Computer boots straight to Ubuntu without BIOS logo or GRUB menu

Question: Not sure why this started happening, but now Ubuntu 16.10 just boots directly to the login screen, without showing the UEFI/BIOS logo or GRUB menu for dual boot. I haven't changed any settings to deal with booting; it was working perfectly before.
Answer: The previous answer about Boot-Repair is unlikely to fix your issue.  Since your Ubuntu is booting we know that you have a working bootloader.
You should be able to press SHIFT during boot to bring up the grub menu, however that can be tricky timing it right sometimes.
Information related to this question can also be found in this previous question, even though they are not using dual-boot the information is identical for grub/grub2:
How to get to the GRUB menu at boot-time?
It is likely that somehow
GRUB_HIDDEN_TIMEOUT and/or GRUB_TIMEOUT got changed in your /etc/default/grub
You can change these to something other than zero (in seconds) and then run:
sudo update-grub
After this your grub menu should show on your next reboot.
Comment: Unfortunately, neither solutions worked. The UEFI logo doesn't show at the start like it usually does, and neither does the GRUB menu. The grub file was intact.
Comment: @fu_______ If `SHIFT` key doesn't work, you can try to press `ESC` key at boot time. This should show you Grub menu.
Comment: I've fixed it...somehow. I downloaded Grub Customizer and saved the configuration and it seemed to fix all my problems. Windows had to fix some disk errors but other than that it solved my problems.
Answer: You should try to boot from live USB. After that, install Boot-Repair with
sudo add-apt-repository ppa:yannubuntu/boot-repair
 sudo apt-get update
 sudo apt-get install -y boot-repair && boot-repair
After that, it should guide you to re-installing GRUB. It should work.
Answer: You note in a comment to another answer that you've gotten it working; however, that comment also notes that Windows complained of disk errors. This makes me think that you may not have disabled Fast Startup and/or Hibernate in Windows. See these pages for instructions on how to do so:

Disabling Fast Startup
Disabling Hibernate

If you don't disable these Windows features, the result can be damage to filesystems shared between Ubuntu and Windows, including the EFI System Partition (ESP), which is where EFI boot loaders live. This could have resulted in GRUB's configuration scripts failing to detect Windows after a GRUB or kernel update; and when that happened, the GRUB timeout and menu-display options might have been altered to prevent the menu from showing up.
This explanation is highly speculative, of course, and there are other things that could cause damage to the ESP's FAT filesystem, such as a system crash, a bug in some tool that accesses the ESP in a low-level way, or user error. Ubuntu's installer doesn't warn you to disable those Windows features, though, so failing to do so is a common mistake.


Why shouldn't I use footnotes in chapter/section titles?

Question: I've seen several times on TeX.SE (example 1, example 2) that I shouldn't use footnotes in titles, with no explanation. However as far as I can tell the technical problems it would cause (the command need to be protected, a short title should be provided for the TOC) appear to be resolvable. So why shouldn't I do it?
Comment: A footnote is some side-remark, while a title is a "main message", combining both looks odd. Did you ever saw a footnote sign e.g. on a book title?
Comment: @UlrikeFischer I'm not talking about the overall title, I'm talking about chapters. My thesis has three chapters, each of which is a published (or to-be-published) article; I'd like to put a footnote with the bibliographical information for each title. Why shouldn't I?
Comment: I know, but book titles and chapter titles are similar. The first is only a bit larger. I find it bad looking -- both aesthetically and logically -- if a large title text has an large superscript asterix or a number to point to some not really important text.
Comment: @UlrikeFischer: I know at least one :-) https://www.discworldemporium.com/wp-content/uploads/2022/06/products-2862.jpg
Comment: What you describe is like the acknowledgements (`\thanks`) in an article, which in LaTeX's `article` class and many publishing style is put as an unnumbered footnote attached to the author's name (not the title). If you need to have per-chapter credits, you could put them as a footnote after the first sentence. But why not just list all the credits together at the beginning. In academic books I often see this on the copyright page ("Chapter 5 is adapted from X article in The Atlantic") and in acknowledgments ("I presented the research in chapter 5 at X conference").
Answer: There are two issues, (a) does it work and (b) is it a good idea.
(a) is the easiest part to handle, the main issue being that you don't want the footnote in the page head or table of contents so (as you mention in the question) you should supply a short from in the optional argument (even if it is just the same as the long form but without the note).
That will then work in the standard classes. If you are using some custom formatting which sets the title in some rotated coloured box surrounded in 100 levels deep boxing levels of tikz positioning then it's possible that the footnote will not be able to automatically migrate to the outer level, but if you knew how to get into that position you should know how to work around the issues, so using \footnotemark where you want the mark and \footnotetext somewhere where it works.
(b) is harder. Use of footnotes at all is very subject dependant. In the mathematical sciences a footnote usually denotes a last minute correction squeezed in at a late stage; so something to be avoided at all costs. So having a footnote in a chapter heading would just be weird. In other subjects footnotes are more common, I've seen biology and law documents where an average page seems to be half footnote text, which looks bizarre to me but is apparently the convention. So the answer to (b) will depend on who you ask.
I can only refer to the famous quotation in the TeXBook

Don't use footnotes in your books, Don.
JILL [KNUTH] (1962)

If you can't get agreement in the Knuth household about when to use footnotes it's probably too much to ask to get global agreement on the same subject.
Comment: I could see an argument for doing this if you had some ironic, humorous, off-hand footnote about your title.  But only *because* combining both looks odd.  In this case where OP wants to give the bibliographic citation, I think it would be better for that to be in the thesis introduction.
Answer:
So why shouldn't I do it?

Because IME many editors regard it as Bad Style. Try to find a more elegant way to explain what you want, like a note under the title block, before the abstract.
As with all æsthetic choices, your mileage may vary. Some disciplines think it is just fine.
Pseudo-footnotes in author lists, however, are an accepted way to signal their affiliation, but the list of institutions needs to go immediately after the title block, not at the foot of the page.


what are Datatypes in SQLite supporting android

Question: Can anyone please tell me the LIST OF DATATYPES in SQLITE supporting ANDROID.
I want to confirm about TIME and DATE datatypes.
Answer: Here is a list of SQLite's data types
TIME and DATE are supported indirectly.
Comment: I think date and time can both only been inserted as string
Comment: Or Using UNIX timestamp and save them as INTEGER, also BOOLEAN is stored as INTEGER 0 or 1
Answer: As said at Datatypes In SQLite Version 3:

Each value stored in an SQLite database (or manipulated by the database engine) has one of the following storage classes:

NULL. The value is a NULL value.
INTEGER. The value is a signed integer, stored in 1, 2, 3, 4, 6, or 8 bytes depending on the magnitude of the value.
REAL. The value is a floating point value, stored as an 8-byte IEEE floating point number.
TEXT. The value is a text string, stored using the database encoding (UTF-8, UTF-16BE or UTF-16LE).
BLOB. The value is a blob of data, stored exactly as it was input.

And for TIME and DATE storage the following is stated:

Date and Time Datatype
SQLite does not have a storage class set aside for storing dates and/or times. Instead, the built-in Date And Time Functions of SQLite are capable of storing dates and times as TEXT, REAL, or INTEGER values:

TEXT as ISO8601 strings ("YYYY-MM-DD HH:MM:SS.SSS").
REAL as Julian day numbers, the number of days since noon in Greenwich on November 24, 4714 B.C. according to the proleptic Gregorian calendar.
INTEGER as Unix Time, the number of seconds since 1970-01-01 00:00:00 UTC.
Answer: First of all check the Sqlite3 version for each android API version: https://stackoverflow.com/a/4377116/737636
I don't think that this is the full list of versions and I think that also OEM can change the SQLite's version installed inside.
Anyway, if you take a quick look you can see that the base version is 3.x
This is the list of supported datatypes by SQLite3 (the same used inside android): http://www.sqlite.org/datatype3.html
and here you are :)


Blinking power button in yellow color on a Dell OptiPlex 7010

Question: We have a Dell OptiPlex 7010 model system with Windows 10 Pro 64 bit and system is working fine. But the problem is whenever I restart the system is not booting and Dell power button is keep on blinking with amber color and when I do the power drain which will start to load the operating system. 8GB RAM installed.
Below are the steps I have done already but still issue not fixed:

Full system diagnostic is done and no error found.
Different RAM has been changed and checked.
RAM replaced in different slot and checked.
Latest Bios version updated.
Answer: The Dell power button is blinking with amber color
You need to look at the blink pattern. It is documented in the Owners Manual and will tell you exactly what is wrong:

Power LED Diagnostics
The power button LED located on the front of the chassis also
  functions as a bicolored diagnostic LED. The diagnostic LED is only
  active and visible during the POST process. Once the operating system
  starts to load, it is no longer visible.
Amber LED blinking scheme – The pattern is 2 or 3 blinks followed by a
  short pause then x number of blinks up to 7. The repeated pattern has
  a long pause inserted in the middle. For example 2,3 = 2 amber blinks,
  short pause, 3 amber blinks followed by long pause then repeats.
...
<code>2,1 system board failure
2,2 system board, PSU or PSU cabling failure
2,3 system board, memory or CPU failure
2,4 coin-cell battery failure
2,5 corrupt BIOS
2,6 CPU configuration failure or CPU failure
2,7 memory modules are detected, but a memory failure
3,1 possible peripheral card or system board failure
3,2 possible USB failure
3,3 no memory modules are detected
3,4 possible system board error
3,5 memory modules are detected, but a memory configuration or compatibility error
3,6 possible system board resource and/or hardware failure
3,7 some other failure with messages on screen
</code>
Answer: Just for followers, for me on my 990 this "did" mean bad psu, I blew out the dust on its intake grill and blew through the thing itself and it started working again FWIW :)
Other options: remove and replug in to motherboard, replace motherboard? :)


Does the SEC apply to overseas investments?

Question: Would a US company issuing private stock outside of the USA have to comply with SEC regulations?
Comment: Does that mean "only issues private shares", or do you mean "is also publicly traded in the US"? The concept of "private shares" needs to be clarified: shares in X can be publicly traded, but that does not preclude X from giving me, personally and privately, shares in X.
Comment: To clarify, the private company takes investment from a foreign source with an agreement to share profits in the future
Answer: There are several ways that a company can be subject to regulation by the Securities and Exchange Commission. The basic threshold is involvement in interstate commerce in the United States, which is a very low one and is probably satisfied simply by incorporating in the United States and having U.S. shareholders.
In general there are also practical reasons to avoid having a large number of investors in a start up company, even when exemptions for securities laws permit you to do so.
A U.S. company that is privately held and does not make a public offering is not required to have its securities registered with the SEC, but is still subject to SEC regulation through Regulation D and Regulation 10b-5.
A foreign company which does not offer to sell shares in the U.S. and does not list its shares for sale in the U.S. will generally not be subject to U.S. regulation by the Securities and Exchange Commission, although it might be deemed to be offering shares for sale in the U.S. if it has any significant number of U.S. shareholders. Obviously, however, foreign companies are subject to the securities laws of the places where they are organized and do business.

