Definition of Multilin DDH

Question: I am on the abbreviation mutlin. DDH, which probably stands for mutliniear Decision Diffie Hellmann. I am currently looking for a definition for this term, but unfortunately cannot find a source. Can anyone here help me further?
Answer: The standard DDH problem is, given $g, g^a, g^b, g^c$, to decide whether $c = ab$. With a bilinear pairing (for example elliptic curve pairings), this is solvable, since
$$e(g^a, g^b) = e(g, g^{ab}).$$
We therefore introduce the bilinear DDH, and it's generalisation - multilinear DDH.
Suppose we have a multilinear map
$$e : \mathbb{G}^\kappa \to \mathbb{G}_T$$
Where $\mathbb{G}^\kappa$ is the product of $\kappa$ copies of group $\mathbb{G}$. Suppose $g$ is a generator of $\mathbb{G}$ and $g_T$ is the corresponding generator of $\mathbb{G}_T$.
The $\kappa$-multilinear DDH problem is: given $g, g^{x_0}, \ldots, g^{x_\kappa}$ (that is, $\kappa+1$ exponentiations in $\mathbb{G}$), and an element $g_T^y$, to decide if
$$y = \prod_i{x_i}.$$
With a bilinear map we can solve $\kappa  = 1$, but don't know of any way to solve for higher $\kappa$. The bilinear DDH is when $\kappa = 2$, and would be solvable using a trilinear map if one existed.
Comment: Thanks for your reply, I'm really amazed how fast you get replies in this forum and how good they are. Can you still give me the source for this so I can quote it?
Answer: In this paper, there is a definition:
In a $n$-linear context $(\mathbb{G}, \mathbb{G}_T)$ with a $n$ linear map which verifies:
$$e(g_1^{a_1},\dots, g_n^{a_n})=e(g_1,\dots, g_n)^{a_1\cdot a_2\dots \cdot a_n} $$
Let $g$ be a public generator of $\mathbb{G}$.
The adversary receives: $\left(g^{a_i}\right)^{n+1}_{i=1}$, and should compute $e(g,\dots, g)^{a_1\cdot a_2\dots \cdot a_n \cdot a_{n+1}} $.
I'm assuming the decisional version is only about distinguish this output from a random element of $\mathbb{G}_T$, even it's not clearly defined in this paper.

Definition of Multilin DDH


What is the meaning of consensual principle for the basis of making law of contract?

Question: What is the meaning of the consensual principle for the basis of making the law of contract?
Answer: A consensual contract is one that requires no formality or further action beyond the consent of the parties
Most contracts are of this type (see What is a contract and what is required for them to be valid?).
Contracts that are required to be in writing (e.g. many real estate related contracts) are not consensual since, after the parties have agreed to contract, a further step is required to make the contract binding.
Answer: To dilate upon Dale M's answer, let's consider contracts that AREN'T consensual. I'll quote English contract law textbooks that explain this better. Note that all these textbooks have new eds. published in Apr 2020.
O'Sullivan & Hilliard's The Law of Contract (2018 8 ed). p 2.

1.6 Similar objections are levelled at the second thread, that contractual obligations are
consensual and based on the intention of the parties. Critics point to the prevalence of
the printed standard form and the frequent insertion of 'implied terms' into contracts,

p. 3

some justified as giving effect to the intentions of the parties but others, like the implied
conditions that goods sold in the course of the seller's business are of satisfactory
quality and fit for their purpose, are inserted by statute regardless of intention.
Contracts are sometimes described as requiring a 'meeting of minds', but even in the
nineteenth century, this was misleading. This is because the parties' intention is judged
objectively—how would their words and actions have appeared to a reasonable person?
So, although questions about the formation, contents, and interpretation of a contract
are resolved by reference to the parties' intentions, this is not really an enquiry into
what the parties subjectively intended or wanted. After all, as Rix LJ pointed out in
Procter & Gamble v Svenska Cellulosa Aktiebolaget SA (2012), 'when it comes to a dispute
the [subjective] question of actual intention is likely to be submerged in wishful
thinking.'

Anson's Law of Contract (2016 30 ed). p 7.

(iv) Implied terms and the standard of 'reasonableness'
Finally, the negative aspect of freedom of contract, that there should be no liability
without consent embodied in a valid contract, sits uneasily with the practice of
implication of terms into the contract, and the use of the standard of 'reasonableness'
as a way of dealing with gaps in the contractual language.28 Terms are implied not
only under statute, but also at common law. Although the basis of such implication
is said to be 'necessity'29 or in the case of custom 'presumed consent',30 in many cases
this is rather artificial, and in truth in many standard transactions the implied terms
are the legal incidents of the transaction,31 from which the parties are, subject to
statute, often free to deviate although in practice many parties will simply be bound
by those implied terms without seeking to negotiate otherwise. Freedom of contract
is also difficult to reconcile with the adoption of the 'objective theory' which provides,
in essence, that a person (A), whose conduct is such that the other party reasonably
believes that A has assented to the terms of a contract, will be bound no matter
what A's real intention is.32 This rule can lead to the imposition of non- consensual
obligations, since what creates the obligation is not consent in fact but acting as if
consent is being given.

Contract Law: Text, Cases, and Materials (2018 8 ed). p 11.

(2) The individualistic ideology
A persistent theme in Market- Individualism is that judges should play a non- interventionist
role with respect to contracts. This distinctive non- interventionism derives from the individualistic
side of the ideology. The essential idea is that parties should enter the market,
choose their fellow- contractors, set their own terms, strike their bargains and stick to them.
The linchpins of this individualistic philosophy are the doctrines of 'freedom of contract' and
'sanctity of contract'.
      The emphasis of freedom of contract is on the parties' freedom of choice. First, the parties
should be free to choose one another as contractual partners (i.e. partner- freedom). Like the
tango, contract takes two. And, ideally the two should consensually choose one another.
Secondly, the parties should be free to choose their own terms (i.e. term- freedom). Contract
is competitive, but the exchange should be consensual. Contract is about unforced choice.
      In practice, of course, freedom of contract has been considerably eroded. Antidiscrimination
statutes restrict partner- freedom; and term- freedom has been restricted by
both the common law (e.g. in its restrictions on illegal contracts) and by statute (e.g. the
Unfair Contract Terms Act 1977— UCTA) [p. 418, Chapter 13, Section 3]. Moreover, the development
of monopolistic enterprises, in the public and private sector alike, has made it
impossible for the weaker party actually to exercise the freedoms in many cases. For example,
if one wants a British family car, a railway ride, telephone services etc. the other contractual
partner is virtually self- selecting. Similarly, where the other side is a standard form
or a standard price contractor, the consumer has no say in setting the terms. Nevertheless,
none of this should obscure the thrust of the principle of freedom of contract, which is that
one should have the freedoms, and that the law should restrict them as little as possible—
indeed, it is consistent with the principle (in a widely held view) that the law should facilitate
the freedoms by striking down monopolies.


PCIE reference clock

Question: I recently completed a PCI-E Gen 1.0 line card design. The line card consisted of 4 Spartan 6 FPGAs sharing one PCIE reference clock. Early on in the design there was a decision to solely use the PCIE reference clock to clock all the internal logic from this PCIE reference clock and hence no on board crystal was used. The FPGAs interface to a bunch of servos on the other side over 10/100MBS interface and the FPGAs used the DCM to divide the PCIE- Reference clock to 25MHz and used for the MII and the PHY interface. There was a lot of Receive Data valid errors on the PHY interface which was attributed to the PCIE reference clock and its inherent jitter of 30KHz to 33KHz(Spread spectrum modulation). We decided to turn off the SSC function in the host software and we stopped seeing the receive error and things are back to normal.
I wanted to ask and see if others had different experience using the PCIE-ref clock for clocking all FPGA logic including the 25MHz PHY interface. I was under the impression that the FPGA's DCM would clean up the PCI- reference clock after it gets in to the PCIE MGT hard macro before it was divided down. This is my first FPGA design and hence would appreciate some insight.
Comment: What was the datasheet spec for the required input clock? What Phy are you using? In my last design, I chose to put an oscillator on the board specifically for the Ethernet phy because the FPGA DCM/PLL would not meet the clock jitter spec.
Comment: @stacey,The PHY is DP83849 and the jitter spec for the TX reference clock is 1.4nS. The FPGA Max input clock period jitter 1nS max. I think the DCM in the FPGA can account for high frequency deterministic jitter of up to 1nS. But i dont see any documented evidence of it keeping track of a clock that is wandering around 30-33KHz increments on purpose. Like I mentioned things are better now that we have turned the spread spectrum functionality off and the clock is fixed at 100MHz +/-300PPM.
Answer: All a DCM can do is add jitter - it's a tapped delay line, so the output clock can jitter as the taps change.  And for the same reason it can't remove any jitter that is already there.
If you want to "clean up" the clock signal, you need to use a PLL - check the datasheet to see if the FPGA PLLs can cope with the spread-spectrum PCIe clock. Or use a dedicated clock cleaning PLL device.

PCIE reference clock


Direct USB-to-USB networking?

Question: Is there an existent method, a driver, or some software that facilitates direct USB-to-USB networking in macOS similar to FireWire-to-Firewire ethernet or the new Thunderbolt-to-Thunderbolt ethernet networking?
Answer: Yes and no. Because of the host-slave relationship USB is built around this requires specific hardware, namely an active cable with electronics. They're cheap enough though, you can get them on Amazon or similar for ~$10 and up.
Just search for "usb bridge cable".
Comment: So there is an existing driver built into macOS that will leverage a cable such as [this](https://www.amazon.com/Mediabridge-USB-3-0-Cable-SuperSpeed/dp/B00D8JFNO8/ref=sr_1_2?ie=UTF8&qid=1479178284&sr=8-2&keywords=usb+bridge+cable) as an ethernet bridge via USB for two Macs?
Comment: No, that's just a cable, it has no electronics. The fact that a company named "Mediabridge" is selling it does not make it a bridge cable...

