mysql database that can be accessed over a vpn?

Question: I have 2 computers. Both with w7 as os. I have installed wampserver on one of them.
I have mysql database on wampserver.
Then I have made a vb.net program to connect to mysql database. I have put the program on both computers.
What I want to do is for those two programs to see the same database that is on one computer. For them to be able to add, delete, update that 1 database.
How do I do that? How do I network the mysql database? Do I also have to install wampserver on the other computer?
What do I do? Please enlighten me.
Comment: i tried googling and I found this stuff, which I can't understand:http://www.nairaland.com/nigeria/topic-127119.0.html
Comment: did you try to connect to your other host? just connect? I dunno what vb use for this, but it must have mysql_connect equivalent. Or your question's just "how to use mysql from vb?"
Comment: No, the program can connect in wamp server where wampserver is installed. I just don't know how to connect it if wampserver is to be accessed on another computer on the lan.
Comment: whats wrong with phpmyadmin?  i see you tagged it but never mentioned anything about it...
Answer: MySQL is a database management system which will respond to ODBC (and equivalent) connections.  Get the name of your computer that is running the MySQL database, and go to http://www.connectionstrings.com/mysql to find an appropriate connection string to use in the  System.Data.OleDBConnection connection string.
Comment: yeah, I already know that. I already have the program. Do I have to install connector.net to the 2nd computer which doesn't have wamp server installed to make this work?
Comment: Yes - the original WAMP server aleady has the required stuff (it came with MySQL), but the 2nd computer needs to have the interface driver (i.e.  Connector/Net) to handle the communication between .NET and MySQL
Comment: I already installed connector/net to the 2nd computer(w/c does not have wampserver installed)I also opened port 80 for both tcp and udp. In the 1st computer(w/wampserveer). Do I also need to open port 80 on the 2nd pc?what do I do next?


Cannot create wifi hotspot

Question: I used connectify to create hotspot in windows. in ubuntu 12.04 i want to create an hotspot, i tried ap-hotspot to create hotspot but it is showing  Your wireless card does not support Access Point mode
how to create a hotspot .
wifi adapter PRO/Wireless 5100 AGN [Shiloh] Network Connection
Answer: You need to check whether your card supports AP mode. To do this execute the following
sudo aptitude install iw
iw list

Look for Supported interface modes:
AP should be listed

Cannot create wifi hotspot


Built-in shelves brackets

Question: I'm planning built-in shelves with a design similar to the pictures belwo.  Building the shelves is straightforward to me but I'm not sure how to do the installation and/or what kinds of "invisible" brackets could be used to secure it to the wall/floor.  Any thoughts on the best approach to achieve a similar look to the photos?
Comment: I'm guessing they are invisibly attaced to the wall
Comment: Fixed or moveable shelves?
Comment: Fixed. Thanks for clarifying.
Comment: Not your question but I would not leave those end shelves to hang in the breeze like that on the far side.  Heavy books will produce a lot of leverage.  I think there should be a final vertical support unless they are somehow invisibly attached to the wall.
Answer: That's an odd set, really.
The ends have to be "floating" to hold any weight, at which point the usual approach is to dispense with the uprights and make the shelves fully "floating."
There are a few ways of floating shelves, all simple in concept but finicky in practice. One is long rods set into the studs (deeper than is safe if you don't know what's in the studs so you can miss wires and pipes) with matching holes in the back of the shelves. Conceptually simple, but if anything is the least bit at the wrong angle, very difficult in practice as the rods and holes won't line up and the shelves won't slide on.
Another approach uses a bracket that screws to the wall surface, and a shelf that's essentially a hollow box that slips over the bracket to hide it.
Comment: I suppose it's possible that the uprights are notched, and the shelves are also notched, so the ends are supported by the un-notched portion of the shelf being held up by the uprights, but even with 2x material that seems a bit of a reach for being loaded with books on the ends and not visibly sagging. And in today's lumber market, I dread to think of the cost of that much framing lumber... (oddly, hardwood does not seem to have shot up as much, so look around for other sources of wood.)
Comment: The uprights could be notched 95% and the shelves 5% ( or uprights  consiting of vertical blocking and a front strip)
Answer: Make sure there is a wood piece at the divider panel to receive the L bracket and nailing/screwing at locations shown in the sketch below.
Also, suggest having a panel at the end, the cantilever seemly too long. Instead of a full-width board, you may consider a stud to make it looks better.
Comment: Your sketch is incorrect on two issues: 1) the vertical dividers extend to the floor, which obviously gives the shelving end support, and 2) there are no "L" brackets. Concealed fasteners give its unique design. The unique design relies on making the vertical supports about twice the distance apart as normal for books , which "cries out" for thicker than normal shelves, which would allow for concealed pegs or in ends and back of shelves.

Built-in shelves brackets


Retrieving a Geocoding Result from ArcGIS Online World Geocoder

Question: I ran the World geocoder service through AGOL on a lot of points (I forgot how expensive it is - dumb dumb mistake, alas!). Worse still, the ArcMap Desktop mxd where I had the results open as I was going through the Tied records and manually matching them, died in the middle of the process, before I saved the mxd. Ack!!! I am trying to figure out if I can get those geocoder results back into Desktop somehow, but I don't know where to look for them. They better be retrievable, given how much they cost me! I have a standalone Standard ArcGIS Desktop license, and an Organization ArcGIS Online account.
Answer: For your MXD, and in the geoprocessing process you have to define the source and destination where the results are stored. So there is someplace on your machine a gdb (usually in your My Documents) where the default.gdb resides.
Comment: Thank you! I did check in my Default gdb, I saw no geocoding results there. The problem is that they were generated by ArcGIS Online, so I wonder if it's possible the temporary file was remotely stored. I found nothing in my ArcGIS Online Content or MyESRI Reports either, sadly.
Comment: You should be able to go to your "Results" view under the geoprocessing menu and see where the last run tasks where processed and stored.
Comment: Good idea! Since that mxd died without being saved, I can't see those Results anymore. A different mxd is using the Default gdb for Scratch Workspace. I don't see any geocoding results in there, sadly. Thank you for thinking this through with me!


What is the byte order for the WiringPi 16 bit I2C reads and writes?

Question: The WiringPi library has APIs called wiringPiI2CWriteReg16 and wiringPiI2CReadReg16 for writing and reading 16 bits of data from a given I2C device at a specific register.  Since the resulting 16 bit value is made up from the read of two 8 bit bytes, is the value Low Byte / High Byte or High Byte / Low Byte?
Answer: By writing an application that reads two 8 bit values at adjacent I2C registers and comparing these against the value returned from a wiringPiI2CReadReg16 I conclude through experimentation that the order of the data returned is Low Byte / High Byte.
Answer: The byte endianess is determined by the I2C device.
It may be low endian or high endian as defined by the I2C device's datasheet.
Comment: Imagine that an I2C device gave the byte value X and address n and Y at address n+1.   Using the WiringPI API, we have a 16 bit read API that doesn't document how it interprets the data.  The data could be read as X<< 8 | Y or Y<<8 | X.   This does not appear related to the intended value of any I2C device ... but rather, what is coded within WiringPI's 16 bit read function?
Comment: Yes, to make sense of I2C words you need to know at least two things 1) the byte order as used by the device, 2) the byte order as presented to you by the client function you are using.