It can make a public offering of securities (not necessarily stock) in the United States. This happens due to Section 5 of the 1933 Securities Act. A public offering is pretty much any offer to sell a security made via a general circulation publication, a mass mailing list, the Internet, or by telephone cold calls, that is not pre-screened to include only "accredited investors".
It can be listed on a U.S. stock exchange (e.g. the NASDAQ or the OTC market). This is due to Section 12(a) of the 1934 Securities Exchange Act. This says:

SEC. 12. (a) It shall be unlawful for any member, broker, or dealer to
  effect any transaction in any security (other than an exempted
  security) on a national securities exchange unless a registration is
  effective as to such security for such exchange in accordance with the
  provisions of this title and the rules and regulations thereunder. The
  provisions of this subsection shall not apply in respect of a security
  futures product traded on a national securities exchange.

It can be a U.S. company with more than a certain number of shareholders and more than a certain asset value, even if it has never made a public offering. This is due to Section 12(g) of the 1934 Securities Exchange Act. This says:

(g)(1) Every issuer which is engaged in interstate commerce, or in a
  business affecting interstate commerce, or whose securities are traded
  by use of the mails or any means or instrumentality of interstate
  commerce shall—
(A) within 120 days after the last day of its first fiscal year ended
  on which the issuer has total assets exceeding $10,000,000 and a class
  of equity security (other than an exempted security) held of record by
  either—
(i) 2,000 persons, or
(ii) 500 persons who are not accredited investors (as such term is
  defined by the Commission), and
(B) in the case of an issuer that is a bank or a bank holding company,
  as such term is defined in section 2 of the Bank Holding Company Act
  of 1956 (12 U.S.C. 1841), not later than 120 days after the last day
  of its first fiscal year ended after the effective date of this
  subsection, on which the issuer has total assets exceeding $10,000,000
  and a class of equity security (other than an exempted security) held
  of record by 2,000 or more persons,
register such security by filing with the Commission a registration
  statement (and such copies thereof as the Commission may require) with
  respect to such security containing such information and documents as
  the Commission may specify comparable to that which is required in an
  application to register a security pursuant to subsection (b) of this
  section.

There are numerous exemptions set forth in Section 12(g)(2). Once you are subject to registration, you can only avoid it by falling below 300 investors, pursuant to Section 12(g)(4).
The Securities and Exchange Commission has considerable authority to tweak these thresholds by regulation.

Private offerings of securities by U.S. companies must comply with Regulation D which creates safe harbors for companies that are not required to register as a publicly held company but are required to make certain private disclosures. Regulation D is a coordinated regulation issued under the 1933 Securities Act and Section 12(e) of the 1934 Securities Exchange Act.
Every sale of securities by a U.S. company or to a U.S. person is subject to the requirements of Regulations 10b-5 under the 1934 Securities Exchange Act. This is a basic anti-fraud requirement.
Some securities exempt under federal law are still subject to state securities regulations which are called "Blue Sky Laws".
Tax law also limits your entity form if you have many owners. An S-corporation is not permitted to have more than 75 owners (with special rules for counting owners). Special rules apply to publicly traded partnerships. And, there are securities, tax and non-tax rules that apply to certain kinds of business trusts such as REITs and mutual funds, to state regulated banks, and to mutual insurance companies.
Answer: If the company is a public US domiciled company the SEC regulates it. In addition, any foreign company that publicly traded its securities in the US is subject to SEC regulation.
Comment: A company doesn't even have to be domiciled in the U.S., if for example, it is listed on a U.S. stock exchange or makes public offers to invest to U.S. persons.


windows 7 and desktop icons

Question: how to put the real webpage icon on desktop of windows 7 with explorer it was simply shortcut on windows xp and in firefox i'd go to view/ page info/media...save the icon in a folder then left  click the icon on the desktop/properties and change the icon ...go and get it in the folder...i'm an icon nut.. i cover my desktop!
Answer: You want to download the favicon.ico file. Most of them are stored in / directory of the website. For example Google is http://www.google.com/favicon.ico. You can save .ico files where ever works for you and then add it to your shortcut.
Comment: ALLRIGHT!!! YESSSSS! IT WORKS....AS YOU SAW IT WAS A SIMILAR WAY IN XP....JUST DIDIN'T WORK IN 7 ...A MILLION THANKS MAN!

windows 7 and desktop icons


Is it "our work and that of others have demonstrated" or "our work and that of others has demonstrated"?

Question: Do I use a plural or singular verb in this sentence? Is one more accurate than the other?
Comment: This is a good place for a passive to get rid of the agent subject phrase, and then extraposition to keep the complement at the end: _It has been demonstrated by our work, and the work of others, that..._
Comment: "Has," but both sound awkward, it took me a minute to figure out what you were trying to say. How about something more like "Prior work has / Prior studies have demonstrated..."
Comment: *Work* is an uncountable noun, so it is singular, despite the *and*, so use *has*.
Comment: I suggest building on Kevin's recommendation: "Prior work—ours and others'—has demonstrated..." It retains all of the information from the original, avoids the awkward singular/plural issue, sounds natural when you read it aloud, and does the job in seven words—one fewer than the original construction requires and five fewer than the passive-voice construction does.
Answer: I think it should be plural-agreeing "have" (unlike others who have commented so far).  Does the subject of the verb refer to more than one thing?  If so, the plural-agreeing form should be used.
Then we need to decide whether "our work and others' work" is just one thing or two things.  If our work and other's work referred to just one thing, as for example in some collaborative effort, it is that single work that would be referred to.  I don't think that is a possible interpretation, here.  Maybe it would be if if the reference was to "our and others' work" (as compared with "our and others' works").
I don't see the relevance of "work" being a mass noun.  I say my work and your work "are the same", not "is the same" (though it's true I could say "our work is the same"). But what does that have to do with whether "work" is a mass noun?  I could say to my wife that my son and her son is late, yet again, though "son" is not a mass noun.
Answer: In your example sentence you have no choice but to use the singular.
For the avoidance of doubt, the plurality or otherwise of 'has/have' has nothing to do with the adjacent 'others'. The verb follows the plurality of the subject, which is 'our work and that of others'. The question is: is that a singular or a plural noun phrase?
Not many speakers would argue that 'our work' + '[work] of others' adds up to two 'works'. Other than in the specialised usage of 'book', 'work' is a mass noun. It doesn't get pluralised. 'You do some work, I do some work; together we do some work' (as opposed to 'I write a work of literature, you write a work of literature, together we write some works of literature'). In the sense of 'work' as 'labour', the subject is singular and the verb should be 'has'.
If the sentence had instead been 'My subscription and that of my parents has/have expired', there would have been more room for debate. I would have gone for the plural, in that the subject is the total of two subscriptions. However, the singular doesn't sound terrible either, because you could rephrase the sentence as 'My subscription has expired, and so has that of my parents.'


How can I fix my 1st gen iPad if I can't restore (error 3194) and have forgotten the password?

Question: I googled to try to solve this, but changing the hostfile didn't work and using tinyumbrella didn't work either. I'm still getting the same error 3194.
Answer: I solved this error by deleting private/etc/hosts and rebooting my computer.
Comment: You generally don't want to entirely delete your hosts file - instead, edit it and remove any lines mentioning `gs.apple.com`.


Manually triggering a low active pin

Question: I have a microcontroller with an active low reset pin. This pin is connected to a timer that periodically triggers a low edge and resets the microcontroller.
What would be the best circuit that would allow me to also manually trigger the reset pin through an external circuit while the pin is connected to the timer? Shorting the two together would obviously not work.
Comment: What kind of timer? Post the link to the datasheet.
Answer: When either the timer output or the GPIO goes low, \$ \overline{\text{INT}}\$ goes low, so the gate is a positive true AND, 1/4 of an HC08 or somesuch.

or, better yet,
Answer: I assume your timer output is a push-pull circuit?
If you can configure it to be open-drain, then you just need a pull-up resistor on the reset line, and use a switch to short the reset input to ground.
If you can't configure the Timer output to be open-drain, then  you just need a series resistor between the timer output (say 1k) and the reset input. Again, the reset switch shorts the reset input to ground.
Answer: SW1 can be pressed in order to generate the reset independent of timer. Whenever the SW1 is pressed first transistor turns on and RST# net will be pulled low to ground. Whenever the Timer is driving low, the first transistor will turn on and  RST# line will go low and a low level voltage will be presented to the MCU RESET# Pin.
When either switch or TimerOutput is low, Q3 will have a base current of about 8 mA (assuming VCC as 5 V, Vbe as 0.8 and diode drop of 0.4 V), which will sufficiently to saturate Q3. Q4 too will be saturated so that, the Vol will be less than 0.1 V.

simulate this circuit – Schematic created using CircuitLab
Comment: What if the switch is pulled up? I expect the final design to be a gpio instead of a switch. The gpio will be pulled down to trigger the reset.
Comment: @Malfunction Sorry for the silly mistake.
Comment: Doesn't work: lowest possible level of ouput will be ca. 1.4V, which is to high for L.
Comment: @Curd pls check now. Sorry for ignoring
Comment: Now logic output level is OK, but still another problem: base resistor R9 is only 100 Ohm, i.e. ca. 36mA at Vcc=5V; might be too much load for TimerOutput. BTW R8 and R9 can be shared.
Comment: @Curd assuming 5 V now, the TimerOutput need to sink about 7.9 mA. Nice correction of sharing the resistors, Else the current will be more when both go low simultaneously.
Comment: You can make the base resistor realy big (several 10kOhm). Q3 will amplifiy the current at least by factor 100 (or saturate). The same is true for R7/Q4
Comment: @Curd the hFE is only about 10 to 20 for selected transistor at saturation. So, i thought of low value for base resistors. Hope it still works

Manually triggering a low active pin


Any Way To Run A New 3" Pipe INSIDE An Existing 6" Pipe & Have The 3" Pipe In A Fixed Position Inside

Question: Is there any way to run a brand new 3" water line pipe through the inside of an existing 6" water line pipe. I want to make sure that the 3" pipe is not rattling around inside the 6" pipe. Any way to keep it in a fixed position as well?
Comment: what distance are we talking about?
Comment: Hello, and welcome to Home Improvement. Why do you want to do this? And, you should probably [take our tour](//diy.stackexchange.com/tour) so you'll know how best to participate here.
Comment: must the 6" pipe remain open?
Answer: There are products called "pipe casing spacers" which can be used for such a purpose.  I'm not meaning to endorse this particular brand but this is an example of such a device:
Pipe Casing Spacer
A drawback, however, is that the original line will be significantly blocked by the combination of the new pipe plus the spacers, so if you need to maintain flow in it, what you get will be quite a bit less than what you currently have.
Comment: That's cool, but looks very expensive and specialized.  OP is going to have to _really_ need the pipe centered to use something as well built as those.
Answer: If the only reason to do this is to prevent "rattling" of the smaller pipe inside, meaning you are NOT going to use the 6" pipe for anything else, then you can just fill the annular space with expanding foam. It would not be important that it is suspended evenly in the larger pipe, it would only matter than it can't move in there.
Answer: Depending on your very vaguely specified needs, there are plastic products made to keep well water lines (typically 1 or 1.25 inch) (roughly) centered in the bore of a drilled well, fairly inexpensively.
Few are sized for a 3 inch pipe, though.