From the description, "Transfer or share data from external hard drives and other peripheral devices with a USB A-Female port to your computer or other Host device."
Comment: In the remarks folks are talking about using it to transfer data between a Mac and PC...  Also see: http://superuser.com/a/945523 -- Now, having said this, it doesn't appear that macOS, or Windows, for that matter, supports networking over USB at the moment...  At least I'm not finding it.
Comment: It's exceedingly unlikely to be an actual bridge cable, since there appears to be no room for the required electronics, and it isn't advertised as having any. Best case scenario it doesn't work, worst case you blow your usb ports.
Comment: Oh yeah, that's yet another type of cable, it's crossed-over so it won't blow your ports but it also won't let you transfer files, sadly.
Comment: It seems as though one will have to have some kind of a kernel extension that would facilitate ethernet transport over the USB crossover, which seems doable, but that no one has yet done... Hmm.
Answer: There's a number of USB to USB direct connection cables out there.  Here's one example from StarTech: https://www.startech.com/Server-Management/KVM-Switches/dual-system-swap-cable-kvm-switch~SVUSB3DSA
Another example from LapLink (look at the USB 3.0 cable):  https://web.laplink.com/cables/
My guess these cables and many more like them are nearly identical, all using the same PL27A1 IC from Prolific Technology:  http://www.prolific.com.tw/us/ShowProduct.aspx?pcid=43
The chip emulates a serial or Ethernet connection but at 5 Gbps.  It's exact behavior depends on how the chip is configured and what software/drivers exist on each computer.  My guess is, given what I've seen from reviews, that most operating systems include support for this chip in providing an Ethernet-like network.
There are cheaper passive USB-A to USB-A crossover cables that meet the USB 3.x spec for data transfer between USB hosts, like this one: https://www.datapro.net/products/usb-3-0-super-speed-a-a-debugging-cable.html
In the description they call it a "debugging cable" since as of today (to my knowledge anyway) that's all any operating system allows it to do.  There's no reason someone can't write drivers for this cable to create a network between two computers.  That is except the large number of similar cables that violate the spec and could open up the authors of such software to all kinds of criticism because people bought the wrong kind of cable and broke their computers.
USB-C to USB-C, and USB-A to USB-C, cables are generally safe to connect two computers since there's been a crackdown, by some big players in the industry, on those selling noncompliant cables.  The only thing I can see stopping these cables from making a network connection is an apparent lack of need for it, at least from the view of the people that make operating systems for desktop computers.  I'm seeing questions over USB to USB connections often enough that I expect this will likely change soon.
In reading things on the new USB4 spec a host to host connection will be supported.  Well, it's technically already supported since it's been in the USB 3.0 spec for some time now.  It's only that this time the feature will be better defined and/or support is mandated.  I have to wonder if this will trickle down to existing USB 3.x devices.  There's not a lot of difference between the old USB 3.x and new USB4.  Like all USB versions before it USB4 requires backward compatibility.
As far as I know every Apple Mac computer with a USB-C port will support "Target Disk Mode" on those ports.  This allows a Mac to boot into a mode that makes the computer a very expensive USB drive.  This is helpful in a number of situations.  This proves that the hardware is there for a Mac to act as a USB device to another computer.  It's just not a feature supported/allowed by the operating system.

Direct USB-to-USB networking?


Sitecore CDP Dashboard Report

Question: I am exploring the Sitecore CDP dashboard. In the Summary Report it is showing number of reports like Guest Type, New Visitors, Converted, Abandoned etc. reports. I want to further drill down the report to see the user details. However I could not find a way to get the details of users. Even if downloading the reports don't contains user info.
Please let me how I can achieve this.
Answer: Most of the prebuilt dashboards have very specific usecases that they are showing data for.  There are a few dashboards that allow you to pull the CSV of the data, however most do not provide access to the raw data including the specific guests that matched in that report.
Sounds like however, you are looking for more of the raw data that you can get inside an S3 bucket of your Sitecore CDP Data Lake.  This is not provided out of the box, however you can reach out to Sitecore support to provide this raw data to you, so that you can create your own dashboards (using Power BI, Tableau or any Business Intelligence software) for your organization, or use this data to explore specific trends.  This data is provided to you on a daily basis and you can move to your own S3 buckets or do with that data as you wish.

You can learn more about the Data Lake export service here in the Sitecore Documentation: https://doc.sitecore.com/cdp/en/developers/sitecore-customer-data-platform--data-model-2-0/using-sitecore-cdp-data-lake-export-service.html
New link: https://doc.sitecore.com/cdp/en/developers/api/data-lake-export-service.html
Answer: Enabling Developer mode gives you all the details about a specific session
To enable it you`d check this and here details about how to use it.
From those docs:
Enable feature flags in Sitecore CDP

To enable a feature flag:

In the navigation pane, hover the mouse over the Sitecore CDP tenant
name, then hover over your name, as shown in the following image.

Click Feature Flags.

Note: The features that you can enable depend on your role. Most user
roles only have access to the debug feature.

To enable a feature, in the Features list, in the Enabled column, turn
on the switch. If you do not immediately have access to
the feature, try logging out and logging back into Sitecore CDP.

Use debug mode in Sitecore CDP

If you enable the debug feature flag, you can view data in a table or
JSON format on certain screens. This is useful when troubleshooting
experiments, because you can access all the data in a guest profile
either in a table or in JSON format.
To use debug mode:

Click the Debug icon next to the data you want to access.
There are multiple screens within the Sitecore CDP application that
this icon displays on.

The following image shows the debug feature that is available on the
guest profile page.

Depending on where you clicked the icon, a window displays that includes data in a table.

The following image shows the window that displays if you click the
Debug icon on the guest profile screen.

To see the data in JSON format, click Show as JSON. A JSON representation of data displays, as shown in the following image.

To download the JSON to a local
destination, click the gray gear icon, then Download. Alternatively, to
copy the JSON to the clipboard, click Copy.

To return to the previous screen, click the close icon.
Comment: Please add a description here as well from the link as the link may expire in the future.

Sitecore CDP Dashboard Report


Map network drive to .local address

Question: I am connecting a network drive to a computer using the map network drive option. If I give it the IP address (192.168.1.x) it will connect fine, but if I give it the .local address (name.local), it ends up saying that it cannot access the device. I would like to use the .local address so that in the case that the ip address gets changed (e.g. 192.168.1.2 to 192.168.1.3), the computer will still know where to connect.
Comment: could possibly be a Master Browser issue. Do you have one always-on machine correctly set to be master browser, or do you just let them all auto-negotiate using random elections?
Answer: This sounds like a DNS issue.  Do you have a server in place that runs DNS?  If not, NetBIOS should find it by just pinging the name instead of typing ".local" after it.  If you do have a server, check the DNS server settings to see if it "knows" about the device.
Answer: Sounds like you don't have DNS set up to be able to handle this. Often, if you're using DHCP to hand out addresses, the consumer-grade router will store a DNS-to-DHCP mapping. So if the router is set up to serve the ".local" DNS domain via DHCP, and I am machine1 and I get 192.168.1.100 from the router, the router will store an entry saying
machine1.local = 192.168.1.100

in its DHCP table, which is often shared by the build-in DNS server. If you're not using DHCP, the router has no idea what your hostname or DNS domain name are.
Alternatively, you could populate the hosts file on the devices (on Windows it's usually C:\Windows\System32\Drivers\etc\hosts and on UNIX flavors it's /etc/hosts).
Comment: I do not think it has anything to do with DNS to be honest. On the same computer, I used putty to access the server command line over ssh, and entering the .local address worked fine. Plus, on a win 10 computer I have, entering the .local address works on both.

Map network drive to .local address


What is the correct "pay to the order of" to get a U.S.passport?

Question: I need to make out a check to submit/pay for my passport application, but I am a little confused about what to write. Do I just put down, "U.S. Department of State," or do I enter the headquarters' mailing address (2201 C Street, Vancouver, Washington)? I don't want to have to rewrite a check!
Comment: You have it right, "U.S. Department of State" according to this [site](https://travel.state.gov/content/passports/en/passports/information/fees.html).
Comment: Don't you pay for the passport application fee at the post office?
Answer: If you're paying for your passport application in the United States, you make your check or money order payable to "U.S. Department of State". The USPS advised me when I went in this year that your name and date of birth should appear on the check/money order (in case it gets separated from your application).
You must also pay a separate $35 execution fee to the agency which accepts your passport application (this fee was $25 prior to April 2018). This must be made payable to the agency accepting your application, but most such agencies will accept debit/credit cards for this payment.
If you're paying for your passport outside the United States, contact the embassy or consulate for payment instructions.
Comment: It used to be the case that when applying for a passport in person at a post office, in addition to the check to the State Department for the Application Fee, you had to write a *separate* check for an Execution Fee, and as I recall the latter was to be made out to the Postal Service.  Has this changed?
Comment: @WGroleau Oh, they charged me separately for the photo. But I got to keep one of the two photos they printed. That'll be useful if I have to apply for a visa anytime soon.
Comment: A lot of US Embassies overseas do not take personal checks, just cash and sometimes credit cards.
Comment: @NateEldredge You pay the $25 Execution Fee when you apply for your first passport at any location.  But yes that fee is made payable to the location accepting your passport, be it the post office or passport office or embassy
Comment: When I got my most recent passport (2012), at the post office, they told me the $25 was for them to take a photo and that I wouldn't have to pay it if I already had an acceptable photo.  (Which I had.)


etymology of phosphatidates and phosphatidic acid

Question: A phosphatidic acid is an ester of a glycerol, two fatty acids and a phosphoric acid.
Who coined this term, and is the word's "phat" part related to english "fat"?
Answer: Apparently, "phosphatidate" is from "phosphate" + "-ide" + "-ate".
The word "phosphatidic acid" is from "phosphate" + "-ide" + "-ic acid".
Phosphate ($\ce{PO4^3-}$) is the anion of phosphoric acid ($\ce{H3PO4}$).
The suffix "-ide" is used in biochemistry to just about anything. It just denotes an element of a group of related compounds.
Now, "-ate" and "-ic acid" denotes the acidic nature of the compounds.

