guess this is a joke, but it's not instantaneous data transfer, and some people could find it confusing.
Comment: sorry mark! was gonna qualify it as false, but i thought it was pretty self-explanatory. For those that don't understand - you have no 'proof' the present is open, only faith that those at the other end did whay they were told, and weren't drunk on egg nog, kissing the undergrad
Comment: haha very nice indeed! (unfortunately i won't be giving a +1 because it's not really a correct answer)
Comment: hah. that's funny and apt for the season !
Comment: @SoulmanZ Got it.  If you edit I can un-downvote.
Comment: hah. nah, i dont mind, it probably deserves to be at the bottom anyway!
Comment: @Soulman: namely, that you can measure spin on the other side along different axis and the result would still be correlated. Unfortunately, this is not possible to do with your classical teddy bear. You need quantum teddy bear for that :-)


How to warm up for a PR attempt

Question: I failed a 120kg bench PR a few weeks ago, after which I went down to 117.5 and succeeded, so I know I was very close. I decided to wait 4 weeks and try again which I'll be doing soon. One vital part of a PR attempt that I've never felt I've managed to perfect if the warm up I.e doing enough set to warm up sufficiently but keeping the intensity low enough to keep fatigue to a minimum. I know of course individual variation will mean there's no one right answer but is there a generally accepted method, specifically for a PR attempt to maximise power output while by meeting the 2 conditions I mentioned above. Essentially what I'm asking is what weight do I use (in term of % of my 1RM), what reps ranges should I use and how many warm up sets do I need?
Answer: There's no definitive answer to this question, just that it should be enough to physically warm you up (especially in a cold climate), gradually increase the loading to the muscles, allow practice of the movement pattern, and re-familiarise you with holding onto heavy weights, while not being so much that it fatigues you and adversely affects your 1RM performance.
This would typically start with a set or sets using just the empty bar, and then over several sets, increase the weight while decreasing the reps, ending with a final warm-up of perhaps one rep at a 3-4RM load. Or, if the test is in the context of powerlifting competition, then you'd probably do something more like warm up to a single at a 4-5RM load, and then perform lifts at predicted 3RM, 2RM and 1RM loads for your three competition attempts, since in that context you need to make three attempts that count, rather than just one.
So for a 120kg bench press attempt, it might look like:

Empty bar for 5-10 reps. (Repeat if cold)
60kg for 5 reps. (Repeat if cold)
80kg for 3 reps.
100kg for 1 rep, then minimum 1 minute rest.
105-110kg for 1 rep, then minimum 3-4 minutes rest.
120kg attempt.

If there are any specific requirements for the PR attempt, such as pausing on the chest or waiting for commands (as would be the case in powerlifting competition), then you should attempt to replicate those as much as possible in at least the 1 rep warm-up sets.
These weights and rep ranges are not at all prescriptive, and should not be scaled to other target weights. Rather they are just chosen as examples using round numbered weights for ease of loading. There's nothing magical about the third warm-up weight being 2/3 of the final attempt weight, and if the final attempt were 130kg instead of 120kg, I'd probably leave everything exactly the same, except maybe bump up the weight on the final two warm-up singles a bit, so that they were heavy enough to be significant for someone who can bench 130kg, but not too fatiguing.
To generalise this to lighter of heavier weights, I'd choose the same starting point of the empty bar (except for deadlifts, where you need a pair of full-size plates on the bar to position it at the correct height), then something like sets of 5 at increments of 40kg/90lbs, reducing the reps when you're at around 2/3 of the target weight, and then a single in each of the ranges of 75-85% 1RM and 85-90% 1RM. If it's a very light attempt, then obviously you'd need smaller increments than 40kg, and may need to reduce the number of warm-up singles. E.g. For a 40kg attempt, you might do 2 sets of 5 with the empty bar, 3 reps at 30kg, 1 rep at 35kg, and then the 40kg attempt.
Comment: Thank you, this method definitely worked, I hit my target PR and even exceeded it with 122.5kg. When using this for a considerably weaker lift like shoulder press, would I need to reduce it accordingly. My expected 1RM for this lift is 80kg so should I use the same method but reduced weight I.e. bar, 40, 60, 70, 75, then 80?
Comment: @Ethan I've edited my answer to generalise this warm-up schedule to lighter or heavier weights.
Answer: Let's say you'd like to hit a particular weight n for a new gym PR. For this situation, I find that first working backwards from n works well with the 91%/96% rule of thumb. (These numbers come from study on successful weight selection attempts on powerlifters[1]).
Then all warm up before that should be selected such that it satisfies the following criteria:

gets you warmed up
gets you in the right groove for the heavier weight
doesn't get you fatigued

I'd like to use very light weights (bar to 55% of 1RM) to get warm. Once the weights are over 55%, I already want to be warm. The warm-ups between 55% and the 91% can be used to grease the groove and get used to the heavier weights.
The number of sets and reps you need to get warm and to get used to the heavier weights depend on many factors. For example:

room temperature
other warm up you did before
type of life
lifting experience
etc.

To put this into practice, for a 120 kg bench:
* 100%    1x 120kg PR
* 96%     1x 115kg (rounded down from 115.2kg) last warm up
* 91%     1x 110kg (rounded up from 109.2kg)
* 56-90%  however many singles you need to grease to the groove
* bar-55% whatever you need to get warmed up

[1] Travis SK, Zourdos MC, Bazyler CD. Weight Selection Attempts of Elite Classic Powerlifters. Percept Mot Skills. 2021 Feb;128(1):507-521. doi: 10.1177/0031512520967608. Epub 2020 Oct 23. PMID: 33095691.

How to warm up for a PR attempt


Can a valid student visa and study permit for Canada be used as a tourist visa?

Question: I have a valid Study Permit and a student visa (S-1 TRV) for Canada, but I no longer attend the university in Canada and have transferred to a college in America. Can I use my student visa to enter Canada as a tourist or do I need to separately apply for a tourist visa?
Comment: Most countries _do not_ permit this.
Comment: When you withdrew, wasn't that reported to the CIC with the effect that the study permit is no longer valid? As is noted by @pnuts link, wouldn't you need to be in status in order to return to Canada (and risk being refused entry at the border)?
Comment: It says you need a tourist visa but then goes on to say that you must reapply for a TRV. So, can my valid S1 TRV be used as a tourist visa?
Comment: The link posted by @pnuts is about traveling internationally *during* your studies.  So it does not seem to be relevant to this question.
Answer: Canada does not have a separate Student Visa. When you apply to study in Canada you are issued a Temporary Residence Visa (a normal visitor visa) and a letter. The letter is exchanged at the border for a Study Permit, which, together with the Temporary Residence Visa allows you to study in Canada.
If you decline to study, or finish your course, you can still use the visitor visa for tourism purposes while it is valid.
Answer: To answer my own question after entering Canada a couple of times on the student visa as a visitor: yes. One border officer said that as long as the Student Visa is valid it can be used a Tourist Visa since they're both a TRV. I did end up getting a tourist visa eventually since I wanted a longer validity period.
Comment: Welcome. A fantastic first question and answer on this site.
Answer: As you are no longer enrolled, your student visa is invalid and cannot be used for travel to Canada; as it only allows travel for the purposes of studies.
If otherwise ineligible, you will have to apply for a normal visit/tourist visa to Canada.
Remember, a visa is not a right to enter the country. It is a permit to request entry based on the premises under which the visa (permission) is granted.
Comment: This may be true in other countries but it is not how the Canadian visa system works.


