What is this photo FX and what app can do this?

Question: Saw this on a twitter's celebrity. How do you achieve this? What phone app could do this? Many thanks
Comment: If you search for "anaglyph effect", you'll likely find a phone app that can do it. For example, [the Google Play store](https://play.google.com/store/search?q=anaglyph%20effect&c=apps&hl=en≷=US) shows several. Just a note, this probably isn't a real 3D anaglyph, but a fake anaglyph effect.
Answer: It has 3 copies of the image in separate layers, moved horizontally a half of the face width apart and colorized differently.

They are combined with layer blending mode ADD - the top layer (=Original) which contains in the middle placed non-colorized version, has it.
In the left there's the red version and in the right there's the cyan version. One gets them by applying Image > Adjust > Hue/Saturation > Colorize in Photoshop:

Alternative: Colorization to red is possible also by closing the green and blue channels of an RGB image. Coloring to cyan happens when one closes the red channel. Photoshop's Hue/Saturation > Colorize gives more control.
The background of the faces are removed and finally a new black background is inserted.
Layer blending mode ADD causes easily clipping. Both colorized versions have a masked curves layer (with the next image layer only -switch ON) to reduce their brightnesses on the areas below the original.
The half-cyan slice on the middle version does not look right, it should be neutral. One can reduce the saturation of the cyan version below the original with another adjustment layer:

In theory this could be made as well in GIMP, Krita, Paint.NET etc... but making it is handy only in programs which have adjustment layers and layer masks. Krita and web service Photopea are such ones available for free.


What is the difference between the processes listed by `ps` and `ps -A`?

Question: What is the difference between the processes listed by ps and ps -A ?
Answer: without the -A, ps will only print the processes belonging to the current session. Think of it like "absolutely everything". On a related note -a does the same thing, but restricting it to the session-owner (username).
Comment: Why does `ps -a` list different processes than `ps x` for GNU ps?  I would think both would restrict it to the current user.
Comment: According to the man page, `ps -a` does not include [session leaders](https://unix.stackexchange.com/questions/18166/what-are-session-leaders-in-ps), whereas `ps x` does.
Answer: The GNU ps command suffers from a severe case of multiple personality disorder.  So it is no wonder that its manual page is confusing.  Perhaps a look at the BSD manuals might help.  After all, this question is tagged unix.
The operation of BSD ps is fairly straightforward when one bears two things in mind:

Processes are selected for display using two basic filters.  Those filters are on by default and command line options turn them off.  Plain unadorned ps is thus filtered through both.
BSD syntax goes back a long way.  Although the modern BSDs use getopt and the convention of options prefixed by a minus sign, the options and behaviour relevant here are much the same as from 30 years ago.

That behaviour is this:

The -a (historically a) option turns off all "selector" filtering.

The various other command line options specify selectors: -U selects by UID, -t by controlling terminal name, -p selects by process ID, and so forth.  All of these are bypassed by -a.
If no selectors are explicitly supplied, the default selector is to only display processes running with the same effective UID as the user who invoked ps.  This default selector is the historic filter that this option turns off.

The -x (historically x) option turns off the restriction that ps only display processes that have a controlling terminal.

Historically, BSD ps did not have an A option.  But modern BSDs implement an -A option, also usable as A, for (a degree of) compatibility with the (later) Single UNIX Specification.  -A is simply the same as using both -a and -x: it turns both restrictions off, leaving one with all processes, unfiltered.
OpenBSD and NetBSD document the -A option, in NetBSD's case explicitly as -a -x.  FreeBSD does not, but a comment in the source code states that it is there as an intentionally undocumented SUSv5 compatibility option.
Further reading

ps 1983-04-13.  4.2BSD Manual pages.
ps OpenBSD Manual pages.
ps 2009-10-22.  NetBSD Manual pages.
ps 2014-08-07.  FreeBSD Manual pages.
Comment: To get the BSD behavior with GNU ps (the default ps in ubuntu), leave off the - in the arguments: `ps a`, `ps x`, `ps ax`, or `ps -A` (that last one you must leave the - on).


Image that looks different on black background and on white

Question:
If you put this image on a black background you see something different than on a white background. Does anyone have a idea how it works?
Comment: It's not a transparent PNG and placing the image over different colored backgrounds does nothing here. There was a trick to alter an image when the cursor highlighted it - only worked in IE. But I can't find info on that now.
Comment: I remember getting insta banned from a forum in my youth for making my avatar have a hidden nekkid lady in it when highlighted. Ah to be young again.
Comment: I might've seen image like this before cause it seems a bit familiar, but I can't remember. One thing I noticed is that the image seems to render differently in different browsers ( Safari showed the other image for me ). Plus saving it to my desktop, I can immediately see the other image and when I open it up in photoshop and adjust the curves I can bring out the other one as well. http://i.stack.imgur.com/rhev7.jpg
Comment: it has to do with how browsers change the colors when the image is selected.
Comment: So the answer is the browser and (possibly) your gamma settings? My Safari showed a worse version of the img above from Joonas where the interlacing was much more apparent. The FF version shows the Batman but more blurred than Joonas's and darker. I'd post a screenshot but I'm lazier than Joonas ;)
Comment: Dithered overlay of one image over another. Every odd pixel is transparent and that image is overlaid over you original as Joonas mentioned. It was a "trend" a few years back which only worked in IE as internet explorer had placed a dithered  mask over the image. That hid one of the images and lets the other one shine through. Your brain fills then in the missing pixels :)  Here's a tutorial http://www.n-sane.net/effects/hidden-image/index.php
Comment: @Ryan that's Mondays...
Answer: I'm going to ignore the actual picture in your question, and just answer the question implied in the title: How to create a PNG image the looks different on a black background than on a white one?
Specifically, the method I'll describe will allow you to combine any two grayscale images A and B into one PNG file C, so that C looks like A on a black background and like B on a white background, with one condition: every pixel of A must be equal to or darker than the corresponding pixel of B.  (If the original images don't quite satisfy that condition, the method I'll describe will force them to, leading to some "ghost image" leakage from one to the other.  Depending on the images, this may or may not be visually apparent.)
The trick, of course, is to make use of the PNG alpha (= opacity) channel.  Specifically, given a pixel with color c and opacity α, the pixel will have the apparent color a = αc on a black background and b = αc + (1−α) on a white background.  (Here, color 0.0 represents black and 1.0 represents white.)  Turning this around, given target pixel colors a and b, we can solve these equations for c and α:

1 − α = b − a
⇒   α = 1 − (b − a) = 1 − b + a,   and
c = a / α

(This, of course, assumes than a ≤ b; otherwise the transparency 1 − α would be negative, which the PNG format, alas, does not support.)
OK, so how can we do these calculations in a graphics editor like the GIMP or Photoshop?  Well, we really just need two operations — subtraction and division — and, as it happens, GIMP (and, if I'm not completely mistaken, Photoshop too) has both of these implemented as layer composition modes.
In fact, the process is essentially the same as the one I described in this answer for reconstructing the alpha channel of an image from versions superimposed on black and white backgrounds; the only difference is that we start from two arbitrary images — for example, these two cat photos from Wikimedia Commons:

I've cropped these images, converted them to grayscale and adjusted their color levels so that they almost — but not quite, just to show what happens — satisfy the relative lightness condition a ≤ b.  (Tip: the "lighten only" and/or "darken only" layer modes are handy for checking this.)
Now, I'll do the following steps:

Open both images as layers of a single picture.
Make a copy of the black cat layer, change its layer mode to "subtract" and merge it down into the white cat layer.  The result will be the transparency of the final image.
Make a copy of the transparency layer we just created, invert it, change its layer mode to "divide" and merge it down to the remaining copy of the black cat layer.  This will be the luminance channel of the final image.
Move the transparency layer to top and make it visible, and use the channels dialog to transfer it to the selection.  Then select the luminance layer and cut the selection out of it.  Finally, delete the transparency layer.
Save the remaining layer as PNG.  It should look like this:

This page has a white background, so you should see the white kitten image (but with some faint "ghosts" of the other image where the a ≤ b condition wasn't satisfied), but you can download the image above and confirm that it looks quite different on a black background.  Or just compare these flattened versions on black, white and purple backgrounds respectively:

The last version shows what's going on: the purple areas are where the PNG above is transparent, allowing the background color to show through.  Conversely, the black and white areas are where the PNG image is opaque, making the color independent of the background.
Answer: OK, I finally got around to looking at the example image, and the trick is pretty simple: gamma correction.
As others have noted, the image is composed of two interleaved pictures: out of every 2 × 2 pixel block, three pixels have RGB values in the range 0 to 210, and show the "red tabby kitten on bed" image, while one pixel has RGB values in the range 214 to 255 and shows a very heavily lightened version of the "batman cat" image.
Here's a small section of the image, scaled up by a factor of 8, with no gamma correction applied.  This is what you'll see if you open the image in a program that doesn't understand PNG gamma correction and zoom in:

As you can see, the light pixels (which contain the "batman cat" image) just pretty much look white.  At normal magnification, they blend in with the other image, which has much higher contrast, and just make it look somewhat lighter.
However, the PNG image also contains a gAMA chunk, which specifies a file gamma value of 0.023.  This is an extremely small gamma value; more typical values would be between 1.0 and 0.45.  When opened in a program that supports PNG gamma correction, this causes the image to be darkened so much that the "kitten on bed" image literally becomes invisible — all its colors are mapped to black — while the colors of the "batman cat" image are mapped to more ordinary values.
For example, here's the same zoomed-in section of the image after gamma correction:

So, to conclude, the appearance of this image does not depend on the color of the background.  Rather, it depends on whether the program you use to view it supports PNG gamma correction (and is willing to apply such an extreme gamma value) or not.
By the way, the gamma correction value used in the image seems a little too extreme: at least on my screen, the "batman cat" image shows up a lot more nicely if you double the gamma.
Answer: The image is two images interlaced. Interlacing is, very basically, where two images are displayed simultaneously by showing a single line (or pixel) of each one in an alternating pattern.
Usually one finds this in TV and video broadcasting since the frame rates mask the interlacing effect.
If one were to take this image and use a deinterlace filter ("even fields"), the batman image would be gone. Also, the batman cat image becomes "primary" when zoomed to 50% or when using the scale feature of photoshop, probably becasue of interpolation/resampling. Curiously, when you commit the scaling change and photoshop re-renders at full quality (it uses a fast render method when manipulating transforms), the other cat becomes "primary" again. The same scaling to 25% eliminates the batman cat as well.
As far as "why" the display changes  depending on software, that is most probably a function of the methods used by various software for rendering images. I checked the header for the file and it looks to be properly formed, but there may be some trick or hack implemented in one of the optional sections (it has 3 optional chunks IIRC) which exploits a rendering bug. Personally, I think it is just the method used interpolate based on the rescaling stuff I mentioned above. Also, there is a gamma section in the header and on my display the image is REALLY dark, so it may just exploit the fact that some pixels are lighter and the rest blend in with a dark background.
Comment: The image appears dark here to me as well, however, saving the file I can only see the other cat and a faint "haze" of the Batman cat as if it's a semi-transparent overlay. Zooming in you can see the interlaced pixels.
Comment: try zooming out. The software used has an impact on the effect as well.
Answer: I ran a test on the PNG you linked to and got the same picture both times so I can't see what you are asking about.
PNGs have an alpha channel which allows the colors to have varying degrees of opacity. This might be the cause of the change with the different colored backgrounds.