And no, it is not related to the English word "fat".
Answer: Phosphatidate can be broken down to "Phosph(orous) + at(e) + id(e) + ate"
-ate means either "having" or "characterised by" and is used in chemistry to denote esp. a salt or ester of an acid.
-ide is a suffix meaning "a compound of " or to denote a group of relate do compounds as in monosaccharide or to indicate that is a binary compound as in Sodium chloride.
-ic also means relating to or characterised by.
Therefore phosphatidate literally means it's a compound containing a moiety characterised by having an ester of phosphorus.

Who coined this term?

According to this site it was 1st used in 1884 by International scientific vocabulary.
Comment: Can it be one of "ate" was for phosphoric acid, and other for fatty acid. and ide for glycerol ( like a monosaccharide)?
Comment: Yes it could be.


What is the source of Higgs field?

Question: Could you please explain me this question.
As we know that every field has a source like electric field has electric charge and magnetic field has magnet etc. So what is the source of the Higgs field?
Comment: The sources for the Higgs field are the rest of fields in the Standard Model, such as the electron field, the $Z$ field, etc.
Comment: @AccidentalFourierTransform Could you elaborate?
Answer: The Higgs  particle is an elementary particle in the standard model of particle physics

All elementary particles have a meaning in the quantum mechanical framework, not the classical that you are discussing.
In quantum field theory, every particle in the table above is a quantized field which covers all spacetime. Creation and annihilation operators define the sources , for example a creation operator creates an electron, acting on the electron field,  or a Higgs particle creation operator acting on the Higgs field creates a Higgs meson.
Answer: Good question. The Lagrangian  of the Standard Model of physics is constructed in such a way that the standard answer to your question is: the Higgs field itself. This is because it contains a huge self-interaction term $\lambda\phi^4$ that causes the ground state of the Higgs to correspond to a constant nonzero value. However, the existence of such a term has not yet been established experimentally and there are other possible explanations for the Higgs vacuum expectation value (which definitely IS established) that do not involve self-interaction.

What is the source of Higgs field?


NSTimer for count down among view controllers

Question: \I have a view controller which starts a count down timer and updates a button label according to it.The problem is that when the user leaves the view and come back i do not want to restart the timer o continue from the paused time.What would be the best way to active it?
Making the timer in app delegate and post notification to the this view controller.Since the other view controllers do not need it they do not need to listen to any notification.The problem is that i have to make sure that the timer has run down to 0 to enable the button os it means when the view is loaded again i need to wait for the notification at least 1 second to make sure that he time is really up.The problem is what to show during this period on the button label.
If the view controller was a tab bar item i would be no problem but it is a pushedv view contller so it is released when the user taps on back.
Comment: @Mert thank you for your responce but i just saved the expiry time in nsuserdefaults and inviewwillaper i read the timestamp of the expiry.If it is before the current time it means tiemr would have already exprired when i was in the background.If it is somehow still forward then i start the timer.Creating Singleton classes is ok but you should watch out.I alrady have like 7 single ton manager classes which must be kept in memory during app run so i would avoid crating additional ones as much as possible.
Comment: It is ok to save expiry date to nsuserdefaults if you have just one. For multiple dates it will be very difficult to manage them from userdefaults. It should not be problem to create a singleton object, which holds some long long values, even if you have 10 more singleton objects...
Answer: Just create a singleton object for managing your timer(s). You can create it from appdelegate.
You can get remaining time from that object anytime you want (firedate - datenow). So you do not need to work with notifications. It is also not good idea to send a notification in every second, which is not used by your view, when it is not allocated. When your view is allocated / comes to screen can ask that object every second the remaining time and updates his label or some subviews etc. You can implement it easily with a NSTimer object.
You can also place a bool variable in that singleton object, which indicates, if firedate reached. Or check remaining time < 1 . So your view check the value of the variable while is allocating/showing then you can decide to activate or deactivate your button.


Do gravitational waves affect light?

Question: Gravity "bends" light, predicted with theory of relativity and subsequently observed: how does gravity and gravitational waves achieve this effect, and shouldn't this effect be present wherever there's gravity, for example shouldn't there be a "shimmering effect" when observing distant stars/galaxies as their emitted light is "bend" this way and that (like heatwave-shimmer)?
Comment: the shimmering of starlight is due to random fluctuations of temperature/pressure in the atmosphere,  causing the light to be slightly refracted in different directions as it travels through the atmosphere.  The gravity wells of stars are by comparison quite stationary, so the bending doesn't change with time, hence no shimmering prior to the atmosphere.
Comment: Instead of gravity "bending light," it's more accurate to say that gravity bends *spacetime*, and light simply follows a geodesic path along the bends.  Thinking of it that way, it makes sense that there should be a "shimmering" effect when light passes near a source of gravitational waves. The difficulty is that this effect is extremely small, and we're only now starting to develop the technology with enough resolution to detect it.
Comment: @DmitryBrant, the effect is extremely small, and the objects that have enough mass to reveal it (i.e., whole galaxies) move raaaaaather slooooooowly through the sky.
Answer: The truth of the matter is that we don't know.  Gravitons are a theory slowly becoming accepted but there is simply little evidence to support them as a viable theory on how gravity works.  Gravity is still just the magical notion that was devised centuries ago.  We only know that gravity is not constant and it is the only force not conforming to the theory of relativity.  We know this because the path and direction of distant galaxies in our expanding universe are not moving in directions that our gravitational math expects.  Something is wrong there.
What some of us are confident of is that light particles are physical things, and like all physical things they adhere to the laws of gravity.  Just as our bodies do.
I think your question is why isn't light reaching us from distant stars shimmering, as might happen when someone holding a flashlight pointed at you moves it from right to left or up and down across your vision.  Even if the star were moving from left to right or up and down (which it isn't) it would still not shimmer - because it's light is emanating at all angles, not a steady stream like a focused flashlight.  And although light from stars are bent by the gravity of other solar systems, planets, and even other galaxies, it stays bent.  It doesn't bounce back to it's original path like a spring.  Therefore when it gets to you it is a steady stream on a fairly straight path from it's last bending point.
The most interesting thing about your question is that light particles are also affected by the gravity of our Sun and Earth as well.  Which means the star's light particles are most likely on a slight bend as they reach our eyes as we stand on this huge gravitational ball called Earth.  I personally believe that planets attract light particles, as well as neutrinos, electro-magnetically.  But that is not a popular theory.
I think you might be interested in learning how our eyes work, the retinas being struck by gazillions of light particles from different bending angles constantly.  Everything we see is due to photons bouncing off of objects and then into our eyes, which have rods and cones that convert them to neural vision.  Although the particles could have bounced 100 times or more off of things before reaching our eyes, they are on a direct and straight path from the last object they bounced off of, giving us that object's distinct color.  It gets a little more complicated than that with the notion of light particles emanating in waves, but it's all relevant.
Comment: agree with you on the single photon ... but it is the next photon, and the next ... that comes in at slightly different angles that would cause the shimmer. I like the slightly skeptic feel of your answer, somehow makes it seem more trustworthy :)
Comment: I would add that everything except quantified observation is only theory.  Theory attempts to explain _why_ those things in the observations occur.  And questions like yours will always be the stimulus to refine theory.  Which is a good thing @slashmais, because no theory is probably precisely accurate.
Answer: Gravity "bends" light, predicted with theory of relativity and subsequently observed: how does gravity and gravitational waves achieve this effect
I conceive the question as one referring to General Relativity (GR) which is an entirely classical theory, and not to a (thus far missing) theory of Quantum Gravity.
The point of view of General Relativity is that free falling reference frames are true inertial frames (ie where Newton's first law holds). Thus, viewed from within a free falling frame, light goes on a straight line:  a person inside a free falling elevator, will see a straight light ray. Then, another person, standing on firm ground on the outside the free falling elevator, will thus see a bend line.
From the above, it is clear that everything that has a gravitational field will bend light rays.  Masses clearly come with a gravitational field.  And because of the equivalence of mass and energy, $E=m c^2$, everything that has energy, has a gravitational field, too.
However, the energy density of a gravitational wave is tiny. So the bending effect will be tiny.
Nevertheless, the "shimmering" you refer to can be observed if the gravitational waves are humungous: this is just what LIGO did with gravitational waves stemming from the merger of black holes.
Comment: that is fine, it is just that the question asks about "shimmering when observing distant stars", and it sounds as if you agree with it.
Comment: Ligo did not observe "shimmering" of light from the stars, but concentrated light from lasers in special geometry.
Comment: @anna:  maybe I expressed myself in a bad way, but I didn't think I said that LIGO observed light *at all*.  LIGO observes gravitational waves, using a earth based laser interferometer. The "shimmering effect" that LIGO did observe is the dynamical distortion of space-time due to a passing gravitational wave.  In the case of a black hole merger, the signal was compared, by the LIGO people, to a tone moving up the scale rather than a shimmering.
Answer:
Gravity "bends" light, predicted with theory of relativity and subsequently observed: how does gravity and gravitational waves achieve this effect,

Physicists are hopeful that gravity will be quantized similar to the other three forces, and that gravitational waves are a confluence of gravitons. There exists the classical general relativity where the photons/light are following the distortion of the stress energy tensor due to a gravitational source of a star, the geodesic, and bend.  In a quantum mechanical view, the photons interact with virtual gravitons and change directions into bending.

and shouldn't this effect be present wherever there's gravity,

wherever there is gravity, in the QM framework there will be a photon graviton interaction, but it will be higher order and the gravitational constant is very very small,

for example shouldn't there be a "shimmering effect" when observing distant stars/galaxies as their emitted light is "bend" this way and that (like heatwave-shimmer)?

