Rabbinical testing in Volozhin

Question: I heard once (can't remember where) that in Volozhin, the students preferred to be tested for the rabbinate by Reb Chaim of Brisk and not the Netziv, because the Netziv didn't give very hard tests. In his view, the true test of a rabbi came if and when he became a communal rabbi. Reb Chaim, on the other hand, would give difficult tests, so if one passed them, that meant that that person was very learned. Does anyone have a source for this?
Comment: Although most rabbinical students today likely want to be very learned, and perhaps also recognized as such, I would guess that almost every one would prefer easier bechinose that they will more likely pass than harder ones which might look more impressive but which they also risk failing. I don't think people have changed all that much, and so the story sounds a bit suspicious.
Answer: While not mentioning R. Chaim, this passage in  Lithuanian Yeshivas of the Nineteenth Century by Prof. Shaul Stampfer confirms the basic concept (though reversed) you have heard:

Many yeshiva graduates were rabbis. This lent support to the belief that the yeshiva's function was to train rabbis, as did the fact that the rosh yeshivah himself ordained rabbis. The more famous the rabbi who granted it the more valuable the ordination certificate; it was only natural that ordination by the rosh yeshivah would be particularly prestigious. Moreover, R. Berlin was known to be very careful about granting ordination; this made his ordination even more valuable, and also enhanced the importance of Volozhin as an institution in which to receive ordination. The following passage, written by a student at Volozhin in the 1880s, reflects this well:

There were some young married yeshiva students who rushed off to R. Naftali Tsevi [Berlin] from other places or yeshivas in order to be ordained by him — and this gave them the title of 'religious rabbi' [in contrast to 'government rabbi']. In truth, every rabbi is authorized to ordain others. But each of them wanted his ordination certificate to bear the signature of a famous authority. The best candidates [for ordination] were prepared to be examined by a more scrupulous examiner. Those who were interested knew in advance that R. Yitshak Elhanan [Spektor] of Kovno was more lenient in his examinations... an ordination endorsed by the name of R. Naftali Tsevi was quite different. The candidates stayed in Volozhin for week after week just in order to get a letter of ordination that he had signed. R. Naftali Tsevi used to test the candidates again and again on their knowledge in all branches of talmudic literature, and the ordinations he issued under his signature were very highly respected by Jewish communities."

(p. 95)
Answer: To get a better sense of this, one might do an informal check of ספר אהלי שם by ר' שמואל נח גאטליב (published in 1912).  It's available here: https://hebrewbooks.org/36602
Typically where the rabbis learned and from whom they received semicha is included in the listings. One could look for rabbis who are mentioned to have spent significant time in Volozhin AND to have received semicha from Rav Chaim Brisker.
Comment: See [this article](https://leimanlibrary.com/texts_of_publications/106.%20Three%20Lists%20of%20Students%20Studying%20at%20the%20Volozhin%20Yeshiva%20in%201879.pdf) from Dr. Leiman containing lists of students at Volozhin around the time R' Chaim first came to the yeshiva. It also discusses the claim that around 60 of the students enrolled were seeking semicha from the Netziv. Finally, it cites the example of R' Avraham Hofenberg, Rav of Vashki, who obtained semicha from both the Netziv and R Chaim ([confirmed](http://www.eilatgordinlevitan.com/kurenets/k_pages/hofenberg.html) by his grandson).

Rabbinical testing in Volozhin


SSL both for web and mail

Question: I have my public web site with EV SSL certificate.
I bought new server and configured postfix and dovecot for mail server.
Can I use my website's SSL certificate for postfix and dovecot for secure connection?
I have .crt and .key files.
Should I convert them into .pem?
They will be on the same domain.
Answer:
Can I use my website's SSL certificate for postfix and dovecot for secure connection?

In general yes. But the names you use to access the mail servers need to be part of the certificate too. So if the certificate only covers www.example.com but you want to use mail.example.com then this will not work.

I have .crt and .key files. Should I convert them into .pem?

PEM is a file format. ".crt" and ".key" are just file names. These files are often in PEM format already but might also be an DER format. If it looks like pure text it is PEM already and no need to convert it. See Wikipedia: PEM for details.
Comment: I added paths of my certificates, but dovecot gives error: imap-login: Error: Failed to initialize SSL server context: Couldn't parse private SSL key: error:06065064:digital envelope routines:EVP_DecryptFinal_ex:bad decrypt, error:0906A065:PEM routines:PEM>. What may be the problem?
Comment: @itdev: Wrong format of key, i.e. likely not a PEM format.
Comment: How can I convert my keys to right format?
Comment: @itdev: All I know from your file is the name, not the content. Assuming that the keys are in DER format you could use something like `openssl rsa -in file.key -inform DER -out key.pem -outform PEM`. But again, I have no idea what your files actually contain.
Comment: I will check your advice and tell it then. My keys are from my website's SSL certificate. That certificate is EV SSL. It uses 2048 bits for securing. All I know - is this.
Comment: @itdev: If you bought a certificate it typically comes with a) instructions or linked vendor documentation and b) maybe different formats. Thus, follow the instructions.

SSL both for web and mail


modal's buttons to be tabbable first once modal is open

Question: I have a list of offers, each of them with a delete button, that once clicked opens a modal with another 2 buttons: delete and cancel.
When the modal has opened, I want to be able to tab to these 2 buttons first, but at the moment tab continues with the other tabbable elements that are after the delete button in the row, eventually hitting the buttons after completing the tab round.
Is there a way of achieving that?

 e.stopPropagation()}>
    <Button tabIndex="0" secondary className={Styles.cancel} onClick={onCancel}>
      {t("abort")}
    </Button>
    <Button tabIndex="0" secondary className={Styles.delete} onClick={onDelete}>
      {t("deleteModal.CTA")}
    </Button>

Answer: Tab continues wherever the focus is set. Your code already includes tabindex="-1" which allows to focus the dialog element after opening it, with .focus() in JavaScript.

If the dialog content includes semantic structures […] then it is advisable to add tabindex="-1" to a static element at the start of the content and initially focus that element.

In your case, the dialog is really simple, and it would be more advisable to focus the primary button instead of the dialog itself.
See Dialog (Modal) from the ARIA Authoring Practices Guide (APG)
Your dialog needs an accessible name. Usually a aria-labelledby attribute is used to refer to the dialog's title element, if you don't have a visual title you can use aria-label.
It is also important that you establish a focus trap in the dialog, meaning that one cannot leave the dialog by means of tab, but only by means of close. Usually the focus wraps around back to the dialog-element.
Browser support for the native <dialog> element recently got better and you might consider using that one.


Does the word "Shiva" appear for Shiva/Rudra in Rigveda?

Question: Does the word "Shiva" appear for Shiva/Rudra in Rig veda?
Comment: Shiva Sankala Shuktam is present in rig veda khilla.
Answer: Yes, Lord Rudra is called Shiva in RigVeda itself:

स्तोमं वो अद्य रुद्राय....येभि: शिवः...॥ (RigVeda 10.92.5)
Address praise today with reverence to Rudra... You are Shiva...

Similarly in RigVeda 10.61.12 also which is the popular "Ayam May Hasto Bhagavan" mantra, we have:

अयं मे हस्तो भगवानयं मे भगवत्तरः ।
अयं मे विश्वभेषजोऽयं शिवाभिमर्शनः ॥१२॥ (RigVeda 10.61.12)
My this hand is the God. My this hand is higher than God!
My this hand holds all the healing. My this hand is the one that touches Shiva!

The Devata of above mantra is "Hasta" hand but the mantra itself is itself talking about power of hand which touches Shiva. This mantra is recited as a part of Sri Rudram itself in traditions. Similarly Shiva Agamas also use this mantra "Ayam May Hasto Bhagavan" as a part of establishing Lord Shiva in hand or purifying hand. For example Karana Agama says:

षडक्षरेण मनुना हस्तं स्वस्य च दक्षिणम् ।
षड्वारमभिमन्त्र्याथ शिवहस्तं विभाव्य च ।
तं शिष्यशिरसि न्यस्य पठेदेकाग्रमानस: ॥
अयं मे हस्त हत्येवं मन्त्रं शिष्यं विलोकयन् ।
भो शिष्येति च सम्बोध्य ब्रूयादेवं विचक्षणः ॥ (Karana Agama 1st Chapter)
After that, he consecrates his right hand six times-with six lettered hymn i.e., Ohm Na Maha Shi Vaaya then while treating his hand as
the hand of Shiva puts it on head of disciple, and with-concentrative
mind, looking at disciple recites hymn of "Ayam may Hasto Bhagavan" and addressing "Oh! Shishya" (Ah! Disciple), preacher says-

Similarly Shiva Sankalpa Sukta which is Khila Sukta of Rigveda also as discussed here glorifies various aspect of Lord Shiva along with word Shiva also:

कैलासशिखराभासा हिमवद्गिरिसंस्थिताः ।
नीलकण्ठं त्रिनेत्रं च तन्मे मनः शिवसंकल्पमस्तु ।। 25 (RigVeda Khila 10.166)
On the peak of Kailash along with daughter of Mountains, who is Nilakantha (blue necked) and Trinetra (three eyed). May my mind do Sankalpa of such Shiva!
Comment: RigVeda 10.61.12 is very powerful mantra. Even listening to it in proper pronunciation is good for healing any sickness faster. @ShivaNamah
Comment: Thank you  @TheLittleNaruto
Comment: Woah...thanks for this
Comment: @TheLittleNaruto can people who are not initiated also listen to it??
Comment: @LSSJBroly yes.
Answer: Yes, the following are two such hymns. But this is not healthy, obsessed in seeking the vocabulary term "Śiva" to determine if Shankara is in Vedas. One has to understand that the entity known or called as "Rudra" or "Shankara" or "Mahadeva" or "Ishana" or "Sarva" and many more is a nameless being that is the most contradictory and polar divinity with a multitude of shapes, sizes and existence. When this "being: emerged, Prajapti just called the child Rudra because the child roared/cried and not because it was a "namakarana saṃskara" (naming ceremony). In another case Agni wept in anguish, so He was also called Rudra. If any divinity shows attributes similar to Rudra they are called Rudratvam and so forth. This is why Rig Veda and all other Vedas termed Rudra as Viṣvarūpam, meaning the cosmic form or the all-encompassing omni-form, and Pururūpam, meaning multiform/multihued, and Virūpam, meaning multifold altering forms and Vahurūpa meaning multi-formed [RV2.33.10,TS4.5.4,TA10.23.1]. The Vedas clearly said, the so-called Rudra is a nameless divinity, He is unknown and only the learned Sages know who he is. Private note: I think this obsession started recently and has become a fashion statement to say Shiva is not in Veda and hence everyone started to search Veda for the Vocabulary term called "Siva" rather than actually trying to explore the Divinity associated with it.