Virtualization Technology Not Supported while Installing HAXM

Question: I want to install HAXM on windows 8.1 (HP pavilion 3515) (I've downloaded the hotfix from here). but the below error occurs during installation  (HyperV is not installed on my laptop). so I installed the "Intel processor utility to check if my laptop supports VT or not, and as you see in the second picture it says "NO". then I checked my bios and I saw "Virtualization technology" item in "configuration manager" tab! and I enabled it, and tried to install HAXM again, but I got the error again and I saw NO in front of "Virtualization technology again.
Now my question is that "is there any way to do so my cup can support it"? or it's a hardware problem?
Answer: It's a hardware problem. If your CPU doesn't support Intel Virtualization Technology then HAXM will not work for you.
Answer: According to ark.intel.com, the T5900 series processors do not support VT-x. Not sure why your bios indicates that you have it.  Perhaps it supports some VT features but not all the VT features that HAXM requires.
Here are some trouble-shooting tips for installing HAXM.
http://software.intel.com/en-us/blogs/2014/03/14/troubleshooting-intel-haxm


h5ai: subfolders are empty

Question: I manage to install h5ai on debian with Apache but I have an issue.
I have unzipped h5ai in my /var/www/files folder and when I go on "http://tmp.domain.com", I can see all the content of my "files" folder. But when I click on one of the sub-folders, h5ai says this folders is empty. Same for every sub-folders of /files...
And when I refresh the page of one of these empty folders (http://tmp.domain.com/nameoffolder), Apache index is shown again but this time with the content of the folder.
I have installed it with Apache and added the next lines at the end of /etc/apache2/sites-available/default :
<VirtualHost *:80>
        DocumentRoot /var/www/files
        ServerName  tmp.domain.com

        <Directory /var/www/torrent>
                Options Indexes

                AuthType Digest
                AuthName "Access restricted"
                AuthDigestDomain "http://tmp.domain.com"
                AuthDigestProvider file
                AuthUserFile /home/tulleb/.htdigest
                Require valid-user

                DirectoryIndex index.html index.php _h5ai/server/php/index.php
        </Directory>

        CustomLog /dev/null "combined"
        ErrorLog ${APACHE_LOG_DIR}/dl-error.log
</VirtualHost>

I tried a chmod -R 777 on the /files folder but it doesn't make any difference.
Do you have any ideas..? Thanks a lot.
Answer: You forgot the leading slash:
DirectoryIndex index.html index.php /_h5ai/server/php/index.php
                                    ↑

Answer: The torrent folder is (as I see it) password protected. Disable it to test h5ai. And check your apache logs dl-error.log, you might have the error there.
Comment: I already tried to comment the lines `AuthType Digest
                AuthName "Access restricted"
                AuthDigestDomain "http://tmp.domain.com"
                AuthDigestProvider file
                AuthUserFile /home/tulleb/.htdigest
                Require valid-user` but it didn't make a change.

h5ai: subfolders are empty


"Green Spire" shrub with no growth

Question: "Green spire" vertical growing shrub not taking root very well, or growing at all. It was planted 4 weeks ago, nutrient rich soil, no problems except mininal morning sun. Ideas?
Comment: A photo would be helpful, but are you saying you only planted it 4 weeks ago? In the ground? Does it look healthy?
Answer: There is a saying regarding woody plants: First year, they sleep; second year, they creep; third year, they leap. In other words No growth the first year, a little growth the second year, and typical growth (or slightly better than typical) the third year. Seeing no growth on any shrub four weeks after planting is typical and normal.

"Green Spire" shrub with no growth


Okay to use essential oil as cat repellent?

Question: We have a cats coming into our yard and pooping in our plant bed. It's getting pretty gross and the neighbors aren't going to do anything about it. I've read that sprays containing citrus essential oils can work as a deterrent. I was considering making a mixture of water and lemon or citronella oil and spraying it on the ground of the plant bed. Should I be concerned about negative effects on the plants? Maybe making the soil too acidic?
Answer: Citronella  plants are toxic to cats and dogs, although their reaction depends on the type of exposure and how long for. Most cats do not like the scent of the plant and will avoid it, though dogs are less selective about it. There are some reports of citronella oil poisoning in cats, but not much testing has been done, see here https://www.canadianveterinarians.net/documents/cats-and-essential-oils, but I wouldn't recommend using the dilute oil  as a spray, given cats' habit of licking their paws.
Cats don't like citrus peels, though you may not like the appearance of lemon and orange peels strewn across your flower beds. The most effective are branches cut from prickly plants laid across the ground (Berberis, holly and the like) where you know they visit, but again, it's not particularly attractive. Inserting lots of short sticks vertically will deter them too. Whatever you use, you must remove any deposits they have previously left behind first I'm afraid.
Cats like loose soil, so if you have recently dug over an area, that's where they will choose to go; as the soil settles back down and becomes firmer, they will be much less likely to use it, so using short sticks or cut prickly branches till the soil does settle should work. I wouldn't recommend gravel or pea shingle, they tend to view that as cat litter because it's easy to scrape around, but larger pebbles do keep them off. Un-neutered tom cats are less fussy about whether they can scrape a surface up, often  just depositing on top or even in un-mown grass.
There are so called cat repellant sprays you can buy, but I never found they worked, and if they did, not for long. They seem not to like mothballs, the old fashioned ones which contain napthalene if you can still find them, but they will need replacing if it rains or after a week or two as the scent fades.
The most effective method of keeping cats off is a sprinkler attached to a movement detector; cats don't like getting wet, so if every time they come along a spray of water gets on them, they will stop coming. Either that or a pebble mulch are the most effective methods.
Comment: Thanks for the info. I've been putting lemon and orange rinds in the bed. We'll see how it goes. Any thoughts on the effect of the citronella on the plants?
Comment: It repels insects, so that might mean you won't get pollinating insects in the area, though it won't harm them.


Do I need statistics in addition to an index?

Question: Suppose I have a table with images that are supposed to go through several steps:
CREATE TABLE images (filename text, extracted bool, cropped bool, resized bool);

INSERT INTO images (filename, extracted, cropped, resized)
VALUES
    ('foo', false, false, false),
    ('bar', true, false, false),
    ('baz', true, true, false),
    ('qux', true, true, true);

At some point I have a query to find all images that are cropped but still need to be resized:
SELECT count(*) FROM images WHERE cropped AND NOT resized;

Now I believe the best way to make that query fast is a partial index:
CREATE INDEX ON images (cropped, resized) WHERE (cropped AND NOT resized);

I'd make it partial because cropped AND NOT resized is a relatively rare state while there might be millions of images that are already fully processed and also millions that are not cropped yet.
My question is now, do I need statistics in addition to the index?
One of these?
CREATE STATISTICS stat1 (dependencies) ON cropped, resized FROM images;
CREATE STATISTICS stat2 (ndistinct) ON cropped, resized FROM images;
CREATE STATISTICS stat3 (mcv) ON cropped, resized FROM images;
ANALYZE images;

I found chapter How the Planner Uses Statistics which I had previously missed (or rather conflated with Statistics Used by the Planner), but it only talks about how statistics are turned into row estimates. What is unclear to me is how indexes are chosen, given that there are apparently no statistics about indexes.
Answer: My experiments showed that currently I don't seem to need those statistics for the index to be used, but it was still unclear why that it. After a dive into the source I think I can answer my own question.
Essentially the decision to use an index is made in btcostestimate() and in turn genericcostestimate().
It helps to remember what kinds of statistics are available for each table:

Number of tuples
For each column: Number of distinct values (sometimes called "cardinality")
For each column: The most common values
For each column: A histogram of the remaining (less common) values
If configured: dependencies stats ("How many values in Column A have only a single value appear in column B.")
If configured: ndistinct stats (Number of unique value combinations in columns A and B.)
If configured: mcv stats (Most common value combinations in columns A and B.)