The photon/light bends when passing very close to  gravitational wells, but  the probability of changing directions randomly ( shimmering) is very very small, not detectable because of the gravitational constant.
The LIGO experiment has enormous spatial accuracy , so it can detect small distortions of space from a passing gravitational wave, which affect the laser light of the experiment. These conditions are not held for general star light . The LIGO experiments gravitational wave is built up by  real gravitons, building up the gravitational wave (of course assuming quantization of gravity analogous to the other three forces). The interactions of light in bending around stars is effected by the exchange of virtual gravitons, the gravitational field. LIGO is a gravitational wave detector.
Comment: In addition to the prior series of inappropriate comments, I also moved some obsolete comments and the ensuing conversation to [a chat room](http://chat.stackexchange.com/rooms/38372/discussion-on-answer-by-anna-v-do-gravitational-waves-affect-light). Feel free to continue the discussion there, if you want.
Comment: @DavidZ  Thanks, I think I have a comunication problem with MBN and he cannot get through to me and I to him/her .  Maybe it is the language barrier, considering that my mother tongue is greek.
Comment: It could be many things. A language barrier is a possibility, though perhaps not the most likely. In any case, it's always up to you whether you care to continue participating in the discussion.
Comment: @annav: There is no communication problem. It is simplly not true that photons follow "distortion of the stress-energy tensor".

Do gravitational waves affect light?


Replacing AZERTY keyboard by a QWERTY keyboard

Question: Is it possible to replace the top case with the  AZERTY keyboard of a 15" MB Pro late 2012 (usb 3.0) by the top case with QWERTY keyboard of a 15" MB Pro Mid-2010. Do they have exactly the same dimensions?
Answer: At one time it was possible to make the change and in the end it became impossible in the official Apple network.
The only solution I could offer you is to buy and replace the keys. I had done it one 2011. ok it's long but I changed the keyboard for 5€... then just change the layout of the language in the keyboard preferences.
Answer: Whilst the topcase-and-keyboard is interchangable, changing the keyboard's nationality layout does not change what the machine knows as its native tongue. Yes, it is possible to change the keyboard layout in the OS to match the new languagr-layout, but that only takes effect once you have logged in - at the login screen, the keyboard will still be using the original language layout.


If else for string types in Business objects

Question: How can I use if...else in Business Objects for strings? My problem is my query returns value in string which can contain duplicates. For duplicates, what I would like to do is find if I have duplicates or not and if found, just use one. I can do it for float type as like this: If [sales_rev]>1000 Then [high_rev] but if I do If [display_name]=['sth'] Then...., I always get an error.
Answer: Try
If [display_name]="sth" Then....
without square brackets and with double quotes
Answer: I'm not too sure based on your question, but it sounds like you're trying to assign a string value instead of compare it. Depending on your language, the most common operator to do boolean logic on strings is "==" while you just have a single "=" in your question so it'd probably be [display_name]==['sth'] based on what you stated.


Swapping Framebuffers or swapping attachments?

Question: I wanted to know what the better approach would be for post-processing: swapping between framebuffers or swapping between textures attached to one framebuffer?
Answer: It's almost always better to change a state's object instead of changing an object's state.
Mostly for readability, as thats how GL is expected to be used.
Even if you have multiple attachments in common between FBOs, I would still suggest binding them to each FBO.
As for performance, its implementation dependent, but framebufferTexture is likely slower due to it having to compute the mapping for the fragment outputs into the texture, while a complete framebuffer already has this mapping computed and just needs to load it.


What about actually identical requests?

Question: Now, it says here that I should not make lots of semantically equivalent requests. It also says that there is a lot of caching. So I was wondering, what about exactly identical requests. Should I wait about a minute in between for them too? Or can I just spam them out as fast as my program can send them, as, like you said, you cache it anyway. Or is it somewhere in between?
Answer:
Or can I just spam them out as fast as my program can send them, as, like you said, you cache it anyway.

No.  The throttle discussion page specifically says:

... we consider > 30 request/sec per IP to be very abusive and thus cut the requests off very harshly.

And even though the underlying data is cached for 60 seconds, the HTML requests are not. You'll still burn one quota slot with each call, cached or not, and risk the wrath of a backoff flag, or worse.

See also: "Is it possible to access data in real-time using the Stack Exchange API?".
Once per minute is a pretty reasonable rate to repeat the same request. Multiple times per minute also risks running out of quota -- especially if you make several calls per batch/poll.

What about actually identical requests?


Curl of magnetic field produced by current carrying wires with infinitesimal small area

Question: Can Magnetic fields produced by  thin current carrying wires with infinitesimal area have curl with a delta function in it ?? As area is Zero current density J definitely becomes infinite at where Wire is located So curl has a delta function in it , Am I right ?Please correct me
Answer: Off hand, I'd think so, for some scenarios: $\vec{r}$ as in cylindrical coordinates,  $\nabla \times \vec{B}=\mu_0\vec{\delta}(\vec{r}-\vec{r_0})J_0$. You could probably find some examples of this researching Green's Method solutions to Maxwell's Equations.


What about the action on this space?

Question: I think it is true that there is no free-action of $(\mathbb{Z}_2)^3$ on product of $\mathbb{S}^{m}$ and  $\mathbb{CP}^n(n$ is odd).  I don't know how to prove it. A detailed proof will be very much helpful.
Comment: This is true for $m=1,n=1$.  Closed $3$-manifolds covered by $\mathbb{R} \times S^{2}$ fall into 5 homeomorphism classes, and $(\mathbb{Z}_{2})^{3}$ does not embed into $\pi_{1}$ or any of them. See page  457 of https://homepages.warwick.ac.uk/~masgar/Teach/2012_MA4J2/geometry.pdf. The case $m=1, n \geq 1$, can probably treated similarly, although it seems quite different to the general case.
Answer: Here's a proof that works for all even $m$ and $n\geq 2$ and shows a stronger result:  There is no freely acting group $G$ containing three non-identity elements $a,b,c$ for which $ab, ac,bc,abc$ are all non-identity elements.  (So, we do not assume $a,b,c$ commute).
I suspect one would have to work much harder for odd $m$, because the corresponding fact about $(\mathbb{Z}_2)^2$ actions on $S^m$ is much harder when $m$ is odd.
Each element in $G$ induces an action on the cohomology ring of $M:=S^m\times\mathbb{C}P^n$.  By Kunneth, the cohomology ring of $M$ is given by $$H^\ast(M)\cong\mathbb{Z}[x,y]/x^2 = y^{n+1} = 0$$ with $|x| = m$ and $|y| = 2$.
Because $H^\ast(M)$ is generated (as a ring) by $x$ and $y$, any automorphism is determined by where it sends $y$ and where it sends $x$.
Lemma:  If $m$ is even and $n\geq 2$, then any such automorphism maps $x$ to $\pm x$ and $y$ to $\pm y$.
Proof:  Assume first $m\geq 4$.  Then $y\in H^2(M)$ generates, so any automorphism must map $y$ to $\pm y$.
Now, we know $x, y^{m/2}\in H^m(M)$.  If $y^{m/2} = 0$, then $x$ generates $H^m(M)$, so $x$ maps to $\pm x$.  If $y^{m/2}\neq 0$, then notice that the equation $0 = (ax + by^{m/2})^2 = 2ab xy^{m/2} + b^2y^m$ forces $a= 0$ or $b=0$ because $xy^{m/2}\neq 0$.  Thus, the are 2 ``lines'' (multiples of $x$ and of $y^{m/2}$) are ring theoretically characterized by squaring to $0$. Since an automorphism must map $x$ to a primitive element, it must be sent to $\pm x$ or $\pm y$.  But $y$ already maps to $\pm y$, and an automorphism is injective, so $x$ must map to $\pm x$.  This concludes the case $m\geq 4$.
If $m = 2$ (and $n\geq 2$) then the only solutions to $0 = (ax + by)^2 = 2abxy + b^2 y^2$ have $b=0$.  Thus, $x\in H^2(M)$ is characterized ring theoretically (up to sign) by being a primitive element which squares to $0$, so $x$ must map to $\pm x$.  Now, the equation $0 = (ax + by)^{n+1} = (n+1) ab^n x y^n$ (using the fact that $y^{n+1} = x^2 = 0$ and that $x$ and $y$ commute) implies $a=0$ or $b=0$.  Since $x$ maps to $\pm x$, then just as in the case of $m\geq 4$, this implies $y$ maps to $\pm y$.
$\square$
Now we use Lefschetz:  if an automorphim of $H^\ast(M)$ maps $x$ to $x$ and $y$ to $y$, then its Lefschetz number is non-trivial, so this map has a fixed point.
  If there are three elements $a,b,c\in G$ with $ab,ac,bc,abc$ all non-identity then we have a contradiction as follows.  There are only three choices of mapping $x\mapsto \pm x$ and $y\mapsto \pm y$ which are not the identity map.  If, say, $a$ and $b$ induced the same map, then $ab$ induces the trivial map on $x$ and $y$.  This implies the non-identity transformation $a\circ b$ on $M$ has fixed points, contradicting freeness.  So, we may assume $a,b,c$ all generated different maps on $x$ and $y$.  Then $abc$ acts as the identity on $H^\ast(M)$, so has fixed point.
All this still leaves the case of $m=2, n=1$, when $M = S^2\times S^2$.  We can still prove the $\mathbb{Z}_2^3$ can't act freely, but it takes more work: there are new automorphisms of $H^2(M)$ coming from swapping $x$ and $y$.
The calculation $0 = (ax + by)^2$ still implies the the two lines (multiples of $x$ and multiples of $y$) are characterized ring theoretically, so $x$ and $y$ map to elements of $\{\pm x, \pm y\}$.  It follows that, up to changing signs, swapping $x$ and $y$ is the only new thing we need to worry about.  Of these new automorphisms, only 2 have 0 Lefschetz number:  $x \mapsto -y, y \mapsto x$ and $x\mapsto y, y\mapsto -x$.  Both of these have order 4, so if every element of $G$ has order $2$, then these can't happen, and so we reduce to the case where $x\mapsto \pm x$ and $y\mapsto \pm y$.
Comment: Does the case $S^{2} \times S^{2}$ not follow from the fact that $\chi(S^{2} \times S^{2}) = 4$?
Comment: @NIck:  You're right!  No group of order greater than $4$ can act freely on $S^2\times S^2$ for that reason.  Much nicer than my arugment - thanks!
Comment: @JasonDeVito Thank you so much. It will be very helpful if you give some hint how to solve it when $m$ is odd ?
Answer: This is not a complete solution, but it puts a relatively strong restriction on the possible $n$ and $m$ for which this can happen.
Suppose that $m$ is even and a group of order $8$ acts freely on $S^{m} \times \mathbb{CP}^{n}$. We will show that $$ n =3 (\mod 4). $$
Firstly, note that since $S^{m} \times \mathbb{CP}^{n}$ is compact the action is automatically properly discontinuous so the quotient is a manifold and the quotient map is a covering map.
Now, $8| \chi (S^{m} \times \mathbb{CP}^{n}) = 2 (n+1)*$  since the quotient has a degree 8 cover by $S^{m} \times \mathbb{CP}^{n}$ (recall that Euler characteristic is multiplicative for covers), therefore $4| (n+1)$ which proves the claim.