I have a 2000 BMW 540i. I have been getting "check coolant level" from the computer recently.

Question: Twice in a week the radiator has been low on fluid. I have added water to it to refill. I was told that I should use coolant/water mix instead of just water as it is most likely evaporating from the radiator. Could someone explain the procedure to bleed any trapped air from the radiator? I have never had to deal with this issue before. Thanks for any help with this!
Answer: this is a serious condition which should not be neglected.
I have been driving a 320I with a similar condition. The solution was to change the radiator and all hoses connected to it.
If you decide to only change a leaking hose, consequences will lead to radiator popping while driving on a highway.  BMW recommended to change the whole thing(radiator,and all hoses) but we did not listen to them and this resulted I n the radiator being blown while driving at high speed on a highway.


MSSQL backups - Can I write a script to check if the DB has been modified before backing up?

Question: Question is in the title really,
I want to know if I can check if the DB has been modified at all since the last backup, as my company change the data very sporadically and we end up with lots of backups all exactly the same.
Answer: You could query the last change to any index.  This includes changes to tables, since tables are stored as a special form of index (clustered or heap):
select  max(last_user_update)
from    sys.dm_db_index_usage_stats ius
join    sys.databases db
on      ius.database_id = db.database_id
where   db.name = 'YourDbName'

These statistics are reset after a server restart.  You can find the server restart time with:
select  create_date
from    sys.databases
where   database_id = 2 -- tempdb

Answer: The easiest way is to keep track of this fact within your schema using a "Last Updated" field.  You can have a "global" value (which would be useful in your case where nothing may have changed, but if anything has you should back it up), and/or a column in each table. You can manage this field with triggers, so none of your apps have to care that the field exists, but you could get false positives with this if your apps ever perform updates to records without actually changing the data.
Then it's simple; set up the backup as a scheduled job, predicated on a query checking any and all "Last Updated" fields to see if they're newer than the last backup date (which you can also put somewhere in your schema for quick reference).


What is faster? Pure rolling or rolling with slipping?

Question: So we know that a ball will slide down an incline when there is no frictional force. Once you say there is static friction it causes a angular momentum which in turn causes a torque and makes the ball roll. Lets say the ball number one does pure rolling, where the static friction is large enough to keep it rolling pure all the way down. Now lets say ball 2 rolls as well, but slips some ways down the incline. What would reach the ground first? The ball rolling and slipping, or the ball purely rolling? If so, why?
Comment: What do you think and what's your reasoning? It's always better, on conceptual questions, to lay out what thinking you have done. As the question stands now, it's a standard homework-style, classroom-style question.
Comment: I think that since the frictional force during rolling is less when kinetic friction appears, which causes a ball thats only rolling to have higher acceleration, thus ball 1 reaches the ground first. Isnt the velocity around the center of mass is lower than the rotational velocity?which then in turn makes the whole system move without rotational velocity, which again makes it move slower when it slips and rolls? I feel like that is logical isnt it? Like when you roll a pool ball it will travel a longer distance, unlike if you tap it a bit or put some more friction on the table
Comment: It is not a homework question, im just trying to understand even if the kinetic friction coefficient is lower, it still will move slower
Comment: @ROLLIN40SCRIPS If it's up to the magicians of this site, whatever you do it's always wrong.
Comment: However, note that if you explain your thoughts in too much detail then the question is classified as a "check my work" question and get closed anyway.
Comment: @gandalf61 OK. That really WASN'T  useful...
Comment: @Gert IMO suggesting that the poster can turn a homework question into a non-homework question by explaining their reasoning is misleading. If this is a homework question then explaining reasoning does not make it less of a homework question.
Comment: @gandalf61 Yeah, keep changing your story. Yaaawwwn.
Comment: Bruh, anyone wanna help me just understand basic physics. It's not that deep guys… Like im just asking a question I just started learning about rotation, i dont even go to school now… Im preparing myself for next year
Answer: The slip scenario caused a faster fall.
Less energy is converted into rotational kinetic energy if there is slipping, and thus (per energy conservation) more is converted into translational kinetic energy, which is a measure of the falling speed.
$$U=K_\mathrm{rot} +K_\mathrm{trans}.$$
Comment: @JohnDarby Sure. I'm assuming an ideal scenario.
Comment: Yes.  Just wanted to clarify that for a given coeff. friction on the plane the body either rolls or slips, so the conditions on the plane for the comparison of rolling or slipping change.
Comment: This assumes a perfectly rigid body; that is, no change in internal energy (heating).  Also for an incline with a specific slope and coefficient of friction the body either rolls or slips.
Comment: Also for an inclined plane with a given coeff. friction the body will roll or it will slip.  You cannot compare rolling vs. slipping for a fixed coeff. of friction; only one will occur.
Comment: @JohnDarby Sure. There is no requirement of constant coefficient of friction in the question. It is merely asking for a comparison of one slope with friction causing rolling all the way down and another similar slope possibly with an iced part or so over which the wheel will slip.
Answer: For the ball (a solid sphere), with slipping, the acceleration of the Center of Mass (CM) of the body is $\ddot x_c = g(sin(\theta) - \mu cos(\theta))$ where $x_c$ is the acceleration of the CM, $\mu$ is the coefficient of friction, and $\theta$ is the angle of the incline. If $\mu$ exceeds ${2 \over 7} tan(\theta)$, the body rolls without slipping.  $\ddot x_c$ is smallest when $\mu$ is this highest value, specifically $\ddot x_c = {5 \over 7} g sin(\theta)$.  For any smaller $\mu$, $\ddot x_c$ will be higher.
The ball either slips or not depending on the values of $\mu$ and $\theta$; it cannot either slip, or roll without slipping, for the same conditions.  For example, for a fixed $\theta$, the ball slips unless $\mu$ is sufficiently high.
So you cannot compare the two cases- slipping and rolling without slipping- for the same conditions; only one will occur.
If you compare an incline with a $\mu$ sufficiently low to cause slipping to  an incline with the same angle of incline but a different surface that has a $\mu$ sufficiently high to cause rolling, the acceleration for the slipping case is larger than for the rolling case and the ball will reach the bottom of the incline sooner for the slipping case.
Note: With no friction $\mu$ is zero and the acceleration is  $\ddot x_c = gsin(\theta)$, the maximum value.
(See the text Analytical Mechanics by Fowles for a detailed derivation of the relationships used above.)
This response assumes a perfectly rigid body for which there are no heating effects.  See Consistent Approach for Calculating Work By Friction for Rigid Body in Planar Motion.
In reality frictional heating occurs, and over a long distance would slow down the case with slipping, so my response is for slipping over a relatively short path down the incline.  Since no body is perfectly rigid, as @Dast points out, if heating (increase in internal energy of the ball) is important the rolling case can be faster.
Additional response to questioner's comment about tape part way on the incline.
Acceleration decreases with increasing $\mu$, and $\mu$ is greater for pure rolling than for rolling with slipping. So if the tape has sufficient $\mu$ to allow pure rolling, and part of the incline has the tape and part has no tape to allow slipping, the ball will reach the bottom of the incline faster than if the incline has the tape all the way and only pure rolling occurs.   Again, this assumes a perfectly rigid body (no heating).
Comment: I see.  Please see my modified response.
Answer: For a reasonably long slope rolling is usually the faster option.
Lets say that we have a slope of a fixed height. We know once as the object has slid or rolled down the slope the potential energy of gravity is lost. This energy ends up split between three places: (1) The linear kinetic energy of the ball. (2) The rotational kinetic energy of the ball and (3) heat from friction.
Once the ball is rolling its surface that touches the slope doesn't rub against it at all. So no energy is lost to this friction. Small friction losses from the non-ideal real world (roughness and air) will remain, but we can probably neglect those and say that once the ball is rolling it doesn't loose energy over time to friction.
So if we compare a rolling ball to a sliding block. The ball has to pay an up-front energy bill to get rolling in the first place (investing in its rotational energy) so has less linear kinetic energy due to this. The block avoids this, but on the other hand it pays in kinetic energy the whole way down the slope as it rubs its way down. Its a one-time payment vs a continuous cost.
For a longer slope, or a higher coefficient of friction the one-time payment (rolling) has the advantage.
Comment: For a short distance I think the heating effects are small, but for a large distance cannot be ignored. If heating effects are small, I think the slipping case is faster (see my response using classical mechanics for a perfectly rigid body).  But for sufficient distances where heating cannot be ignored, I agree with you.  A detailed calculation including  thermodynamics (the first law) is needed.
Answer:
you have  those two equations:
$$m\,\ddot s=m\,g\sin(\alpha)-F_r\\
I_\theta\,\ddot\theta=F_r\,r$$
in case of pure rolling is $~\ddot s=r\,\ddot\theta~$ and in case of partial rolling $~F_r=\mu\,m\,g\cos(\alpha)~$
pure rolling
$$\ddot s=\frac{m\,g\,r^2\,\sin(\alpha)}{m\,r^2+I_\theta}
\quad \Rightarrow\\
s_1(t)=\frac{m\,g\,r^2\,\sin(\alpha)}{m\,r^2+I_\theta}\,t^2\quad (1)$$
partial rolling
$$\ddot s=g\,(\sin(\alpha)-\mu\,\cos(\alpha))
\quad \Rightarrow\\
s_2(t)=g\,(\sin(\alpha)-\mu\,\cos(\alpha))\,t^2\quad (2)$$
hence (with $~I_\theta=m\,r^2$)
$$\frac{\dot s_1}{\dot s_2}= \frac 12\,\frac{\sin(\alpha)}{\sin(\alpha)-\mu\,\cos(\alpha)}< 1$$
so  partial rolling is faster then pure rolling
Comment: This assumes a perfectly rigid body; no "heating" (no change in internal energy of the body).  Also for an incline with a specific slope and coefficient of friction the body either rolls or slips.