स्तोमं वो अद्य रुद्राय शिक्वसे क्षयद्वीराय नमसा दिदिष्टन येभिः शिवः
स्ववाँ॑ एवयावभिर्दिवः सिषक्ति स्वयशा निकामभिः We sing our devotional
praise (स्तोमं) for you (वो) oh Rudra (रुद्राय) we show/shower
(दिदिष्टन) our salutations (नमसा) oh intelligent one (शिक्व) oh ruler
of men/beings/heroes/brace (क्षयद्वीरा), who is Śiva (येभिः – शिवः)
with our devotion may we unite with you (सिषक्ति) oh majestic glorious
one (स्वयशा) our protector (स्ववान् / स्ववाँ॑), who is swift in
delivering that protection (एवयाव = swift, एवयावभिर्दिवः
swift-moving/walking), oh granted of our wishes (निकाम). ~Rig Veda
10.92.9
कैलासशिखरे रम्ये शंकरस्य शुभे गृहे । देवतास्तत्र मोदन्ति तन्मे मनः
शिवसंकल्पमस्तु ।। Rig Veda 24 The splendid (रम्ये) summits (शिखर) of
mount Kailash (कैलास) is the auspicious residence (शुभे गृहे) of the
Lord Shankara (शंकरस्य). All divinities (देवता) are stationed (स्तत्र)
delighted in His presence (मोदन) with the inner self-absorbed into Him
(तन्मे मनः) may that be my auspicious resolve – many my resolve be of
Śiva (शिव-संकल्प-मस्तु). ~Rig Veda Bhaskāla Śākhā Khilani 33 (Śivā
Sankalpa Sukta)

Source: LINK-1 LINK-2
Comment: Thanks  namaḥ Śivāya


What is the meaning of Brahmarakshasa?

Question: What is the meaning of Brahmarakshasa and how is it different from Narasura, Rakshasa, Asura etc?
Is Ravana a Brahmarakshasa?
Answer: A partial answer.

When does one become a BrahmarAkshasa?

Samyogam patitaira gatvA parasya eva cha yoshitam | Apahritya cha
  viprasvam bhavati brahmarAkshasah ||  .....
Manu Smriti 12.60. He who has associated with outcasts, he who has
  approached the wives of other men, and he who has  stolen the property
  of a Brahmana become Brahmarakshasas.

So, this is talking about afterlife.
Answer: Brahm Rakshas is actually the
spirit of a Brahmin , a dead
scholar of high birth, who has
done evil things in his life or
has misused his knowledge, who
has to suffer as a Brahm Rakshas
after his or her death. The earth-
bound duties of such a scholar
would be to disperse or impart
knowledge to good students. If
he did not do so, he would turn
into a Brahma Rakshas after
death which is a very fierce
demonic spirit. The word
Brahm means Brahmin and
Rakshas , a demon. As per
ancient Hindu texts they are
powerful demon spirit , who have
lot of powers and only few in this
world can fight and over-come
them or give them salvation from
this form of life. It would still
retain its high level of learning.
But it would eat human
beings. They have the
knowledge of their past lives and
vedas and puranas . In other
words, they have qualities of
both Brahmin and Rakshas

source:—https://en.m.wikipedia.org/wiki/Brahmarakshasa

your another question about ravana ravana was a rakshasa .rakshasas were born to a great sage, Vishrava (or Vesamuni) and his
wife, the daitya princess Kaikesi so all children birn to kaikasi were rakshasas and Ilavida the another wife of sage vishwara was kubera.

source:— https://en.m.wikipedia.org/wiki/Ravana
source:— https://en.m.wikipedia.org/wiki/Kubera

What is the meaning of Brahmarakshasa?


Failing to determine common ground due to excessive insight into some field/topic

Question: Is there a term for the situation where a person knows so much about a topic or field, that she has a hard time talking about the topic with laymen because she fails to comprehend the limits of the layman's understanding of the topic? In other words, she fails to distinguish between what is common knowledge and what isn't, thereby failing to explain the non-obvious, which in turn leads to a complete lack of understanding from the layman.
Comment: Steven Pinker famously called this ["the curse of knowledge"](https://www.psychologicalscience.org/observer/the-curse-of-knowledge-pinker-describes-a-key-cause-of-bad-writing); one of our members [summarized the theory on Meta](https://english.meta.stackexchange.com/a/9800/55623) pithily: *Pinker means that experts know what they know and can no longer imagine not knowing it. They thus cannot tell when they've left out explication crucial to the understanding of non-experts*. In the first link, Pinker concludes *I think the curse of knowledge is the chief contributor to opaque writing*.
Comment: Thanks Dan Bron
Answer: 'Blinding someone with science' is also used in a metaphorically broadened sense outside the science domain.

blind with science (British & Australian)

If you blind someone with science, you confuse them by using technical language that they are not likely to understand.

[Collins Cobuild Idioms Dictionary]
You're also not on the same wavelength as your audience.

be on the same wavelength [phrase]

to understand the way that another person thinks because you often have the same ideas and opinions as they do

[Macmillan]
Answer: Someone who

"...fails to comprehend the limits of the layman's understanding of the topic..."

and who

"...fails to distinguish between what is common knowledge and what isn't..."

may be said to be living in an ivory tower


Display macOS window full screen on secondary monitor using Cocoa

Question: I'm working on a Cocoa Mac app where I need to display a window/view on a secondary monitor, full-screen.
I know how to create a window that could be dragged onto the secondary monitor, but I was wanting to programmatically create the window and make it full screen on the external monitor.
Answer: First, determine which screen you want to use by iterating over [NSScreen screens].
Create a full screen window with:
NSScreen *screen = /* from [NSScreen screens] */
NSRect screenRect = [screen frame];
NSWindow *window = [[NSWindow alloc] initWithContentRect:screenRect
    styleMask:NSBorderlessWindowMask
    backing:NSBackingStoreBuffered
    defer:NO
    screen:screen];
[window setLevel: CGShieldingWindowLevel()];

You might want to google CGDisplayCapture() as well.
Comment: Thank you, that worked great.  Since I was trying to output to the secondary screen I just used [window setLevel: NSStatusWindowLevel];
Answer: You can call the enterFullScreenMode:withOptions: method of NSView to acheieve the desired behaviour.
See Apple's documentation.
Read here and here for the options that can be supplied to this method.
You can use [NSScreen screens] to get the list of available screens. See here for details.
Comment: Thank you for the help.  For what I was trying to do I think the first solution is going to work best for me, but I'll have to keep that View full screen method in mind.
Answer: The full screen window animations are choppy and don't look good in my opinion.  The fullscreen view is much smoother.
Try this:
- (void)toggleMyViewFullScreen:(id)sender
{
    if (myView.inFullScreenMode) {
      [myView exitFullScreenModeWithOptions:nil];
    } else {
      NSApplicationPresentationOptions options =
          NSApplicationPresentationHideDock |
          NSApplicationPresentationHideMenuBar;

      [myView enterFullScreenMode:[NSScreen mainScreen] withOptions:@{
             NSFullScreenModeApplicationPresentationOptions : @(options) }];
                                                                                }];
    }
}

You can connect this to the fullscreen menu item in the Window menu (after inserting that into your nib) but be sure to change the action that the menu item fires to your toggleMyViewFullScreen: .  Or your can invoke toggleMyViewFullScreen programmatically or when your app loads.


How long would it take to brute force an 11 character single-case alphanumeric password?

Question: My previous home wifi router's WPA2 password was permanently fixed to FZ4HBCKHGC8.
How long would it take to crack via brute force?
Or more pragmatically:
How long would it take to exhaust all possible iterations of this 36 character set {A..Z}+{0..9}?
Comment: Yes, cause if your WPA2 has been permanently fixed to `FZ4HBCKHGC8`, it now takes no time to crack it.
Comment: @M'vy has succinctly captured a deep truth of password cracking here. If you don't understand why, ponder it until you do.
Comment: Ponder further. :)
Comment: I don't get the pondering meme, can someone explain?
Comment: @user1306322 They are commenting on the fact that the password that the OP posted now requires only 1 attempt to crack because the fact that they posted it on this site means that it is now public, operating under the assumption that the password shown here is not a placeholder password and that someone can correctly locate the user's router somewhere in the world, given their user information here.  There is no meme, just a joke.
Comment: oh, I thought OP wouldn't be so silly as to post the actual hardcoded password, and so I thought the joke was somewhere else. Also, it's a meme to call regular jokes and pretty much anything a "meme". Thought I should explain in turn :p Sorry, this is just how it is.
Comment: @M'vy As snarky as that is, that is an answer not a comment.  No, really.  It takes 1 try to crack a password fixed to `FZ4HBCKHGC8`.
Comment: Is this pondering meme from Breaking Bad scene?
Comment: @user1306322 OP said "previous" so since he's not using that router anymore, he may not care anymore.
Answer: The speed of WPA2, and the speed of modern GPUs, are essential to this answer.
A reasonable prosumer-sized (~US$5K) GPU cracking rig with 6 GTX 1080s can try around 2 million hashes per second - but there are 36^11 candidates to try!
For demo purposes, this is an actual attack, using the example WPA2 hash from the hashcat website:
$ hashcat -a 3 -m 2500 -2 ?u?d hashcat-wpa2.hccapx ?2?2?2?2?2?2?2?2?2?2?2
hashcat (v4.1.0) starting...

OpenCL Platform #1: NVIDIA Corporation
======================================
* Device #1: GeForce GTX 1080, 2029/8119 MB allocatable, 20MCU
* Device #2: GeForce GTX 1080, 2029/8119 MB allocatable, 20MCU
* Device #3: GeForce GTX 1080, 2029/8119 MB allocatable, 20MCU
* Device #4: GeForce GTX 1080, 2029/8119 MB allocatable, 20MCU
* Device #5: GeForce GTX 1080, 2029/8119 MB allocatable, 20MCU
* Device #6: GeForce GTX 1080, 2029/8119 MB allocatable, 20MCU

[...]