Does "poco a poco accel. fino al misura 25" sound like proper Italian tempo mark?

Question: Is ""poco a poco accel. fino al misura 25" correct tempo mark in Italian for "step by step accelerate until measure 25"?
Comment: There's no harm in being specific, but you can also just say "poco a poco accel.," and then in m 25 "a tempo"
Answer: Musical directions are not always correct Italian, and don't really need to be as long as the meaning is clear. But a much more usual way of notating something like this is write just "poco a poco accel." and then at M.25 add "a tempo" or give a new tempo indication. Musicians don't read measure numbers while they are playing.
Comment: Thank you for the suggestion.
Answer: Beyond the obvious "accelerando", there are various ways to note a gradual increase in tempo; "di piu in piu animato" (ex. Saint-Saens, Danse Bacchanale, lit. "more and more animated"); "cresc. e stretto poco a poco" (Grieg, In the Hall of the Mountain King, "stretto poco a poco" being literally "gradually tighter" or "narrower", i.e. referring to the time interval between the notes).
As others have pointed out, if the accelerando is not until the end of a piece (in which case one can simply append "al fine"), the end point is more practically indicated by a new tempo marking (coupling it with a rehearsal mark  is definitively a good idea) or other mark that makes this clear, such as an extension line for the accelerando. In a case like this, I would very likely pencil in a mark on the sheet music if it were not already present
In any case, although that is a bit moot, and ignoring that musical Italian is not always proper Italian (for ex. "fine" is also feminine), "misura" is grammatically feminine (see the dictionary) and thus, to be strictly correct, it would be "alla misura".
Comment: Thank you for such a detailed answer. I actually translated from English to Italian via ChatGPT. I guess, it does not know enough about music Italian :-)
Answer: Assuming the music wouldn't be performed by an Italian-only orchestra, it may be most effective to simply write that in English.
Comment: All classically trained musicians worldwide learn to understand 'musical Italian' but not all of them understand English.
Comment: First thought: "well, I want everything in the same language", but then, the title is in English, so it is already two languages :-)
Comment: I do think tempo indications in English is more and more commonplace these days.


Is jenkins-x the successor of jenkins?

Question: I strongly doubt that jenkins-x is the successor of Jenkins. To me it looks like a tool that is able to create a ci/cd pipeline for new apps.
Answer: Jenkins X is a CI / CD platform for Kubernetes.
This project is a sub-project of Jenkins.
see below
here
Comment: Ok, but is it identical to the original jenkins?
Answer: After evaluating and trying to build our current monolith, my conclusion is that jenkins-x should be used if one starts from scratch and wants to deploy a spring boot app in the cloud automatically. If that is not the case and one would like to run an existing monolith in the cloud then I suggest to use the stable/chart jenkins.

Is jenkins-x the successor of jenkins?


Matte effect in illustrator

Question: I'm new to graphic design.
Does anyone know how to create a the matte shape like this from the picture below!
Answer: It is possible using Illustrator and using the photoshop effect built in illustrator..
please follow the following steps.

place you image in your artboard.
copy [CTRL+C] the placed image and past it exactly on the old on by pressing [CTRL+F] - now you have two images over each other.
draw a circle over the desire spot.
by the black arrow select the circle and the image that you have just pasted over the old one. and make a clipping mask by going to object>clipping mask>make or press [CTRL+7]
now we are ready to make the effect to the circle which appear exactly on the old image spot. be careful not to move it. go to effect>blur>Gaussian blur and set a desire value preferable to be between 3 to 7
after doing the effect group the circle with the old placed image by selecting them both and press [CRTL+G]

if you want to make the matte effect you can add another circle with a white color having 50% opacity and put over or under the mask circle.

that's it.

Matte effect in illustrator


What happens when a blackbody absorbs light that does not have a blackbody spectrum?

Question: If radiation with a blackbody spectrum with temperature $T$ is incident on a blackbody, the blackbody heats up to the temperature $T$. What happens if the spectrum of incoming radiation is not that of a blackbody? What temperature will the blackbody reach?
Comment: Your assumption about blackbody absorption is wrong. To quote wikipedia (where you should have looked before asking) "A black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence"
Comment: A blackbody in thermal equilibrium at temperature $T$ emits radiation that resembles a blackbody spectrum. I do not understand what do you mean by blackbody spectrum being incident upon a blackbody.
Comment: @SRS for example, one can shine monochromatic light on a blackbody
Comment: Sorry, that should have been obvious. Silly me
Answer: An object absorbs some fraction of the incoming radiation. If the object is indeed an ideal black body, all the power is absorbed, and none is reflected.
The body will heat up, and emit a black body spectrum corresponding to its temperature. The emitted power is given by $P=A\sigma T^4$ (this is known as the Stefan–Boltzmann law).
When the absorbed power is equal to the emitted power, the temperature of the blackbody stops changing. By equating the two expressions, you can solve for temperature. So, if the incoming power is $P_\text{in}$, then the steady-state temperature of the blackbody is
$$T_\text{steady state} = \left( \frac{P_\text{in}}{A \sigma} \right)^{1/4} \, .$$
Note that a real body (not an ideal black body) has a finite, wavelength-dependent reflectivity.
In other words, real objects don't absorb all of the incoming radiation.
Answer: I think you're confusing it with the equilibrium state where the incident radiation is exactly the same as the emitted radiation. An ideal blackbody absorbs all radiation and re-emits it with the spectrum given by Planck's Law. You consider this equilibrium state only to calculate the conversion factor between the energy density and the radiated power (Wikipedia and related question).
In your case, the spectrum is irrelevant. The temperature of the blackbody will reach an equilibrium at $T = (I/\sigma)^{1/4}$, where $I$ is the total power absorbed per unit surface area of the blackbody (for all frequencies) and $\sigma$ is the Stefan–Boltzmann constant. This is known as the Stefan–Boltzmann law.


Mycelium - more than one unused Bitcoin HD account

Question: Why can't I have more than one unused Bitcoin HD Account? What if I want to use a Bitcoin mixer that allows for 5 destination addresses?
Answer: You can have multiple addresses on one account.
Accounts are not a feature of the Bitcoin protocol, they are a feature created by some wallets to help you organise your money.


2D Galaxy Spectrum taken by long-slit-spectrograph

Question: I have the 2D data (shape(1125,2058) taken by a long slit spectrograph, which looks as follows:

The y-axis corresponds to the distance from the center, where the center is at the most bright green part (y_index about 595). The x-axis corresponds to the wavelengths. The data shows the long-slit-spectra of an elliptical galaxy. And here is where I am getting a bit confused, I expected the red end of the spectrum (i.e. longer wavelengths) to be more bright, since ellipticals have a red continuum, due to their old stellar population. In the plot, the spectrum is clearly more bright on the left, does that mean that the wavelength goes from right to left? Of course I also extracted the wavelength array from my fits file, if I include it into the plot, it runs from left to right. But I guess since my wavelength range does not extend into the red part of the spectrum only until yellow (570 nm) this resolves the problem?
After some changes one can see the absortpion lines as well as the rotation of the galaxy.
Comment: Your wavelength range is not very large, and in particular doesn't include the Balmer break and the 4000 Å break, so I'm not sure it's so surprising that it looks flat-ish. From ~4800 Å to a Mg line at ~5200 Å the spectrum actually typically does decline a little, but after that it should increase again, I think. But perhaps extract the 1D spectrum? I think that would make it easier to see.
Comment: And yes, I understand that you're puzzled as to why the spectrum seems to decrease toward longer wavelengths, and I agree that it should stay flat or increase somewhat. But if you look at other elliptical galaxy spectra, you'll see that it's not unusual to decrease in the first half of your range. It should then increase again, which is indeed puzzling, but see Peter Erwin's answer. And then perhaps try and extract the 1D spectrum which might be more easily interpreted.
Comment: to my understanding the balmer break and the 4000 $\AA$ break in elliptical galaxies is roughly at the same wavelength right? the 1d spectrum from the center looks pretty much as expected for the wavelength range
Comment: Was just wondering why there is more blue light then yellow light in the central region.
Comment: Yes, the two breaks partially overlap (but evolve differently with age of the stellar population).
Comment: Why would you assume that a 2D image like this has had any instrumental effects removed (e.g. the wavelength response of the grating)?
Answer: In the plot, the spectrum is clearly more bright on the left
One possibility is that this was taken with a blue-sensitive CCD, and/or the diffraction grating has a higher efficiency in the blue. In general, the recorded brightness is the product of the object's intrinsic brightness (in photons/sec per wavelength interval) and the spectral response of the system (telescope mirrors/lenses + grating + filters + CCD).
I would be grateful about tips on the colormap settings, so one can see the absorption lines and therefore also the rotation better.
You don't say how you are generating these displays, though I'm guessing it's via the Matplotlib library in Python. So, in case you're not already displaying the data with a logarithmic scaling, try that (i.e., display np.log10(data) instead of data. You could try other scalings as well (e.g., square root).
A second point is that you appear to be using the default "viridis" colormap of Matplotlib. This is a terrible colormap if what you want to see are faint variations in your data. I would suggest something like a rainbow color map, maybe the "turbo" color map which is a bit better than the older "jet" colormap.
Comment: I updated my plot with a different colormap and displayed the data logarithmically


Can you use the same key to provide a signature and a MAC?