since Euler characteristic is multiplicative on products.
Answer: Here is the $m$ odd case with one major gap.  I don't know how to fill it - does anyone else?
We will assume $m \geq 3$ since $m = 1$ has been handled in the comments by Nick L.
Consider the principal $S^1$ bundle $N\xrightarrow{\pi} M$ where $N:=S^m\times S^{2n+1}$ and $M:= S^m\times \mathbb{C}P^n$.
Lemma:  Let $X$ be a reasonable topological space and consider a map $f:X\rightarrow M$.  This map has a lift $\tilde{f}:X\rightarrow N$ iff $f^\ast:H^2(M)\rightarrow H^2(X)$ is the $0$ map.
Proof:  Because $H^2(N) = 0$, if $f = \pi\circ \tilde{f}$, then $f^\ast$ factors through the $0$ map on $H^2$.  So the condition on $H^2$ is definitely necessary.
Sufficieincy is more fun.  The bundle $\pi$ is classified by a map $\phi:M\rightarrow BS^1 = \mathbb{C}P^2$, which also classifies $H^2(M)$.  If $f^\ast$ is $0$ on $H^2$, it then follows that the pull back bundle $f^\ast N$ is a trivial bundle, $f^\ast N\cong X\times S^1$.
Of course, $f^\ast N = \{(x,n):f(x) = \pi(n)\}$ comes equipped with a natural map into $N$: projection onto the second factor.
Then $\tilde{f}$ is the composition $X\rightarrow X\times S^1\cong f^\ast N\rightarrow N$.
$\square$
We will use this lemma in the case where $X = N\times G$, with $G$ a finite group acting freely on $M$.
Then the map $f: X\rightarrow M$ given by $f(n,g)= g\ast\pi(n)$ induces the $0$ map on $H^2$, because $H^2(X) = 0$.  By the lemma, there is a lift $\tilde{f}:X\rightarrow N$.

Is this lift a group action?

I don't know the answer.  But for the rest of the post, I will assume it is a group action.  Then note that the projection map is equivariant.  Indeed, writing $g \tilde{f}(n)$ for $\tilde{f}(n,g)$, the fact that $\tilde{f}$ is a lift means $\pi(g\tilde{f}(n)) = g\pi(n)$.
Proposition:  This group action is free.
Proof: Suppose $g$ fixes $n$.  Then $\pi(n) = \pi(g n) = g \pi(n)$ so $g$ fixes $\pi(n)$.  Since the action of $G$ on $M$ is assumed to be free, $n$ is the identity element.$\square$
So we have a free action of $G$ on $S^m\times S^{2n+1}$.
If $G$ contains $(\mathbb{Z}_p)^3$, we are done, by an Annals paper of Adem and Smith which can be accessed here.
Comment: Thank you for your nice idea. I have some problem related to group action on a space which uses spectral sequence. May I contact you through mail. If you can give some idea it will be very helpful.
Comment: @Sneha:  I prefer if you would just ask a question on MSE and the direct my attention to it (say, by commenting here.)  I should also add that I am comfortable with compact connected Lie groups, and less so with finite groups.

What about the action on this space?


Paper Sizes on Android 5.0

Question: On the Android print screen, I only see American paper sizes (Letter, Legal etc.). Is there a way to add or switch to ISO sizes (A4, A5 etc.)?
Comment: For everything else, I have my phone set to American English with metric units.
Answer: The issue is linked to the language you've set in your Android device.
Most probably you set the language as "English (United-States)" and with these settings you only get the American format in you "Paper Size" dropdown.
Therefore, if you want to have the ISO size but still keep your phone in English, you just have to set the language of it in "English (United Kingdom)" and your problem will be solved.
Manipulation :
Settings > Language & input > Language > English (United Kingdom)
Close your apps and try again, in "Paper Size" you should now see the ISO formats.
Answer: I believe it's up to the developer to provide whichever print job attributes they want to support for their app. The ISO, Japanese, and other media sizes are defined in Android's API here and I think it would be up to the developer to use them.
The developer can modify the print dialog for the user via a custom print document adapter that can include a set of default print attributes that are appropriate for what's being printed, and I bet this is where they'd include the ISO sizes.
Unless someone has a better solution (and who knows-- there could simply be an option somewhere to change based on location or language that I just don't know about), your best bet may be to contact the developer of the app you're using and ask them to add the ISO sizing attributes to the list of default print attributes.
Comment: Hm, according to that documentation, the app only sets the default. I did try changing the language of the phone to German and now it shows only ISO sizes, so it seems to be linked directly to the language.
Comment: Interesting.  So does that work as a solution then?  I think the developer can make changes to include those ISO formats as well...
Comment: Well, it will get me through the printing process, though I think it's rediculous that I have to change the language every time I want to. :/

Paper Sizes on Android 5.0


Re adding myself to a Group

Question: I was added to a group but a few months later removed myself from the group via my phone. Punching too many "yes" answers I hit the button that said "prevent anyone from adding you to this group again". So now I want to be in that group again and I don't know how....
Answer: I was in a similar situation once. The way I solved this problem was by asking the group admin to copy and send me the full URL that appears when s/he opens the facebook group page (excluding strings like &view=doc if any). Example: https://www.facebook.com/groups/298596911542862/ After I got the URL and pasted it in my browser, an "Ask to join" button appeared which let me request the group admin to add me in the group. After the admin accepted, I was back in again. I hope this solves your problem.

Re adding myself to a Group


Custom ordering/z-index of icons in Leaflet layer

Question: I have a single layer (points) with a bunch of markers. Each point has an attribute that is regularly updated. I understand by Leaflet sorts the layers such that those lower in latitude are on top for aesthetic reasons, but I would like to reorder them such that in areas of clustering, those with higher values in the attribute are on top. Is there any way to do this without having to split up each marker/point into an individual layer and forcing draw order that way?
Answer: Until somebody finds a better solution, here what I would do ...
As you noticed, leaflet is using pixel position to set zIndex (in Marker.js)
pos = this._map._latLngToNewLayerPoint(this._latlng, opt.zoom, opt.center).round();
this._zIndex = pos.y + this.options.zIndexOffset;

What I suggest is to undo leaflet zIndex using setZIndexOffset()
Say you want to set zIndex = 100, you would do
var pos = map.latLngToLayerPoint(marker.getLatLng()).round();
marker.setZIndexOffset(100 - pos.y);

There is a bit of a glitch: you have to do it every time the map is zoomed :(
Here is a JSFiddle example (comment the code in adjustZindex() to see the difference)
Comment: It does look like Leaflet redoes z-indexes every time it's zoomed anyways (at least playing around with it and checking a marker's z-index on chrome's dev tools). So I think this a perfectly fine option, thanks!
Comment: thanks a lot for sharing the source of the problem. With this knowledge we can also produce alternative workarounds like "zIndexOffset: this.leafletMap.getSize().y*(index+1)x" which works like a charm for me without the on-zoom problem
Answer: The current version of leaflet (as of early 2017) has a zIndexOffset property with default = 0 that can be set to atler the z-index of a marker. http://leafletjs.com/reference.html#marker-zindexoffset
Works fine with zooming and everthing.
Answer: This plugin will do the trick: leaflet.forceZIndex.js
// Force zIndex of Leaflet
(function(global){
  var MarkerMixin = {
    _updateZIndex: function (offset) {
      this._icon.style.zIndex = this.options.forceZIndex ? (this.options.forceZIndex + (this.options.zIndexOffset || 0)) : (this._zIndex + offset);
    },
    setForceZIndex: function(forceZIndex) {
      this.options.forceZIndex = forceZIndex ? forceZIndex : null;
    }
  };
  if (global) global.include(MarkerMixin);
})(L.Marker);