For each index Postgres determines which conditions can be checked using the index (the "Index Conds" or "indexQuals"). Based on those, genericcostestimate() (using clauselist_selectivity()) calculates a selectivity for the index, taking into account those extended statistics. This was actually reflected in my experiments in that I got better row estimates with extended statistics of the mcv kind:

The difference in actual time is down to caching.
The predicate of the partial index would also be taken into account but only if it introduces additional restrictions, so it's not relevant here.
So what I think how the index got chosen is this: First the index predicate was checked to see if the index is even usable. Then a selectivity was calculated for the particular conditions, which without the extended statistics was indeed be a bit wrong. But then further down, when the actual cost is calculated, it's so low because the index is so small that even with the wrong row estimate the cost is very low.
So the answer is yes, theoretically extended statistics are still needed for good row estimates but also no, the index is still chosen without extended statistics because it is so small.
Answer: You are substantially over-thinking this.  Your query is very simple, and there are only a few ways it could be executed.  Whether it will return 7000 rows or 2000 rows, it doesn't matter because either way the index will appear to be better than the meager alternatives.
If you really do want to run a wider variety of queries which have more opportunities to make the wrong planner choice, it might be important to include the extended statistics of the mcv variety.
Your two examples are utterly mismatched.  The table of counts in your question would lead to vastly different row estimates than are shown in your answer.  It would give around 5,000,000 with no extended MCV stats, and around 1 with the extended stats.  Certainly not 6872 vs 1782.
Comment: While I disagree that I'm overthinking this (I have encountered [cases where Postgres didn't optimize what would have been trivial to optimize](https://dba.stackexchange.com/questions/310530/index-not-used-for-date-range-condition).) your assessment is correct, Postgres chooses the smallest index that can do the job. I removed that confusing table again, I shouldn't have posted that, it was from a different test case and was just to illustrate the ranges involved.
Answer: your query does not sufficiently filter the table. so the postgres query optimizer doesn't use this index and chooses the sequential table scan (or table full scan).
You can change the table design and query design. Or, you should aggreed sequential scan.


Fundamental domains of measure preserving actions

Question: Suppose a finite group $G$ acts on a standard probability space $(X, \mu)$ by measure-preserving actions (i.e. $\mu(g(A)) = \mu(A)$ for all $g \in G$ and $A \subset X$ measurable).  In addition, suppose that for $g \in G$ and $g$ not the identity then $\mu(\{x:g(x) = x\})=0$.  I am wondering if it is always possible to find a fundamental domain $D$ of the action of $G$, i.e. is there a measurable $D \subset X$ such that $G$ is the disjoint union (up to measure 0) of the sets $g(D)$?
Comment: Do you know the answer in the case where $(X,\mu)$ is the unit interval with Lebesgue measure?
Comment: The measure preserving property is not relevant, because you can always replace mu(A) by sum_g mu(gA)
Comment: Mike Hartglass, will you please include the specification of your measure space in the question?  I think many of us don't use the terminology "standard probability space" to mean [0,1] with Lebesgue measure.
Comment: no I do not.  In fact by standard probability space I meant [0, 1] in Lebesgue measure
Answer: I believe the answer is yes. You can first show, that once you have a set $A$ of non-zero measure, such that $g(A) = A$ for any $g \in G$, there is some $B \subset A$ of non-zero measure, such that $g(B) \cap h(B) = \emptyset$ for $g \ne h$. If this is shown, then you can prove your statement by Zorn-like argument, showing that the set $A$ of maximal measure, such that $g(A) \cap h(A) = \emptyset$ will be the fundamental domain.
Here is a hint how to show the first claim. Take $g \in G$, $g \ne 1$, and a $G$-invariant set $A$. Since $g(x) \ne x$, there is some $B_0 \subset A$ of non-zero measure such that $g(B_0) \ne B_0$, set $B_1 = B_0 \setminus (g(B_0))$. Then $g(B_1) \cap B_1 = \emptyset$. Now do the same thing for other $h \in G$ starting from $B_1$.
Alternatively, once can pass to a Stone space and use continuity to get this statement, though this requires more machinery.
Comment: I use separability of $X$. There is a theorem, usually attributed to von Neumann, that every measure preserving automorphisms of a separable measure algebra has a unique point realization on a standard Lebesgue space. From this existence of $B_0$ follows. By $g(B_0) \ne B_0$ I mean $\mu(g(B_0) \setminus B_0) \ne 0$, so $B_1$ has positive measure.
Comment: The second paragraph has some gaps. What properties of $X$ are you using to construct $B_0$, since such a set doesn't have to exist for a more general probability space? You also want $B_1$ to have positive measure. How do you ensure that?
Answer: Yes, take a look at the old papers by V. Rohlin:
MR0030584 (11,18f) Rohlin, V. A. On the fundamental ideas of measure theory. (Russian) Mat. Sbornik N.S. 25(67), (1949). 107--150
MR0030710 (11,40b) Rohlin, V. A. Selected topics from the metric theory of dynamical systems. (Russian) Uspehi Matem. Nauk (N.S.) 4, (1949). no. 2(30), 57--128.
Answer: Here is another way to find a fundamental domain.
First identify $X$ with $[0,1]$. You want to pick a single point in each orbit of the action. Just take the smallest one.
Let be more specific.
Consider the set $A$ of points $x$ for which the number of points in the orbit of $x$ is equal to the cardinal of $G$. Your assumption insures that this set is of full measure. Take some Borel subset $B$ in $A$ of full measure in $A$, such that $G$ acts on $B$ through Borel transformations.
The fundamental domain $D$ is then defined as the image of $B$ by the map $x \rightarrow  \min\lbrace\ gx\ |\ g\in G\ \rbrace$.
Now the image of a Borel set by a Borel map is always measurable. Restricting again to a Borel subset of full measure, we get  a Borel fundamental domain for the action.
Answer: No, you need some other condition, since you aren't guaranteed to have many measurable sets.
For example, a probability pace consisting of a single atom (which need not be a set with one element) with measure $1$ has no measurable subsets of probability $|G|^{-1}$. You can let the space be  $(-1/2,0) \cup (0,1/2)$, let $C_2$ act by reversing signs, and let only the countable and cocountable subsets be measurable, with measures $0$ and $1$, respectively. Then no set has measure $1/2$.
Ok, assume $(X,\mu)$ has no atoms. You still aren't guaranteed that there are enough measurable sets. Let the space be $(-1/2,0) \cup (0,1/2)$, let $C_2$ act by reversing sign, and let the measurable sets be  $A\cup-A$ where $A\subset(0,1/2)$ is Lebesgue measurable. Then all measurable sets are fixed by the action of $C_2$, so there can't be a fundamental domain. As far as measure theory is concerned, this is still the trivial action, even though nothing is fixed by the nontrivial element.
You need some condition like that any set of positive measure contains subsets of positive measure which are moved by the action, whose images are measure-disjoint. Then (at least using some level of choice, but perhaps this isn't necessary) you can build a fundamental domain as a maximal set which has measure $0$ intersection with its images by nontrivial elements of $G$.
Comment: By Standard probability space, I meant the standard diffuse non-atomic Borel probability space.  This is [0, 1] in Lebesgue measure.