How to hide Firefox titlebar in Gnome Shell?

Question: I am running Gnome-shell 3.2 in Ubuntu 11.10. Is there a way to hide the firefox title bar and make it like Chrome's ? I would need the min/max buttons in tab-bar. There was a tutorial in webupd8, but that was with compiz.
Comment: Related: [On Ubuntu 18.04 can I hide the Firefox title bar AND still see the close/minimize/maximize icons?](https://askubuntu.com/q/1039360/480481)
Answer:
Notice: HTitle is now discontinued, you can find an archive on Wayback Machine for historical reference.

You could use HTitle Firefox extension to hide title bar.
This can be used only with GNOME 3.
Comment: The addon site currently says it was discontinued. :(
Comment: @LiveWireBT, I don't think that extension is no longer required since Firefox UI in full screen mode has been changed.
Answer: I wondered how to do this as well and found that I do not need the title bar in any window if it is full screen in order to safe some pixels, so I did some changes to my metacity-theme-3.xml
/usr/share/themes/Adwaita/metacity-1/metacity-theme-3.xml
<code><frame_geometry name="max" has_title="false" title_scale="medium" parent="normal" rounded_top_left="false" rounded_top_right="false">
    <distance name="left_width" value="0" />
    <distance name="right_width" value="0" />
    <distance name="left_titlebar_edge" value="0"/>
    <distance name="right_titlebar_edge" value="0"/>
    <distance name="title_vertical_pad" value="0"/> <!--
                            This needs to be 1 less then the
                            title_vertical_pad on normal state
                            or you'll have bigger buttons                               -->
    <border name="title_border" left="0" right="0" top="0" bottom="0"/>
    <border name="button_border" left="0" right="0" top="0" bottom="0"/>
    <distance name="bottom_height" value="0" />
</frame_geometry>
</code>
With the "Menu bar" unticked, it just looks like Chrome. :)
There is also a Gnome 3 extension for this: Pixel Saver.
PS: I can also recommend the GNOME 3 theme for a consistent look and feel (scroll bars etc).
Comment: That **Pixel Saver** extension was what I was looking for, very good!
Comment: But what about the minimize and close button
Comment: I never minimize windows (I just move them to other workspaces) and I close windows using shortcuts such as `Ctrl+Q`.
Comment: Pixel Saver was exactly what I was looking for in Gnome Shell.  I use Super+up/down/left/right to maximize/unmaximize/move to left or right half of the screen. Alt+F4 to quit. No need for minimize/maximize buttons (for me - I understand it doesn't answer OP's question)
Answer: I found a solution:
Install this extension:
https://addons.mozilla.org/en/firefox/addon/hide-caption-titlebar-plus-sma/

It'll show that is not compatible with our version, still click Add to Firefox and install.

All credits to WebUpd8:
http://www.webupd8.org/2011/03/firefox-4-get-tabs-in-title-bar-like.html
Comment: This addon is now listed as compatible for Linux. You can also easily disable the orange menu button and the windows like close buttons if you prefer.
Answer: As of September 2019, this functionality is built into Firefox:

Right click empty space on the tab bar
Select "Customize..."
At the bottom of the window there is a "Title Bar" checkbox. Uncheck it to hide the title bar.
Answer: As the global menu extension in gnome-shell doesn't work with firefox, you have to make do with the firefox extensions.
There is an extension called "Mozilla Labs: Prospector - LessChrome HD 7" which shows the navigation bar only on mouse-over. It works quite well (works also very well with unity).
You can get it here: https://addons.mozilla.org/en-US/firefox/addon/prospector-lessChrome-HD/?src=userprofile
Additionally, in View -> Toolbars, unmark "Menu Bar" and mark "Tabs on top"
Your Firefox window should now look like this:
Comment: Thanks for replying, but it still doesn't answer my question to hide the title-bar. I have Hide Caption Title Bar extension installed, but doesnt seem to work in Ubuntu 11.10.
Comment: I am also facing this problem..Does anybody has any suggestions for fixing this?
Answer: There is a good workaround here.  It provides a keyboard shortcut for a fullscreen mode in which the Firefox bars are preserved.
Steps:

Open 'System Settings' (under the user menu top-right on the screen)
Select 'Keyboard'
Click on the 'shortcuts' tab
Select 'Windows' group of shortcuts
Click twice on 'Toggle fullscreen mode'
Type a shortcut (I use Shift+F11)

It's not perfect, but it does 90% of the job.
Answer: Settings -> Window Manager Tweaks -> Accessibility -> Check "Hide title of windows when maximized"


Magento checkbox layered navigation

Question: One of my client need to select multiple values in magento layered navigation using checkboxes for product weight options . How can I do this by coding without using any extensions ? I'm using magento 1.9
Comment: What have you already tried?
Answer: You need to:
1. rewrite catalog/layer/view block to show check boxes
2. rewrite the catalog/layer/model/filter/item to modify URLs
3. rewrite catalog/layer/model/filter/attribute to create the new query, based on multiple attribute options.
Comment: hi yes I have tried those and able to resolve it , thanks for the input
Comment: @Amasty what code needs to be changed, can you please tell us?

Magento checkbox layered navigation


To retrieve deleted Gmail mails

Question: Every six months I used to delete the mails in my inbox but now I need access to one of the emails I deleted a year back.
Is there any way to recover the mail?
Comment: You are aware that if there is a chance you need an email the proper action is archive not delete?
Answer: Any email you delete goes to your Trash. Just go to Trash just below Spam on the left vertical menu. However, you you have emptied your Trash, then it now depends on Gmail.
If Gmail has not deleted the messages from the server, you can recover deleted emails after deleting them from your trash folder. You must open your gmail then click the drop down arrow on the left from your search box at the top of the screen. it will then ask you several question such as to: or from, subject, key words or date within. Click on the blue search button on the bottom left of the search box and all of your deleted emails will be revealed. http://www.itpro.co.uk/email-clients/26076/how-to-recover-deleted-emails-in-gmail-2.
Gmail Help link:https://support.google.com/mail/answer/7015314?hl=en&visit_id=1-636441828911162408-3017547055&rd=1
Answer: If you know exactly which email you need, the easiest option to get a fresh copy is to contact the sender and ask for that email to be resent/forwarded to you. It's unlikely you'll be able to get Google to restore a single email from their backup archives.
The other question is whether you've downloaded the email to one of your computers (e.g., Outlook, Mail, Thunderbird, or another client). If so, that's an option for you to obtain the email (or restore from backup).

To retrieve deleted Gmail mails


What exactly is source control?

Question: So I'm using Visual Studio 2012, and whenever I create a new project I always leave out the "Add to source control" option, mainly because I don't what it is, what it's purpose is, or if it would be beneficial to me to use it.
I'm developing a rather large library on my own. I plan on making it open source. Is this what source control is for? Reading up online did not help me unfortunately, I'm still desperately confused. Is source control the same as version control too, by the way?
Some clarification would be greatly appreciated.
Thanks!
Alex
Answer: Quite simply, source control is a repository where your source code is stored. It's purpose is to provide a storage spot that is separate to the copy that you are currently working with, so you can make changes locally and then submit them back to your source control repository.
Think of it as being like a library - you can get code out, you can return (modified) code to it. It also gets more complex than that - you can use source control to handle multiple versions of the same source, and to merge changes from one copy to another. Additionally most (all) source control systems should provide a mechanism for tracing changes to code (i.e. audit tracking), and a means to "roll back" or revert to a previous version.
There are a variety of source control systems, some of which you install locally and some which you install centrally and access remotely. In all cases the underlying philosophies are mostly the same, although there are some differences around terminology and implementation of features. Source control is considered to be a system that is separate to your editor, your editor then provides a means to specify which source control provider to use and the location of the source control repository.
Answer: Source control (otherwise known as revision control: http://en.wikipedia.org/wiki/Revision_control) is a way in which you can store and manage changes and revisions to your code-base.
Google: SVN, Git, Mercurial and Google Code, GitHub, Bitbucket.
I personally never start a propject (personal or otherwise without it). Also, I like BitBucket (www.Bitbucket.com) for managing my projects. It's free for a single user and they support Git and Mercurial.
Answer: I have found this

Git Source Control Provider is a plug-in that integrates git with
  Visual Studio

According to this, seems like source control = version control
Answer: You were asked this when starting a project because it's much easier to setup before you have many messy pieces. Github is a form of version control. An organized place for code to grow.
Version control really shines if you're collaboratively working on a project with many different pieces to it. Source or version control is the process of taking the many different parts that are being worked on simultaneously and bringing them together as a final project.
If you're working on your own it still may be beneficial to practice with, as still holds some value in allowing a process of documentation, and modular pieces being brought together for a final project.
This wonderful post explains many of the benefits of using version control for your pet projects.