To use:
var aMarker = L.marker([lat, lon], {
    icon: icon,
    title: title,
    forceZIndex: <Value> // This is forceZIndex value
})

forceZIndex declaration will assure that ZIndex will be always set from aMarker.options.forceZIndex
Update it somewhere to re-force ZIndex value
aMarker.setForceZIndex(<New Value>)

Or setForceZIndex(null) to automatical zIndex state:
aMarker.setForceZIndex(null);

By the end of the day, if no forceZIndex option declared, Marker will work with normal behaviour.
Answer: The only solutions currently available are hacky and may break in the future, if they decide to use something else than pos.y or refactor the code heavily.
Here's another simple hacky solution which is easier to use and doesn't require updating the value every time zoom changes. It makes the zIndexOffset reflect the actual zIndex:
L.Marker.prototype.__setPos = L.Marker.prototype._setPos;
L.Marker.prototype._setPos = function ()
{
    L.Marker.prototype.__setPos.apply(this, arguments);
    this._zIndex = this.options.zIndexOffset;
    this._resetZIndex();
};

Then if you want to set zIndex to 100, just use:
marker.setZIndexOffset(100);

Comment: May i just am not an expert, but i don't get where I must put your code to get it Working. May you add a bit more of explonation?
Answer: As stated in https://gis.stackexchange.com/a/115954/35248 the source problem, is that markers have a default zIndex which is build by the y pixel position.
With this knowledge, we can simply overgo this setting the zIndexOffset over the scale of y by using:
let myMarker = Leaflet.marker(coord, {
    zIndexOffset: this.leafletMap.getSize().y*myWishedIndex
})

But here we have also small backdraws:

we have a (strongly) reduced amount of zIndexes so this solution is not usable for high resolution maps with a lot of markers
we may run in to a problem if we use frameworks around the map which allow rescaling of the map-container without reinitiating the map Markers


How do I hold a package in landscape?

Question: I have a held package in Ubuntu, and Landscape recognizes it -- this is great.  But how do I hold a new package from the Landscape UI?
Answer: This is a very common question, and the UI is a bit tricky.  I took some screenshots.
First:

Computers -> Select Computers -> Packages -> Find Package to Hold

You should end up on a screen like this:

Expand the row of the package you want to hold:

Click the little package icon:

Then apply changes to generate the activity that will be sent to the selected computer.
Answer: You need to click the package icon in a search results page to set/unset a hold on that package. There are also some interesting details about holding kernel updates, with the kernel metapackage playing a key role.
See Mastering Package Holds for coverage of this functionality from both Landscape and the shell.

How do I hold a package in landscape?


How can i make a field read only when i selected a picklist value

Question: This article was old one but I want to know step by step. Consider there is picklist status in case object. When I add extra picklist values to status, It should be visible to only particular users. How can I do this. could any one help me out for this issue.
Comment: Which article? You mention an article, but forgot the link.
Comment: Hi peter please find the below link
Comment: https://success.salesforce.com/answers?id=90630000000gilKAAQ
Comment: See my answer below. It is still record types you want to use, but you don't need separate page layouts, just the record type picklist filter. The online help for this is pretty good and you should find the step-by-step you want in there. If you have specific problems making it work, be sure to ask a new question.
Answer: In standard point-and-click configuration the way you allow one user to see some picklist values and another to see different ones is with record types. You then set the record type accessibility to be exclusive for each user's profile.
From the docs:

Record types allow you to offer different business processes, picklist values, and page layouts to different users.
...
Create record types for opportunities to differentiate your regular sales deals from your professional services engagements and offer different picklist values for each.

You can also make it so that for one user with one profile, the record type allows you to assign a different page layout.
This is pretty quick and easy to do, so if you can get these to work for you, go for it. Record types work for most standard objects and all custom objects, but check the considerations section in the link I put above.
If you can make it work, the step-by-step instructions for creating and modifying record types are shown in the links in the docs above.
After record types it is a huge cliff to fall off to implement in Visualforce and Apex. But this requires basically re-implementing your entire edit page in Visualforce, just to write some filter logic into the picklist you want.


Remove DVD drive from device manager - command line

Question: Is there a way through command line to uninstall the entry under DVD/CD-ROM drives in device manager? Somehow someone applied a GPO on our top level domain and it's affected a number of things, I've resolved the other issues but when users try to access their DVD drive it says Access is Denied. Currently as a workaround if we manually go into device manager and right click and remove the DVD drive and reboot the machine it reinstalls it and then when a user tries to access it, it is fine. I'm wondering if there's a command I can push out to all machines to take care of this without having to run through that manual process on each machine.
Answer: There is a tool from Microsoft called devcon.exe that can manipulate devices from the commandline like you're looking for. If you need the x64 version, which I assume you do since this is 2013, you have to jump through a couple of hoops, but it's available and can do what you need it to.


Any suggestions about which questions to answer?

Question: Greetings fellow stackoverflowers.  I'm the only programmer in my company, so I thought stackoverflow might be a "fun" way to get some camaraderie.  Also, I like to be helpful.  I've been using the system actively for about two weeks, asking the occasional question and answering many (rep now=145).  The "problem" I am facing is that I find my mileage varies a lot: sometimes people are very grateful and responsive to help.  But a couple of times I have given very detailed answers, only to be ignored: the questioner just vanishes without up-voting or accepting my answer.  That makes me feel like I'm wasting my time--and I feel "stupid" posting "please accept my answer" comments.  Do y'all have any suggestions about how to discern which questioners are to be taken seriously?  Or any other comments about how best to participate as someone who is more of an answerer than an asker?  I am not so much interested in maximizing my rep as minimizing my frustration, and maximizing my benefit to serious users.
Comment: If you think 5K is high... http://stackoverflow.com/users?tab=reputation&filter=all
Comment: Here are some ['coherent strategies'](http://meta.stackexchange.com/questions/30910/how-do-you-increase-your-stack-overflow-reputation-in-a-realistic-manner/30919) you may find interesting.
Comment: See also: [How do you know when to answer a question?](http://meta.stackexchange.com/questions/73758/how-do-you-know-when-to-answer-a-question)
Answer: People who put effort into their questions will generally take care of them. Ways to recognize:

proper spelling
well formatted
includes all the info you need but not more
tends to be longer

If you just want to get reputation there are strategies for that (search here on meta) but they're not too fun. Usually it involves answering easy questions in popular tags very quickly.
Comment: Good point: there's more to life than reputation, just as there's more to life than money. :-)
Answer: Honestly, you kind of answered your own question:

Also, I like to be helpful.

Sometimes very good answers go without upvotes or acceptance.  Conversely, sometimes a lot of rep is dished out when half a dozen people quickly give a simple answer to a simple question.  There's really no rhyme or reason.
I guess if I were to define my "strategy" it would be that I just like to answer questions.  Sure, getting rep is fun.  But the real value is in the activity itself.  By answering a question, I take a step to refresh (or even further) my own knowledge of a subject.  Often I get just as much out of it as the person asking the question.
Though I don't have any stellar answers by measure of upvoting, I like to think I've written some good content on Stack Overflow and have contributed to the community.  And, in doing so, have contributed to my own career growth.  (I came close to the Unsung Hero badge once, but I think I'm pretty far now.)  Maybe I just tend to answer less popular questions?  I don't know.  But the main thing is to just keep doing it.  Answer stuff that's interesting to you.  As I said, it helps the person answering as much as it helps the person asking.
Keep in mind also that these questions and answers are saved for posterity.  Sure, maybe the person who asked the question has gone on their merry way without any gratitude.  But that question and its answer (an answer you provided) is now there for all to see.  It's not uncommon for people to search for things on Google and find what they need on Stack Overflow.  That answer may help someone else days, weeks, months down the road.
If you enjoy contributing to the community, then by all means keep doing so.  There's no shortage of appreciation, even if any given question doesn't indicate as much.  If nobody else has said it to you yet... Thank you.  Thank you for contributing.
Comment: Thanks for your answer. Very encouraging. After one month, I have reached the conclusion that the best "strategy" for me to pursue, considering my own temperament, is simply to ignore rep. I just don't like nagging people about acceptance, or "rushing" to answer quickly to get the jump on other potential answerers. I just monitor my tags of interest via rss, and answer when spirit moves. We'll see where that takes me. :-)
Comment: I just got the Unsung Hero badge.  I guess you had me pegged. :-)
Answer: The bottom line seems to me how much satisfaction it gives you. Speaking for myself, I use the Stack Overflow sites chiefly to ask questions. Merely posing them (usually through the blinding anguish of coding-block) forces me to clarify my thoughts and sometimes that gets me to a good enough answer on my own. But the sites have been an enormous help (the tex." subsite especially, in my case). The answers by experienced hands to my direct questions and the larger resource of accumulated answers are both terribly, terribly useful. So before answering, I ask myself: "Can I contribute in a worthy way to that resource?"
Now three other matters you've raised:

Whether to write answers? I find I only answer questions that interest me, and those tend to be general (as here) or within my very technical and narrow fields of expertise. In the interest of sharing, I've sometimes looked for questions on the latter, but there aren't too many.
Who deserves an answer? You ask "how to discern which questioners are to be taken seriously": I think the primary sign is clarity. But there is also merit in clarifying an unclear question.
Coveting reputation points. The rep-quatloos are of limited interest to me. I grant you, it's nice to get useful privileges on the site for having them, but apart from that they're unimportant. And I'll wager you have better things to do with your time than gather imaginary points, too.