I am curious if the answer is true even in this case
Comment: Ah, thanks for the clarification. I think the condition I gave in the last paragraph is still relevant for that.


Lemon curd bonbons

Question: When trying to cap my lemon curd bonbons the chocolate would not set up/seal?Any help is appreciated. Wondering if consistency of curd is not firm enough? Maybe the water portion in it from the fresh lemon
Juice?
Comment: Welcome! Can you add a few more details, what exactly are you making, is there for example a recipe that would help the community understand what's happening?
Comment: As Stephie said, more information would be helpful. A picture can help us understand how you're trying to seal them with chocolate - this may be a language issue, but in English, I've only seen "bonbon" refer to hard sugar candy, not to something that contains lemon curd and chocolate (which may be normal in French).
Answer: Lemon curd does not set the way that ganache does. The fat in ganache (cocoa butter) crystalizes, and this is how ganache sets. But lemon curd is a type of gel made from the (denatured) proteins in the egg yolk, and the phospholipids in the yolk also help emulsify the curd. It is a very different mechanism, and they are not going to have the same consistency.
Typically, bonbons (aka pralines) are made with a ganache center or some other center that does not need to be refrigerated so it is shelf stable. If you're actually making lemon curd, that is probably not the best idea for a bonbon as it needs to be kept refrigerated. You might make a lemon filling that is not curd. Sugar is highly hygroscopic and hydrophilic, and the high amount of sugar in bonbon fillings typically is designed to prevent bacterial growth when stored at room temperature. (Sugar greatly lowers the Aw.)

Lemon curd bonbons


Stream iOS camera live video to other device over Wifi

Question: Is there an app that can capture live video on one iOS device and broadcast it over wifi so all my other iOS devices can play it? Streaming it over the internet would be even better...
Answer: TJUNKS Video Camera
You can stream to other devices. This app does a lot, with video including uploading and sharing video on facebook, twitter etc but to stream video all the devices will need this app.
It also works across platform.
As far as personal experience goes I have only used it once to test it quickly, it workked accross a home WiFi network, other than curiosity I had no need for this.
Comment: I mean live streaming of what the device is capturing. The video does not need to be saved and it shouldn't be accessible to other users. I don't see how facebook is anything like this.
Comment: did you click the link? facebook is just another tool this app has after live streaming.
Comment: I'm sorry, I thought that was a post title style. Not used to this site. Thanks, I'll try that...
Answer: You may be interested in this incredibly awesome little app called CollabraCam. I've never used it, but the idea is just brilliant. One iOS device serves as the "director," and receives video from up to 4 other iOS devices. The Director can send requests to the cameramen (pan up, shift left, zoom in), and I think can record video by switching between different live feeds.
CollabraCam is obviously designed for making movies with multiple camera operators, but it sounds like it might suit your simpler purpose of just broadcasting a live camera from one iOS device to another.
I should mention that I have absolutely no experience with this thing, I just think it's incredibly cool.
(By the way, the wording of your question ("capture live video") is a little confusing. I'm not sure if you want to capture video and stream it afterwards, or if you want to view a live feed with/without recording it. In any event, I think readers who arrive at your question because they are curious about streaming live video from one device to another will find CollabraCam interesting).
UPDATE:
After answering this post I became curious and bought CollabraCam. I thought it would be fun to mount my iPhone on an RC car, and stream video back to my iPad. Turns out CollabraCam does work for live streaming, but not very fluidly. The video feed staggers a lot, and is definitely too slow to use for steering an RC car.
I didn't try anything to speed up the system -- I think I could create an ad hoc network with a laptop, and thereby get faster WiFi streaming than going over the home router...
But the short message is, CollabraCam does work for live streaming, but maybe there are better solutions out there.
Comment: Looks like a nice app although not what I was looking for. I'm not really looking for a capturing method, just a way to stream live video like if it was a webcam...
Answer: I think I have the perfect solution for you,
I found an application called Friendsee!
You can stream live video with great quality to all of your iOS friends, and directly to your facebook wall or twitter account (with 3G/Wifi connection), its really simple and fast, I think it's the fastest way to broadcast live stream with iphone
Good luck & Enjoy!


Kubeflow: Not able to build Makefile - recipe for target 'presubmit' failed

Question: I want to build kubeflow from source and deploy it in a container.
I used the setup instructions from over here:
https://www.kubeflow.org/docs/started/getting-started/
My resulting docker container looks like this:
<code>FROM ubuntu:16.04

ARG DOCKER_HOST=tcp://localhost:2375

RUN apt-get update
RUN apt-get install sudo
RUN sudo apt-get install -y curl gnupg apt-transport-https ca-certificates software-properties-common curl git bash make python wget autoconf

COPY . /usr/src/app
WORKDIR /usr/src/app
RUN ls

RUN wget https://dl.google.com/go/go1.11.1.linux-amd64.tar.gz && tar xvfz go1.11.1.linux-amd64.tar.gz && sudo chown -R root:root ./go && sudo mv go /usr/local && . env/.profile && . env/.bash_profile

RUN cd .. && chmod -R +x app && cd app
RUN make && sudo make install

RUN bash

RUN ./scripts/kfctl.sh init kfapp_config --platform none
RUN ./scripts/kfctl.sh generate k8s
RUN ./scripts/kfctl.sh apply k8s
</code>
Everythings works except the "RUN make && sudo make install" part.
It fails with:
<code>(...)
Boilerplate header is wrong for: /usr/src/app/bootstrap/cmd/bootstrap/app/options/options.go
Boilerplate header is wrong for: /usr/src/app/bootstrap/version/version.go
Makefile:17: recipe for target 'presubmit' failed
make: *** [presubmit] Error 1
The command '/bin/sh -c make && sudo make install' returned a non-zero code: 2
ERROR: Job failed: error executing remote command: command terminated with non-zero exit code: Error executing in Docker Container: 1
</code>
If I skip the making of the source, I encounter the following error at the part "RUN ./scripts/kfctl.sh init kfapp_config --platform none":
<code>(...)
+ check_install ks
+ which ks
+ echo 'You don'\''t have ks installed. Please install ks.'
You don't have ks installed. Please install ks.
+ exit 1
The command '/bin/sh -c ./scripts/kfctl.sh init kfapp_config --platform none' returned a non-zero code: 1
ERROR: Job failed: error executing remote command: command terminated with non-zero exit code: Error executing in Docker Container: 1
</code>
Does anyone has an idea on how to address this issue?
Thanks a lot :)
Answer: You shouldn't have to build Kubeflow in order to deploy. Based on your question it looks like you want to deploy Kubeflow in a single machine/local deployment. Kubeflow runs on top of Kubernetes. If you want to try it locally, the easiest way to do that is using Minikube (https://www.kubeflow.org/docs/started/getting-started-minikube/) or MicroK8s (https://www.kubeflow.org/docs/started/getting-started-multipass/). You'd need to have a reasonable sized machine/laptop and a way to run VMs.