Well, there are a number of good reasons:
1. It's good to be in the habit.  Sure, you may be working alone.  But in
the future you may not be.  Or your "weekend hobby project" might turn
into a popular project with many developers.  If anything like that
happens, being in the habit of using source code control will stand
you in good stead.
2. It protects your code.  Since your code is stored
in on a server apart from your development machine, you have a backup.
And then, you can even backup the code on the server.  Sure, you can
zip it all up any time you want, but you don't get all the other
benefits I'm listing here.
3. It can save your butt.  Sometimes, you
might accidently delete something.  You might make mistakes and change
code that you didn't want changed.  You might start off on some crazy
idea when you are feeling a bit saucy, and then regret it.  Source
control can save you from all of these by making it a piece of cake to
revert to any previous state.  It's like a really powerful "undo"
feature.
Answer: Wikipedia says:

Revision control, also known as version control and source control
  (and an aspect of software configuration management), is the
  management of changes to documents, computer programs, large web
  sites, and other collections of information. Changes are usually
  identified by a number or letter code, termed the "revision number",
  "revision level", or simply "revision". For example, an initial set of
  files is "revision 1". When the first change is made, the resulting
  set is "revision 2", and so on. Each revision is associated with a
  timestamp and the person making the change. Revisions can be compared,
  restored, and with some types of files, merged.

In simpler terms, source control allows multiple people to work on the same project without everyone clobbering everyone else's changes.  Changes to the same file are merged together, and conflicts can be resolved.
It also allows you to keep version history.  You can roll back a file to the way it existed at any point in time, as well as maintain different branches of the program; a way to organize groups of changes.
Answer: "Source control" refers to the concept of storing all files that make up the source of an application in an (usually online) repository - where one can manage with fine detail exactly what has changed in the source code between versions, manage issues, involve other people with the project, and generally provide a platform to collaboratively manage a project.
The term is usually synonymous with version control - although you can have a repository and not explicitly name versions. Version control in particular just refers to labeling major versions of your project with a hierarchical numbering scheme (1.2.4 etc.)
There are several different tools that implement different kinds of source control. For example, Git is a source control tool that sets lets you manage a project. Github is a web-site that repositories managed with Git are usually stored on. Mercurial is yet another source control tool.
Typically, source control can be complicated to understand, and usually requires significant time studying tutorials. A lot of concepts are unfamiliar to solo programmers who have only worked on small projects. Changes to source code are managed through commits, and different branches of the project can be worked on by multiple people. Changes from separate branches can be merged, and the project can be forked by another user entirely (at least for Git).
Making your library open source is a good idea, and it's not too hard to give it a home on github. I would look into some resources to find out how to set it up.
Comment: "Version control in particular just refers to labeling major versions of your project with a hierarchical numbering scheme (1.2.4 etc.)"  I strongly disagree.  Version control is source control, period.  If you want to talk about the versions, you can talk about the changeset or commit IDs, they need not be released versions.
Answer: As Si puts it, source (or version) control will save your life 20 times over.
Source control keeps a detailed history of your code. This enables you and any collaborators to:

Review the history of any code
See changes between two versions
Revert any code you just goofed up
Prevent you from losing any code
Discover which changes caused bugs / crashes
Blame whoever messes up your code
Keep an up-to-date backup of all of your code

Even if you're not working collaboratively, source control is so useful when you accidentally delete your code. I speak from experience.
I would recommend using Git or Mercurial, as they allow branching and offer a more isolated workspace for collaborators to experiment without fear of harming any code.
For online collaboration, Github and Bitbucket are great and integrate well with Git and Mercurial, respectively. They also allow you to track collaborators' changes, choose and merge changes into a central repository, and keep a detailed list of issues.

What exactly is source control?


Do movement restrictions end if country of origin gets added to green list?

Question: If at the time of my arrival in Ireland, the Green List does not contain the country I started my travels from and I start to restrict my movements, but before my 14 days restriction are over my country of origin is put on the list and it goes into effect, do my requirements to restrict my movements end the day the updated list goes into effect, or do only new arrivals from that  time on not require movement restrictions any more?
Comment: Not sure about Ireland, but in most countries you continue to be bound by the regulations that were in force when you arrived.  This makes more sense when you think about it in the other direction: if a country is removed from the green list, they don't slap you in quarantine halfway through.
Comment: @lambshaanxy Actually, at the very beginning of this pandemic (before a lockdown or any contraining legal framework was in place in my country of residence), I have actually been advised to self-isolate after the fact (returning from a country 4-5 days before it was considered dangerous, being asked to spend the last 3-4 days of a 14-day quarantine at home). Intuitively, this does make sense because prevalence or positivity rate on a given date reflects the circulation of the virus some time in the past and you could be most contagious a few days after any potential contact.
Comment: But if we ignore this and look at it from a purely legal point of view, what makes you think that in most countries you would be bound by the regulation that were in force when you arrived? It's more difficult to do anything if you are not testing and registering people at the border but on a theoretical level it's certainly possible for the regulations I am familiar with (it's phrased as "having been in area XYZ in the last X days" so if it kicks in before X days after you returned, you're formally covered).
Comment: @Relaxed This is policy in both Australia and Singapore, where quarantine is strictly enforced and changes in regulations take effect at a point in time for new arrivals.   Before the point, old rules apply (and continue to apply), after it, new ones do.  See eg. https://www.gov.sg/article/updates-to-border-measures-for-low-risk-countries-from-1-sep
Comment: @lambshaanxy I also found some European countries where this is explicitly handled in this way (Germany, the UK).
Answer: As it stands (based on information I got from phone calls with the HSE and Irish Citizens Information) one is bound by the rules that were in place on entering the country, unless any official source would explicitly state otherwise - which so far has not been the case.
In short, if you're subject to restrictions of movements when you arrive, they stay in place even if the country is cleared in the meantime. Similarly, you won't have to go into movement restrictions if you came from a country on the Green List that is taken down from there within two weeks after your arrival.


WPF Toolkit's DataGrid sort, change reflect to the data source?

Question: When I sort one of the columns (click the header column) in DataGrid, does the change on UI (the row order) reflected back to the data source ? I use List object as source.
Answer: No, the UI sort has no effect on the data.
There is another way to sort directly the data (actually, the representation of the data), but connecting it to the normal UI functionality (clicking on a header to sort) might be a bit more involved and also has several other downsides: wrap the collection in an ICollectionView (ListCollectionView for instance) and use SortDescriptions or custom sort predicates to sort the items; the ICollectionView will stand between the source collection and the UI and you can get the sorted items from it.
Also, you can sort the source collection directly as well (either manually or by using a sortable collection), but again, you need to connect the datagrid headers to the sorting code, which means something along the lines of intercepting the click events of the headers and drawing appropriate glyphs for sort hints etc.


Is there anything such as gravitational field-lines in GR similar to the electric/magnetic field lines in electromagnetism?

Question: I sometimes mistake space-time curvature for gravitational field lines. Do geodesics in some ways represent $g$-field lines? Why is not it traditional to show $g$-field lines around a massive object in general relativity the same as we do for $E$ or $B$ field lines around an electrical charge or a magnet in electromagnetics?
Answer: I would encourage anyone to draw any lines that help either them or others to get a good understanding. But the reason field lines are so much used in electromagnetism, and much less used in gravity, is because they nicely capture a mathematical property of a field whose divergence is zero (on a flat spacetime background). The zero divergence of the field translates to the field lines being continuous, and their spacing then expresses the field strength.
There are aspects of gravity that are a bit like this. In the weak field limit and for simple cases (e.g. a static case or one with steady rotation) one can express gravitational influences using a pair of fields analogous to electric and magnetic fields. But mostly in GR we are interested in stronger effects where the equations are more complicated and non-linear. Then there is nothing quite so convenient as the field lines of electromagnetism. But there are plenty of things one can do. Drawing sets of light cones can help in getting an impression of a region of spacetime. Or one could draw a selection of null geodesics. I like to add to this a set of timelike lines marked off by proper time along them. But such diagrams can be stretched and squeezed, twisted and distorted in all sorts of ways merely by changing the coordinates being used to plot them, so they have to be interpreted with care. If one merely changes coordinates then the diagram changes but the spacetime does not.
Answer: Probably not just as you ask, but there are interesting ideas for visulazations in
Visualizing spacetime curvature via frame-drag vortexes and tidal tendexes. II. Stationary black holes by David A. Nichols, Robert Owen, Fan Zhang, Aaron Zimmerman, Jeandrew Brink, Yanbei Chen, Jeffrey D. Kaplan, Geoffrey Lovelace, Keith D. Matthews, Mark A. Scheel, Kip S. Thorne.


Multi-Connection bluetooth application

Question: I am developing a J2ME MIDP 2.0 game that can can be played over bluetooth,
it is similar to tic tac toe where players exchange turns until the game is over,
the problem is that i want to give the both players the ability to restart the game or quit
at any time but i only have one connection for transferring game related data, and i wonder if i could have two separate connections between the two devices at the same time, one for eg. game management (restart, quit, etc ..) and the other for game data to separate the logic.
or you could put it this way : is it possible to have multiple bluetooth connections between two MIDP devices at the same time ?
Thanks in advance
Answer: I think for bluetooth.Via j2me mobile application the Midp device is connected at most only one  device and at most at a time only one connection is enabled.So there is no chance of going to multiple connections.

Multi-Connection bluetooth application


What are good alternatives to Mac Address Book for Windows?

Question: I used Address Book on Mac and I really liked the software. It's simple, easy and it has all the things I needed in an address book. Now, I recently switched to Windows 7 and I am looking for a software which works like Address Book. Light weight with Speed. Even, a Website or a Cloud App which syncs contacts from my system to the cloud is OK.
Answer: Since Vista contacts are built-in, if you have MS Office you might also want to consider MS Outlook.
Another alternative is Free Address Book.

Free Address Book is an address book wherewith you can note the addresses, phone numbers, mobile, fax, company name, country, city, website address and email addresses of your contacts and partners.
You can easy print list of contacts and details of selected contact.
  Phone Book is absolutely free!
Organize your contacts easy and fast!
Comment: I checked out Free Address Book and it looks like a good option. However, is there someway where I can transfer Mac Address Book contacts to Free Address book?
Comment: @Ctroy: Seems it only allows to import CSV files for now. I suppose Mac exports iCard files so I don't think you can, I'm sorry...
Answer: Google Contacts.
Plus Android and addressbook nirvana arrives regardless of computing platform.
Answer: What do you use for email?
Thunderbird and the Contacts extension work for me.  And I access my contacts on my phone as well with Google.