Alexa audio streaming skill http

Question: Can somebody explain how get http url to https url. I wont to put SHOUTcast stream radio url (http://live.narodni.hr:8063/stream) in Alexa skill but can't figure out how. :(
Skill work great with https url. but I want specific radio to use in Alexa skill
Please help
Answer: Use of HTTP vs HTTPS (secured HTTP) is determined by the owner of the service.
Unless the person that owns the service has a HTTPS endpoint, you'll be stuck running on HTTP.
Normally when you make a request a HTTP endpoint, it'll default to port 80 whereas HTTPS endpoints default to port 443.
From your URL above, it looks like the owner of live.narodni.hr opened up port 8063 and used that port to listen for HTTP requests. If you can find if and where they have a HTTPS port, then that should work.
You may want to to just try https://live.narodni.hr/stream which will use port 443 which is equivalent to https://live.narodni.hr:443/stream. For testing, some people like to use port 8443 too (8080 for http), so you can try https://live.narodni.hr:8443/stream.
TL;DR - it's up to the service owner to enable/support an HTTPS endpoint. Find that endpoint and you'll be good.

Alexa audio streaming skill http


Is a process matrix of rank $1$ unique?

Question: It is said that when an unknown process is unitary, its $\chi$ matrix is rank-$1$ and possesses only one positive eigenvalue. See eg  https://arxiv.org/abs/2306.07867. So when the process matrix has rank $1$, is the matrix unique? If the process matrix is not unique, what is the relationship between these rank-$1$ process matrices?
Answer: TL;DR: The elements of the process matrix with respect to an operator basis $E_i$ are just the coefficients in the expansion of the channel, viewed as an operator on $\mathcal{H}\otimes\mathcal{H}$, in the basis $E_i\otimes\overline{E_j}$. Therefore, the process matrix depends on the chosen operator basis, but is otherwise unique. This is true irrespective of its rank.
Definition
The process matrix of a channel $\mathcal{E}(\rho)=\sum_kK_k\rho K_k^\dagger$ with respect to an operator basis $E_i$ can be obtained by expressing the Kraus operators as a linear combination $K_k=\sum_ia_{ki}E_i$ of the basis elements $E_i$ and collecting the terms $E_i\rho E_j^\dagger$ from all Kraus operators
$$
\mathcal{E}(\rho)=\sum_{ijk}a_{ki}\overline{a_{kj}}E_i\rho E_j^\dagger=\sum_{ij}\chi_{ij}E_i\rho E_j^\dagger\tag1
$$
where $\chi_{ij}:=\sum_ka_{ki}\overline{a_{kj}}$ is called the process matrix.
Uniqueness
Now, $E_i$ is a basis in the Hilbert space $L(\mathcal{H})$, so $E_i\otimes\overline{E_j}$ is a basis in the Hilbert space $L(\mathcal{H}\otimes\mathcal{H})$. By vectorizing and suppressing $\rho$ in $(1)$, we obtain an equality between a representation$^1$ of $\mathcal{E}$ as an operator on $\mathcal{H}\otimes\mathcal{H}$ on the LHS and a linear combination of the basis elements $E_i\otimes\overline{E_j}$ with $\chi_{ij}$ as the coefficients on the RHS. The representation of $\mathcal{E}$ on the LHS is unique$^2$. The coefficients in the linear combination on the RHS depend on the basis, but are otherwise unique. Thus, the process matrix of a channel depends on the chosen operator basis $E_i$, but is otherwise$^3$ unique.
Rank
The definition $\chi_{ij}:=\sum_ka_{ki}\overline{a_{kj}}$ implies that the rank of $\chi_{ij}$ is the Choi rank of $\mathcal{E}$. In particular, the rank of the process matrix of a unitary channel is necessarily one.

$^1$ This is sometimes denoted with $K(\mathcal{E})$ and called the natural representation of the channel $\mathcal{E}$. Intuitively, it is what $\mathcal{E}$ becomes when we forget that it acts on operators and pretend that it is merely a linear operator on vectors (stripped of their own operatorial structure by vectorization).
$^2$ Since vectorization is a bijection.
$^3$ Sometimes folks implicitly assume that $E_i$ is the Pauli basis. Obviously, under this more restrictive definition, the process matrix is unique without any qualification.
Comment: Thank you, I see. But there can be lots of unitary processes, right? Can you give some examples of unitary process?
Comment: Yes, there are infinitely many unitary processes. Every quantum gate is an example.
Comment: It is said that if the map is unitary, there is only one term of Kraus operators, namely, $\mathcal{E}(\rho)=U\rho U^{\dagger}$. So if there are infinitely many unitary processes, it means that it is common we don't need the sum of the Kraus representation. And when it comes to the noise channel, we can use the sum. Therefore, the Kraus representation can be used to represent every quantum channel, and when the channel is quantum gate, it is only one term. Right?
Comment: Yes, that's right.
Comment: Ok. Thanks. : )

Is a process matrix of rank $1$ unique?