Basic understanding of op amp circuit

Question: I am wondering if my understanding of the following circuit is correct. Specifically it is about the 0V at the non-inverting input. Is it correct to say that, since there is ground at R7 || R8, there will always be 0V at the non-inverting input and when voltage is turned on, the op-amp will try to adjust the voltage at the inverting input to 0V, in order to create zero voltage difference? Which means, no matter what resistor combination there is in the upper branch (non-inverting input), it will not affect the circuit?
Comment: All you need to know is : V+=V-, no current flows through V+ and V-, and ouput is seen as a voltage dependant voltage source :Vout = A(V+ - V-) where A is about infinite. You need to do the nodal analysis using these rules.
Comment: Yes, the opamp will always *try* to keep V+ = V-
Comment: Additionally, these resistors may also add thermal noise to the circuit. The larger the resistance, the larger the thermal noise.
Comment: @Pigrew : true, but you have to give a magnitude to evaluate this. Thermal noise voltage of a resistor is v = sqrt(4kTRB). Noise of a 2k resistor in audio bandwidth (22kHz) is about -120dBu. This is much less then the noise added by most generic opamps by theirselfs.
Answer: The resistors at the non-inverting input are for the bias current compensation. In a non-ideal opamp the bias currents flowing into the inputs are non-zero.
Parallel resistance of R7 and R8 is nearly equal to the parallel resistance of R2 and R3 (1485 Ohms).
The matching in this circuit is not correct, because the output impedance (R1) of the source is not taken into account.
(R1 + R2) || R3 = 1973 Ohms. A better choice for R7 and R8 would be 3900 Ohms resistors.
For more information on the input stage and bias current compensation take a look at this app note.
Comment: Your basic understanding of the circuit is correct. The opamp has a large open-loop gain (> 1e6). It amplifies the voltage difference between the inputs. Due to the feedback loop, the output voltage participates on the inverting-input voltage. This forms a control loop and tries to keep the voltages in equilibrium. In an ideal opamp with infinity open loop gain, the input voltage difference gets zero as you descriped it. In non-ideal opamps with limited open loop gain a small voltage difference will remain.
Comment: Yes, thanks for that. Could you also give me some feedback to my attempt of explaining the circuit behaviour? Then I would gladely check your answer as the solution.
Answer: First, the parallel combination of R7 and R8 is equivalent to a single resistor. In this case, Req=R7/2=R8/2.
Second, an ideal OpAmp draws no current on its input pins. If no current flows through Req, the voltage across Req is zero and Vp is at ground potential. There is no need for the OpAmp to regulate anything or even to be operating in the linear region for this to happen!
This is how the simulator is getting Vp=0. If the OpAmp is ideal, Vp will be zero by the same argument regardless of the value of Req (i.e. the values of R7 and R8). With real OpAmps you may have to account for bias currents depending on the specific problem at hand.

Basic understanding of op amp circuit


Is Microsoft.Practices.Composite earlier version of Microsoft.Practices.Prism?

Question: In my project I use some kind of SDK libraries written by external team. These libraries are using Prism. For some reasons we had to rollback to previous version of their SDK and now build is failed trying to find reference to Microsoft.Practices.Composite.dll. Am I right that this is how Prism library was called in earlier version or is it something completely different?
Answer: Microsoft.Practices.Composite is from Prism 2.x
The currently version of Prism is 4.1!
Source: http://msdn.microsoft.com/en-us/library/microsoft.practices.composite.aspx
Answer: If you have problems with the upgrade/rollback, you might take a look into:
http://msdn.microsoft.com/en-us/library/ff921073%28v=PandP.40%29.aspx
and, more specifically about the SDK's assemblies:
http://msdn.microsoft.com/en-us/library/ff921144(v=pandp.40).aspx#AssemblyRef
While these documents talk about the upgrade process, it should help you understanding what should be taken care of during a rollback.
Answer: You are correct. Pre v4, Prism's dlls included Microsoft.Practices.Composite.dll but, as of v4, this has been rolled into Microsoft.Practices.Prism.dll along with some other functionality.
The documented list of changes is this:

The Composite Application Library was renamed to the Prism Library.
The Composite and Composite.Presentation portions of the namespaces were removed and the Composite and Composite.Presentation assemblies collapsed into a single assembly named Microsoft.Practices.Prism.
The Microsoft.Practices.Prism libraries for Silverlight and WPF now register the Microsoft.Practices.Prism.Regions, Microsoft.Practices.Prism.Commands, and Microsoft.Practices.Prism.ViewModel namespaces with the http://www.codeplex.com/prism xmlns definition.
Several reusable user interface (UI)–based behaviors were extracted into the Prism.Interactivity assembly, including the interaction request behavior.
You can now use MEF as the dependency injection container. This functionality required two new projects in the Prism Library solutions: Prism.MefExtensions.Desktop and Prism.MefExtensions.Silverlight. These projects create a new assembly, Microsoft.Practices.Prism.MefExtensions.dll, in the respective Desktop and Silverlight folders. Also included in the solutions are new unit test projects for the new functionality.

Source here.


Dual-monitor – dialog boxes appear on "wrong" screen

Question: At home I work regularly between two monitors – I have a Dell monitor connected to my MacBook Pro. It's not something I've done a whole lot until recently so I was trying to find a solution to the problem that when I'm working in an application on the connected monitor, how can I always have the dialog boxes for that application open on that monitor, and vice versa? It's pretty annoying and definitely slows down work a bit, especially if an application open on the other monitor is in full screen.
Comment: I have scoured the web again and only find people pleading for an answer. Someone has to have an idea what is wrong, no?
Answer: If you open System Preferences - Display, then click on Arrangement, you will see a white bar above one of the displays. This is where the menu bar is and will be the primary display and probably the screen that the dialogue boxes are opening on. To change just drag it across to the other screen in the Arrange window and this should then make the dialogue box appear on that screen.
Comment: The problem is that it is never "solved". It happens to multiple software systems, so it is a Mac/Windows problem with dual monitors, not the problem of a specific system. Also, it is erratic. That is, the first click drops down the menu on the secondary monitor (where it should), the second one drops it back to the primary (laptop) monitor and also in very small size. Extremely annoying, and questions all over the web. No solutions found to date.
Comment: OP's not talking about Notification Center pop-ups. OP means dialog boxes within apps, which do sometimes appear on the wrong screen in multi-monitor setups.
Comment: I tried this and it solved the problem with the dialogs


One to Many outer join using KTable in Kafka

Question: I have a table contact published to topic#1 with key as contact.id. Also a subtable is published to topic#2 with key as subtable.id. Now, I need to do a join like
select * from contact c  outer join subtable st on c.id = st.id;

There is a 1..n relationship between the tables. How to perform a aggregation with/without lambda here.
Answer: Kafka Streams API does not support 1:n KTable-KTable joins yet (version 1.0 or earlier).
The only non-primary key join supported is a KStream-GlobalKTable join.
More details about join can be found in this blog post: https://www.confluent.io/blog/crossing-streams-joins-apache-kafka/
Also, there is a JIRA for 1:n joins: https://issues.apache.org/jira/browse/KAFKA-3705


Side by side subtitles

Question: Is there a video player in Linux that can display subtitles from a srt file correctly for side by side 3D video?
Answer: Yes, Bino. It can display the videos in even/odd rows (or other options for different 3D displays) so you don't actually need to set the display to 3D mode, and it can display subtitles. At least... it supposedly can show subtitles, I can't seem to get a .srt file working with it at the moment, but it shows options for subtitle display.