Combine ASP.Net MVC with WebForms

Question: Is it possible to create a MVC root application (Portal with masterpages and themes) and add a couple of WebForms based subprojects (we already have an existing WebForms application that we would like to integrate into the Portal)?
How would you centralize navigation (sitemaps, url routing)?
How would you share the masterpages?
How would you refer to resources (~ issues etc.)?
Answer: Combining web forms with MVC is entirely possible. See this blog post by Scott Hanselman for an introduction.

Sharing master pages: see this StackOverflow question
routing: In ASP.Net 4.0, routing has been enabled for web forms page routes (scottgu's blog)
Comment: Great post, all the links I used when doing the same thing.
Answer: For reference I managed to accomplish this task yesterday - add MVC functionality to an existing WebForms website, that is - so thought I would write in case it is of any use to anybody.
People say: "the best way is to rewrite your web forms application as MVC" but there is often no business case for this when the application is well established and has grew over time to be quite a monster (that still works well I might add). In my case, we had been asked to implement a new function into this application - and having been developing in MVC on other projects recently - I wanted to build this new function in MVC. This is what I did:

Create a new ASP.NET MVC 4 Application and selected the Empty Template (with Razor)
Once loaded, right-clicked References > Browse and then navigated to the Bin folder of my existing website. I then added in the DLL's I was using in the existing site. For reference, I am using AjaxControlToolkit so decided to add in the latest one from NuGet. This should then auto-update the web.config file
I then opened up Windows Explorer and navigated to the root folder of my existing website. Using Visual Studio, I then replicated the top-level folder structure of the existing site (minus the Bin folder) resulting in the 2 sites having the same top-level folders (but the new project having the MVC App_Start, Controllers, Models and Views folders also, but not Bin obviously).
Once setup, I copied the sub-folders and files from the corresponding top-level folder in Windows Explorer and pasted them into the solution. If you do it the way, the project allows you to copy in sub-folders and content directly, adding existing item via visual studio only copies in files thus saving time. Please note: for the root folder, I did NOT copy in web.config or Global.asax, but copied in all other files (like default.aspx, etc)
Once all content was copied, I opened up web.config in the new project and web.config in the existing site. I copied into the new web.config anything specific to the existing website (appSettings, connectionStrings, namespaces, membership info, etc). The same applied to the Global.asax files.
I attempted a build and fixed anything I'd missed.
Once everything was right (and I had shot down a few Yellow screens of death) I noticed that I got the page not found error when loading default.aspx. This is down to routing, here is the changes I made to the Routing info (located in Global.ascx MVC 3, or App_Start > RouteConfig.cs MVC 4):
public static void RegisterRoutes(RouteCollection routes)
{
    routes.IgnoreRoute("{resource}.axd/{*pathInfo}");

    // This informs MVC Routing Engine to send any requests for .aspx page to the WebForms engine
    routes.IgnoreRoute("{resource}.aspx/{*pathInfo}");
    routes.IgnoreRoute("{resource}.aspx");

    // Important change here: removed 'controller = "Home"' from defaults. This is the key for loading
    // root default page. With default setting removed '/' is NOT matched, so default.aspx will load instead
    routes.MapRoute(
        name: "Default",
        url: "{controller}/{action}/{id}",
        defaults: new { action = "Index", id = UrlParameter.Optional }
    );
}

With all these steps taken, the site loaded and ran as it always did, but I now have a ASP.NET WebForms and MVC hybrid. I am sure there are probably easier ways of accomplishing this task, but I tried other methods and failed. This way took me around 20 minutes and I was up and running... well sat and coding :)
Comment: This was quick and helpful; easy.  I was also able to use standard Add dialog to add more webform pages to the MVC project.  This was a must in case I wanted to add both older webform pages, but also add MVC Razor pages.
Answer: Great answer above.  The other thing you could consider is migrating from ASP.NET to ASP.NET MVC, in which case this post might help you migrate the project files.
Answer: I was too stucked in this situation, after all i got a solution which worked, first of all copy web.config in your views folder.
After that open Route.config(Routeconfig) file, there you will find one method RegisterRoutes, keep it as it is and create another method named RegisterRoutes2(), and copy paste the below method code from the mvc application:
public static void RegisterRoutes2(RouteCollection routes)
{...}

in some cases method name in mvc app would be RegisterRoutes.
If above code does not work, copy paste it from your machine's mvc application.
Then, go to global.asax file, ther you will find RegisterRoutes() has been called, simple now , also call RegisterRoutes2().
And you are done, restart your application, or clean build, it will run.

Combine ASP.Net MVC with WebForms


Geometric proof of Robinson-Schensted-Knuth correspondence?

Question: Famous Robinson Schensted Knuth correspondence gives a correspondence between the matrices with non-negative integer entries and pair of semi standard tableaux. The proof that I have seen is highly combinatorial e.g. in Knuth's paper [Permutations, matrices, and generalized young tableaux]. Does there exist a geometrical proof of this correspondence?
Answer: This depends on the meaning of the word "geometric".  If you are thinking of RSK and want the geometry in the way the algorithm is presented (in the case of permutations only), Viennot's paper Une forme geometrique de la correspondance de Robinson–Schensted mentioned by PeterR is your answer.  If you are thinking of geometry of flag varieties, you might like Steinberg's theorem (see van Leeuwen's thesis A Robinson–Schensted algorithm in the geometry of flags for Classical Groups and his J. Algebra paper An Application of Hopf Algebra techniques to Representations of Finite Classical Groups.  Finally, if you want RSK to be a map between integer points in polytopes, there are several versions of that, going back to Gansner in 1981 (you can find a description of that and references in the second half of the paper Hook length formula and geometric combinatorics of mine).
Comment: Note that there's now an open access link to Steinberg's 1988 paper: http://www.sciencedirect.com/science/article/pii/0021869388901779
Comment: Thanks a lot for references. Let me look at these and find out, if I get the answer of what I am looking for.
Answer: This is more a comment on some of the answers given than an answer to the question itself (which isn't too clear in the first place: a correspondence is not something to prove, and why should one want a geometric proof of a statement (whichever) that is fundamentally combinatorial, unless it's just the right brain hemisphere that wants to join in the game?)
I have said this before, in print long ago, but since many people seem not to be aware I'll say it again: while a beautiful presentation, Viennot's description with shadows (ombres) is just a restatement of the "graph-theoretical viewpoint" that is given in Knuth's original paper Permutations, matrices, and generalized Young tableaux (section 4), in a somewhat more geometric language. In fact I think the approach is really poset-theoretic: a permutation gives rise to a poset by embedding the $1$'s of permutation matrix into $\mathbb N^2$ with the product partial ordering. The graph that Knuth considers is basically the restricted partial ordering, viewed as a directed graph (there is a minor complication for general integer matrices instead of permutation matrices, as they give rise to points with multiplicity). Viennot's partition into shadows is just the partition of the poset into anti-chains according to the "height" of its elements: the length of the longest chain ending in the element (which description is given in the Knuth paper). Once the shadows are defined, the construction proceeds by iterative formation of derived (and ever smaller) subsets of $\mathbb N^2$/posets/digraphs in exactly the same way. Fulton's matrix-ball construction is again a reformulation of Viennot's construction, taking care to adapt to the general integer-matrix context of Knuth's paper; otherwise there is nothing new here either.
Personally I find the iterative nature of these constructions limit their transparency, even though it does not interfere with their proof of the symmetry of the RS / RSK correspondence. A similar "geometric" approach, but which really adds insight with respect to Knuth's paper is Fomin's growth-diagram approach (which can be given for the full RSK correspondence, even though it is often presented for permutations only). One has to consider all points of (the relevant part of) $\mathbb N^2$ rather than just those occurring in the poset, and one has to attach partitions to each of them according to a somewhat subtle rule, but then the construction produces in one swoop (without iteration) the full pair of tableaux.
Comment: @darij grinberg: For the Knuth correspondence, since the matrix can have entries larger than one, the "growth" of shapes can correspondingly be by more than one square; the natural condition is growth by horizontal strips. The original reference is S. Fomin, "Schur operators and Knuth correspondences", J. Combin. Theory, Ser. A
72,(1995), pp. 277–292, though I might unmodestly suggest that my discussion in section 2.2 of "Spin-preserving Knuth correspondences for ribbon tableaux" http://www.combinatorics.org/Volume_12/Abstracts/v12i1r10.html is maybe more accessible (you can skip section 1).
Comment: You speak of "Fomin's growth-diagram approach (which can be given for the full RSK correspondence, even though it is often presented for permutations only).". Is there a good way to extend growth diagrams to the non-standard case without the (rather artificial IMHO) concept of standardization?
Answer: Do you just want a proof that some bijection exists, or do you want a proof that the RSK bijection itself is a bijection? If the former, this is equivalent to the Cauchy identity in the theory of symmetric functions, for which many proofs have been given. For instance, it is equivalent to decomposing the symmetric algebra of $V\otimes W$ (where $V$ and $W$ are finite-dimensional complex vector spaces) into irreducible representations of $\operatorname{GL}(V)\times \operatorname{GL}(W)$, for which a direct argument is possible.
Answer: There is a generalization of Viennot's "ombres" mentioned in some of the other answers, called the matrix-ball construction, and which realizes the full RSK correspondence.
It is presented in Fulton's book "Young tableaux: with applications to representation theory and geometry", and as far as I remember  was devised by Fulton itself.
Answer: See "Une forme geometrique de la correspondance de Robinson–Schensted" by Viennot if you can read French. Otherwise, if I recall correctly it is in Sagan's book on the symmetric group.
Answer: Not involved approach to RSK is in
Danilov, V. I.; Koshevoy, G. A. The octahedron recurrence and RSK-correspondence. Sém. Lothar. Combin. 54A (2005/07), Art. B54An, 16 pp. (electronic)


Jesus Christ Superstar shooting places guide?