Question: Apologies if this is a stupid question, but I'd be grateful if someone could provide an answer. I'm assuming that you need to use different keys to provide these functions, is that correct?
Comment: What exactly do you mean by a signature? And you should specify the exact schemes you want to use. With common terminology a signature uses asymmetric crypto, a MAC symmetric. So you need different kinds of keys for that.
Answer: The usual meaning of signature (in cryptography) is this:

Bob has a pair of keys, a public and a private one. The public one is known to Alice, too. (And to any possible attackers, too.)
Bob wants to send a message to Alice.
Bob does some operation with the message and his private key (he signs the message), and obtains a signature.
Bob sends message and signature to Alice.
Alice does some operation with message, signature and the public key and can verify these fit together, i.e. that the message really was signed by Bob.

The usual meaning of a MAC (message authentication code) is this:

Bob and Alice have a shared secret key.
Bob wants to send a message to Alice.
Bob does some operation with the message and the secret key, and obtains a MAC.
Bob sends message and MAC to Alice.
Alice does some operation with message, MAC and the secret key and can verify these fit together, i.e. that the message really was sent by Bob. (Actually, as she knows the key, too, she can only verify that "the message was sent by Bob or Alice", and has to know by some other means that this was not her own message.)

How could they use the same key to produce a MAC and a signature? We could use one of the signature keys for the MAC:

If this would be Bob's private key, Alice would need to know it too for verification, and then it wouldn't be private anymore. (And Bob would be able to forge Alice's signatures, too.)
If they used the public key, then anyone knowing it would be able to forge the MAC.

Both are not good ideas, normally.
There might be a way to do something similar, though. For example, if both Alice and Bob have a key pair, they might be able to calculate a shared secret from both (e.g. using the Diffie-Hellman key exchange if these are discrete-log-type key pairs), and use this shared result for a MAC. At the same time, Bob might use his private key to sign the message.
Answer: As Paŭlo Ebermann said, digital signatures involve a keypair ( a private key and a public key ), and there's no way to directly use either key for a MAC.
There are several ways to indirectly use public key cryptography to generate something that could be used as a MAC key -- but I don't see any reason to ever do that.
Both the MAC (message authentication code) and a public-key digital signature have the same function: they allow the receiver to confirm that the message really did come from the person he thought it came from.
Once you've confirmed that the digital signature attached to a file is valid, you already know that the message is authentic and it came from the person who knows the private key used to create that public key. I don't see any point of also having an additional MAC.


Points in the Stone Cech compactification are intersection of open sets

Question: Let $\beta \mathbb{N}$ be the Stone Cech compactification of the natural numbers and let $ x\in \beta \mathbb{N}$. Is it true that there exists a sequence of open sets $\{U_n\}_{n=1}^\infty$ in $\beta \mathbb{N}$ such that $\bigcap_{n=1}^\infty U_n = \{x\}$? What about if $ x$ is in the corona?
Comment: Any $G_\delta$ point of a compact Hausdorff space has a countable local base, and this certainly doesn't happen at any point of the remainder in $\beta\mathbb{N}$. In fact, any nonempty closed $G_\delta$ subset contained in the remainder has cardinality $\geq2^\mathfrak{c}$.
Answer: Yes if the point is from $\mathbb{N}$ (it is isolated).
No if the point is in $\beta\mathbb{N}\setminus\mathbb{N}$ because in that subspace every nonempty $G_\delta$-set has nonempty interior, see this paper for more elementary properties of $\beta\mathbb{N}$.


Major private labs that did physics research in the USA

Question: I've been reading about Bell Labs in the USA and find the idea of private labs to be interesting. What are some other major private labs that have existed in the US that did/do physics research?
Comment: During the 1950s through the 1970s (my guess), I believe a lot of the better known engineering-type companies had such labs --- Xerox, DuPont, General Electric, RCA, Westinghouse, IBM are a few names that come to mind.
Answer: The GE research lab in Niskayuna (Schenectady) NY - variously called GE CR&D (corporate research and development) or GRC (global research center) has a long history of innovation - from the early invention of ductile tungsten (first for the light bulb, and then resulting in the Coolidge tube which is still the foundation of virtually all medical X-ray tubes today), through jet engines, magnetrons, laser diodes, superconductivity (for which Ivar Giaever shared the Nobel prize), thin films (for which Langmuir received the Nobel prize), amorphous silicon X-ray detectors, MRI, wind turbines, locomotives, remote sensing, robotics, plastics, ceramics, ...
It is one of the few large private labs that is "still standing".
Other labs included the Xerox Palo Alto Research Center (PARC), Kodak research center, the Philips Natuurkundig Laboratorium (now High Tech Campus Eindhoven - no longer purely for Philips, and not in the US), Siemens (multiple locations - very diverse), and many others.
None of these are "just physics" labs - they usually thrive because they are cross-disciplinary.
Comment: RCA, Westinghouse, IBM, Draper, Boeing ... The list goes on and on...


Heater cartridge with 7.2 ohms - 12 or 24 V?

Question: I have a heater cartridge here that measures 7.2 Ω.

At 12 V that would amount to 20 W
At 24 V that would amount to 80 W

The cartridge itself is 2 cm long
Could it handle both voltages or is this clearly an under-powered 12 V cartridge?
Comment: Just give it 15V, what controller board are you using?
Comment: The question is what voltage, and/or continuous power level, is the cartridge rated for?  You're going to dump roughly the same total energy to reach a given temperature setpoint regardless of input voltage. ("Roughly" because there is heat loss from the hotend to air)
Comment: any way I can figure that out?
Comment: The same total energy given a set temperature, but a higher wattage will have a higher max temp and be able to heat up quicker. Regardless I don't see how we can determine the rated voltage with the information provided. We would be better off looking at the source of the heater cartridge, from a pre assembled printer, printer kit, preassembled hotend, or sourced individually. If it was sourced with the rest of the printer what voltage of the PSU?
Comment: @user1282931 as my answer explains, you might need to supply photos - both possible types exist.
Answer: e3D Heater Cartridges are documented to be around 4.8 Ω for 12 V & 30 W, 3.6 Ω for 12 V & 40 W, 19.2 Ω for 24 V 30 W and 14.4 Ω for 24 V 40 W.
7.2 Ω is a value quite far away from these values - about double of what the 12 V versions are listed and about a third/half of a 24 V heater cartridge. So it is not a cartridge that is similar to those. I have contacted e3D about updating the specs of their High Precision Heater Cartridges to include the nominal resistances to get more references, but that information is pending.
However, we don't need that. A variant of Ohm's law can help us estimate what this cartridge could be: Power is the square of Voltage (U) divided by Resistance.
$P=\frac {U^2} R = \frac {144\ \text V^2}{7.2\ \Omega} = 20\ \text W$
$P=\frac {U^2} R = \frac {576\ \text V^2}{7.2\ \Omega} = 80\ \text W$
Mounted in a 12 V Machine this should act work at 20 W, in a 24 V it would work at 80 W. It does not say, however, if it is designed to get powered with 12 V or 24 V. It doesn't help, that both types exist: I was able to find 12 V 20 W heater cartridges just as well as 24 V 80 W heater cartridges. If it is an underpowered 20 W one and mounted in a 24 V machine, you risk burning it though, but not the other way round.
To differentiate what exactly you have, a photo of the heater cartridge would be needed - it should bear at least a marking of what it's nominal wattage or voltage is.


Help! Why is my image divided into these small squares/boxes?

Question: Help! Why is my image divided into these small squares/boxes? The file is saved as a PDF file, and when I open it in Ai, it does this when I use the Selection tool and hover over the image.
Answer: This may just be a result of how the PDF was saved. I always find PDF files given to me have images sliced, words separated, strangely placed clipping masks etc. and generally there isn't much that can be done about it.
If it is just a bitmap image then you should just be able to open it in Photoshop and save it to another format. If it is a vector image though, you may have to manually fix it by deleting any clipping masks and using pathfinder to join all the shaped together again.
This is all based on the assumption that the image has actually been split, it may just be something overlaying the image as Kaylee said. Without seeing the actual file it's hard to tell.
Comment: The argument against a grid overlay is that there are some cells that are not the same size. I note that the two that are larger are all white. I suspect some optimization was done where the blank white areas were split out in order to be able to compress them more: each cell being an arbitrary compression. A one-color white square might compress better with RLE for example. This would mean your assertion about save-as settings is probably correct. Googling "PDF image split" comes up with old complaints pertaining to Acrobat V4 doing this(?)
Comment: could be related to transparency or effects though
Comment: Very good point. The all white merged cells do pretty much rule out an overlay, and make it being something to do with optimization during save seem a lot more likely.
Answer: It looks like the rectangular grid tool. You should be able to select and delete it.


What was character of Chmielnitski's rise?

Question: What was the character of Chmielnitski's rise? It has started as a private conflict between him and a polish nobel, Aleksander Czapliński. But what was after? And what was the reason to fight when Chmielnitski ruled all the Ukraine?
Comment: Who?  It is very useful to link proper nouns to some source for those of us who aren't immediately familiar with the individual. H:SE should be a learning resource for all of us, not just for the person asking the question.  It would also be useful to summarize the conflict a bit more clearly than you've done here.
Comment: Please modify your question to include the clarifications in your comments on the answer.  The goal is to create a Q&A that is of general reference use.
Answer: Bogdan Khmelnytsky was arguably the "William Wallace" of the Ukraine. He not only took on the one noble, but the whole Polish noble "Establishment," to the point where even the King of Poland (a personal friend), wouldn't defend him.
So he basically started a war between his Cossacks and the Poles. He was initially successful but the Poles were unforgiving; they wanted his head and continued the war, which started to turn in their favor. This drove him and Ukraine into the arms of Russia, and helped set the stage for Poland's decline and Russia's rise.
Comment: It's a very useful answer but the question was about the character after the first phase of the rise.
Comment: @enedil:The sequence of events is something like the following. 1) He got into a fight with the one noble. 2) the other Polish nobles supported the first noble. 3) He incited the Cossacks against the Polish nobles and started a war. 4) He won early, and started to lose. 5) He tried to recoup his losses by allying with Russia ("out of the frying pan into a fire"). The "character of the rising is that a private quarrel "morphed" into a multinational war.