Side by side subtitles


how come square root of 2 times itself equals 2

Question: since square root of $2$ is a irrational number, which we know or assume its decimal part is not a finite number, or doesn't terminate, how come we say that this infinite number (not in terms of being a large number) times itself is equals $2$.  It is like saying $0.928041\dots$ times itself is some finite number (like $1.64$ not going $3$ dots in the end).
Comment: If it helps, you can represent $2$ by $2.000\ldots$ They're two ways of saying the same thing.
Comment: Isn't that the definition of the square root? "A number which produces a specified quantity when multiplied by itself."
Comment: Note that $0.9999\ldots\cdot0.9999\ldots=1\cdot1=1$.
Answer: It's by definition. $\sqrt 2$ is the unique positive number that, when squared, results in $2$. Are you also surprised that $\frac13+\frac13+\frac13=1$ even though you are adding a number repeatedly to itself that has a decimal representation that is non-terminating? (Similar to above, $\frac13$ is defined as the unique number that, when multiplied by $3$, results in $1$)
Do not confuse a number with the representation of a number by a decimal expansion.
Comment: The last sentence is probably good advice. But. It is possible (and maybe even traditional) to construct the real numbers as decimals. I remember that when I first taught analysis (c. 1965) there had been set the previous year as a finals examination question "Explain briefly how to construct the real numbers by Dedekind Cuts, Cauchy Sequences, Decimals. Prove that these yield isomorphic ordered fields."
Answer: There is no such thing as an infinite number. If you mean that the decimal expansion of $\sqrt2$ never ends, then you are right. But that also applies to the decimal expansion of $\frac13$; however, I suspect that you see no problem with the assertion that $3\times\frac13=1$.
The fact that the decimal expansion of $\sqrt2$ never ends is something that concerns the specific way that we use to represent numbers, which was developed centuries after square roots started to be studied. It is not something about the number $\sqrt2$ itself, which is defined as the only real number greater than or equal to $0$ whose square is $2$.
After having defined it this way, we can try to compute its decimal expansion. For instance, $1^2<2<2^2$, and therefore $1<\sqrt2<2$. Furthermore, $\left(\frac75\right)^2<2<\left(\frac85\right)^2$ and therefore $\frac75<\sqrt2<\frac85(1.4<\sqrt2<1.6)$.
Comment: That is just not something you can disagree with. One can use calculus to prove $1/3=0.333...$. That is not a matter of opinion.
Comment: You are mixing two things that you should not mix: the **number** $\frac13$ and its **decimal representation** $0.3333333\ldots$. We *define* $\frac13$ in such a way that $3\times\frac13=1$. And when we compute its decimal representation we get $0.3333333\ldots$
Comment: derivation of this "a=0.3333... then 10a=3.333.... then a=1/3 " is weird too. how come you multiply something that never ends and then subtract it with  another thing that also never ends.  So I don't really agree  on 0.3333... = 1/3.
Comment: @ozguncan $0.333...$ can be expressed as the infinite sum $\sum_{n=1}^{\infty}3(1/10)^n$. This is a geometric sum that equals exactly $1/3$
Answer: In the real world, you only want to be accurate up to a point.  Within an inch, or a millimetre, or much less if you are building a computer chip.
Two people might agree that two numbers are close enough to being equal, but a third might disagree.
If two numbers are so close that everyone agrees they are close enough - so they agree to every decimal point - then we say they are equal.  So $1/3=0.333333333...$


Does there exist anything more general (extension) than operator of QM?

Question: In classical mechanics we simply have quantities which are simply scalar like momentum, energy of system but when we transit to QM it's an ad-hoc principle, at least to me, that we'll be dealing with quantities which are operator and hence commutativity can't be assumed a priori. Now I want to know if there exist anything which are kind of extension of operator since scalar in mechanics are scaled identity operator. By extension I have analogy of Real, Complex, Quaternion, Octonion in mind, the preceding thing is a substructure of the next.
Comment: Operators in quantum mechanics like momentum and energy can be viewed as a kind of measurement. There are [POVMs](https://en.wikipedia.org/wiki/POVM) in quantum mechanics, which are more general measurements than operators, but unfortunately these can't be made to correspond to physical quantities like momentum, energy, position, spin (which I think is what you're really asking about).
Answer: All operators met in textbook QM are linear. So, if one rephrases slightly your question
Does there exist anything more general than linear operators in QM?,
then the answer suggests itself: non-linear operators.
As it happens, non-linear operators do play an important role in quantum field theory, where interactions can change the number of particles.
Answer: A possible extension of operators will be operator-valued distributions.
Functions contain scalars as the constant functions, distributions are a generalization of functions, but there is a wide class of distributions whose codomains are not a field, it is an algebra of operators instead.
The context for this generalization is vector bundles over a manifold.
This article will be helpful for you.


How can I calculate the potential values of the final value from the value of standard deviation?

Question: If I have a list of 99 values, say $x_1$ to $x_{99}$, and I have a value for the standard deviation of an $x_{1}$ to $x_{100}$ data set, containing these 99 values + 1 other, can I calculate the potential values for the extra data point? Is there a simple algebraic solution to this that I am missing?
Comment: Hint: see the definitions of mean, variance and standard deviaton. Then write them as applied to this case.
Comment: looking... haha
Comment: so let's say you have: (N-1)std^2 = sum(xi-sum(xi)/N)^2 (i=1 to 3)
Comment: I'll help you get started. So let's first look at the $99$ data points. I'll assume that you have calculated their mean $\mu_{99}$ and variance $\sigma_{99}^2$. Now we assume that $\sigma^2_{100}$ is also given. The formula for $\sigma_{100}^2$ is
$$
\sigma_{100}^2 = \frac{1}{100} \sum_{i=1}^{100} (x_i - \mu_{100})^2 = \frac{1}{100} \sum_{i=1}^{99} (x_i - \mu_{100})^2  + \frac{1}{100}  (x_{100}- \mu_{100})^2
$$
On the other hand, we know that $\mu_{100} = \frac{99\mu_{99} + x_{100}}{100}$ ...
Comment: never used LaTex before, can I post a picture on here to show my working? see if you can point me in the right direction? my assumption was to put mu 100 into the first equation for sigma squared 100 and then expand the squared brackets and re-arrange so that you have two sums, one can be reduced down simply to sigma squared 99
Comment: you can simplify further by using the sum of squares formula too... still not getting there though
Comment: Hey @MattiP. I am still looking at this, so any further assistance would help
Answer: We are basically comparing two data sets, one with $99$ entries and another with $100$ entries. In the latter, the first $99$ are the same as in the first set and the last value is "unknown". Additionally, we know

the mean of the first $99$ values, $\mu_{99}$
the variance of smaller set, $\sigma^2_{99}$
the variance of larger set, $\sigma^2_{100}$