Question: I have been up to Avdat today and I think I recognize it being used in a few scenes of the 1973 movie version of Jesus Christ Superstar, but of course it's really hard -- obviously they "dressed" the ruins. Is there any information about the specific locations where they filmed the movie?
Answer: A Songfacts interview with Ted Neeley offers the following shooting locations;

"Trial Before Caiaphas" was shot on the site of the Herodian
  fortresses near Bethlehem.
"The Temple" and "The Crucifixion" scenes were shot at Avdat.
Ein 'Bokek provided the location for "Pilate's Dream."
"What's the Buzz" truly was under the Dead Sea... in a cave [The Beit Guvrin Bell Cave]. The same
  cave served as Neeley's "prison" after "The Arrest."
For "King Herod's Song," the crew set up shop on the Dead Sea not far
  from Masada, where Herod actually once had a summer palace.
An ancient Roman ampitheater in Beit 'Shaan, tunneled through with
  catacombs, served as the location for "Trial Before Pilate" and
  "Superstar."

Note that I've corrected a couple of the place name spellings for ease of reference.
Comment: This book about [Jesus in Cinema](https://books.google.co.uk/books?id=CnsgxAc0L-sC&pg=PA63&lpg=PA63&dq=Hosanna+avdat&source=bl&ots=rEL967zDEj&sig=BEMa5BEgi27rH1TNPRmzkP8LPII&hl=en&sa=X&redir_esc=y#v=onepage&q=Hosanna%20avdat&f=false) would certainly suggest that that's the case
Comment: Great! It'd be even better if it included details at Avdat... also, I suspect but not 100% that Hosanna was also shot at Avdat, simply by comparing a certain shot with what I saw today.


Latex warning package pgfheaps

Question: I have a latex beamer file for a presentation, however, running the file creats the following warning and no pdf is created at all
package pgfheaps warning. this package is obsolete. on input line 16.
What does this mean please?
Thanks a lot for your help.
Sanaa Moussa
Comment: it is not a generic warning, the file itself has been edited by its author to say `\PackageWarning{pgfheaps}{This package is obsolete.}`  usually if package authors do something like this they make the warning say what to use instead but not here. As this is part of the base pgf distribution I assume pgf itself has suitable facilities now so this little extension package was given this warning, but I can not tell you  what to use instead so not posting this as an answer.
Answer: PGF has been restructured quite a bit since 2006 and the functionality this piece of library offers is already available with regular commands and node placement on curves. Your file must be really really old to have this library mentioned. Or your beamer version is exceedingly old that still uses pgfheaps. But still it is not the cause of no PDF file produced error.
However, if it is that old, many packages could be conflicting anyways. Best is to try it on a new version of beamer and see whether it flies or not.
Comment: Thank you very much for your answer. I haven't used beamer years ago, that's why I know you are right telling it is very old. My question now, if you please, how to get the latest verion of beamer?
Comment: @user402254 If you update your TeX distro (TeXLive or MikTeX) then you would get a recent version of everything)

Latex warning package pgfheaps


Ubuntu 18 scale problem

Question: Recently i have upgraded from Ubuntu 16 to Ubuntu 18. Usually i use Mate desktop and the upgrade was made in Mate session via command line.
The problem is, that when i log into Ubuntu desktop it is as if scaled to 200% i can see only one quarter of the desktop and i can reach other corners with mouse.
I have checked the scale option in settings: it is 100%, so the problem is not here.
When i log back to Mate, it is all right.
I have rebooted several times and ran "sudo dpkg --configure -a" (if it matters).
Comment: That sounds like a resolution problem not a scale problem. Scale tends to be the size of text and icons whereas resolution will affect the size/shape of the desktop. Do you have another monitor/tv you could connect to to see if the issue persists?? I'd check your resolution settings in gnome and see if changing it to a lower resolution and back again will solve the problem.
Answer: It was zoom (in Universal access) enabled.

Ubuntu 18 scale problem


Excel - Get value of a column in a row where match found in another column

Question:
I need to track where value in J column starts appearing zero. In following table, its 7th row. From this row, I need to get value of D column to appear in some designated Cell outside this table. In this case it would be 11 y, 2 m.
I used offset function, that gives me first occurrence of zero in J column. But I'm not sure how to lookup different column (D) from the same row to fetch it's value.
My formula: =OFFSET(J1,MATCH(0,J:J,0)-1,0)
Here actual data starts at first row; hence 1 is used above. It correctly returns first zero (0) occurrence in J column.
Answer: Assuming the column is sorted so that the first zero is the only zero that matters, one of many solutions is Index Match.
=Index(D:D,Match(0,J:J,0))
Index returns the value of nth position in a list.  Match returns the position of the of the first match in a list; the 0 at the end is a parameter to specify an exact match.  When both lists are the same length, such as two columns in a table, then you use Match to find the index and use Index to find the value of an arbitrary column.
So, in this example, Match returns 7.  The 7th position in Column D is 11y, 2m
In the very near future (or right now for some people) you will be able to use XLookup to do the same thing.
Comment: That worked, thanks. I was not aware of `Index` function.
Comment: @benjamin54 Happy to help!


How to fix gap between drywall and brick inside the home

Question: I have a gap between brick framing that holds my microwave oven and oven and the rear dry wall.  This is due to house shifting over the years.  The former owner filled it will ugly rubbery filler that cracked and fell mostly out.  I have removed that and cleaned up the gap.  Now the hard part.  I want this to look as as aesthetically pleasing as possible as it's in my kitchen. The gap is about 5.5 inches deep before I hit a wood frame that is probably holding the ovens. The width of the gap varies from 1/4 to 1" inch.  In a perfect world it would be nice to fill with mortar so it looks like it never separated from the wall.  Not sure how they would hold up over time. I guess an alternative would be to fill with something and paint it the wall color so it looks seamless.  Due to the ovens being there I am concerned about using any flammable filler.  Looking for suggestions on how to make this look it's best.  Appreciate your input! Thanks!

Comment: Sounds like the previous fellow caulked it.  He may not have cleaned surfaces properly or may have used the wrong kind of caulk if it separated without additional settling forces.  If well done it is not typically an eyesore.  More rigid fillers will crack or separate similarly if the building settles more.  Another option is to look at the wall and whether you can add a piece of molding to cover the gap, and this could be either very easy or very hard to tie into the wainscoting below.  If you can find the same shape molding as the crown of the wainscoting, it looks not too hard.
Comment: I agree with the molding idea.  Trying to fill a crack that big and have it hold over time is probably not a long term solution.
Comment: It appears that there is caulk between the wainscoting and the brick. For consistency purposes, that's what should be done above, too.
Comment: Caulk will look like caulk (and a lot of it). I'd be tuck-pointing mortar so it looks like it originally did.
Answer: Use a spray expanding foam filler in the larger gaps then shape the foam so it is slightly recessed.   Then use a paintable flexible caulk.   Paint it if you want.
The foam will provide a better backing for the caulk.
Hopefully the drywall edges are all supported.  If any of those gaps is just a layer of drywall with a large void behind that not even expanding foam can reasonably fill, you may need to repair that .
The brick should not be getting hot.  Not at all.  If it's getting so hot you think it might ignite things, stop using the oven immediately.  Figure out what's wrong.    For reference it is normal to install wall ovens in painted wood cabinets without expecting any heat  damage to the wood or even the paint.
Comment: Instead of wasting can after can of spray foam filling a 5" deep gap, I'd suggest simply using foam backer rod. My local big-box store sells it in 3/4" diameter, and I'm sure that with a small additional amount of effort, one could also find it in 1" diameter. Wedge some of that in then caulk over the surface. Otherwise, I believe this is exactly the right answer - +1.


Convert GPT disk to MBR disk in-place?

Question: Is there any way to reliably convert a disk from GPT to MBR in-place?
(i.e. without having to backup/restore any partitions.)
Answer: This conversion can be done with my gdisk program, The actual GPT-to-MBR conversion isn't all that complex (to the end user), although there are some significant caveats, such as:

If there are more than four partitions, some of them will have to be logical partitions under MBR, and those have spacing requirements that may not be met in the original. Thus, it may not be possible to convert all the partitions without resizing some of them.
Some GPT data, such as partition names, will be lost.
If the disk is over 2TiB (assuming 512-byte sectors), partitions that extend beyond the 2TiB mark won't be converted. MBR has a 2TiB limit on partition sizes and start points, which is one of the main reason it's being put out to pasture; it's just not useful on larger disks today.
If the disk is a boot disk, you may need to re-install your boot loader. This will be a real pain with Windows, but is likely to be easier with Linux.