Session..........: hashcat
Status...........: Running
Hash.Type........: WPA/WPA2
Hash.Target......: 8381533406003807685881523 (AP:ae:f5:0f:22:80:1c STA:98:7b:dc:f9:f9:50)
Time.Started.....: Tue Apr 24 06:51:26 2018 (54 secs)
Time.Estimated...: Sun Oct 11 02:45:49 4105 (2087 years, 168 days)
Guess.Mask.......: ?2?2?2?2?2?2?2?2?2?2?2 [11]
Guess.Charset....: -1 Undefined, -2 ?u?d, -3 Undefined, -4 Undefined
Guess.Queue......: 1/1 (100.00%)
Speed.Dev.#1.....:   336.2 kH/s (6.89ms)
Speed.Dev.#2.....:   330.8 kH/s (7.03ms)
Speed.Dev.#3.....:   332.0 kH/s (6.96ms)
Speed.Dev.#4.....:   331.1 kH/s (6.97ms)
Speed.Dev.#5.....:   334.2 kH/s (6.90ms)
Speed.Dev.#6.....:   333.8 kH/s (6.90ms)
Speed.Dev.#*.....:  1998.0 kH/s
Recovered........: 0/1 (0.00%) Digests, 0/1 (0.00%) Salts
Progress.........: 108544000/131621703842267136 (0.00%)
Rejected.........: 0/108544000 (0.00%)
Restore.Point....: 2539520/3656158440062976 (0.00%)
Candidates.#1....: 82TCFESS123 -> 8MXVZONANDA
Candidates.#2....: 9JGXQW12345 -> 9O3QWESS123
Candidates.#3....: 9BBZPANANDA -> 93M1YONANDA
Candidates.#4....: 96RCZONANDA -> 9WMXQW12345
Candidates.#5....: 5S3O3123456 -> 59QC6678999
Candidates.#6....: 40QC6678999 -> 4CUZPANANDA

But don't feel too reassured by the "2087 years" estimate. Fixed passwords often do not require brute force to be cracked. Many permanently fixed WPA2 passphrases are algorithmically generated, and many of those algorithms are either known, or discoverable by reverse-engineering the device's firmware.
Comment: @R.. Normally an attack like this captures some information for offline brute forcing.
Comment: How is your rig going to "try around 2 million hashes per second"? Where does it get a test function to tell it whether a guess is correct or not?
Comment: @R.. There's no hashed password in the handshake, nor device present, cracking WPA2 basically consists on creating keys and testing against the MIC in the 2nd or 3rd packet of the four way handshake. So... **once captured** the handshake you don't need the AP, nor the Supplicant ("Victim"/Station). The test function part it's not that hard, just input X quantity of "passphrases" pass them through the PBKDF2 function (4096 times hashed each pass) and mean the time by X.
Comment: @JohnV.: I'd really like to see how you think that happens.
Comment: @R.. ... Are you asking how people capture WPA2 handshakes in order to crack them? Tools to do this have been around for a loooong time. The algorithm for WPA2 is public, and tools like hashcat generate candidates, hash them, and compare them to the original hash - at speed. This isn't a new thing. Maybe we're not understanding your point?
Comment: I forgot the hashed password was part of the handshake. This does however presuppose there is some device present which knows the password and can authenticate with the access point, no? If not, I don't see how you're going to get the hashed password or any means of brute-forcing it. So I wouldn't so much call this brute-forcing but intercepting.
Comment: @R.. It's both. You definitely have to intercept it first, so you to have to have physical proximity. But then you *also* have to crack it (bruteforce is only one approach, and less useful for a slow hash like WPA2; wordlists, masks, etc. would come first). Both interception and attack are necessary, but neither is sufficient. :) You might be interested in ZerBea's hcxtools - it has many features that help with both aspects. https://github.com/ZerBea/hcxtools
Comment: Well, strictly speaking, it's hashcat that helps more with the attack side. :) The two projects work closely together.
Comment: What is ?2?2?2?2?2?2?2?2?2?2?2 and ?u?d in the command above...Maybe someone knows...or can point me to the necessary part of the documentation. Thanks
Comment: See the output of `hashcat --help`
Answer:
Your password is 11 characters long and has 542,950,367,897,600
  combinations. It takes 10,534.62 hours or 438.94 days to crack your
  password on computer that tries 25,769,803,776 passwords per hour.
  This is based on a typical PC processor in 2007 and that the processor
  is under 10% load.

source: https://tmedweb.tulane.edu/content_open/bfcalc.php

source: https://www.grc.com/haystack.htm

source:https://asecuritysite.com/encryption/passes
Comment: You assume the router will be used to authenticate the password. It won't. The attacker will capture the handshake used for authentication and brute force it offline.
Comment: _"This is based on a typical PC processor in 2007"_  Given the processing speed increases in the past decade I don't find this data particularly useful for current cracking estimates.
Comment: Stuffing three answers that disagree into one "answer" is not helpful.  Which one is right?  Possibly none of them are computer using the same assumptions as the question?
Comment: … especially since this is exactly the kind of operation that hugely benefits from the embarrassing parallelism of GPGPU, let alone FPGAs and ASICs.
Comment: For asecuritysite, you have [a-zA-Z] checked, not [a-z0-9]
Comment: Helpful sources but this assumes the router will accept attempts at a really high rate.  I would be surprised if a home router actually accepts password attempts at 1 million per second.
Comment: @JimmyJames me too ! I answered just his "pragmatically" question. After all in a real life scenario no need to bruteforce the router... it's all about cracking the handshake.
Comment: One hundred billion guesses per second?
Comment: These images don't seem to indicate what algorithm was used to encrypt the password, which greatly affects speed. WPA is on the slower side. My 2020 graphics card does not come close to the speeds indicated in this answer. If you assume MD5 instead of WPA, then it does.


How do we get to another stronghold in minecraft without using cheats?

Question: so me and my boyfriend were playing minecraft and threw several eye of enders which led us to a stronghold that only consisted of one room, so we tried to mine all around it up and down and couldn't find any ender portal near it or in it and after doing research online we found that our best choice was to head to the second stronghold in the world out of the 3 and see which one has the ender portal, but the eye of ender only leads to the same one no matter how far we go, so we can't use that as a guide to any of the others. Is there any easy way that we can get to a different stronghold without having to use cheats? We are playing on xbox 1 and playstation.
Answer: If you know seed - search tool (choose right edition)
It is unlikely that there is no portal in stronghold, maybe check again digging up and down around spot that ender eye marked.
Otherwise another one can be 5000 blocks away in any direction, so you have to keep trying - like 2000 blocks away, which can be really long journey.
Comment: We looked it up and other people have had the same problem of finding a stronghold and it only consisting of 1 or 2 rooms and nothing else around it. It is rare, but the main thing that was directed to do was to travel to a different stronghold, so that's why we're trying to do that. Haven't found it yet but thanks for the website!


Nav2 Custom planner plugin does not exist

Question: I followed this nav2 getting started guide without issue: I am now following this nav2 custom planner plugin tutorial
I have created a new plugin nav2_straightline_planner in my ws/src directory. And I have confirmed there are no typos between the sample code and mine. When trying to run
$ ros2 launch nav2_bringup tb3_simulation_launch.py params_file:=/ws/src/navigation2_tutorials/nav2_straightline_planner/nav2_params.yaml
I get the following error:
Failed to create global planner. Exception: According to the loaded plugin descriptions the class nav2_straightline_planner/StraightLine with base class type nav2_core::GlobalPlanner does not exist. Declared types are  nav2_navfn_planner/NavfnPlanner
I have checked my CMAKELists has  pluginlib_export_plugin_description_file(nav2_core global_planner_plugin.xml)
my global_planner_plugin.xml is the following
<library path="nav2_straightline_planner">
    <class name="nav2_straightline_planner/StraightLine" type="nav2_straightline_planner::StraightLine" base_class_type="nav2_core::GlobalPlanner">
      <description>This is an example plugin which produces straight path.</description>
    </class>
</library>

And in my params.xml I have the following:
 planner_server:
  ros__parameters:
    plugins: ["GridBased"]
    use_sim_time: True
    GridBased:
      plugin: "nav2_straightline_planner/StraightLine"
      interpolation_resolution: 0.1

Answer: Did you re-source your workspace's install directory?


Existence of creation and annihilation operators

Question: In a multiple particle Hilbert space (any space of any multi-particle system), is it sufficient to define creation and annihilation operators by their action (e.g. mapping an n-particle state to an n+1-particle state) or does one have to do anything else, like "proving existence". Intuitively, one can apriori say it exists if you can write it down, but I donno.
Answer: If the action is given, the existence is obvious (unless the definition is faulty). But one would usually want to verify that all creation operators commute,
all annihilation operators commute, and the commutator of a creation operator and an annihilation operator is a c-number.
Moreover, for a field theory, one would usually want to verify that this c-number transforms covariantly under Poincare transformations, and vanishes at spacelike separation. This applies for free field theories. For interactive field theories, there are no intrinsic c/a operators, as their properties are destroyed by renormalization. But one can associate (according to Haag--Ruelle theory) to each bound state a family of creation and annihilation operators parameterized by momentum, describing the free asymptotic motion in a scattering process.
Answer: Providing the action of an operator on the vectors of a basis is exactly the same as "writing it down" since it extends to all other vectors by linearity. This is the same as writing the matrix of a linear operator between vector spaces. Thus creation and annihilation operators automatically exist when they are defined.
Comment: This is true, modulo infrared problems as in QED or more sever infrared problems such as in QCD. The demonstration is simply from the noninteracting nature of plane waves.
Comment: This is true in the fermionic case where the operators are bounded but not in the bosonic case. On rather defines the operators on vectors with finite particle number which are just dense in the Hilbert (Fock) space. I refer to the second volume of the books of Reed, Simon.
Comment: Yes, that is correct, and it is fundamentally due to bosonic operators being unbounded - just like position and momentum in the SHO. Whatever mathematical difficulties one encounters for boson creation and annihilation has an exact counterpart in the SHO formalism.
Answer: For a theory with a mass gap, it is sufficient to define the creation and annihilation operators by their action, but these operators will not have local properties in general, they won't be the Fourier transform of fields that obey microcausality or come from a local Lagrangian.
When particles are spread out over all space, they have no interaction. Their S-matrix element is
$$ S = I + i A $$
Where I is the identity, so has delta-functions making the incoming k's equal to the outgoing k's, while A has only one overall delta function for energy momentum conservation (A might be the identity on some of the particles and not the identity on others, but this is still fewer delta functions). This means that two infinite plane waves have delta-function scattering in the forward direction, but only a smooth distribution in the off-forward direction. The interpretation is just that as you make the beam of particles more tenuous, the number of collisions goes to zero.
This argument fails to a certain extent in infrared divergent theories (like quantum electrodynamics), because you need to make a nonlocal field with every charged particle. In this case, you need to prove existence, since it isn't clear that the same Hilbert space includes both a zero electron state and a one electron state with an infinite range field (although the zero electron state must contain electron-positron states which are net neutral, and so have a shorter range field).
But in any relativistic field theory with a mass gap, if you have a vacuum and a one particle state, you can define creation and annihilation operators which make n-particle (necessarily noninteracting) states. Then you can define a quantum field from these operators. This quantum field will not have local interactions in general, since the particles can be H-atoms, or pions, they don't have to be pointlike photons.
The issue with constructing quantum field theories is to make sure that the field for the elementary particles is local, and this sometimes requires different degrees of freedom, like quarks, which are not asymptotic. In this case, you can't define a quark creation operator, because the quark's infinite range strong field is an infinite mass string, and is definitely outside the Hilbert space of the theory.