Fridge is not cooling

Question: When I was cleaning my fridge to remove the ice from the fridge, Accidentally my I made a hole in the bottom plate of the fridger and some liquid start leaking, I guess it's a coolant. After that my fridge is not cooling. Can you please help me with the resolution ?
Comment: Yes, buy a new fridge.
Answer: Sounds like you've hit one of the refrigerant lines, and let the magic juice out. This is a very common cause of refrigerator failure. Unfortunately, if it's even possible to repair the damage, it'll likely cost more for the repair than for a new fridge.  Of course that completely depends on the make and model of the unit, and how accessible the damaged section of tubing is.
Next time you want to defrost the freezer, follow the manufacturer's defrost procedure. Stabbing at the ice with pointy bits, is never a good idea.
Answer: With the freon line punctured a HVAC service call will be needed to advise if they are able to seal the line, pump it down to a vacuum and recharge. It would best to not run the fridge because air is getting in the system and the compressor life will be shortened if you do try to have it repaired. If it is a nice fridge it may be worth the cost if it is a low end model it may be cheaper to replace it.

Fridge is not cooling


where do combinatorics play a role in computer science?

Question: I am graduated from computer science. I need some advice in Combinatorics material related to master fields of computer science. I already know that discrete mathematics have a key role in some concepts which I need a little more elaboration. In which fields could I trace combinatorics concepts and theorems and techniques?
Comment: Combinatorial optimization.
Comment: Analysis of algorithms.  For example, see "Analysis of Algorithms" by Sedgewick and Flajolet, or "Concrete Mathematics" by Graham, Knuth, and Patashnik.
Comment: If you never heard about complexity theory, read about it, it is quite pleasant.
Comment: Honestly this question surprises me. Combinatorics is used pretty much everywhere in CS. You have a degree in computer science, what did you specialize in? I'm certain that no matter what you specialized in, combinatorics is used there.
Comment: @StellaBiderman I know It's surprising But combinatorics topics and methods aren't recognized since they are very vast; from inclusion–exclusion principle to well-known theorems like Ramsey theorem Hall theorem or even Regularity lemma) I was interested in some(including setsystems properties for example vc dimension or epsilon net theorem) and I find the other type of problems very hard to work with(like VC density and other topics in complexity) this post was for some clarifying purposes to finding the next step of combinatorics methods in cs.
Answer: Graph Theory is a part of Combinatorics that is extremely important in computer science. Roughly, it is the study of objects and connections between them.
Algorithms and Data Structures uses combinatorial techniques for analysis and design extensively. Several Data Structures use graph theoretic ideas to efficiently store and retrieve data. Randomized Algorithms and Approximation Algorithms require combinatorial methods for analysis. They are often faster or more "useful" than deterministic ones.
Complexity Theory, the study of the complexity of resources used by computational procedures, use combinatorial problems and ideas as examples of several sorts of complexity classes. Discrete Optimization, in particular, is a class of problems which are of relevance to both theoreticians and practical purpooses. Some of them represent problems which are too hard for detereministic procedures and give way to randomized and approximate algorithms.
Other interesting Computer Science fields where combinatorics is important include Coding Theory(Error Correcting Codes over Communication Channels), Computational Geometry, etc.

Also see: Why is it important to study combinatorics?


What does m mean in kernel configuration file?

Question: CONFIG_UNIX=m

I know what y and n stand for,but what about m?
Comment: Not programming-related - belongs on http://superuser.com
Comment: Stack Overflow is a site for programming and development questions. This question appears to be off-topic because it is not about programming or development. See [What topics can I ask about here](http://stackoverflow.com/help/on-topic) in the Help Center. Perhaps [Super User](http://superuser.com/) or [Unix & Linux Stack Exchange](http://unix.stackexchange.com/) would be a better place to ask. Also see [Where do I post questions about Dev Ops?](http://meta.stackexchange.com/q/134306).
Answer: I assume, this refers to the same as the (y,n,m) prompt when running make config; in that case it would be "module".
Note that compiling Unix domain sockets (CONFIG_UNIX) as module is probably not a good idea. A lot of system components and programs depend on them, and some services might fail to start up if the module has not been loaded at that time.
Most functionality in the Linux kernel can either be compiled in ("y") or left out ("n"), and much of it can also be compiled as a loadable module. This makes sense when you don't know for certain whether you will need some feature in the future.
If you compile it as module and it turns out that it is needed, it will work, but until then it will not bloat the kernel.
It does not, however, really make sense to configure Unix domain sockets as a module, because they are needed almost everywhere (e.g. udev will fail to launch at startup).
If you know you will need something anyway, that should be "y", not "m"
Comment: Most functionality in the Linux kernel can either be compiled in ("y") or left out ("n"), and much of it can also be compiled as a loadable module. This makes sense when you don't know exactly if you will need some feature in the future. If you compile it as module and it turns out that it is needed, it will work, but until then it will not bloat the kernel. For unix domain sockets, it does not _really_ make sense to configure as module, because it is needed almost everywhere (e.g. udev will fail to launch at startup). If you know you will need something anyway, that should be "y", not "m".
Answer: y  =  yes (always installed)
m  =  loadable module (can install and uninstall as you wish)
n  =  no (never installed)

see: https://www.linuxquestions.org/questions/linux-general-1/boot-config%2A-273853/
Answer: Read the below excerpts from "Understanding Linux Kernel ":
some Linux code must necessarily be linked statically, which means that either the corresponding component is included in the kernel or it is not compiled at all. This happens typically when the component requires a modification to some data structure or function statically linked in the kernel.
For example, suppose the component has to introduce new fields into the process descriptor. Linking a module cannot change an already defined data structure such as task_struct because, even if the module uses its modified version of the data structure, all statically linked code continues to see the old version. Data corruption easily occurs. A partial solution to the problem consists of "statically" adding the new fields to the process descriptor, thus making them available to the kernel component no matter how it has been linked. However, if the kernel component is never used, such extra fields replicated in every process descriptor are a waste of memory. If the new kernel component increases the size of the process descriptor a lot, one would get better system performance by adding the required fields in the data structure only if the component is statically linked to the kernel.
As a second example, consider a kernel component that has to replace statically linked code. It's pretty clear that no such component can be compiled as a module, because the kernel cannot change the machine code already in RAM when linking the module. For instance, it is not possible to link a module that changes the way page frames are allocated, because the Buddy system functions are always statically linked to the kernel.


Surveys answers reports

Question: Our company want to certificate emploees knowledge and for this reason created several surveys (about 10). The task is to collect data from this surveys and determine how much correct answers each users have. At first i was thinking about workflow, which collect users answers (only correct one) from all surveys and store in separate list. But the problem is that sharepoint can't export to excel filtered list and create view for individual user is not possible, because there be more than 8000 users (our company is chain-store). And now i have some doubts will it be a good user experience for survey administrator to filter list for more than 1.2kk items to validate user knowledge (8000 users, 10 surveys, more than 15 questions each).
Maybe you can suggest some good approach to collect data across multiple surveys, like using SSRS or develop custom web part to display all results for certain user, or maybe some OOtB feature will be useful?
We are using SharePoint 2013 Enterprise
Answer: Combining the lists may be best achieved using SSIS. Once you have a single data source you can access the list from Excel and design a dashboard in Excel and deploy it to SharePoint. Alternatively you could use PowerView to create a dashboard. You should also consider using PowerBI which would suit any 'power' users.
Figure out how to combine the data first, and test it, Then design the UI. You'll need Excel Services running on your SharePoint instance to get a nice interactive experience. Use PowerPivot in your Excel Workbook for the data, and use slicers and suchlike to filter and give a nice UX.
The advantage of using Excel is that you can empower your users to create their own interactive reports. SSRS would take not give the same interactive UX. Once you have the Power Pivot Workbook loaded it can also be used as a report source for SSRS.
I can't remember, but you may need Pivot Table services running in Share Point as well. There are quite a few options here, this is just one approach. Have fun!

Surveys answers reports


Any suggestions on filling gaps in wood flooring? - Pine Tongue Groove

Question: Any suggestions on filling gaps in wood flooring?
Pine tongue groove was used that is usually used for wall or ceiling.
Have no idea why they put this on the floor?
I was thinking cutting wood strips and gluing them in then sand.
Better idea may be to rip it up and put a real floor down.
Impossible to clean for the owner.
Comment: Personally I would do one of 3 things, 1, live with the floor as-is and just accept that the grooves may get dirty 2, use carpeting or lots of rugs 3, replace the entire floor.
Comment: I don't think there's any reasonable way you can fill these. There are ways you can fill them but the amount of effort required would be immense, not to mention the amount of filler required! And I don't think you could guarantee the filler would be durable enough with foot traffic and wood movement. Wood strips seem like a viable alternative until you factor in making them to the correct profile in the first place (a not-inconsiderable task) and then cleaning every groove thoroughly beforehand so that they'll glue in properly. Just don't think the amount of work required is justified.
Answer: You cannot fill these gaps in.  Any wood floor needs to expand and contract across its width and the tongue and groove profile of the individual strips allow the floor to absorb changes in width without buckling. All wood floors will have a similar approach for the same reason.  I will grant you that pine makes for a floor that is very susceptible to scratches and dents, and the exposed screw is singularly unattractive!  If you want invisible seams consider a plastic laminate flooring in a wood pattern.


How to set up a connection to a MySQL-server within Dreamweaver CS4?

Question: If you have credentials to login in via SFTP to a server that is running PHP and you can access MySQL with SSH, how do you combine these things within Dreamweaver CS4?
Comment: I'd almost think this would be a question for Stackoverflow?
Comment: Not really about programming, more about using software.
Answer: You start by opening the Databases panel (top menu: Window->Databases). You'll need to set up a site. Then you should be able to create a database connection to associate with the site.
Answer: the problem is dreamweaver leans towards setting up the database connection from a local file or database...which was a simple job using the database (+)add connection and pointing it to either the file or database. Unfortunately when I tried to go live and my remote host is different than the location of my remote hosted database it becomes really a pain. I can transfer and sync my files to the remote host but get a "testing server not defined error" for some unknown reason. My database definition is correct pointing to mysql2.hostingcompany.com but the connection isn't ever made to the remote database.
if i set it back to the local one complete access to all tables. user name and password are correct for my hostingcompany.com database.
I can't set the remote to point to the database and i can't change the testing server to point to it either without getting a failure. so what is the right way to set up a remote database connection?


What does F#'s `box` keyword do and where is it documented?

Question: The Null Values article in the F# Language Reference show an example that uses it, but it does not explain what it does exactly.

You can use the following code to check if an arbitrary value is
null.
match box value with
| null -> printf "The value is null."
| _ -> printf "The value is not null."