What do you call this instance of break in noise?

Question: You're in the gym waiting for the school principal to make a speech and everyone is talking, the whole place humming loudly. Suddenly it goes quiet because you all thought the principal has arrived, yet no one told you to be silent. Everybody else notices the others being quite and follows suit. Everyone is silent for a moment, but after they realize no one arrived and no one is enforcing the silence, the noise starts again. This has happened to me in classrooms and at seminars.
Comment: *there was a sudden hush*
Comment: In OP's *specific* context, a ***premature** hush* (a "false alarm", or "false start", but idiomatically there's no such thing as a "false **stop**").
Answer:
lull - a temporary calm, quiet, or stillness
Comment: A *lull* sounds more like a coincidental break in conversation.
Answer:
The air left the room

... is a metaphor often used to describe the sudden quiet when everybody's attention is drawn to someone or something.  It sounds like you are describing the same kind of quiet, even though the someone is only imagined.
For example:

The air left the room for a moment, until we realized that the principal had not, in fact, arrived.


Bipolar Junction Transistors

Question: If the switching frequency is increased for BJT what happens to the power dissipation? Will it increase, decrease, or remains the same and whats the logical reason for it?
Answer: To add on the @Barry answer, switching losses also depend on time at which you switch, ie. The amount of current or voltage across the switch. This is why in high efficiency applications, power electronics designers choose to employ what is classed as zero voltage switching or zero current switching.
To better understand this we can refer to the graph below:

Whenever you give a command (ON or OFF) it takes time for the device to achieve its final state. This is to do with movement and recombination of charges within the device.
So lets assume there is a certain voltage across the switching device, and the switch is Turned on. The voltage takes time to drop and similarly the current takes time to achieve its final value. Using \$P=VI\$ we can determine the switching loss.
The model above is highly simplified, in reality the curves are non linear.
Answer: Your question is very general so only a general answer can be given.  If you assume the BJT dissipates a certain amount of power every time it switches states, then the average power it dissipates is proportional to the number of switches per unit time.  Therefore, if you increase the switching frequency, the power dissipation will increase.

Bipolar Junction Transistors


Vista 32-Bit Home Premium will not let me install any updates

Question: Windows update says that it could not search for new updates and gives the error code 80070643. From what I have read it has something to do with Office 2003 and I dont even have it installed. I tried downloading and installing vista sp2 but it fails as well giving me error: ERROR_EVT_CONFIGURATION_ERROR(0x89973aa2). Can someone please help me resolve this issue?
Comment: Can you make sure BITS, background intelligent transfer service is running & set it to run automatically.
Comment: Do you happen to have ZoneAlarm installed on your computer?
Comment: BITS is running and set to auto. No I do not have zonealarm installed.
Answer: You might want to read up on this article about the Windows Update error 80070643. If you were installing Microsoft Office 2003 updates when you received the error, follow the instructions in the "Microsoft Office 2003" section of the article. Otherwise, follow the instructions in the "Windows .NET Framework" section.
Good luck, and hope this helps some.
Comment: In the article you give, they says it could be a .net framework issue too. As he don't have installed office 2003, I think it's more relevant...
Comment: I have tried both of these and neither of them seem to work. Any other suggestions, I am trying to avoid re-imaging this laptop. I know theres got to be a way.
Answer: Windows Update error 80070643 could also be a .NET framework problem... Follow the link gived by Chris (Windows Update error 80070643)
If that doesn't help, you could try Fix WU, this utility will re-register a total of 114 .dll,  ocx and .ax files which are required for the proper functioning of Windows Updates.


How does digital signatures provide integrity?