Java Application + Subpixel Rendering

Question: How i can enable subpixel rendering in java applications like NetBeans?
Answer: If I interpret this blog posting correctly, then subpixel rendering should automatically be available with Java 6 runtimes, unless you are using a Mac. If so, this posting has instructions how to re-enable it there.
I'm running Eclipse with a Sun Java 6 runtime and did not have to manually enable subpixel rendering, but that may be something different as Eclipse uses SWT graphics library while NetBeans uses Swing.

Java Application + Subpixel Rendering


Speed up API Post requests than curl?

Question: I am trying to make API post requests of JSON to a web server. This task needs to be asynchronous with least overhead. I am currently using CURL with infinite looping in bash and used '&' to make it asynch, which is slow for my task. How do I boost and reduce asynchronous API post task from CURL or without it? Are there any alternative tools that I can implement for this task?
Answer: Partial answer:
In particular when using https, opening the connection takes a measureable amount of time. If you do many requests, re-using the connection will make it faster.
Putting curl into the background with & just opens connections in parallel, it doesn't re-use an existing connection.
I do not know a tool that can re-use connections from the commandline.
However, you can do it e.g. with libcurl, see e.g. this answer, and many other HTTP client libraries.

libcurl is a C library, so you need to write a C program to perform your API calls, instead of a shell script.
If C is not your thing, pick any other programming language you prefer, as written above.
In particular with libcurl, have a look at the example program under the link provided. Also read the libcurl docs and example programs (google). With the easy libcurl interface, the trick is to get a single handle, and then keep using that handle for all API calls - it will reuse the open connection.
Real concurrent operation will open several connections in parallel. Which may or may not make it faster - the bottleneck is usually the network, and not processing.
Comment: Since I am new to these concepts. Can you elaborate a little if I can implement asynchronous api post requests with libraries like libcurl?

Speed up API Post requests than curl?


Fantasy novel with female protagonist and mages who can be drained of power by a cut?

Question: I believe it was a standalone, but possibly 2 books. The main character was a female. The mages in the story exploded when they died unless they were completely drained of power first.  They could be drained by making a small cut and then the other mage touching the cut and draining them (to the extent it would kill them). A large portion of the book was main character's training to use magic at a school where thy would basically put up magical shields and duel each other. I believe I read it in late 90's to early 2000.
Comment: Almost certainly Trudi Canavan's [*Black Magician* trilogy](http://www.trudicanavan.com/books/the-black-magician-trilogy/), possibly conflated with some elements of her [*Magician's Apprentice*](http://www.trudicanavan.com/books/the-magicians-apprentice/), whose plot is basically identical.
Comment: @CearonO'Flynn I would've posted an answer if I hadn't repcapped for the day, but you're welcome to claim the glory :-)
Comment: @randal'thor - argh you beat me by 5 minutes :( that'll teach me for trying to work at the same time as typing an answer and find evidence to back it up
Comment: @randal'thor thank you. most gracious
Answer: Trudi Canavan The Black Magician Trilogy published between 2001 and 2003.
Magicians were able to drain magical power from other magicians through a cut using black magic, they could be drained to death.
I don't remember any of them exploding though Cat'r'pillar informs me they do if they still have power and this rings a faint bell now its been mentioned.
The books follow Sonea a young girl who learns she has magical ability as she is trained in the arts of magic. A lot of dueling, throwing magic at the shields goes on during her training.
She ends up falling in love with the with a fellow magician who teaches her Black Magic.
Comment: They explode when they have power remaining when they die, and it's released all at once or something like that.
Comment: It just so happens that I reread the books very recently, or I wouldn't've remembered either :)
Comment: @Cat'r'pillar cool, I can easily believe I forgot that fact it has been ages since I read them, have so many books in my head since then including all the ASOIAF books, Peter Hamilton's tomes and a bunch of Malazzan books, they all take up some serious memory space. It was the draining magic and throwing spells at each others shields that triggered my memory.


Mount permissions problems

Question: when I tried to mount and other ssd in my elementary os I had get this problem:
All the files with no permission, seems like they are being mounted as root, but why?
I trid to follow this another topic: NAS share mount permission problems , but without success.
this is my fstab file:
<code>  GNU nano 4.8                                                                                etc/fstab
# /etc/fstab: static file system information.
#
# Use 'blkid' to print the universally unique identifier for a
# device; this may be used with UUID= as a more robust way to name devices
# that works even if disks are added and removed. See fstab(5).
#
# <file system>  <mount point>  <type>  <options>  <dump>  <pass>
PARTUUID=89a21891-8b81-4bb6-b362-5798978f99f1  /boot/efi  vfat  umask=0077,uid=1000,gid=1000  0       0
UUID=c3261c9f-cb24-4dc4-a2a2-bd7df74cca76  /  ext4  noatime,errors=remount-ro  0  0
/dev/disk/by-uuid/290B3424686AD77F /mnt/arquivos/290B3424686AD77F auto nodev,gid=1000,uid=1000,nounix,nofail,x-gvfs-show,\040 0 0
</code>
Answer: Can't comment (weird Stack rules...) so here is a question:
why do you have backward slash in your fstab line?:

gid=1000,uid=1000,nounix,nofail,x-gvfs-show,\040 0 0

Mount permissions problems


21 points on circumference of a circle must have at least 100 pairs separated by 120+ degrees.

Question: Prove that at least 100 of the arcs determined by the pairs of these points subtend an angle not exceeding 120 degrees at the center. How do I prove this? Induction? Help please. Thanks.
Answer: Form a graph with 21 vertices corresponding to the points and edges between pairs of points more than 120 degrees apart.
This graph is triangle-free.
The number of edges in a triangle-free graph on 21 vertices is (uniquely) maximized for the bipartite graph on 10 + 11 vertices which has 110 edges.  Therefore, there are at least ${21 \choose 2} - 110 = 210 - 110 = 100$ pairs of vertices not joined by an edge. Those pairs correspond to pairs of points at most 120 degrees apart.
This also explains what the maximal arrangements of points look like.  All of them can be formed by taking two disjoint closed 120 degree arcs on the circle, and placing $10$ and $11$ points respectively inside the two open intervals that are the complement of those arcs.
Answer: Since the angle subtended by any 2 adjacent points is $360^\circ/21$, we need to have a 7 point gap to subtent an angle of $360^\circ/21 * 7 = 120^\circ$.
Hence, 8 adjacent points subtend $120^\circ$, and any smaller number of points will subtend $<120^\circ$. So, we can pick $8, 7, \ldots 2$ adjacent points and they will all subtend the angle we want.
The number of $n$-adjacent points that exist on a circle will be $p - n + 1$, where $p$ is the total number of points.
Computing $S = \sum_{i=2}^8 p - i + 1 = \sum_{i=2}^8 21 - i + 1 = 97$ (if one does the math)
I'm not sure what I'm missing - however, I do believe that if we are to take the smaller angle when the arc angle crosses $180^\circ$ (example, if I cut out a $300^\circ$ arc, do we take the arc angle to be $60^\circ$? In that case, the complement of all arcs that have arc angles greater than $360^\circ - 120^\circ = 140^\circ$ will also need to be computed. This will definitely push the number of arcs to over a hundered
Comment: What is S equal to? I don't get 2nd part below the summation. Can you explain?
Comment: The points do not have to be equally spaced or located at some vertices of a regular 21-gon.


Engine swap & ecu

Question: So I was thinking about putting a vw 2.0t into my 1999 eclipse gs. Would I have to get a standalone ecu or can I wire in the engines ecu in place of the eclipses oem. I'd be able to figure everything else out on my own but electrical is my downfall
Comment: you are going to need an ECU for the VW engine.
Comment: Welcome to Motor Vehicle Maintenance & Repair!
Answer: There are a few different ways to go with what you are trying to do.

Piggyback the original ECU and use a VW ECU to run the engine. If you want the rest of the vehicle to work, you'll need the original ECU to display activity to the stock dash.

Replace the original ECU with the VW ECU, then modify the gauges to accept the input of the VW ECU. This may entail complete change out of the all gauges. This is your standalone solution.

Replace the original ECU with an aftermarket ECU as well as gauges.

In all three situations you'll need either the VW ECU or an aftermarket one to do the deed. Someone may have already done the swap you're talking about. Do your research. If it has been done and someone has recorded what needs to be done, it'll give you a leg up. Don't reinvent the wheel if you don't need to. If you go this route, you'd want to look at the Eclipse forums as well as YouTube for the information. Particulars would be well beyond the scope of this site.
As an example, my son blew up the engine in his Impreza. It was a 2.2L 4-cyl boxer style (EJ22). He had heard you could replace it with the 3.3L 6-cyl engine out of a Subaru SVX (EG33). I looked around on the internet and found someone who had done it and followed his example. It gave me what I needed to retrofit and what to buy to make it happen. Ultimately it was a fun project and there was very little guessing to make it work.