I went through the entire PDF version of the language reference, but there is not trace of it ever being introduced.
Answer: box and unbox are operators documented in the F# Core API Reference; it tersely states that they box / unbox (respectively) a strongly typed value.

box a returns value a wrapped in a .NET System.Object instance
type Lofa = | A | B

match      A  with         | :? Lofa -> A //=> error FS0016
match (box A) with         | :? Lofa -> A //ok
match (A :> obj)           | :? Lofa -> A //ok
match (A :> System.Object) | :? Lofa -> A //ok

unbox<a_type> a will try to unwrap value a and cast it into an a_type or throw an exception otherwise

For completeness' sake, there is also a tryUnbox operator, that can be called the same way as unbox, but it returns the boxed value wrapped in an Option.
Example use cases
1. Type detection during runtime
The F# for Fun and Profit article Built-in .NET types1 has an entire section dedicated to box and unbox which ends with the following example:
[1]: That is, types of the Common Language Infrastructure (CLI).

Let's say that you want to have a function that matches based on the
type of the parameter, using the :? operator:
let detectType v =
    match v with
        | :? int -> printfn "this is an int"
        | _ -> printfn "something else"

Unfortunately, this code will fail to compile, with the following
error:
// error  FS0008: This  runtime coercion  or type  test
// from type  'a to int involves  an indeterminate type
// based on  information prior  to this  program point.
// Runtime type  tests are  not allowed on  some types.
// Further type annotations are needed.

The message tells you the problem: runtime type tests are not allowed on some types.
The answer is to "box" the value which forces it into a reference
type, and then you can type check it:
let detectTypeBoxed v =
    match box v with      // used "box v"
        | :? int -> printfn "this is an int"
        | _ -> printfn "something else"

//test
detectTypeBoxed 1
detectTypeBoxed 3.14

2. Interoperability with non-F# .NET projects
The Functional Adam post What is box in F#? nicely complements the example in the question:

The Problem - The variable was in Int, it was required, and if it was not provided, our application was supposed to return an error
code. When the parameter was not provided, WebAPI would set the
required parameter to null even though Int cannot be null. This
was very confusing, and, to make matters worse, we were not able to
make the required parameter a Nullable<Int> with WebAPI.
The Solution - Box is the solution! From our searching, by boxing the int we could check if it is null even though it's not nullable
(confusing). But what does the box function do? According to [the
docs], the box function in F# "boxes a strongly typed value".
Ok... Thanks documentation.
That's not all very helpful, but with further digging in [the Boxing
and Unboxing article of the C# Programming Guide], the answer is

Boxing is the process of converting a value type to the type object or to any interface type implemented by this value type.
When the common language runtime (CLR) boxes a value type,
it wraps the value inside a System.Object instance and stores it on the managed heap. Unboxing extracts the value type from
the object. Boxing is implicit; unboxing is explicit. The concept of
boxing and unboxing underlies the C# unified view of the type system
in which a value of any type can be treated as an object.

So, boxing a value wraps the value type inside a System.Object
and therefore we can check if the object is null. Then we have to
unbox it back to an Int after the null check.


azure cloud costs

Question: I was looking for any basic costs for a simple site.
this site accepts an input from the customer. This field is a product number much like a ISBN or UCC.
We have a database which has this number along with name and price. we have a program in C# which displays the price and name.
my question is, what is the cost to have this database in the Azure cloud, and the small app I described. there would also be a user table and some smaller tables. how does the costing work regarding the database tables, and app code?
There is only one lookup per data entry, straight forward, enter the code, if it exists, display the price and name.
I dont know if this is unclear or ambiguous question i apologize if it is, but am trying to determine costs of using the ac.
Answer: You should look at the calculator.

100MB SQL Database: $5.00/mo (this can be much cheaper if you convert it to table storage).
Web Site

Free (you cannot use a custom DNS): $0.00
Shared (allows you to use custom DNS): $9.36/mo

Egress Bandwidth (10GB/mo to USA/EU): $0.60/mo

azure cloud costs


How are higher dimensional sphere packings used in signal transmission?

Question: How are higher dimensional sphere packings (HDSPs) related to the error-correcting codes used by cell phones, space probes and the Internet for signal communication through noisy channels? HDSPs strike me as super abstract objects and I'm having a hard time even wrapping my head around their possible concrete applications.
Comment: In what context did sphere packing come up?  Please reference where you've read or heard about it.
Comment: Was reading an article published onQuanta on HDSPs and that's literally the context i.e. it's casually mentioned in a single sentence that HDSps are used thus.
Comment: That's not a reference to an article as references go, @user51309 .
Comment: https://www.quantamagazine.org/sphere-packing-solved-in-higher-dimensions-20160330/?utm_content=buffer15425&utm_medium=social&utm_source=facebook.com&utm_campaign=buffer
Answer: Does the sphere radius indicate the acceptable deviation away from nominal magnitude and phase? If the total energy (signal + noise + ISI + deterministic trash) moves outside that radius, the energy is interpreted as a different symbol.
Answer: The math is way over my head for the proof of \$E_8\$ and \$E_{24}\$ which defines how to arrange spheres to occupy the most space.
It uses d-dimensional Euclidean space,  Fourier transforms, Schwartz function, holomorphic congruent subgroups,  Eisenstein series, Fourier expansions, Eigenvalues, Modular Forms, Poincare´ series with Bessel Functions.

Although some residuals were rounded off, the best computer tests could not find an error of more than \$10^{-28}\$ in \$E_{24}\$ .

Although based on the Leech Lattice, the inspiration for the proof used the \$G_{24}\$ 24 bit [Golay code] used for error correction such that 3 errors could be corrected in 12 bits of data in 24 bit words. 2
The symmetry of this lattice model may lead to more beautiful revelations in nature using this with Quantum Theory, not just how to stack oranges better in the supermarket.


Can I mention my current grad school in my PS for PhD application to other schools?

Question: I am currently a master's student at upenn, and I am trying to apply for phd in other schools. Would it be okay if I explicitly mention upenn in personal statement because I was trying to talk about a specific lab I've been working in here
Comment: Can you clarify why you think it might not be okay?
Comment: Well, personal information is pretty much what they are looking for. Knowing UPenn, your lab likely has a good reputation so that reflects well on you and your abilities.
Comment: I thought it could give away too much personal info and potentially bias a reviewer's decision?
Comment: @Lleryomof Are you applying to schools that have ablind application process somehow?
Answer: Yes! You should absolutely mention your lab, the names of the people you work with and the university at which you are studying/researching. I would find it extremely odd if a research statement or CV did not include any of those details.
Networking is extremely important in academia. In fact, I think it's the single most important component to a successful career, or at least to getting a position (PhD/postdoc/permanent). Some people realise this too late, and have a lot of problems getting a postdoc after their PhD, for example.
In a comment you said you are worried about giving away too much personal information, and biasing a reviewer's opinion. Unfortunately, it is very difficult to conduct an academic career anonymously. People need to know who you are and what you are working on. It's like word association: they need to see the name Lleryomof and instantly think Mating habits of three-toed sloths... or whatever your research topic may be. As you get further on in your career, this association will become automatic, as you write papers, give talks, and visit other institutions (i.e. networking).
Regarding biasing a reviewer's opinion -- well, you want to do that, don't you? In your favour! That's the goal of any application -- convince the person hiring that you're the best person for the job. One great way to do that in this case is tell them about all the cool research you have been doing during your Master's degree at an internationally renowned university, UPenn.
Make sure you get a lab mate (maybe a friendly PhD student or postdoc) and your Master's supervisor to look over your application and give you feedback before you submit it.


Does momentum change when Kinetic energy changes?

Question: $K.E=\frac{1}{2}mv^2$ , $P=mv$ thus there is a relation between them. I really can't understand why there is no decrease in momentum when kinetic energy is decreased in an inelastic collision. I am only a high school passed student(ready for college). Should I leave this question until I learn "Lagrangian" and "Noether's Theorem"? Because I literally read most of the answers about these kinds of questions and spend almost half a day on it, BUT I do not get it. Can anyone please answer this question? please help me.
Comment: You really don't need Lagrangian to understand these. The Lagrangian is just another equivalent formulation to newton's laws.
Comment: Also I think it will be good if you saw this stack:

[https://physics.stackexchange.com/questions/92051/how-can-momentum-but-not-energy-be-conserved-in-an-inelastic-collision](inellastic collisions)
Comment: Yes kinetic energy is not a linear function of momentum but a quadratic function of it. ( you only get proportionality if you have a linear function)
Comment: It means that we cannot say that kinetic energy is  always proportional to momentum, but K.E is directly proportional to energy right?
Answer:
$K.E=\frac{1}{2}mv^2$, $P=mv$ thus there is a relation between them.
I really cant understand why there is no decrease in momentum
when kinetic energy is decreased in inelastic collision.

It doesn't make sense to consider a single body during a collision
(i.e. during interaction with a second body).
You need to look at both bodies together.
Therefore you need to consider the total momentum of both bodies.
$$\vec{P}=m_1\vec{v}_1+m_2\vec{v}_2 \tag{1}$$
It is this quantity which is conserved during the collision.
Likewise you need to consider the total kinetic energy of both bodies.
$$E_\text{kin}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$
It is this quantity which is conserved during an elastic collision.
So there are many possible ways how the velocities $\vec{v}_1$ and $\vec{v}_2$
can change to make the total kinetic energy $E_\text{kin}$ decrease while still
preserving the total momentum $\vec{P}$.

Should i leave this question until i learn "Lagrangian" and "Noether's Theorem" ?