Question: Digital signature provides authentication by encrypting the text with the sender's private key and decrypting the text with the sender's public key. But how do we add integrity to it so that the text sent by the sender does not vary at the receiving end?
Comment: That's not how signatures work. [Here's](https://security.stackexchange.com/q/183347/151903) a similar recent question, though there's probably a better duplicate if you dig.
Comment: You use the words `data` and `text`. Are you referring to two different things or the same things?
Comment: https://searchsecurity.techtarget.com/definition/digital-signature has a good explanation. The use of the hash is key in digital signatures.
Comment: @NeilSmithline Not sure I'd call that explanation good. `The reason for encrypting the hash instead of the entire message or document is that a hash function can convert an arbitrary input into a fixed length value, which is usually much shorter.` It's misleading people into thinking that the purpose of the hash is to produce a shorter value to "encrypt", which misleads people into thinking the hash might not be necessary when signing smaller data such as in the question I linked. It's also perpetuating the common misconception of signing as "encryption".
Comment: @AndrolGenhald - Fair enough. I picked it because it clearly discussed the hash something the OP, seems to have missed.
Answer: Your question implies a slight misunderstanding. First, to clear that up: Digital signatures do not encrypt the original data. You can choose to both sign and encrypt the original data, but those are two totally distinct processes. I will address signed and encrypted messages at the end. The original data is called the plaintext below, which is the official term.
A standard digital signature works by hashing the plaintext first. For message signing, only the hash is encrypted. Since both the sender and the recipient can hash the plaintext, they should get the same result. A digital signature requires the sender to append the hash to message, and he encrypts the hash with his private key.
Since the sender encrypted the hash with his private key, the recipient knows that only the sender's public key will allow him to decrypt a hash that matches the plaintext. This is the essential concept behind signing.
In order to verify the plaintext, the recipient decrypts the hash attached to the message. Then he computes a hash for the plaintext on his own. If his computed hash matches the decrypted hash, then he knows the message has not been changed. If his computed hash does not match the decrypted hash, then the message cannot be trusted.
If the sender wants to encrypt the entire message, he can do that as well. He would simply encrypt the plaintext with the recipient's public key (the encrypted data is called ciphertext). Then he calculates a hash for the ciphertext, and he sends the ciphertext with an encrypted copy of the hash, just like a regular signed message. The process is similar for the recipient, except this time the recipient will use the hash to verify the ciphertext. If it matches, he will then decrypt the ciphertext into plaintext.
Comment: The hash is not encrypted. It is signed using public-key cryptography, but cryptography!=encryption. The fact that you use one kind of key (private) to "encrypt" the hash and another kind (public) to actually encrypt messages is a bit confusing to explain if you say that the hash is "encrypted" just as if it were a message to actually encrypt. Even more confusing if you consider that encryption (in public-key schemes) is done with the *receiver*'s key, while signing obviously requires the *sender*'s key. IMO it's better to say that the hash is signed/verified than encrypted/decrypted.
Comment: I take some issues with the terms "encrypted/decrypted" in the context of digital signatures, also because signing and encrypting are different processes as you so well explained.
Comment: I'm not sure where you have a problem. The hash is generated based on the plaintext, but that hash is encrypted afterward by the signer/sender. Digital signatures require both hashing and asymmetric cryptography. I can edit if there is a point where the answer is ambiguous or uses the wrong term.
Comment: I agree, technically, it should be sign/verify for most of that post. I wanted to simplify as much as possible though. Primarily, to convey that the hash provides the guarantee. The original question doesn't seem to understand that the signed message includes more than just the plaintext/ciphertext.
Comment: In RSA cryptography, the underlying function used to sign the hash is more or less the same as the function used to encrypt the session key when encrypting message, except with public and private key reversed and session key replaced with the message hash. This is why people often speak of signing as encrypting using private key, but strictly speaking this is just an implementation detail of RSA cryptography, so using the same terminology can be confusing.
Comment: -1 for `For message signing, only the hash is encrypted.`. This is incorrect as others have pointed out. Please don't keep spreading this myth.