OFDM , cyclic prefix

Question: If OFDM did not have a cyclic prefix or zero padding would have equalization of the individual sub-carriers been possible in the frequency domain?
Answer: Since the goal of the CP is to remove or prevent ISI and ISCI, you would need a form of DFE with feedback from several neighbouring subcarriers of the previous symbol.
That can be expanded into a multi-symbol MMSE with feedback.
So yes it's possible.
Comment: @alikhalil, there's a lot in your comment that depends on semantics, so I won't answer it literally, but maybe this helps: the CP enables ISI-free matched filtering for each sub-carrier. If there were no or negligible ISI, you wouldn't need a CP. The FEQ generally takes care of gain and phase rotation, not interference. The cyclic-convolution is not a requirement for FEQ, but a way to take advantage of a CP where needed.
Comment: Inspite of the fact that frequency domain equalization relies on cyclic convolution which is ensured by the cyclic prefix?
Comment: Interesting,  so does that mean that the one tab channel inversion or  zero forcing FEQ per sub-carrier  can be done with a formula corresponding to linear convolution?
Comment: @alikhalil FFT + FEQ, for one subcarrier, = I&D matched filter + inversion. The linear convolution happens in the I&D matched filter. Cycl. convolution is just a fancy name for shifting the I&D window into the CP.
Answer: To extend on P2000's answer:

So yes it's possible.

and he even tells you what kind of equalizer might be worth looking into.
However, in almost all cases (there's a few sparse signal speciality channels out there), you do OFDM to avoid having such a complicated equalizer.
So, when using a long (and thus, complex) equalizer, why still use OFDM? If you've done your equalizer right, your channel is flat afterwards. If you do (classical) OFDM right, you don't need the equalizer.
There's a few corner cases (and those are very real) that could benefit from a combination, but seriously, reading your questions, you're not designing one of these. You're trying to understand what OFDM is about. And OFDM is mostly about not needing an equalizer.
Comment: Agreed. I was never tempted to design an OFDM system without CP. I even avoid channel shortening TDEQ where possible, and pay a bit extra overhead for the CP.

OFDM , cyclic prefix


Enable binary logging MySQL

Question: How exactly can I enable binary logging on MySQL server. Conf.file? Server Startup Parameter? If possible give me an example.
Answer: You can start the server with the <code>--log-bin</code> option.
The MySQL documentation will cover the details if you need more information.
Comment: I examine the documentation and I try different combinations of --log-bin with mysqld but I am not sure I proceed. So could someone please give an example
Comment: Executing mysqld with the --log-bin option should be enough. Is that not working?
Comment: Here is what I did.
I stop the service with net stop mysql
Then on command line I type mysqld --log-bin
Then nothing happened. Just cursor goes to new line and waits for something.
Then I try to start to service from MySQL Administrator but it gave me error. "Pluging FEDERATED is disabled' and service did not start.
As a summary I think it is not working.

Enable binary logging MySQL


Is installing malicious add-ons via drive-by attacks the only way to add new toolbars to browsers?

Question: Some attacks on web browsers consist in installing new toolbars on the browser of the user (which is just a small aspect of browser-hijacking. My question: is installing (malicious) add-ons (via drive-by download attack) the only way to add new toolbars to browsers such as firefox, internet explorer and google chrome ?
Comment: Your title and body do not ask the same question.
Comment: @Stephane Thank you for your remark, but for me i think i expressed the same idea (English is not my mother language, i learnt it by myself)
Comment: I assume you mean aside from the user installing the toolbar him/herself.
Answer: Simple answer: No
Any attack which compromises your machine could add toolbars to your browser.
However, this is a very different type of attack, and to be honest, any attacker who has full control of your machine is likely to do worse things, like making you part of a botnet, adding keystroke loggers etc.
Comment: Could you list me one attack that leads to install any given random toolbar to a browser ?
Comment: Irrelevant. You asked if it is the only way, and the answer is most definitely no. In my answer I explain why you wouldn't bother, as adding a toolbar is low value compared to total compromise.


Eating and Needing the Bathroom

Question: Is one allowed to eat if he needs to use the bathroom?  Does it make a difference if he made a Bracha yet?
Answer: The needing of the bathroom is divided in categories by Halacha. In Shulchan Aruch Orach Chayim siman 92, you can find all those halachot.

If one needs to bathroom and cannot contain himself the time needed to walk a distance of one parsa (72 minuts), he is not allowed to continue the  prayer , blessings  or learn Tora. But if one feels a need before starting, he cannot start to pray.

If one is able to contain himself for this amount of time, it's right to continue the Tefila and to go in bathroom afterward,  it's better than interrupt the prayer. (Poskim in name of Teshuvot Harashba).

Concerning eating, if one needs to use the bathroom, I did not find anything in Hilchot Derech Erets Baseuda (OC 168-171).

See  Seder  Hayom (סדר בדיקת עצמו קודם האכילה ונטילת ידיים"") that says to check oneself before the meal and which who reports the Gemara in Masechet Shabbat 41a,

One who has a call of nature yet eats is like an oven which is heated up on top of its ashes, and that is the beginning of perspiration odour.

Note regarding prayer. Further I learned in Shulchan Aruch OC 92 in Magen Avraham sk 1 that (see Bait Chadash on Tur in the same siman before the reading of Magen Avraham) accordind to the Rif (and Rambam) if he can refrain himself up to the end of prayer, and the time of tefila is finishing, he can pray lechatechilla. The MA stated as the Rif. So, for birkat hamazon this is obviously the same if the time is ending.
Comment: Is that last quote from Seder Hayom?
Comment: @anonymous the quote is from the gemara shabbat 41b
Comment: Does this mean that one should not bentch until he goes to the bathroom, even if he can wait more than 72 minutes?
Comment: @anonymous if I understand right yes.
Comment: Do these Halachos also apply to learning Torah?
Comment: @anonymous yes,  see again the answer I have made it in bold characters

Eating and Needing the Bathroom


How do I break formation

Question: After pressing CTRL + a number to put several units in a formation, how do I get an individual unit out of that formation again?
Answer: I am not aware of an immediate one click/input way to do this, if there is one, I'd like to know it as well. I always just do it by:
1) Press the number to select the formation
2) SHIFT + click on unit you want to remove
3) CTRL + formation number again
Answer: By pressing the G key the unit is immediately ungrouped. In addition this is useful for getting the last unit out of a group.
Comment: Ooh, that's good to know, I wonder if it also works in other strategy games, I'll have to try it out later.

How do I break formation


How to import MXF recordings on FCP?

Question: I have a problem with my MXF files when I was about to import them into FCP7. It refuse to work at all. I tried couple of plug-ins and applications but nothing worked. How could I solve this problem?
Answer: FCP does not support the MXF wrapper type by default - only MOV files will import. There's a third-party plugin which provides MXF support - MXF4MAC - but I'm not sure which version you would need, and it may depend on the video format of your files.
However, if you don't have that plugin, you (probably*) do not need to transcode your input files - just re-wrapping them should suffice.
Background: MOV and MXF are both 'wrapper' formats, rather than video/audio formats as such. They can be 'unwrapped', which would leave you with a number of separate video and audio files, which FCP would be able to import.
Any transcode (re-encoding the video/audio essence) will cause some degradation in the image/sound quality, so should be avoided. It's also hard work to transcode video, so normally takes a long time. A re-wrap can use the same video/audio essence, but merely package it different. Think of it on the same lines as taking a RAR archive, extracting the files from it, and then putting them into a new ZIP archive - the files are completely unchanged, but the format they're delivered in has changed.
One (free) way to rewrap files would be to use command-line tool 'ffmpeg' with the 'copy' pseudo-codec specified. Without knowledge of your input file, OS version in use (the tool is included on most Linux distributions, but is available for other platforms - take your pick here), and ffmpeg version available, I can't give a precise command-line, but it will along the lines of:
ffmpeg -i MyFile.mov -c:a copy -c:v copy MyFile.mxf
Hope this helps - if you need any further details, we'd need some more information about the input file - the easiest way to get this information is probably MediaInfo (again, free, available on all platforms), which can give all the internal video/audio formats in an easy to access way.
* depending on the actual video format within the MXF file - FCP supports most video formats.

How to import MXF recordings on FCP?


how can i make a .jar executable + I'm on ubuntu Studio 13.10

Question: how do i make my mine craft executable.Ive tried running it in terminal and many forum answers. when I get to mine craft and i go into properties. and i look for allow to execute it doesn't appear i have no option to make it executable. I suppose i must
do it in the terminal yet i do not have the extra time for reading this whole thing Java -h
please help me get a quick easy or hard answer. <----- Original
Ive done what someone in the comments said and i got my computer to recognize it as a .jar
but it still won't  work i tried a very very helpful forum and learned some stuff but still nothing please help me guys i am stuck
Comment: Possible duplicate of [How to create a shortcut for Minecraft like in Windows?](http://askubuntu.com/questions/566153/how-to-create-a-shortcut-for-minecraft-like-in-windows)
Comment: a simple web search for 'minecraft ubuntu' gives several results with step by step instructions. If you want to use this site as a resource you must be willing to do some work and research on your own first.
Comment: look i have searched every corner of the inter net i have even used unrelated forum topics and changed some of the terminal properties i don't want to search any more
Answer: I always run it via. command line:
java -jar /path/to/minecraft.jar

You could create a short script called start_minecraft.sh with the following contents:
#/bin/bash
/usr/bin/java -jar /path/to/minecraft.jar

and chmod +x the file.
Comment: i think you're the best person ever for answering but it wont let me it says unable to access /path/to/minecraft.jar help
Comment: You'll have to figure out where the minecraft.jar file is stored and update the path in my example. This would be the file you downloaded from Mojang (with a similar, but possibly not the exact same name).
Comment: can you help me find this name it isn,t neccesarily easy to find
Answer: Try creating a Launcher using the commands, it's easy and simple!
Right-Click
Click Create launcher and fill in the command and directory, also change "minecraft.jar" to "Minecraft.jar" as that's what it's called for 1.9
Comment: Welcome to Ask Ubuntu! I recommend [edit]ing this answer to expand it with specific details about how to do this. (See also [How do I write a good answer?](/help/how-to-answer) for general advice about what sorts of answers are considered most valuable on Ask Ubuntu.)