Engine swap & ecu


Single word for someone who does not eat mammal's meat

Question: Is there a single descriptor for someone who eats anything (chicken, fish, insects, veggie burger,...) but just not any mammals (from the conviction that they are "to close to us" to serve as food).  Not a compound something like pollo-pesco-ovo-etc-arian, but a single word?
Comment: Did you mean '**too** close to us'?
Comment: yes, sorry!  "too close" in the sense all mammals descend from a common ancestor so it comes close to cannibalism to eat mammals
Comment: by the way, what would be the ethymology of the word "mafist" - I can't really recongnize any root.
Comment: I've become a non mammal eater about 6 months ago.It really is frustrating not having an official term for this lifestyle. Could not find Mafist in any dictionary.
Answer:
Is there a single descriptor for someone who eats anything (chicken, fish, insects, veggie burger,...) but just not any mammals

Someone who does not eat mammals'meat is called mafist. Mafists are sometimes called pesco-pollo-tarians.
Pescetarian is someone who does not eat meat but eats fish.

Pescetarians eat the meat of fish and other non-mammalian sea creatures, but not of birds or reptiles, nor of mammals. Mafists avoid only the meat of mammals. They eat poultry and fish. (I guess they might also eat reptiles. [Yahoo Answers]

Also read:
Mafism — Scribd
Comment: @HotLicks Wow, I always thought they were pesky Presbyterians.
Comment: Gee, I always thought they were pesky politicians.
Answer: Mafism or mafist seems to be a correct answer, since the you asked for a single word. However, this word doesn't seem to be in common use; it doesn't seem to appear in any dictionary, doesn't have any Google Trends data. I have never heard it before now. So if you choose to use this word you will likely have to clarify the meaning.
What I have heard many people say is that they don't eat red meat. This isn't one word, but it commonly used. Red meat is a synonym for mammal meat, while white meat is used for the meat of other animals, usually fish and birds.
Also there are many different reasons one might choose a certain diet such as this or vegetarianism. Not wanting to eat something that is "too similar to us" is a possible reason, but health reasons or ecological reasons are common.
Comment: "White meat" is ambiguous, because in common usage it includes pork.
Comment: This is a very sensible answer. Single word "labels" often do more harm than good.


Geometry Nodes: How to make tube from curve

Question: Given a curve node's output (in this case, a quadrilateral Bezier), how can I get a tube shape? I'm looking to make a rope from the curve.
Answer: You can use a Curve to Mesh node and link another curve (for instance, a Curve Circle primitive) into the Profile Curve socket to sweep it along.
Comment: hi, what if i have a static curve object in the scene and i want to do use it as profile it in the geometry node? can this be done?
Comment: @adrianli You should be able to do that with an Object Info node. Select your curve from the scene in the dropdown, and link the Geometry output to the Profile Curve input on Curve to Mesh.
Answer: With Curve to Mesh you get the desired mesh, but no mapping coordinates to be able to use your shader.
If you want/need to add seamless shading/texturing to your mesh generated with Curve to Mesh, follow this post: Node "Curve to Mesh" with UVs in Blender 3.1?


Is there an explicit irrational number which is not known to be either algebraic or transcendental?

Question: There are many numbers which are not able to be classified as being rational, algebraic irrational, or transcendental. Is there an explicit number which is known to be irrational but not known to be either algebraic or transcendental?
Comment: $e+\pi$ or $e\cdot\pi$.
Comment: Erdős proved that the Erdős–Borwein constant $\sum_{n = 1}^{\infty} \frac{1}{2^n - 1} = 1.60669\!\ldots$ is irrational, and to my knowledge whether it's algebraic remains open; the comments below this old answer suggest that's the case: http://math.stackexchange.com/a/266638/155629 . https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Borwein_constant
Comment: @studentforever: I don't know that either of those numbers has the indicated property. (Of course at most one of them is rational, but I can't see how that helps here.)
Comment: I would think numbers like $.121121112\dots$ would do the job, no?  Liouville doesn't obviously give us transcendence (at least it isn't obvious to me).
Comment: @lulu I believe that number is known to be transcendental.  It's the sum of a rational number ($\frac19$) and a theta-value at a rational argument that I'm pretty sure is known not to be algebraic.
Comment: @StevenStadnicki  Ah...let's see. Certainly a mistake to use only $1's$...makes it too easy to subtract $\frac 19$.   Is my example better if we cycle through the digits?   $.192293339\dots$?  Surely there's a simple way to write a non-periodic decimal to which we can't apply Liouville or the like...
Answer: Maybe the best-known example is Apery's constant,
$$\zeta(3) = \sum_{n = 1}^{\infty} \frac{1}{n^3} = 1.20205\!\ldots ,$$
which Apery proved was irrational a few decades ago; this result is known as Apery's Theorem.
By contrast, $\zeta(2) = \sum_{n = 1}^{\infty} \frac{1}{n^2}$ has value $\frac{\pi^2}{6}$, which is transcendental because $\pi$ is.

Apéry, Roger (1979), Irrationalité de $\zeta(2)$ et $\zeta(3)$, Astérisque (61), 11–13.
Comment: @Jake I think that's true, but I don't think it's known for any odd $m > 3$ whether $\zeta(m)$ is even rational. It is known, perhaps a little oddly, that at least one of the first four such numbers is irrational. (See Zudilin, Wadim (2001), *One of the numbers $\zeta(5)$, $\zeta(7)$, $\zeta(9)$, $\zeta(11)$ is irrational*, Russ. Math. Surv. 56 **(4)**, 774–776.)
Comment: And $\zeta(2n+1)$ in general for $n\geq 1$ is not known either, I think.
Comment: Ah yes. I was being hasty. Thank you :)
Comment: Wow, I thought I knew the Greek symbols by sight, but I had to look up that weird salamander-looking thing to find out it's a lower-case Zeta.
Comment: @MasonWheeler: in maths undergrad terms that's "other squiggle", with "squiggle" being $\xi$.
Comment: Shouldn't the summation in Riemann's Zeta function start at n=1?
Comment: @ViníciusBarros Yes, of course, thanks for pointing it out (I've fixed the typo).
Answer: The most famous have been answered. Let us be a little  less constructive. At least one of $\zeta(5)$, $\zeta(7)$, $\zeta(9)$, $\zeta(11)$ is irrational, a result due to V. V. Zudilin, Communications of Moscow Mathematical Society (2001), and their true nature (algebraic and transcendental) seems unknown at the present time. This result improves the irrationality of one of the nine numbers $\zeta(5)$, $\zeta(7)$, $\ldots$ $\zeta(21)$.
Comment: Fair enough. :)
Comment: Is it known that these numbers are all irrational?
Comment: @Charles not by me, but at least of them (explicitely) is irrational, and unknown to be algebraic or transcendental. This is why I called my answer non constructive
Comment: @Charles glad you liked it, pfiou...
Answer: $$ \frac{\ln\pi}{\pi}, $$
This example is irrational because $$\frac{\ln\pi}{\pi}=\frac{p}{q} \implies \pi^{q}=\left(e^{\pi}\right)^{p}$$
But this is impossible because Nesterenko
demonstrated that $e^{\pi}$ and $\pi$ are algebraically independent.
As far as I know, there is not a demonstration that $\ln(\pi)/\pi$ is transcendental which doesn't invoke the far from proven Schanuel's Conjecture.
Another nice example might be $t$ such that $2^t+3^t=6$. This is perfectly explicit despite us not being able to write down $t$ in terms of other constants. It's also nice and constructive in a way that should allow for making more examples. The solution to an exponential integer equation should be demonstrably irrational but not necessarily demonstrably transcendental (But again assuming Schanuel's Conjecture one can make progress).
Comment: Great example, Mason!
Comment: ${\ln(\pi)}/{\ln\Gamma(\frac{22}{3})}$ has the same property of being clearly irrational but not necessarily demonstrably transcendental because $\pi$ and $\Gamma(22/3)$ are algebraically independent according to [this.](https://en.wikipedia.org/wiki/Gamma_function). $\{a,b\}$ has transcendence degree $2$ implies $\ln(a) /\ln(b)$ irrational(but not transcendental). The question becomes: Did the theory that allowed for proving $a$, $b$ algebraic also provide an an inroad for proving $\ln a / \ln b$ transcendental.


Creating a Card Stack with Geomery Nodes

Question: I'm trying to create something like a stack of cards using Geo Nodes. Following a tutorial on YouTube, I managed to arrange the cards along a curve using the Resample Curve node. Now, I want to give the cards a random rotation around the Z-axis to make the stack look more natural. I've attempted to use the Rotate Instances node, but it didn't work, and I'm not quite sure how to achieve what I have in mind.
The nodes I've used.

Card stack I'd like to create
Comment: noise texture (scaled up or down) + euler rotation (on Z) and plug that in the instance on point rotation, most likely
Comment: @lemon, thank you for your response! However, I'm still having trouble understanding how to constrain the rotation to the Z-axis. I attempted to incorporate those geonodes, but now it's rotating on all axes.
Comment: UPD: i've done rotation with combine XYZ node. Now I would like to break the curve into segments and rotate these segments relative to each other. Could you please advise on how to achieve this result?
Comment: make segments rouding for instance the curve point indexes, or the curve height (Z).
Comment: just an idea: you can also control the rotation using the curve tilt (+ what we said above)
Answer: Perhaps you could get a nice result with just one Noise Texture with the right settings, but I find it easier to reason about multiple separate noise textures:

First a noise texture giving a basic profile to the stack, low scale for smoothness (actually very low, because I use Index as coordinate, and indices tend to be bigger than position components) and no roughness/detail for the same reason; consider some minimal distortion for uneven profile.
Then noise with slightly bigger scale to add some more chaos with local shifts of the stack; plus some distortion, so the shifts vary in size.
Add them together:

Voronoi texture is good for sudden boundary changes, but you could achieve similar by posterizing (snapping) a normal noise texture. This simulates big shifts of the stack. If you want to instead simulate joining separate stacks together, you need to use the voronoi cell position as the base for the coordinates for all other noises!
A white noise can be added to slightly move all cards randomly, however another noise with much bigger scale and/or roughness works more realistically, because in reality cards will tend to line up.