What we want to know, is the value of the last entry on the larger set, the unknown value $x_{100}$. We can start by considering how to calculate means and variances. It's easy to see that the mean of the larger set is just
$$\tag{1}
\mu_{100} = \frac{99 \mu_{99}+x_{100}}{100}
$$
The variances are
$$\tag{2}
\sigma^2_{99} = \frac{1}{99}\sum_{i=1}^{99} \left(x_i - \mu_{99} \right)^2
$$
$$\tag{3}
\sigma^2_{100} = \frac{1}{100}\sum_{i=1}^{100} \left(x_i - \mu_{100} \right)^2
$$
Let's try to write the latter in terms of the former. Splitting the sum and using Equation ($1$), we get
$$
\begin{split}
\sigma^2_{100} =& \frac{1}{100}\sum_{i=1}^{99} \left(x_i - \mu_{100} \right)^2 + \frac{1}{100} \left(x_{100} - \mu_{100} \right)^2\\
=& \frac{1}{100}\sum_{i=1}^{99} \left(x_i - \frac{99 \mu_{99}+x_{100}}{100} \right)^2 + \frac{1}{100} \left(x_{100} - \frac{99 \mu_{99}+x_{100}}{100} \right)^2\\
=& \frac{1}{100}\sum_{i=1}^{99} \left(\frac{100x_i -99 \mu_{99}-x_{100}}{100} \right)^2 + \frac{1}{100} \left(\frac{100x_{100}-99 \mu_{99}-x_{100}}{100} \right)^2\\
=& \frac{1}{100}\sum_{i=1}^{99} \left(\frac{100x_i -100 \mu_{99} + \mu_{99}-x_{100}}{100} \right)^2 + \frac{1}{100} \left(\frac{99x_{100}-99 \mu_{99}}{100} \right)^2\\
=& \frac{1}{100}\sum_{i=1}^{99} \left([x_i -\mu_{99}]+\left[\frac{\mu_{99}-x_{100}}{100}\right] \right)^2 + \frac{99^2}{100} \left(\frac{x_{100}- \mu_{99}}{100} \right)^2\\
\end{split}
$$
Let's look at the first term. We have
$$
\begin{split}
\sum_{i=1}^{99}  \left([x_i -\mu_{99}]+\left[\frac{\mu_{99}-x_{100}}{100}\right] \right)^2 =&  \sum_{i=1}^{99} \left\{ [x_i -\mu_{99}]^2 + 2[x_i -\mu_{99}]\left[\frac{\mu_{99}-x_{100}}{100}\right]  + \left[\frac{\mu_{99}-x_{100}}{100}\right]^2 \right\} \\
=&  \underbrace{\sum_{i=1}^{99} [x_i -\mu_{99}]^2}_{=99\cdot \sigma^2_{99}} + 2\underbrace{\sum_{i=1}^{99}[x_i -\mu_{99}]\left[\frac{\mu_{99}-x_{100}}{100}\right]}_{=0}  + \underbrace{\sum_{i=1}^{99}\left[\frac{\mu_{99}-x_{100}}{100}\right]^2}_{=99\cdot \left[\frac{\mu_{99}-x_{100}}{100}\right]^2}  \\
=& 99 \sigma^2_{99} + 99\cdot \left[\frac{\mu_{99}-x_{100}}{100}\right]^2
\end{split}
$$
Plugging this back in, we have the equation
$$
\sigma^2_{100} = \frac{99}{100} \sigma^2_{99} + \frac{99}{100}\cdot \left(\frac{x_{100}- \mu_{99}}{100}\right)^2 + \frac{99^2}{100} \left(\frac{x_{100}- \mu_{99}}{100} \right)^2
$$
Phew. Can you continue from here?
Comment: That helped massively! Thanks. One question, why do you use N instead of N-1 when calculating standard deviation? so 1/N instead of 1/(N-1)?
Comment: I assume that it's the **population** standard deviation. A **sample** standard deviation would use $\frac{1}{N-1}$.


Flashlights and Capacitors

Question: Why capacitors are used in flashlights of cameras instead of attaching the light directly to the battery? How this leads more light?
Answer: There are 3 main reasons for using a capacitor.
First it stores the energy, so it can deliver a pulse of energy that is far larger than the battery can. Remember it may take several seconds of battery energy to fully charge the flash capacitor. Then the capacitor releases all that in less than a millisecond ($10^{-3}s$) or even just a few microseconds, so the flash bulb gets a massive jolt of energy.
Secondly, the flash capacitor stores the energy at a much higher voltage: we're talking about up to 1000V (typically around 300V), instead of the 6V from 4 AA cells.
Finally, the capacitor is designed so it can deliver extremely high currents, again higher than the battery can deliver by itself.
Finally, the charge can stay on that capacitor for a very long time. Never touch a charged flash capacitor. The energy stored in them can be lethal!
Comment: The charging circuit first converts the 6V DC to AC. Then it uses a transformer to increase the voltage to whatever the flash needs to fire, rectifies this and charges the capacitor. This high voltage, and the lerge current is what gives the flash its power. The same system (but in reverse) is used in powerpacks. First the AC mains is rectified. The resulting DC is converted to high frequency AC. A transformer is used to bring the voltage down, and then it is rectified again. In both cases, the use of high frequencies means a tiny transformer can be used.
Comment: Thank you hdhondt and freude for helpful answers. I am doing some calculations but I can't get to this conclusions. The capacitor $C$ charged with a battery $\varepsilon$ will release the charge like $\varepsilon C e^{-t/RC}$. $R$ is the resistance of light. The power $P=I^2R=(\varepsilon^2/R)e^{-2t/RC}$ which is not more than the $\varepsilon-R$ circuit power which is $P=\varepsilon^2/R$ !
Comment: About the second reason, how is that done? $300V$ from $6V$?
Answer: Because we don't need lighting, but flash, that means we have to release much energy is a short time interval. Capacitors could accumulated and release energy fast.

Flashlights and Capacitors


Run Adobe Fireworks commands from Command Line

Question: I've written few Fireworks commands (jsfw files) and Flash commands that I use for my image processing. Is it possible to run them from the command line without launching the whole UI app for Fireworks or Flash?
Answer: I know Keith Peters had a chapter in Extending Flash MX 2004 about calling jsfl scripts from the command line, but it will open up Flash, as in the whole UI. I don't see how you can do that without opening the UI.
If you want to do some image processing from the command line I suggest giving ImageMagick a try.
Goodluck,
George
Answer: The JSFL API can't be accessed outside the IDE-s, so you must run Flash or Fireworks in order to be able to run the commands.
Cheers,
Rob


Could Mycoon as a company name be insulting?