To understand momentum conservation (1) you don't need Lagrangian
mechanics or Noether's theorem. Newton's mechanics is just enough.
According to Newton's third law (actio = reactio) you have
$$\vec{F}_{2\to 1}=-\vec{F}_{1\to 2}$$
By applying Newton's second law ($\vec{F}=m\frac{\Delta \vec{v}}{\Delta t}$) to these two forces
you further get
$$m_1\frac{\Delta \vec{v}_1}{\Delta t}=-m_2\frac{\Delta \vec{v}_2}{\Delta t}$$
Rearranging this you get
$$\frac{\Delta(m_1\vec{v}_1+m_2\vec{v}_2)}{\Delta t}=\vec{0}$$
and hence
$$m_1\vec{v}_1+m_2\vec{v}_2=\text{const}$$
which is just the above mentioned conservation of total momentum.
Answer: Just to add a bit of more mathematical flavour to the answers, here's why momentum can stay constant, yet kinetic energy may decrease. First we set up some ground rules.
You mention that momentum is $p=mv$ and kinetic energy is $\frac12mv^2$ and that there should be a relation between them. And yes, there is. Basic algebra tells you $K=\frac{p^2}{2m}$. Well and good.
Now, consider 2 bodies colliding, or rather just interacting. In this picture, the total momentum, at say time $t=0$ is $p_1+p_2$. As others have shown, Newton's third law guarantees that $p_1+p_2$ is constant at all times. If so, then $$\frac{\partial}{\partial t}(p_1+p_2)=0$$ where by $\frac{\partial}{\partial t}$ I mean the derivative with respect to time treating all other variables like space etc as constants. This mathematical statement contains the same information as the one written in words just before that. But this will come in handy.
Now, what about the total kinetic energy? At time $t=0$, it's $K_1+K_2=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}$. To make life simpler, we assume $m_1=\frac12=m_2$. This won't change much. Hence $K_1+K_2=p_1^2+p_2^2$. It would help us if we write this as $K_1+K_2=(p_1+p_2)^2-2p_1p_2$ Then, how does the kinetic energy vary in time? To find that, we take a time derivative of the kinetic energy, and get
$$\begin{align*}\frac{\partial}{\partial t}(K_1+K_2)&=\frac{\partial}{\partial t}(p_1^2+p_2^2)\\&=\frac{\partial}{\partial t}\{(p_1+p_2)^2-2p_1p_2\}\\&=2(p_1+p_2)\frac{\partial}{\partial t}(p_1+p_2)-2\frac{\partial}{\partial t}(p_1p_2)\\&=-2\frac{\partial}{\partial t}(p_1p_2)\end{align*}$$
where the last inequality follows because of our momentum conservation equation.
The leftover term is not necessarily 0. Check this by taking a collision with two balls which stick after impact and replace the partial derivative with differences.
An elastic collision is defined as a collision where this leftover term is $0$ and an inelastic collision as one where it's not.
As to where the lost kinetic energy goes, it goes into the internal energy of the bodies themselves, like vibrational motion, heat etc.
Also, this did not require Lagrange, right? There's the answer to this question as well.
Answer: Actually there is no relationship between Kinetic Energy and Momentum (for a generalized system of particles). You can have a non-zero kinetic energy at the same time with momentum being zero.
You can see this from the following equations:
$$\mathcal E = \frac 12 \sum m_i v_i^2$$
$$\mathbf P = \sum m_i \mathbf v_i$$
Now if $v_i \neq 0$ then $\mathcal E$ will never be zero, whereas $\mathbf P$ can easily be zero.
But if you were talking about some special cases then yes there do exist some relationship between Kinetic Energy and Momentum (which you can easily derive yourself).
Answer: The relation between momentum and kinetic energy for a single particle:
Momentum is defined as:
$$\vec{p} = m \vec{v}$$
So, we can write velocity as:
$$ \frac{\vec{p}}{m} = \vec{v}$$
Kinetic energy is defined as:
$$ K = \frac{1}{2} mv^2$$
Using the previous equation,
$$ K = \frac{p^2}{2m}$$
So, it's very easy to see that it can be said that kinetic energy is a function of momentum and mass for a single particle.

In an inelastic collision, the momentum is conserved but the kinetic energy is not. You may think otherwise from the previous equation I wrote because it says that kinetic energy is a direct function of momentum.
When we have a collision what happens is the total momentum is split up between the different bodies
such that if we summed momentum of every single body then the total momentum is the same.
The simple idea is that in an inelastic collision, the total momentum redistributes among the objects of collision in a way that when you take the difference of kinetic energy between the final and initial state, it gives a negative number. The lost energy goes into heat and vibrational energy.

Example:
Consider two clay balls body $A$ and $B$ with the same mass m. Each momentum velocity $v$ and collide head-on undergoing inelastic collision. Let velocity after collision be $v'$
The initial kinetic energy is:
$$ K = \frac{mv^2}{2} + \frac{mv^2}{2} = mv^2$$
Now after collision, they become one body with a velocity of zero (Refer). By momentum conservation for before and after collision:
$$ mv - mv = (m+m) v'$$
Hence,
$$ v'=0$$
Putting this into the kinetic energy post-collision is:
$$ K' = \frac{ (2m) (0)^2}{2} = 0$$
So we can see that the kinetic energy $ mv^2$ was lost completely. This energy went into deforming the clay and vibrational energies as Feynman has said in a quote that I put in the references (*).
The heart of the matter is that for an individual particle, we can relate it's kinetic energy and momentum but for a system of particles the two are not directly related.

Derivation the loss in energy:
Let two bodies $A$ and $B$ with mass $m_a$ and $m_b$ respectively. Let their initial momentums be $\vec{p_a} $ and $ \vec{p_b}$ and after collision in which they stick let their momentum be $\vec{p_{ab}}$ then their loss in kinetic energy.
$$K_{i} = \frac{ (\vec{p_a})^2}{2m_a} + \frac{ (\vec{p_b})^2}{2m_b}$$
The final kinetic energy is given as:
$$ K_{f} = \frac{ (\vec{p_a} +\vec{p_b})^2}{2(m_b +m_a)}$$
Note the momentum in the final must be equal to the initial momentum of $p_a +p_b$ due to the momentum conservation.
Hence,
$$ K_f = \frac{ (\vec{p_a})^2 + ( \vec{p_b})^2 + 2 \vec{p_a} \cdot \vec{p_b}}{2(m_b +m_a)}$$
Now consider the difference of kinetic energy between final and initial states:
$$ K_f - K_i = \frac{ \vec{p_a} \cdot \vec{p_b} }{m_a + m_b} - [ \frac{(m_a \vec{p_b})^2 + (m_b \vec{p_a})^2}{2(m_a + m_b)(m_a m_b)}]$$
$$ K_f - K_i =-\bigg[ \frac{(m_a \vec{p_b})^2 + (m_b \vec{p_a})^2 - 2m_a m_b \vec{p_a} \cdot
 \vec{p_b}}{2(m_a + m_b)(m_a m_b)} \bigg]$$
Or,
$$ K_f -K_i = - \bigg[ \frac{|m_a \vec{p_b} - m_b \vec{p_a}|^2}{2(m_a + m_b)(m_a m_b)} \bigg] $$
Since $m_a m_b$ is a strictly positive quantity, we can move it into the modulus :
$$ K_f - K_i =\frac{m_a m_b}{2(m_a +m_b)} (|\frac{\vec{p_b}}{m_b} - \frac{\vec{p_a}}{m_a}|)^2$$
Since we can switch terms in the square modulus,
$$ K_f -K_i = \frac{m_a m_b}{2(m_a +m_b)} (|\frac{\vec{p_a}}{m_a} - \frac{\vec{p_b}}{m_b}|)^2$$
Now, we can say that kinetic energy change has been reduced in the final state. This absolute value of the amount which is lost is taken as $K_{loss}$ and is given as:
$$ K_{loss} =  \frac{m_a m_b}{2(m_a +m_b)} (|\frac{p_a}{m_a} - \frac{p_b}{m_b}|)^2$$
q.e.d

Further reading:
HC-Verma: Concept's of Physics
Feynman lectures (under energy and momentum of chapter-10)
To understand these ideas in more depth, see the answer by Ron Maimon here
Answer: You've run into a mathematical issue, not a physical one.
Kinetic energy $K=\frac 12 mv^2$ and momentum $p=mv$ may be related, but not uniquely. They share not one but two parameters, both $m$ and $v$, and their relations to them are not the same (some are linear and others squared). Therefore they are not directly related.
Imagine doubling $p$.

Did you do that by doubling $m$? Then $K$ also doubles.
Or did you double $v$? then $K$ quadruples.
Or did you maybe change both $m$ and $v$? Maybe you increased $v$ a lot but also lowered $m$, so it together equated to a doubled $p$? Then maybe this combination causes some entirely different change in $K$.
And maybe even, if they are changed just right, $K$ doesn't change at all.

All this is possible if you find the right combination of changes in the parameters.
In short: When two properties are related via multiple parameters, then they are not necessarily directly related. A change in one does not directly or necessarily correspond to a change in the other. The change may be "absorbed" in the internal changes of those parameters.
Answer: Total energy and total momentum of two interacting particles are conserved. In an inelastic collision between them, the kinetic energy is not conserved, but total momentum is still conserved.
In order that the kinetic energy is not conserved some degree of freedom other than velocity must be present. The particle(s) should have internal energy, for example a temperature, vibration or rotation, dissociation - breaking up.
Answer: Your misunderstanding is in writing $p=mv$ and forgetting that $v$ should really be a vector, i.e. the correct relation is $\vec p=m\vec v$.  Momentum can be positive or negative.  On the other hand, kinetic energy $mv^2/2$ is necessarily non-negative.
Imagine a system where $m_1=m_2=m$ and the two particles approach each other with the same speed but different direction, along the $x$-axis.  To simplify: $v_2=-v_1$ in 1d.  Both particles have the same (positive) kinetic energy $\frac{1}{2}mv^2$ but opposite momentum $p_2=-p_1$: the momentum has a direction (and thus a sign) but the kinetic energy does not (it is a scalar quantity).  This sign is important as it indicates the direction of motion.
The net momentum before collision is $P=p_1+p_2=0$ since $p_2=-p_1$.  The net kinetic energy is the sum of two positive terms so it positive and certainly non-$0$.
After the collision, many $v'_2=-v'_1$ will conserve the total momentum but not necessarily the total kinetic energy.  Say: $v_1'=v_1/2$ and $v_2'=-v_1/2$ will do the trick.  You can verify that the total kinetic is decreased by $4$ after the collision.  Note that both momenta after collision have decreased in magnitude, i.e. $ p_1'< p_1$ in magnitude, and also $p_2'<p_2$ in magnitude, but the net momentum is still $0$ because one momentum is negative and the other is positive.
The point here is that, because momentum is a vector, it's possible to combine momenta in more than one way to add to $0$ and conserve total momentum because momentum is a signed quantity.
Answer: The question you're asking is why is momentum always conserved even during collisions where kinetic energy is not conserved. Consider Newton's third law of motion which states that during any interaction, the force one body exerts on another is equal to, but opposite in direction, to that exerted by the second body on the first, so that for the rate of change of total momentum
$$\frac{dP}{dt} = 0$$
so that for a two-body collision
$$m_1 v_1 + m_2 v_2 = constant$$ and
$$F = \frac{dp}{dt} $$
which is Newton's second law. So momentum must be conserved.
Also, kinetic energy can be transferred to other forms of energy, such as heat and sound during collisions. This cannot happen for momentum in this fashion since momentum is not "a form of energy".
Comment: your equations will look a bit nicer if you encase them in two dollars on both sides. I have done it for you now.
Comment: Thanks again Buraian. Very kind of you my friend!