The gdisk documentation covers the process in detail. The bigger question, though, is: why? There are legitimate reasons to want to do this conversion, but there are also a lot of cases when such a conversion would be ill-conceived. Knowing why you want to do the conversion would enable me to better advice you about whether converting to MBR makes sense or if there's some better way to achieve your ultimate goal.
Comment: +1 I appreciate the answer. The part that concerns me (and that you didn't mention) is what happens if my computer came with Windows 8 preinstalled -- how does the process change (assuming my computer supports legacy mode), and does that invalidate my license? The "why" is basically that I want to be able to boot non-UEFI-compatible OSes (e.g. old Linuxes, etc.), which the UEFI Windows boot manager doesn't do.
Comment: @Mehrdad That same C't magazine spend 20 pages on just those issues. The GPT->MBR conversion story was one of those articles. If you can read Dutch I can recommend it. It is very thorough and explains in great detail. Issue 2013-10.
Comment: @Rod Smith Great program, your gdisk !
Comment: @Tonny: lol I can't read Dutch, is there an English version?
Comment: Mehrdad, see my fourth bullet point: "If the disk is a boot disk, you may need to re-install your boot loader. This will be a real pain with Windows, but is likely to be easier with Linux." If Windows is booting from the disk, it will no longer boot from the disk after such a conversion -- at least, not until you re-install a boot loader.
Comment: @Mehrdad Unfortunately no. C't is published in German and has a Dutch and (I believe) a Russian version. The Dutch and Russian versions translate most of the in-depth articles of the German one (with some changes e.g. they leave out products from reviews if they are not available in their country) and add localized stuff.
Comment: RodSmith: Reinstalling the boot loader is fine, I've reinstalled it a million times so I can live with that... but @Tonny made it sound a lot scarier with regard to the licensing etc. If that's the only issue regarding Windows I might actually give this a try. :) Thanks!
Answer: MiniTool Partition Wizard can do this.
I just tested in both directions (MBR <-> GPT) and it did the job in ~10 seconds.
Comment: I can confirm that. You just need to set system partition as inactive if Partition Wizards tells you that it cannot modify boot drive.
Comment: https://www.partitionwizard.com/free-partition-manager.html
Answer: It is possible with a Linux tool called "gdisk"...
But the process is very complicated, dangerous and a straight conversion is not always possible.
(It highly depends on the original partition layout. Not everything converts from GPT to MBR, because MBR is more limited in features.)
If it is your boot-disk then afterwards you will have to mess with BCDBOOT/BCDBOOT to repair the Windows boot-process.
(Or do complicated chroot stuff to get Linux booting again.)
Example: The Dutch computer magazine C't had an article on this last month. It took them 3 full pages just to describe the process. (I won't repeat that here.)
Unless you are an expert in these matters (since you had to ask you are NOT) it is probably better if you don't even attempt this.
I would say that the full backup/restore process is easier and less dangerous.
(You will have to make a full backup anyway just to be safe before you do something like this. Might as well use it for the restore then.)
P.S. If you have a (pre-installed) Windows 8 computer that came with GPT as standard to process is even more complicated and it is even possible it will invalidate your (pre-installed) Windows license.
(It is possible the manufacturer setup the UEFI BIOS/Secure boot such that it will ONLY work on GPT.)

Convert GPT disk to MBR disk in-place?


Raspberry Pi 3 can't boot up directly to the desktop

Question: I've got a Raspberry pi 3 with a Retropie installed on it.
I want to be able to boot it directly to the desktop, without having the need to type startx every time, as I prefer to boot the Retro machine from the Terminal on the desktop whenever I want to play.
I tried to configure it manually in the boot options under the raspi-config settings, so I chose Desktop Auto-login - desktop GUI, automatically logged in as 'pi' user, but it changed nothing.
So then I found the file
 /lib/systemd/system/getty@.service and saw that the relevant line was already written as
ExecStart=-/sbin/getty/  --noclear -a root %I $TERM
Important note: at the beginning of the boot the Pixel desktop appears for 3 seconds and then it immediately goes to the CLI.
What am I doing wrong/ what can I do more in order to change it?
Answer: I got the same problem on my pi 3 and solved it according to archlinux wiki:
reinstall the gdk-pixbuf2 and lightdm package and reboot.


Do Revenue Sharing Agreements Carry on when the company is sold?

Question: I've had an agreement with the Virginia company I work for to receive 1% of the generated revenue on a monthly basis. Of course that was after a long time of me being with the company (the bulk of my career). After enjoying the benefits of this agreement for about half a year, the company was sold (merged) for an undisclosed amount. The speculation is that it was sold for about 5M+. Am I entitled to get my share for this final transaction (about 50k+) or is that not considered a revenue and I'm out for good?
Neither the old owner nor the new one said a word of course.
Answer: That is going to depend on the exact wording of the agreement, but it is probable that it defines "revenue" as revenue from ordinary transactions, and excludes a sale of the entire firm.
The jurisdiction you are in (country and quite possibly state/province) may also matter.
Comment: I see more promise in trying to get 1% of the future revenue from the the portion of the new company that does the business that the old company did. Buying a company from its owners is noting like revenue or profits capital gains or anything else to the *company* itself.
Comment: I strongly suspect in that case that funds from a purchase of the company would not be considered revenue **to** the company. They are income, or more likely capital gain, to the former owner. Your revenue sharing agreement might possibly bind th new company in some way, but quite likely not..
Comment: @yazanpro  the only suggestion that I can give is that you might want to consult a local lawyer. many will give free or low-cost initial consultations, and that might help you decided if this is worth perusing. 50k is big enough that if there is a significant change of winning it would pay to hire a lawyer. Evaluating that chance beyond the above answer would be giving specific legal advise, which we don't do on this site, even those who are actual lawyers, as some posters are.
Comment: Thank you for the answer. It was a verbal agreement, which is accepted in Virginia (that's where the company was based). There was no mention/discussion of any nuance regarding the word "Revenue". It was just whatever the company receives into its bank account that month.
Comment: I guess the question is: do you think the situation is worth pursuing from a legal standpoint, or just don't bother?
Answer: I see more promise in trying to get 1% of the future revenue from the the portion of the new company that does the business that the old company did.
Buying a company from its owners is nothing like revenue or profits or capital gains or anything else to the company itself. I hope whomever you had the verbal agreement with to acknowledge that verbal contract but there should be internal company records of the payments to you and knowledge of people in accounting, etc. to back up the fact that they had been calculating any paying you per that formula.


How to iptables REJECT outgoing traffic?

Question: So I have a few docker containers running and each of them configured to use a certain network interface for outgoing traffic.
Let's say for an example that there is a docker network called docker_1 , using interface eth1 and I have marked its outgoing traffic with 0x0c3f (--set-mark 0x0c3f).
Now, if eth1 is down the container will default to use eth0 instead for outgoing traffic, this is unwanted behavior. I would like for the container to never use eth0 even if eth1 is unavailable.
I have tried running:
<code>iptables -t filter -I OUTPUT -p tcp -o eth0 -m mark --mark 0x0c3f -j REJECT
</code>
It does not work.
Does anyone have any idea of what is not working properly?
Thanks.
Answer: The problem is that you need to use the FORWARD - not output chain. Try changing "OUTPUT" to "FORWARD"


What goes wrong doing quantum mechanics on a nonseparable Hilbert space?

Question: What kind of things can go wrong if we try to do quantum mechanics on a nonseparable Hilbert space? I have heard that usual mathematical manipulations that we take for granted will no longer hold. What I am looking for is either a general high-level argument explaining why things would go wrong or some specific examples where it is so.
Answer: The main thing that goes bad in nonseparable Hibert space is the loss of Stone-von Neumann theorem. Loosely speaking, the Stove-von Neumann theorem assures us that Schroedinger representation of the canonical commutations rules is irreducibile, and it is unique modulo unitary equivalence. Hence Schroendinger and Heisenberg pictures are physically equivalent, and all pictures obtained by Schroedinger's one by means of the adjoint action of the group of the unitary operators on the Hilbert space is physically equivalent to that of Schroedinger.
Remark. Separability of the Hilbert space is part of the conlusion of the Stone-von Neumann theorem, not of hypotesis. In fact, there is no serious physical request for the separability of the Hilbert space. All that is required in hypotesis is that Weyl commutation rules can be irreducibility represented on some Hilbert space, which a posteriori turns out to be separable. This is a physical request, being the direct formalization of the Heisenberg-Born-Jordan commutation rules extrapolated from expertiments.
Remark. One may note that also in QFT Stone-von Neumann falls. This is true, but for a different reason: one of the hypotesis is that we are dealing with finite degree of freedom, while specifying the status of a field at a given time requires to give the value of an infinite number of parameters. This is effectively a phyisically different situation, mathematically encoded in the different quality of the statements.
The best known example of a theory endowed with a nonseparable Hilbert space is Loop Quantum Gravity. In order to see how nonseparable character plays his role, it is convenient to consider its "first-order approximation" (in a sense), the so-called polymer quantum mechanics. In this context, it is particulary simple to see that usual physical quantities, e.g. the spectrum of the harmonic oscillator, gain a correction of order $\hbar^2$ or smaller. Hence, such a model provides a method to verify its correctness (or not), but corrections are so small that this will be hard to do in practise. Obviously, so far LQG and related models are not physical theories, so this point is purely accademic. I am not aware of any example of fully meaning physical theory having a nonseparable Hilbert space.
Remark. On separable Hilbert spaces, orthonormal bases can be explicitely constructed by means of Grahm-Schmidt algorithm. Thanks to the Zorn's Lemma, each vector space has an algebraic (Hamel) basis, so we can affirm that even a nonseparable Hilbert space has an algebraic basis but we can't construct it explicitely. (in the context of Banach spaces, this is the case for $\ell^\infty(\mathbb N)$.)
Personal considerations. Separability means the existence of a dense countable subset. At the moment, this doesn't sound to me like a true physical necessity. It is more like the consequence of more fundamental facts. This is different from the hypotesis of being Hausdorff, which is the mathematical transcription of the fact that we cannot erase completely experimental errors, but we can make them smaller and smaller, at least in principle.
Comment: Good point: Stone-von-Neumann distinguishes precisely the separable case!
Comment: Some remark: Zorn's lemma assures also an orthonormal basis!
Answer: I'm normally loath to just quote Wikipedia, but it does have some relevant things to say:
The article on separable spaces tells us that

every continuous function on a separable space whose image is a subset of a Hausdorff space is determined by its values on the countable dense subset.

(cf densely defined operator) and that

many theorems that can be proved for nonseparable spaces have constructive proofs only for separable spaces. Such constructive proofs can be turned into algorithms for use in numerical analysis [...]. A famous example of a theorem of this sort is the Hahn–Banach theorem.

Furthermore, the article on Hilbert spaces contains the following:

A Hilbert space is separable if and only if it admits a countable orthonormal basis.