The setup then becomes:

And the stacks (the 2nd with all offsets scaled down for a more subtle effect):

One non-realistic thing here (the continuity across voronoi boundaries could be considered another) is that all cards centers stick to $xy = 0$. You could use the same noise textures (or transform them e.g. using other noises) to also offset the cards slightly.
Comment: I used Blender 4.0 for this answer, but you should have similar results with 3.4.1.
Comment: Thank you so much for the effort and the detailed explanation! It's an excellent solution and incredibly helpful; I truly appreciate it. I'm marking your response as the answer because it provides the most comprehensive solution.

I do have one remaining question: for some reason, in my Blender, the Voronoi Texture node has fewer input fields than shown in your screenshot. I only have Scale and Randomness, whereas in your screenshot, there's also Detail, Roughness, and Lacunarity. Could this be due to the Blender version? Or do I need a specific add-on? I'm using Blender 3.4.1.
Answer: Basically, you need to rotate and translate the instances using a noise texture. I found that tweaking the pivot of the rotation helps. The result looked good, but was missing something.
In the reference, it seems that the cards tend to clump together; a group of them roughly rotate together by the same amount. To replicate that, I posterized the noise using a Color Ramp (because it seems that the posterize node only exists in the Compositor). I ended up mixing the original noise with the posterized version.
A clamped noise texture produces values between 0.0 and 1.0. Typically, we subtract 0.5 and multiply by 2.0 to move the values to the -1.0 to 1.0 range. Then, we can scale the values up or down according to our needs. I combined the latter two steps in my setup.
Note that the rotation should only be around the z-axis, so I multiplied the x and y components of the noise output by zero. On the other hand, the translation should be limited to the x-y plane. Multiplying the z-component by zero does the trick.
The rest is to tweak the values to get something that looks good.
A screenshot showing the final node setup and the result is attached below.

There is an issue with my setup. I shouldn't have used the Factor output from the noise as a vector. This will cause all components to be the same.
Comment: You're welcome!
Comment: Thank you! I'll try to play around with these Color Ramp node and see what comes out of it.
Answer: You can play with the Voronoi texture node to get different results. All you really need from my node tree is the Set Position, Combine XYZ and Voronoi Texture nodes.


host:port for testnet and mainnet

Question: I am testing my Dapp and need host:port for testnet and mainnet.
Please can you let me know where can I find these urls?
Also, what if the url (host:port) i provide and it goes down?
Answer: I would suggest you create an account at https://infura.io. They're providing the infrastructure for what you're looking for.
Answer: One way is to run your own node with geth and geth --testnet. This way you can connect to them from your app. geth also allows custom post numbers with --port

host:port for testnet and mainnet


rotation of conic sections

Question: In the discriminant test of conic sections(rotations), why we're checking with $B^2-4AC$.
How $B^2-4AC=B'^2-4A'C'$, where $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is changed to $A'x^2+C'y^2+D'x+E'y+F'=0$ using rotations by angle alpha.
Comment: You need to do the replacements of $x,y$ by the rotation, and then find the new $A',B'$ etc, and just compute $B'^2-A'C'$. If you have rotated so as to eliminate the $xy$ term, the $B'$ will be zero here.
Answer: We can find from here, the condition for the general 2nd degree equation to represent a pair of straight lines.
If that condition is not satisfied we can apply the following logic.
A conic is the locus of a point $(x,y)$ which maintains constant ratio(called eccentricity$(e)$) of the distances from a fixed point(called focus $(h,k)$ (say),) and a fixed line(called directrix) $lx+my+n=0$(say).
So, $$e=\frac{(x-h)^2+(y-k)^2}{\frac{lh+mk+n}{\sqrt{l^2+m^2}}}$$
On squaring and rearrangement we get,
$$\{m^2+(1-e^2)l^2\}x^2-2lme^2xy+\{l^2+(1-e)^2m^2\}y^2+(..)x+(..)y+(..)=0$$
Comparing with the original equation,
   $A=m^2+(1-e^2)l^2,B=-2lme^2, C=l^2+(1-e)^2m^2$
So, $B^2-4AC=4(e^2-1)(l^2+m^2)^2$
Now,
for ellipse $0\le e<1, B^2-4AC=4(e^2-1)(l^2+m^2)^2<0$
for parabola $e=1, B^2-4AC=4(e^2-1)(l^2+m^2)^2=0$
for hyperbola $e>1, B^2-4AC=4(e^2-1)(l^2+m^2)^2>0$
for circle, $A=C\implies m^2+(1-e^2)l^2=l^2+(1-e)^2m^2\implies e=0$,the $xy$ must be absent in the general equation and the focus co0incides with the centre.

Using Rotation of axes,
$A'=A\cos^2\alpha+B\sin\alpha \cos\alpha +C\sin^2\alpha$
$\implies 2A'=(A+C)+(A-C)\cos2\alpha+B\sin2\alpha$
$B'=B(\cos^2\alpha-\sin^2\alpha)-2(A-C)\sin\alpha \cos\alpha=B\cos2\alpha-(A-C)\sin2\alpha$
$C'=A\sin^2\alpha-B\sin\alpha \cos\alpha +C\cos^2\alpha$
$\implies 2C'=A+C-\{B\sin2\alpha+(A-C)cos2\alpha\}$
$2A'\cdot 2C'-B'^2$
$=(A+C)^2-\{B\sin2\alpha+(A-C)cos2\alpha\}^2-\{B\cos2\alpha-(A-C)\sin2\alpha\}^2$
$=(A+C)^2-B^2\{\sin^22\alpha+\cos^22\alpha\}-2B(A-C)\{2\sin2\alpha\cos2\alpha-2\sin2\alpha\cos2\alpha\}-(A-C)^2\{\sin^22\alpha+\cos^22\alpha\}$
$=(A+C)^2-(A-C)^2-B^2$
$$\implies 4A'C'-B'^2=4AC-B^2$$ (This can be established by eliminating $\sin2\alpha, \cos2\alpha$ from the three equations.)
Answer: For any rotation and translation we need three invariants to be conserved:
$ A+ C, B^2- 4 A C $ , and another determinant of triple product comprising b $ A,B,C,D,E,F.$ The second second one is associated with constant/conserved radius.

rotation of conic sections


Giving the wall thickness of hollow irregular pipe. Solidify not working well with the problem

Question: I want to add the wall thickness around 1.5mm-2mm to that irregular pipe
 For me important is to have the lumen and therefore the inner surface not changed. I tried with solidify, I played with setting, but all the time I am narrowing the pipe, or some of irregularities are appearing inside the lumen. Its usually cracking when the number of thickness exceeds the offset
.
Any ideas how to create a wall to mesh like this preserving inner surface intact?
Comment: Hard to tell without looking at your topology, but looks like you should cleanly model the inner shell as a new cylinder object. You can then connect it to the existing outer shell manually
Comment: Just googled lumen, Interesting.
Answer: I don't think the solidify modifier was meant to handle such complex geometry. If you apply the modifier, you can repair any irregularities by hand.
The other idea would be to use the Shrinkwrap Modifier. It's ability to take the closest surface point as reference makes it pretty invulnerable to small holes.
Below is the result of a Cylinder with just 8 sides. I sandwiched the Shrinkwrap Modifier between two subsurfs. The first one decides how many verts the shrinker gets and the second is for smoothing. That way you can tweak how much inner surface structure leaks through.

All that's left is to apply the modifiers, join the meshes and stitch them together.


Azure Read Replica in Elastic pool performance

Question: We have a Azure Database with a read replica. However the read replica seems to be effected by the load of the primary database. The primary database is in an elastic pool.
Does the read replica make use of the same elastic pool?
Comment: You will need to provide more details. How are getting a read replica? Just by deploying premium/business-critical or via active geo-replication/failover group? What kind of effect on read replica?
Comment: Business critical read replica. When we run heavy loads that take up a lot of cpu the read replica starts timing out.
Answer: Read replica is not using resources from the same elastic pool as the primary (read/write) node.

As part of High Availability architecture, each single database,
elastic pool database, and managed instance in the Premium and
Business Critical service tier is automatically provisioned with a
primary read-write replica and several secondary read-only replicas.
The secondary replicas are provisioned with the same compute size as
the primary replica. The read scale-out feature allows you to offload
read-only workloads using the compute capacity of one of the read-only
replicas, instead of running them on the read-write replica. This way,
some read-only workloads can be isolated from the read-write
workloads, and will not affect their performance. The feature is
intended for the applications that include logically separated
read-only workloads, such as analytics. In the Premium and Business
Critical service tiers, applications could gain performance benefits
using this additional capacity at no extra cost.

I suggest you monitor your resources of the read-only replica. Especially to find out if you are hitting the maximum limit of resource for that database that is getting time out. I will start here sys.dm_db_resource_stats (Azure SQL Database).
Here is a list of DMV's you can use to Monitoring and troubleshooting read-only replicas.
Read this Q&A Azure Elastic Pool Database SQL Connections are Timing Out - DTU Purchasing model answered by David Browne from Microsoft. Even though this question is talking about the primary node, you will experience the same if your read-only replica uses up all allocated resources.
Ref: Use read-only replicas to offload read-only query workloads


Retreive column names of a list and (list items) and export to excel using powershell script

Question: I need to retrieve  column names  of a list and export to Excel using PowerShell script. Can someone assist me to write a script for this? Thanks in advance.
Comment: I think this question is answered here http://sharepoint.stackexchange.com/questions/83641/export-all-the-columns-from-sharepoint-list-via-powershell/83658#83658
Answer: try with the following code,
<code>Add-PsSnapin Microsoft.SharePoint.PowerShell -ErrorAction SilentlyContinue -WarningAction SilentlyContinue | Out-Null
    $site = new-object Microsoft.SharePoint.SPSite("YourSiteURL")
    $web = $site.RootWeb
    $list = $web.Lists["ListName"]
    $view = $list.Views["All Items"]
    $items = $list.GetItems($view)
    $items | %{ select-object -input $_ -prop @{Name='Title';expression={$_.Title;}}, @{Name='ColumnName1';expression={$_["ColumnName1"];}}; } | Export-Csv -Path c:\test.csv -NoType
    Remove-PsSnapin Microsoft.SharePoint.PowerShell
</code>
In the above script build your select object by giving the required columns names from the SPListItem properties before passing it to Export-CSV.


Why do you need to pull HDG knob a320 for selected mode?