Cayley graphs of finitely generated groups

Question: Let $\approx$ be the binary relation on the class of finitely generated groups
such that $G \approx H$ iff $G$ and $H$ have isomorphic (unlabeled nondirected)
Cayley graphs with respect to suitably chosen finite generating sets. Is $\approx$ an equivalence relation?
Comment: @Jacques:  The point is that the generating set for H used to show that G \approx H and H \approx I need not be the same, so it is not obvious that G \approx I.
Comment: That it is reflexive and symmetric is completely trivial.  What is your doubt about transitivity?  Maybe if you expanded on that point in your question, someone here could help with 'bridge the gap' between your understanding and a resolution.
Comment: I don't see any particular reason why it should be transitive.  Of course, the 'usual' equivalence relation that looks a little bit like this is quasi-isomorphism.  Does this question have a context?
Comment: Yes, for some years, I have been trying to prove that the quasi-isometry relation for finitely generated groups is strictly harder than the isomorphism relation (in the sense of the theory of Borel equivalence relations). I can prove that the quasi-isometry relation for arbitrary connected degree 4 graphs is strictly harder than the isomorphism relation for f.g groups ... but the quasi-isometry relation for f.g. groups is much harder to work with. So it seems reasonable to consider the above "equivalence relation"  as a first approximation ... except that I doubt that it is transitive!
Answer: The answer is no, as expected. The following proof is "joint work" with L. Scheele.
Consider $G=\mathbb{Z}$, $K=D_\infty$ and $H:=\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. Then $G \approx K$ and $K\approx H$, but $G \not\approx H.$
Indeed, the Cayley graph associated to $\{-1,1\}$ for G and the Cayley graph associated to $\{s,t\}$ where $D_\infty=\langle s,t: s^2=t^2=1 \rangle$ are clearly isometric.
Similarly, the Cayley graph associated to $\{s,st,ts\}$ for $K$ and the graph associated to $\{(0,1),(-1,0),(1,0)\}$ for $H$ are isometric.
However, let $S$ be some symmetric generating set for $G$. Then $S_k$, the set of vertices which have distance precisely $k\geq 1$ from the identity, has even cardinality because $S_k$ is invariant under the mapping $x \mapsto -x$ and doesn't contain 0.
Now let $T$ be some symmetric generating set for $H$. Let $k_0$ be the distance of $(0,1)$ from the identity in the associated graph. Then $T_{k_0}$, the set of vertices which have distance precisely $k_0$ from the identity, has odd cardinality. Indeed, it is invariant under the mapping $(x,y) \mapsto (-x,-y)$ since $T$ is assumed to be symmetric. But $(0,1)$ is a fixed point.
Comment: Very elegant! I had thought about these groups but was unable to make any progress on the problem of showing that $G \not \approx H$.


Is AES a group?

Question: The question I'm wondering is whether the AES cipher is a closed cipher (which is equivalent to AES being a group). And this question interests me due to the lack of understanding of whether it is possible to encrypt using the AES algorithm with two different keys and hope for increased strength, as for example in 3DES. But for DES this is possible because DES does not form a group (and there are public publications to confirm this). So, can you give me a detailed answer to my question with obligatory confirmation from the scientific literature?
Comment: Compared to DES, [Why is double encryption that's equivalent to single encryption no better than single encryption?](https://crypto.stackexchange.com/q/75120/18298). the blocksize is too big to test, imho
Comment: @kelalaka, With the general principle and DES the situation is clear. What about AES specifically?
Comment: Just for context (because it took me a bit): A block cipher can be considered a subset of the group of permutations (of the set of all blocks the right size), and the main question here is whether the composition of two encryptions can be written as a single encryption. [If that's the case, we'd automatically get the other properties of a group, I think, due to the finiteness – i.e. a key with a "null" encryption, and inverse elements.]
Comment: @Paŭlo Ebermann it is interesting thought, thanks for you time
Answer: See the paper DES is not a group by Campbell and Wiener.
TL;DR There are computational proofs that DES is not a group. The point is to carry out the types of computations that established DES is not a group is not feasible for AES due to the much larger space involved and its better design. Maybe someone will discover a new property, but it is unknown as of now.
There are no weak keys or special short cycle properties that are currently known [Coppersmith used properties of encrypting alternatively all 1 and all 0 vectors with DES, see below].
Just mathematical properties were not enough without randomness assumptions as Kaliski et al found out:

Kaliski, Rivest, and Sherman developed novel cycling tests which gave evidence that the
set of DES permutations is not closed [3]. However, their work relied upon randomness
assumptions about either DES itself or a pseudo-random function which was
used in cycling experiments. Because of the randomness assumptions, it is difficult to use
the results of their cycling tests to make any claims about the probability that DES is not
closed.

Afterwards, DES was shown not to be a group computationally:

We have developed our own DES cycling experiments which provide evidence that DES
is not closed; this evidence does not rely upon randomness assumptions

and

Don Coppersmith has developed an approach to finding a lower bound on the size of the
subgroup generated by the DES permutations [ 11. He has shown this lower bound to be
greater than the number of DES permutations, providing conclusive proof that DES is not
closed.

Coppersmith himself described his approach in a 1992 sci.crypt posting.
Comment: This doesn't rule out that some looked and did not find evidence, however,I've not seen a result on this.
Comment: @kelalaka, I agree, didn't mean to imply that
Comment: @kodlu, In this case, the proof is that DES is not a consequence. Where is the information about whether the protocol is AES?
Comment: @Ss1996, I'm sorry I don't understand your exact question. If you are asking "this proof shows DES is not a group, where is the proof that AES is" then as my answer says, the state space is much larger and no weak key and like shortcuts are known; and no one has demonstrated it is not a group. It would be unexpected that it was a group, but it is open right now.
Comment: @kodlu I have seen evidence that the AES round function forms a group. But, as I understand it, this does not automatically make the entire AES cipher a group. I'm right? Second question: If AES has not been proven to form a group, is it correct to assume that using Triple AES does not (proven) increase crypto currency as in 3DES?
Comment: @kodlu Sorry. *Cryptographic Strength instead crypto currency

Is AES a group?


What is the difference between the concept and treatment of measurement error in psychometry and in statistics?

Question: There is some confusion with respect to the measurement error. What is the definition in statistics and definition in psychometry ?  The statistics does not seem to recognize the measurement error popularly called construct bias in psychometry.
Comment: Your question title is about measurement error; your final question is about bias. The two are not the same. You also ask simultaneously: what is the difference, and is there a difference? That said, here goes. The central idea of bias is that a procedure is wrong on average, yielding a mean that is higher or lower than the true or correct value, known somehow. That could be true of a measurement procedure, but it can apply where measurement error is not the question. That's what I understand from statistics; psychometrics may have a different notion, but I doubt it.
Comment: I am at a loss to know what kind of answer you seek. But the implication that statisticians ignore measurement error is contradicted by a substantial literature by statisticians. Go to www.amazon.com and search for "measurement error models" to bring up several major works.
Comment: @Nick The psychometry presumes that generally there is an error in Scale for measurement of say, fear. And hence, pursues the analysis. For example, error (measurement) variance is deducted from observed variance for arriving at true variance. The statistician presumes inertia of large numbers for mean and variance. There is no need for dealing with measurement error.
Answer: For measurement error there really isn't a difference in the definitions. Psychometry defines "true score" as "measured score" + "error" and this is the same thing as the statistical definition. The confusion may come from different terminology; that developed because psychometry deals with tests while statistics can deal with almost anything.
"Bias" is a bit more complex. @NickCox gave the definition in statistics. In psychometry, it is used (at least some of the time) in a slightly different way, again due to the specialized nature of the subject. A test is biased for/against a group if its predictions work differently in another setting. So, e.g. if we are using SAT scores to predict college GPA, bias would be that one group gets lower/higher GPA with the same SAT score.
In statistics, a scale could be biased against everyone - e.g. if my scale estimates everyone's weight as 5 pounds less than the actual value, that's bias. In the psychometrics definition, that can't be bias.
BUT psychometricians often use "bias" in the statistical sense as well.


Mechanism for ring contraction from 2-methylenecyclobutanol to 1-methylcyclopropanecarbaldehyde

Question:
I am stuck between two paths. I was going to protonate the alcohol to give a better leaving group, leaving a 2 degree carbocation and then go on with a hydride shift. But, I'm not sure what that may turn out to look like, and when exactly the ring is supposed to contract. Or, I was going to have the alkene attack the acid, but after that I wouldn't know what to do with the alcohol.
Comment: How might acid react with this compound? Perhaps start there...
Answer: Mechanistically, both of those protonation options (at the alcohol or the alkene) are reasonable. Without any other information, I would probably try both and then see if either gets closer to the product.
However, two things suggest to me protonation of the alkene:
1) As suggested, protonating the alcohol generates a good leaving group (water). However, the product has a C-O double bond, so breaking the C-O single bond appears to be unproductive. That's not to say it's impossible, but it leads me to think about trying something else first.
2) The product has a methyl group, but the starting material does not. Somewhere in the mechanism, that methyl group has to be formed. If protonation of the alkene is the first step, then a methyl is formed, which I see immediately as productive. Again, that's not to say that this protonation is definitely correct, but at least something productive has happened.
One last tip before I show my suggested mechanism. For any mechanism that is difficult to draw, but especially rearrangements, it can be very helpful to draw in the hydrogens. In this case, drawing in the hydrogens just about gives away an exact mapping of the carbons from the starting material to product. With a mapping, it becomes easier to determine all of the bonds broken and formed, which is essentially a checklist of all of the goals of the mechanism.
I'll leave it to you to add the curved arrows, but the intermediates are shown below. Protonation of the alkene in 1 gives carbocation 2, which contains the methyl group seen in the product 4. Next is the ring contraction, sliding a C-C bond over the the carbocation and generating new carbocation 3. Compound 3 might appear to be less stable than 2, but it turns out to be quite stable due to resonance structure 3', which is fully octet. Deprotonation gives 4.
Answer: Contraction only takes place when the oh group is axial right? If the oh group is eqitorial then then hydride shift can take place to give a planar carbonation through pinacol-pinacolone rearrangement to give more stable 4 membered ring structure.
Comment: Hi Dhruv, you should consider supporting your answer with more sources/information. Leading with a request for confirmation makes it seem like you aren't sure of your own answer.