In case of field theory, it states:

Even in quantum field theory, most of the Hilbert spaces are in fact separable, as stipulated by the Wightman axioms. However, it is sometimes argued that non-separable Hilbert spaces are also important in quantum field theory, roughly because the systems in the theory possess an infinite number of degrees of freedom and any infinite Hilbert tensor product (of spaces of dimension greater than one) is non-separable. For instance, a bosonic field can be naturally thought of as an element of a tensor product whose factors represent harmonic oscillators at each point of space. From this perspective, the natural state space of a boson might seem to be a non-separable space. However, it is only a small separable subspace of the full tensor product that can contain physically meaningful fields (on which the observables can be defined).
Comment: Probably compactness, metrizablility, and Hahn-Banach are most important.  To understand those pretty much requires at least a masters in math though...
Comment: This answer doesn't really address what is wrong with nonseparable spaces.
Comment: @SudipPaul: quote 1 implies that in general, you can no longer define operators on just a countable subset; quote 2 implies that Hahn-Banach becomes non-constructive: Zorn's lemma will tell you that there is a linear extension, but you have no way to actually get it; quote 3 implies that there's no (countable!) Hilbert basis, which is often assumed in QM when you treat infinite spaces like finite ones; quote 4 implies that restriction to separable spaces doesn't have a known down-side as far as physics go
Comment: Again, none of these quotes address what is wrong with doing physics on a nonseparable space. Is there some physical result that holds only for the separable case? The reason I asked this question because in LQG you generically get nonseparable Hilbert spaces and this fact is sometime used to criticize LQG. I wanted to know why that should be a problem.
Comment: According to Arnold Neumaier's answer of http://physics.stackexchange.com/q/29740/, "Due to superselection sectors, the Hilbert space of QED is already nonseparable"
Answer: ''In this case quantum might become classical !''
To complete the above remarks, classical physics can be seen as being the limit of quantum behaviour in some particular cases of non separable Hilbert spaces.
As a matter of facts, the quantum description of a macroscopic system involves the huge tensor product of the N elementary Hilbert spaces of each elementary component of the macroscopic system. If -- as usual in statistical physics -- macroscopic is modeled by the limit where $N \rightarrow \infty$ then the macro Hilbert space becomes non-separable.
Now there is a theorem by von Neumann (1939) that shows that this non-separable space breaks down into the uncountable direct sum of separable subspaces. Each separable subspace is defined by a reference state and all the states obtained by changing the state of a finite number of elements in this reference state. The reference state can be seen as a vacuum and the modifications as states of a microscopic quantum system.
With this theorem, one can show that any operator on the macroscopic system becomes block-diagonal -- having no element between different separable/microscopic subspaces (i.e between  macroscopically different states), and especially density matrices become block diagonal, providing a decoherence mechanism without considering environment influence à la Zeh-Zurek.
You can have a look at arxiv:2209.01463 for more details.


The Bochner integral about a semigroup of bounded linear operators on a Banach space

Question: Let $T(t)$ be a semigroup of bounded linear operators on a Banach space $X$. When does the following hold
$$
\int_0^t T(s)x ds = \Big(\int_0^t T(s) ds\Big)x, \quad x \in X \, ,
$$
where $ t \in (0,1)$?
Comment: Almost never. The operator valued function is in general not measurable (only if the generator is bounded).
Answer: Following ideas of John von Neumann,
J. von Neumann, Über einen Satz von Herrn M. H. Stone, Ann. Math. (2) 33, 567-573 (1932). ZBL0005.16402,
it can be shown that if a function satisfies the semigroup law and is measurable, then it is already continuous. Hence your identity is only true if the generator is bounded.
See also
Engel, Klaus-Jochen; Nagel, Rainer, One-parameter semigroups for linear evolution equations, Graduate Texts in Mathematics. 194. Berlin: Springer. xxi, 586 p. (2000). ZBL0952.47036., Exercise I.1.7.(2).
Comment: Thanks. If $T(t)$ is a  uniformly contiouous semigroup, then the identity is true. Right?
Comment: Yes. But then, the identity of the right-hand side is also just a Riemann integral of a continuous function. In that case, the generator is a bounded operator, which is almost never satisfied in relevant applications.


Termination from academic program when I was unable to pay the fee and write the exam because i was being Covid +ve

Question: I'm a student of IIT Dhanbad from India, I was terminated from the academic program because I was unable to pay the fee and write the exam. The payment date was 1-7 April and the exam date was 19-24 April. Starting from April month my whole family and I was being COVID +ve. I was tested on 13 April and got Covid +ve my mother and father were in critical condition due to being COVID +ve. in that situation it was not possible to pay the fee and write the exam. I was requested to give a chance to pay the fee but they didn't reply. I had also request for make up exam but they said you are not applicable for it because you didn't pay the fee.
I want to know how can take legal action against them?
Answer: You can't
I'm sorry to hear about your circumstances but they are, unfortunately, yours. The fact that you or a family member is sick or has otherwise suffered misadventure does not give you an excuse for avoiding your contractual obligations; in this case, to pay the fee by the due date. In fact, as I read it, the due date for payment had come and gone before you became ill.


Substitute for bell peppers in meat dishes

Question: I recently became addicted to the taste of fried bell peppers (green specifically because that's what's cheapest) as an ingredient in meat dishes. However, in my area the price has doubled so I can't buy as many.
What are some things (if any) that can replicate the flavor (and the lack of spiciness) of bell peppers? Specifically I like to add them in beef and make sandwiches with cheese and onions.
Comment: To run on from the above comment, in the UK you can by frozen peppers cut in small strips.  Not as nice as fresh, granted, but *much* cheaper.
Comment: I find that bell peppers freeze pretty well so when they go on sale you can stock up, that is if you don't already do that.
Comment: Do you mean roasted/fried peppers? Raw green peppers don't taste like much, just watery and crunchy like yossarian said.
Comment: How are you preparing the bell peppers?  or even, what meat dish?  that might help give us ideas for suggestions.
Comment: @Joe Sorry, I'm frying them and like to add them in for beef cheese steak sandwiches. I'll edit my question.
Comment: @Kryptic : blah ... frying ... bell peppers have a fair bit of sugars in them, which aren't in a lot of vegetables ... let me go update my answer.
Answer: Others are saying that bell peppers have no taste -- but I'd disagree.  The green ones have a kind of grassy quality to them.
Obviously, there's wheat grass, but I never have that around.  There might be some herbs that could pull it off (eg, flat leaf parsley; maybe taragon, but also gives a licorice note)
You might try some dark greens ... spinich, kale, swiss chard.  They've got a touch of other stronger flavors to 'em, but if you cook them down first, you might be able to tone them down some.  (I guess you might describe it as a 'spiciness' ... it can be a sharp flavor.
... for the texture, though, my first thought was bok-choy.  It's actually a cabage, and it has some dark green leaves, but the thicker white parts have that watery/crunchy quality that other people have mentioned about bell peppers, but there's that sharp cabbage note that might keep it from fitting your requirements.
update : okay, as the issue is specifically fried peppers, you're going to be bringing out the sweeter notes of the peppers, which you're not going to get from a cabbage.  Sweet onions might give some similar qualities (once fried well), as the cooking will mellow the oniony qualities, but the sugars will caramelize.
You can also try looking for jarred peppers -- they're usually packed in oil or a vinegar solution after being roasted.  There are 'hot' and 'sweet' varieties, where the hot ones also have some hot peppers in there.  The non-vinegared ones sweet ones might work for you.
Also, remember that bell peppers, like most vegetables are a seasonal item -- if you're in the north, it's the end of winter, so any peppers are being being grown in hot houses, or brought in from the southern hemisphere, which jacks the prices up.
Comment: ooh I agree with trying bok choy. It might not give you exactly what you are looking for, but I definelty feel the same sort of kindred with bok choy that you feel for bell peppers. Asian markets are the best place to find cheap produce, in face you may even find cheaper bell peppers there than your local supermarket.
Comment: Bok choy rules, however if you find yourself going to an asian store to get it, be sure to check the price of bell peppers as well. Around here (US), bells are often half price at these stores compared to the local supermarkets. Peppers are also usually cheaper at latin American stores.
Answer: I find that bell peppers are a predominately watery and crunchy. As such, I'd look for other vegetables that have that characteristic. Depending on how much you are cooking them, you could use carrots, red or white onions, or celery. I think those are all pretty close. To get a little further away from the texture and flavor, you could also try broccoli, asparagus, or green beans; although I'm not sure any of those will be cheaper than bell peppers.
Depending on your climate, peppers are actually quite easy to grow too (at least in my experience). If price is a significant issue, you could try some of your own gardening.
Comment: The great thing about peppers -- there's no need to ever buy seeds ... every time you buy a pepper to eat, you get enough to plant more peppers than you need for a small garden.
Answer: Nothing will ever replace the taste of fried bell peppers. And I mean nothing, if the price double then you pay double. Simple as that.
Comment: I agree, a very definite taste and hard to replicate. *Particularly* when fried.
Answer: Watery and crunchy could also describe water chestnuts, or to go a bit further afield, bamboo shoots or even baby corn. These all tend to be inexpensive, since they are purchased canned.
Answer: Can't eat peppers any more (red, green, yellow etc) in chili I substitute with green squash.  No one yet has noticed.
Comment: This is a great suggestion.
Answer: I haven't found any thing that tastes like a bell pepper. So, I collected the seeds from one I purchased at the grocery store (well over 50 seeds) Bell pepper grow well in pots even on my tiny deck. I picked and froze/dehydrated 15 peppers from just one plant! I saved those seeds too.  It only cost me a slight increase in the water bill. This year I planted three. Dehydrating changes the texture but saves the flavor and space.


Accounts limits on Cross-Program Invocation

Question: Can a Solana program invoke another on-chain program via CPI with more than 35 accounts addresses?
Answer: No. Currently the limit of a single transaction is 1232 bytes. One address is 32 bytes, which means 35 accounts is 1120 bytes. You also require some bytes for instruction data, program data reaching the limit quite fast. Depending on the complexity of the instruction the hard cap is around 35 - 37 accounts at this moment.
CPI requires all addresses to be part of the transaction. Therefore its nearly impossible to have more than 35 accounts in one tx.
There is an architectural proposal made to solve this issue, for more info see here:
https://docs.solana.com/proposals/transactions-v2
Comment: "CPI requires all addresses to be part of the transaction". Does that include read-only accounts as well or I can get them from a data-account when the transaction is being executed? Those read-only accounts are those that are needed for the CPI.
Comment: Yes, it doesnt matter if you need to write or only read data from accounts. Any account
 which you interact with needs to be added to the TX.