Question: If you are flying in nav mode, then atc tells you turn to a different heading, as soon as you change the heading knob the aircraft starts turning, then why do i need to pull it?
Comment: What does the FMA say when your rotate the heading knob without pulling?
Comment: The FMA remains in NAV mode until you pull the knob out.
Answer: When you are in NAV mode the heading indication on the FCU is normally dashed (---*). Then, if you just rotate the knob it first synchronizes the heading to the current heading, then displays the selected heading value and you can pre-select it in the FCU. But it remains in NAV mode and there still is the managed dot after the selected heading value in the FCU (e.g. 270*) until you pull the knob out. When you pull the knob it goes to HDG or TRK mode and the managed dot in the FCU disappears (e.g. just 270 without dot) and only then does the aircraft initiate the turn towards the selected heading.

If you do not pull the knob then the heading value will disappear after a while (45s or so) and the FCU heading will be dashed again.

Once you are in HDG or TRK mode you can just rotate the selected heading knob and the aircraft will immediately turn. You don't need to pull it again.

It's the same for the selected airspeed and selected vertical speed. You can rotate the knob to pre-select a value but you have to confirm the change to selected mode by pulling the knob.
Comment: @SachinChaudhary Are you talking about a simulator or the real aircraft? I'm fairly certain it won't turn in NAV mode when rotating the knob.
Comment: But , why can't I just rotate knob in managed mode and then fly my desired heading without pulling? what happens if i continue to follow atc radar vectors but don't pull it but keep on changing the knob ?
Comment: it starts turning as soon as i change the knob, even without pulling it
Comment: FCOM is available online if you want to read the real documentation: https://www.smartcockpit.com/aircraft-ressources/A320-Autoflight.html


Slice an array out of a Google Charts DataTable object

Question: I have a Google Charts DataTable object and I'd like to access one column as an array. I know that there is the getValue(rowIndex, columnIndex) function to access individual fields, but is there also a way to get the entire column?
Answer: A simple for-loop can do:
var my_slice = []
for (var i=0; i < data.getNumberOfRows(); i++){
    my_slice.push(data.getValue(i, 4));
}

data is my Google Chart DataTable object and 4 is the ID of the column that I want to access.


(UK freelancer) Do I still need to put a client's address in the invoice if they are not a business/company?

Question: According to gov.uk my invoice must include "the company name and address of the customer you're invoicing". However my customer does not have a company, I'm simply invoicing them as an individual. Do I still need to include their address?
Comment: Assuming that it won't hurt to include it, why not just include it and remove all ambiguity?
Answer: Yes.
Even if you invoice a private person (no company involved), then the invoiced person & their address who pays bill must be named on your invoice.
There are two tax and one criminal reason.
a) in your accounts you are showing a true record of source of payment to you.
b) This keep your tax accounts in order.
c) If "No person" & "no company" is shown on your invoice, the source of the funds is unclear and you could be into tax evasion or obtaining funds from criminal sources. This gives you a big problem with both HMRC and police if it comes into an investigation.
Cash receipts in a shop are not invoices but tally cash in with goods out to unnamed persons. That is why cash economy is truly "untraceable".
Answer: In a Danish context, the client address is required as the inclusion (or not) of sales tax on the invoice depends on location (among other factors).
Domestic clients must pay an additional sales tax, while foreign businesses may be invoiced without added sales tax.
Rules in your jurisdiction may differ. In addition, I could imagine the authorities could extend the thinking behind e.g. banks having to 'know their clients' to your business.
For this reason I would always include an address - as it seems the authorities seem to prefer/demand a paper trail of where money originates from and where it ends. Not being able or willing to explain who or where your clients are, could be considered shady and cause needless problems.


How to use SSRS to retrieve HTML snippets with embedded images the easy way

Question: I have a web-app that retrieves reports from our SSRS server dynamically - we hit a URL and out pops some HTML4.0 which I stuff into a div for the user to view. I recently tried adding a report that has an embedded image (in the RDL itself) and the image doesn't make it through. What does make it through is an IMG SRC reference back to the SSRS box but we do not allow end users to hit the SSRS box directly... users query the web-app and the web-app interacts with the SSRS service. There is an option to render in MHTML (note that we typically use rs:command=RenderHTML with rs:format=HTML4.0) - the blob returned appears to be valid MIME but does not seem friendly for stuffing into a DIV... am I missing something obvious? My next step is to parse the MIME, swizzle the references and stuff the whole thing back into the page but I feel like this is the hardway.
What is the easy way to retrieve HTML snippet reports out of SSRS with embedded images?
Answer: I have faced the same issue.  Depending on how you implement you Reporting Services, you can get around this without having to write additional code, but you will need to be able to setup kerberos authentication to SSRS.  If you don't do this, you will need to capture the Image Stream from the SSRS Webservice.  This is done by using the RenderStream method of the webservice.  When you render an HTML report it will return an array of streamId strings.  These get passed into the RenderStream method, which in give you a byte[] in return.  You will need to swap out the sources in the HTML with new paths to the local files.  Then you will need to manage your file system.  As you do this, the number of stored images could get out of control, depending on the number of users.  So you may have to manage clean up of the images.


Full Control and Limited Access security, huh?

Question: If I look at the site permissions of a site I can see that some people have both Full Control and Limited Access.  The two seem to contradict each other.  How can someone have Full Control yet also have Limited Access?
Comment: How to resolve this then? I am currently an owner but i am seeing Full control and Limited access.
Answer: It's just how SharePoint displays permissions.  It means that the user has permissions to an object with custom permissions.  At the same time, the user is a member of a site group or possesses individual permissions that provide full control permissions.  SharePoint will provide the user with the least-restrictive permission set available for that object.
Answer: Just because a user has Full Control permissions does not guarantee blanket access. Only Site Collection Admin has that. Full Control users and groups can still be subject to actions that result in the Limited Access designation.
Answer: Limited Access means a SharePoint user or group has permissions to an item or document or page or library within a Site, but does not have permissions to the entire Site.
This is normally used when you want to share content that resides within a site with users but not allow to browse the site.
See the following Microsoft TechNet article  for further help on the subject.


What is the halving thickness of diferent materials against high energy gamma rays?

Question: By high energy gamma rays I mean 10+ MeV to 100 MeV. Is there a way to calculate it ?
Also, at these high energies, do heavier elements like Lead for example still have the advantage over lighter ones like Aluminum ?
Comment: Note that particle physicist at least talk about "radiation length" which represent (roughly) the length over which the intensity is reduced by a factor of $e$ rather than 2. There is a moderately complete discussion in the Particle Data Book published by the [particle data group](http://pdg.lbl.gov/).
Comment: I have tried this : http://www.wolframalpha.com/input/?i=What+is+the+maximum+electron+radiation+1+cm+of+Lead+can+block and the result for 1 cm of Lead was 29 MeV, so what does that mean ? how much will 1 cm of lead reduce the intensity of a 29 MeV gamma ray by ?
Comment: I think a more useful subject for your question would be "What is the attenuation length of diferent materials for high energy photons?"
Answer: In general it can be difficult to calculate the interaction of photons with atoms as a function of energy as different interaction mechanisms turn on and off: pair production turns on at 1.022 MeV, Rayleigh scattering turns on at low E as a function of atomic/molecular size, etc. An important question to ask yourself is "how precisely do I need to know the attenuation length of a photon in a given material?" Note that attenuation length is generally a more convenient quantity for calculations than halving thickness, and it corresponds to the distance at which $1/e \simeq 63\%$ of photons have been absorbed.
In many materials the mean free path of 10-100 MeV photons is roughly flat, as pair production is the dominant mechanism by which photons interact in this energy range. You can look up the attenuation lengths of various materials as a function of energy at this NIST page. You'll probably want to select units of $\frac{cm^2}{g}$. Whatever number you get from the table you will multiply by the density of your material---which has units of  $\frac{g}{cm^3}$ ---the resulting number will be in units of $\frac{1}{cm}$ and is 1/attenuation length. Also, to answer your second question, in general you'll find that the increasing density of a material corresponds to shorter attenuation lengths, since the more charged particles you pack into a tighter volume corresponds to more photons stopped via pair production in that volume. In general heavier elements correspond to denser materials.
For a more precise analytical treatment of attenuation length, you may want to consult the literature. Looking in the PDG, they reference a paper by Y.S. Tsai in Rev. Mod. Phys. 46, 815 (1974). Good luck!
P.S.: Note that if you're asking this because you're trying to protect yourself from radioactive materials, you should contact a radiation safety expert as radiation protection can be much more complicated than just stopping gamma rays (material handling, neutron radiation, etc.).
Comment: @AbanobEbrahim: it's not a stupid question, and the answer is "not necessarily". As I stated in my answer, different photon interaction processes turn on and off as energy changes. Towards the lower end of your energy range, pair production is still "turning on", meaning it isn't converting the lower energy photons to $e^{+}e^{-}$ pairs as efficiently as it converts higher energy photons.
Comment: @Kevinismus, I am having a hard time using NIST as it gives the results in a weird format and I can't get a number. If you could only give me the attenuation length for say Lead, that would be greatly helpful.
Comment: I have also come across this : http://www.laradioactivite.com/en/site/images/Long_attenuation_en.jpg. Does it mean that 1 cm of lead will absorb 63% of the incident gamma rays energy if they are 1200 MeV (according to the image) ?
Comment: @AbanobEbrahim: The attenuation length as a function of energy for lead (density ~ $11.3 \frac{g}{cm^3}$) goes from about 2 cm to 1 cm for E = 10 MeV to 100 MeV. And yes, you're reading that graph correctly.
Comment: @Kevinismus, I am sorry if this is a stupid question, but shouldn't the attenuation length increase with the energy ?
I mean the more energetic the gamma ray, the more it will penetrate in a material and therefore you need more of the material to stop 63% of it.


iBook on iPad: turn pages with voice command

Question: I bought an iPad for somebody who is paralysed; primarily as a book reader for its ability to enlarge the text- but also to surf the net, etc. with a family member or caregiver using the touchpad. I want to know if there is a voice command so he is able to turn pages himself, rather than needing to rely on anyone else.
Comment: There isn't a voice recognition option in Accessibility for iPad (1 or 2). Have you explored AssistiveTouch feature in Settings app under the General -> Accessibility ? Hope this helps.
Comment: I am not aware of any voice recognition in iOS, unfortunately.
Comment: You might want to look thru the App Store products searching for Voice Command, accessibility and assistive devices. The leader in voice activation stuff on the mac is Dragon. They have an iPhone app called Dragon Go, that is free. I am in the market too as my eyesight is failing a bit. So I would welcome knowing what else you find.
Answer: No - without some external hardware that would somehow touch the screen for you, voice input is for Siri / VoiceOver and not general accessibility controls.
You might look to a program like Instapaper where you can scroll the view by tilting the device but iBooks doesn't have any accelerometer based controls or voice controls at present. It's not that it couldn't be added, just that it hasn't.
Answer: I am visually impaired and have been to multiple VA Blind centers, to make a long story short the VA is issueing apple products, including iOS products, not only to the visually impaired but also Vets who suffer from TBI, Spinal trauma... This came up this past Feb. If the individual has head movement a stylus can be attached to an extension and used with the mouth, not only does this allow the person to turn pages but also will allow them to open apps and move through app screens, and possibly the ability to use the home button.