Do bonuses on one weapon affect other weapons in Nier Automata?

Question: For instance, when I have a weapon giving me "critical+" in the heavy attack slot; is it applied to my weapon in the light attack slot when I strike only with light attacks ?
Answer: Throughout my research, it has noted a few times (here, here) that only non-combat specific weapons effect will apply regardless of which weapon is active. This is limited to two effects:

Robot Discount
Discount

Fortunately, these two effects do stack! So you'll get major discounts at some shots.
For combat specific weapon effects, this reddit has researched this a little bit. The users on that community have stated that they did not see a difference in weapon speed when they have two weapons with Attack Speed Up equipped as opposed to only having one weapon equipped.
Comment: what about cypress stick and chests effect?
Comment: @arghtype I didn't find anything on that one. Lemme see what I can find
Answer: The endurance effect of type 3 weapons and the beast weapons also works passively. This is easily seen by equipping them, pausing the game and checking your defense stat increase on the status menu.


Positive and Negative works

Question: When we displace a body from ground to a height we do positive work against gravity and gravity too does work but negative. Do they cancel each other? Or does the external force do extra work?
Answer: Since the body is at rest at the beginning and at the end of the motion, the total change in kinetic energy is zero. This means that the total work done on the body is zero, which means that the work done by the lifting force and gravity canceled each other out. In other words,
$$\Delta K = W$$
or the change in kinetic energy of a body is equal to the work done on it.
Now, sometimes it is useful to separate work done by conservative forces (like gravity) and non-conservative forces, so
\begin{align}
\Delta K &= W_{cons} + W_{non-cons} \\
         &= F_{cons}\Delta x + W_{non-cons} \\
         &= -\Delta U + W_{non-cons}
\end{align}
where the subscript $cons$ means conservative and $non-cons$ means non-conservative, with $F_{cons}$ referring to the sum of all conservative forces. These forces act through a distance $\Delta x$. The symbol $\Delta U$ refers to a change in potential energy, which is equal to the negative of the work done by a conservative force. If we define the total mechanical energy of a body as the sum of kinetic and potential energy
$$E = K + U$$
then we can write
$$\Delta E = \Delta K + \Delta U = W_{non-cons}$$
So, the total change in mechanical energy of a body is equal to the work done by non-conservative forces. If all of the forces on a body are conservative, then total mechanical energy is conserved.
Comment: A better term for K+U is mechanical energy rather than total energy.
Comment: @BillN True. Fixed.
Comment: @MarkH A body alone cannot be a store of gravitational potential energy. It the body and the Earth whose gravitational potential energy changes.
Answer: The answer given by Mark is perfectly correct but I do have an another point of view, u can consider the whole earth and the body that is taken up, as one system. Hence in this system the gravity is an internal force. Therefore here this work done by gravity should be compensated by the work done by us. Therefore we do positive work and they do negative work, hence canceling each other's work.

Positive and Negative works


Drupal on OSX 10.7 (Apache,PHP) - Clean URLs cannot be enabled

Question: I'm using the following:

OSX 10.7 Lion
PHP Version 5.3.26
Drupal 7.23
Apache/2.2.24

I'm trying to get 'Clean URLs' to work, but no matter what I do, I always see this message: "Clean URLs cannot be enabled".
I have mod_rewrite loaded in Apache, it appears on the 'Loaded Modules' list when viewing phpinfo().
I have AllowOverride set to 'All' in my httpd.conf file, under both "Directory /" and "Directory /Library/WebServer/Documents". This seems to be allowing me to use .htaccess files just fine.
I have a .htaccess file in the root of the website I'm working on, and it is being used because if i put something invalid in there, I get an error in the browser.
Here is the contents of my .htaccess file:
<code>Options +FollowSymLinks
RewriteEngine on
RewriteBase /
RewriteCond %{REQUEST_FILENAME} !-f
RewriteCond %{REQUEST_FILENAME} !-d
RewriteRule ^(.*)$ /index.php?q=$1 [L,QSA]
</code>
What am I doing wrong..? have I missed something..?
What else do I need to do to enable 'Clean URLs' in Drupal..?
Thanks!
Answer: Are you using the default Drupal 7 htaccess file?
https://github.com/drupal/drupal/blob/7.x/.htaccess
That could be it.
Otherwise you could take a look here https://www.drupal.org/node/566880


Are fan games legal?

Question: "Fan games" are video games made by fans using a popular intellectual property, without license. Are they protected by "fair use" or are the IP holders just allowing it?
Answer: Short answer
Fan games are not protected by "fair use".
The IP holders probably have a solid copyright and trademark infringement case.
The IP holders may have simply not taken legal action to enforce their rights yet. But keep in mind that this doesn't prevent them from doing so within the statute of limitations for doing so. In many countries, the statute of limitations is three years for copyright and trademark infringement from discovery of the infringement.
Why might IP holders refrain from bringing lawsuits?
There are various reasons that the IP holders may not have sued. More than one of these reasons may be present at the same time and influence the ultimate decision to not seek to legally shut down fan games.
The IP holder may not know there is a serious problem
The IP holder may not be aware of the activity.
The ability to make good IP does not always overlap with having good monitoring of gray area fan activity on the Internet. Corporate IP holders may have simply not tasked employees with monitoring this activity, and individual IP creators may be do busy creating new work to be bothered with what their fans are doing.
Even if the IP holder may have some dim sense that some sort of unlicensed fan games are out there, they may not have a sense of how big they are and whether it is something worth pursuing legally.
These lawsuits are hard to bring and may not result in much money collected if they are won
Identifying defendants to name in an infringement lawsuit based upon conduct by pseudonymous people on the Internet is hard, sometimes fails to identify those involved, and takes expensive investigative work to event attempt. '
Fans are often minors or young people who have few assets from which a judgment could be collected if the IP holders won, so the lawsuits may not be economic to bring (this also makes it hard to hire IP lawyers to sue on a contingent fee basis). The IP holder might not be able to afford a lawyer on a non-contingent fee basis for reasons unrelated to the merits of the lawsuit.
The economics and P.R. of allowing fan games may make sense
The IP holder may have decided that shutting down the games is a bad public relations move vis-a-vis the fan base. The grassroots backlash that legal efforts to enforce one's rights in contexts like this can create is called the Streisand effect.
Similarly, the IP holder may have decided that the games aren't hurting their royalty income (or might even be enhancing their royalty income by generating buzz). There have been some attempts by professional economists to assess the economic impact of fan fiction, fan games, and unlicensed derivative works of copyrighted and trademarked material. Some of these studies have found it actually helps sales of the primary work that fans buy, and others have found that these works have little impact on sales of the primary work.
In something of the same vein, sometimes IP holders refrain from taking action out of professional respect for the fan works (some IP creators got started in fan works), or because the fan works stoke their ego in a way completely divorced from their corporate minions whose praise for their work is tainted by money and may not be as sincere.
IP holders could be waiting to enforce their rights in a global sweep
The IP holders could issue "takedown notices" to firms that host these games, to stop them from being used, even if the names of the people who make them and use them can't be determined.
But the IP holder and other similarly situated IP holders may be biding time to issue takedown notices to and sue everyone in the fan game marketplace at once in one big sweep in order to avoid "wack-a-mole" fan game operator tactics where one site is shut down and another other already in existence or newly created thrives resulting in little net benefit to shutting it down. There have been several of these sweeps by coalitions of IP holders since fan fiction and fan games on the Internet became a thing.
Often sweeps like this are done only in advance of a corporate decision of the IP holder to expand into a new product line, like licensed fan games, and intellectual property right enforcement against the derivative works isn't pursued until the IP holder had a plan to make money from and fill the void that the unlicensed fan made derivative works were filling.
For example, unauthorized fan translations of manga and anime used to be a big thing on the Internet before those translations were available commercially, and the IP holders mostly just let those infringements go because the translations didn't impair sales of the primary works in their original language. But when IP holders secured commercial contracts to translate, subscript and dub their works for sale abroad, the IP holders did a sweep of enforcement to shut down the fan translation sites (possibly as a requirement of the deal with the firms that were doing the translations on a paid and licensed basis).
The IP holders could be covertly licensing the games as a marketing ploy
It is also conceivable, although unlikely, that some fan games are licensed and that this fact simply hasn't been made public. It isn't entirely unheard of for the marketing arm of an IP holder to use the "guerrilla marketing" tactic of setting up a fan game or fan website that is secretly sponsored by the IP holder itself, while creating the appearance that it is illicit activity, in order to entice young fans who find petty transgressions against the law thrilling in order to create interest in their product.
A fictional example of this kind of guerrilla marketing tactic can be found at this Webtoons link (this is a pretty big spoiler in this fictional work, so don't click on the link unless you are willing to be spoiled).

Are fan games legal?