Does LPG (cooking gas) react with water?

Question: If cooking gas (or LPG) was exposed to some water in the gasline system, what would be the resultant by-product, and would this induce flame into the cylinder for explosion?
Comment: Everything that comes out of a refinery has been exposed to water and contains residual water in the ppm range or lower. A menial job in a refinery is to occasionally open valves on the storage tanks to drain water ( that has separated from the product) into the refinery sewer system. And corrosion inhibitor is added to products going into pipelines to prevent corrosion of the steel pipe by water continuing to separate from the hydrocarbons
Comment: @làntèrn While your comment might be useful for OP, it is inaccurate. Even quite low solubility may be enough; not to say that substances can react on interfaces between the phases. Also in specific conditions H2O and  hydrocarbons may react - in vapor phase, at very high temperature, when O2 isn't present.
Comment: LPG is a mixture of hydrocarbon gases (propane and butane) and the mixture will have non-polar character. Water is polar. Many chemistry students learn that non-polar can not react well with polar substances (and vice versa), thus you will not observe like chemical explosion like you wrote. If they are not react, you might want to check the fact about their solubility. From there, we know that the mixtures are slightly soluble in water, it means that there is tendency that the water will evaporate (and the evaporation does depend on the temperature of the system).
Comment: @Mithoron Well, the OP doesn't tell about the temperature and I know the temperature does impact on interfacial phenomena of the mixtures. I do learn it but given the OP information, it is not suffice enough to explain the outcome of the system. Perhaps it is not too good to elevate one thing to some extents.
Answer: No. It produces water after combustion, so it does not react with it.


Hausman test with or without covariates

Question: I am using panel data and I would like to determine whether I can use the Random Effects (RE) model instead of Fixed Effects (FE) to estimate one coefficient of interest.
When I use the Hausman test comparing FE and RE, I have to reject the null hypothesis (meaning that the RE model is not OK). However, the difference between the coefficient of interest estimated by FE and RE is not statistically significant.
So my question is: Can I justify using the RE model only based on this fact? After all, the null hypothesis of the Hausman test must be rejected only because the estimation of some other covariates (control variables) significantly differ between the RE and FE approaches. But in my case, these variables are not of interest and I do not seek for consistent estimators.
Comment: So if you don't care about consistency (why would you not?), why do you perform the Hausman test?
Comment: I care about consistency for the variable of interest but not for the control variables. My sense is that my variable of interest is not correlated with time-invariant unobservables (as the Hausman test on this very variable suggests) but some control variables are. So, is it correct to say that, if I use a RE model, the estimation of the variable of interest will be consistent while estimations of control variable will not (but this does not matter to me as I do not claim to have the right coefficients on these variables). Thanks a lot for your help.
Comment: Having unbalance data might be the reason. If you do have unbalance data, you can directly use random effect. Read more about this in Wooldridge's book.
Answer: The choice between FE and RE models depends on the focus of the statistical inference. The FE model is an appropriate specification if we are focusing on a specific set of $N$ individuals (say, $N$ firms or $N$ OECD countries, or $N$ American states) and our inference is restricted to the behavior of this set of individuals. The RE model is an appropriate specifiction if we are drawing $N$ individuals randomly from a large population and are trying to make inferences about that population (see Baltagi, Econometric Analysis of Panel Data, 2008, §§2.2-3). The Hausman test can't say anything about your focus.
The Hausman test is asymptotically equivalent to a standard Wald test for the omission of $\tilde{\mathbf{X}}$, a matrix of deviations from individual means (see Baltagi, 2008, §4.3). In other words, given the model
$$y_{it}=\mathbf{x}_{it}\boldsymbol{\beta}+\mu_i+u_{it}\tag{1}$$
one can split $\mathbf{x}_{it}$:
$$y_{it}=(\bar{\mathbf{x}}_i+\tilde{\mathbf{x}}_{it})'\boldsymbol{\beta}+\mu_i+u_{it}\tag{2}$$
where $\bar{\mathbf{x}}_i$ is the vector of individual time-invariant means for the $i$th individual and $\tilde{\mathbf{x}}_{it}=\mathbf{x}_{it}-\bar{\mathbf{x}}_i$.
Further, one can give separate parameters $\boldsymbol{\beta}_1$ to the individual means and $\boldsymbol{\beta}_2$ to the deviation variables:
$$y_{it}=\bar{\mathbf{x}}_i'\boldsymbol{\beta}_1+\tilde{\mathbf{x}}_{it}'\boldsymbol{\beta}_2+\mu_i+u_{it}\tag{3}$$
$\boldsymbol{\beta}_1$ is a between regression coefficient, while $\boldsymbol{\beta}_1$ is the within (FE) regression coefficient. The Hausman test is based on $\hat{\boldsymbol{\beta}}_{RE}-\hat{\boldsymbol{\beta}}_{FE}$, but can equivalently be based on $\hat{\boldsymbol{\beta}}_1-\hat{\boldsymbol{\beta}}_2$ (see Baltagi, 2008, §4.3).
As to correlation, some variables in $\mathbf{X}$ may be correlated with $\boldsymbol{\mu}$, but $\tilde{\mathbf{x}}_{it}$ is orthogonal to $\mathbf{1}\mu_i$ ($\mathbf{1}$ is a vector of ones) for all $i$. Thus:

under a FE framework, the time-invariant terms $\bar{\mathbf{x}}_i'\boldsymbol{\beta}_1$ and $\mu_i$ are swept out and one gets unbiased and consistent estimates for $\boldsymbol{\beta}_2$;
under a RE framework,

if one estimates models $(1)$ or $(2)$ the implicit assumption $\hat{\boldsymbol{\beta}}_1=\hat{\boldsymbol{\beta}}_2$ doesn't hold and the Hausman test fails;
if one estimates model $(3)$, then $\hat{\boldsymbol{\beta}}_2$ is unbiased and consistent (it is exactly identical to $\hat{\boldsymbol{\beta}}_{FE}$) and one can ignore or suppress $\bar{\mathbf{x}}_i'\boldsymbol{\beta}_1$, or use $\bar{\mathbf{x}}_i'\boldsymbol{\beta}_1+\mu_i$ to model the random intercept (see, e.g., Snijders and Bosker, Multilevel Analysis, 2012, chap. 4).

In brief, you can estimate a RE model which passes the Hausman test by just splitting your $\mathbf{X}$ matrix into its individual time-invariant means $\bar{\mathbf{X}}$ and the within-individual time-varying deviations $\tilde{\mathbf{X}}$ (see here for a simple example, additional details and references). I'd say that such a coherent approach would be better than an eventual and questionable 'mixture' of consistent and inconsistent estimates.
Comment: Isn't the fundamental distinction that FE allows for endogeneity of all regressors with the individual effect, whereas RE assumes exogeneity?
Comment: It is worth adding that the hausman test requires the full set of assumption. It is often difficult (in observational studying) to justify this.

Hausman test with or without covariates


Can't SSH into Pi 3 after enabling in config

Question: Just loaded a new image of PIXEL for my Pi 3 . I enabled SSH via raspi-config but was unable to access using SSH (via PuTTY). Tried putting ssh file in /boot/ directory and still didn't work. Any other suggestions on how to access my pi headless?
Comment: Given your description, the answers could even start with: did you connect the power to the Raspberry Pi? You provided zero details.
Comment: Did you reboot?
Comment: Is your pi connected to internet?
Comment: Can you check the status of ssh using the command `sudo service ssh status`. If there are any errors like  error: `Could not load host key:` then clean up the files using the command `sudo rm -f /etc/ssh/ssh_host*`. Then perform `sudo service ssh restart`. That should get you going fine.
Comment: go to /etc/ssh/sshd_config and edit mode and put 'PasswordAuthentication yes'
Answer: Since you provide zero details, let's get started:
Firstly, make sure that the IP on the Pi is correct, if static. If it's dynamic, search for it with an "IP Scanner" until you find a device named like raspberry.

Once found, try to connect to port 22 (SSH).

If it throws an error, put the card in the PC and be sure that the ssh file is in the main directory of the card, since Windows only shows the boot partition.
Also make sure, that the file is only named ssh - nothing more, nothing less. Under Windows, when you create an "empty" text file, it puts .txt as default file extension and hides it from you, so you wil only see ssh, but actually it is an .txt.

To prevent this from happening, search in Windows for "Folder Options" and in the list, disable the tick box named: "Hide file extension on known file types".

After that, go into the directory again and remove the .txt on the end of the file and confirm the warning dialog.
Answer: Make sure that you have assigned a static ip to your pi. If you are not sure about it then then scan your network using fing(or other relevant apps) on your android phone and get ip address of raspberry pi.
Assign a static ip to your pi from DHCP settings of your router if you haven't done.
If you have enabled SSH before loading new image revisit raspi-config. Make sure you have selected SSH on your putty.Hope it helps!
Comment: A static IP is not required.


What is this three-engine beauty?

Question: What is this three-engine beauty?
Answer: It appears to be a Westland IV/Wessex
I don't know why it is marked P-1, but the elevator is distinctive.
Answer: P-1 is a temporary civilian registration issued to the company to use on prototypes. Each company was assigned a letter, so any time you see a British civilian aircraft with a letter-number code like this, it is a prototype. The best known are probably E.2 to E.3 used for the de Havilland Albatross airliners, as seen below:
Comment: Hi Mike, the question asked it was about the type of airplane in the photograph, not so much the number on the side. That's an interesting addendum though.
Comment: well, the sole answer was unsure of the importance of the number, and my answer added that it was a prototype Wessex, not just any Wessex.

What is this three-engine beauty?