Question: We would like to name our e-commerce startup Mycoon. Everybody in the team is happy with Mycoon (mainly because it resembles to tycoon and we just love racoons) but I've just found it out on urban dictionary that coon can be insulting. We are based in Europe but we expect customers from the US as well. Could Mycoon insult anyone or people would associate to the racoon as we intend (with a racoon in our logo)?
Comment: Just as coment as I'm not native AE: I like the name and I suppose it's "different enough" from *coon*, especially with the logo.
Comment: I think that this is difficult to answer from a language perspective. Certainly some folks will not find it offensive and some will. In the Southern US, I think it would be likely to be understood as the slur, but I don't have any objective evidence to support my opinion. You may want to try a name that works with the beginning of raccoon instead of the end. A better link for those of us at work that describes the etymology of the slur is https://en.wikipedia.org/wiki/Raccoon#Etymology
Answer: Potentially, since "coon" is a (rather outdated) racial slur for African Americans.
It could evoke "my n****r" or the literal possession of a "coon" which of course would be a very bad association to make.
That said, I rather doubt most people under the age of 35 would bat an eye or consider it anything other than a raccoon reference. In keeping w/a startup approach, evaluate the name w/your target audience to determine their impressions & adjust as needed.
If you were a U.S. company based in a southern U.S. state I would advise against it. A European startup? If there are variations on using this term in the UK or elsewhere around Europe, I'm not familiar with them. There are always plenty of other rude words people invent to disparage one another. It will likely be clear that this isn't your intent. Of course, there's always a chance of making somebody angry on the internet no matter what you do. At some point negative words from the past deserve/need to have their meaning readjusted - you'll have to decide how much of a risk you'll be taking with the name.
Comment: I think whether anyone younger would understand it as a slur would depend on who their parents are. Also, it can be applied to any dark skinned minority depending on what you precede coon with based on what I've overhead. In my opinion, this is more of a business risk question than a language question because there's no question that it's still in use in some areas as a slur.
Comment: True - I suppose there's a distinction between "understanding" it as a slur, and a willingness to use/interpret it as such. If there are variations in the UK or elsewhere I'm not familiar with them. Will edit a bit.
Comment: I think it is uniquely American based on the etymology. I don't associate with folks that use racial slurs, and I work a company that doesn't tolerate any type of disparagement in the workplace, so I really can't say how widespread the usage is, but I have overheard it being used more than once, so I don't think it's completely archaic at least in the rural South.
Comment: Everyone in the US knows what "coon" means as a racial slur, even if they're not racist creeps. And ColleenV, I have lived many years in the South and Northeast, and that term is widely and actively used. Adding "my" in front of it raises some historical concerns, although I think people would consider the name an unfortunate choice rather than an intentional slur. That's all I can tell you as a native US English speaker.


Is there an app for at least chat?

Question: I use Stack Exchange almost exclusively from my (Android) phone, where chat is basically unusable. A conversation on one of my questions got moved to chat. Is there an app or any other way to participate besides the main web UI?
Comment: There used to be... https://stackapps.com/q/4659/32479 .. You could nose about on StackApps to see if there is a current version.  If an app exists, that is where you'll find it.
Answer: I agree that the chat UI is not very mobile friendly. But as far as I know, there is no longer an official stackexchange app. The old app didn't support chat anyway IIRC.
Perhaps there is a third party solution.

Is there an app for at least chat?


Terminal engaged while using gedit or firefox

Question: When I launch gedit or firefox from a terminal, the terminal is occupied(by occupied I mean no other commands can be run from the same terminal window) untill I press ctrl+c but it kills gedit or firefox. Any way other than opening a new terminal window?
Answer: As per http://www.linuxandlife.com/2012/07/how-to-run-multiple-commands-in-terminal.html follow the command to launch the app with an ampersand, e.g.:
firefox &
gedit &
Following the command with the ampersand launches the app then returns cursor control to the terminal window immediately so you are free to run another command within that window.
Comment: That will not work, the commands will hang until you exit firefox
Comment: I don't want to run multiple commands...it is that when i use the command gedit the terminal is occupied and can't take any other command until is press ctrl+c and when i press ctrl+c the program gedit gets killed as well. So i wanted to know is there any other ways of doing that without opening another terminal window...??
Comment: No, not && , just a single &. If you use && or || , that is something else. See http://stackoverflow.com/questions/4510640/command-line-what-is-the-purpose-of
Answer: You put the applications into the background with the & ...
firefox &

If you want to re-direct error messages and close the terminal leaving the applications running,
nohup firefox > /dev/null 2>&1 &

Comment: yes...nohup worked.


What can increase natural reach for a druid?

Question: I was wondering which items and feats would help my druid (level 15) increase his tripping ability. I plan on doing a reach trip build in wild shape and tree shape form.
Answer: Feats
Lunge and Monkey Lunge are the easiest way to increase your reach.
For tripping, we have several good options:

Improved Trip and Greater Trip are both solid choices for any build focused on trip;
Fury's Call will allow you to add your Dex bonus to your CMB;
Tanden Trip is also an interesting option if you plan on tripping with an ally;
Wolf Trip will allow you to pick where the target will fall prone;
Drag Down will grant a free trip attempt when you are tripped.
Tripping Strike will allow you to make a trip attempt whenever you score a critical hit.
Riptide Attack might be interesting if you desire to drag enemies aswell.

Items
Both Longarm Bracers and Pliant Gloves do talk about arms when increasing your reach, so depending on what you agree with your GM, your tree forms may or may not have arms per se.
The Gauntlets of Skilled Maneuvers will grant +2 on the chosen combat maneuver, a Brown Thorny Ioun Stone will grant +2 competence bonus on trips. Other than that, most trip enhancing magic items are magic weapons or weapon ehnacements.
However, a Wild Adhesive Armor will grant a +2 enhancement bonus on trip checks when using natural attacks. The Wild enchantment is required or otherwise your armor bonuses will cease to function while polymorphed.
Comment: Right and how about for trip? Improved trip, greater trip, fury's fall, are there any others? I only know of the three. Are there ant traits that will help tripping specifically?
Comment: It's worth noting that Lunge/Monkey Lunge only apply during your turn, not for Threatening opponents during their turns.


Is lorentz force attractive, repulsive or both?

Question: Current carrying wire is placed near permanent magnet so wire experience a force, the Lorentz force but what is the nature of this magnetic interaction between wire field and magnet field? Is this a repulsive force? attractive force or both at the same time? (wire is being pushed from one side and pulled from another)
Answer: You no doubt are happy with the idea that the north pole of one magnet repels the north pole of another magnet.
The two forces on the magnets are a Newton third law pair - the force on magnet $1$ due to magnet $2$ is equal in magnitude and opposite in direction to the force on magnet $2$ due to magnet $1$.
The forces of interaction are due to the magnetic field between the magnets.
In your case the force on the current carrying wire wire due to the magnet is equal in magnitude and opposite in direction to the force on the magnet due to the current carrying wire with the magnetic field providing the interaction between the wire and the magnet.
The force on the magnet is usually not noticed as the magnet is "fixed" (eg via frictional forces) to the ground.


Principle of friction force

Question: When two equal and opposite forces act on a body at rest or in motion we say it is rest or in its state of motion. But when a body kept on a table starts moving uniformly due to a force applied, it overcomes force of friction.Does it really overcome the force of friction or is just equal to it.   ie for force F =  ū.N if it's so then why would it start to move,why not stationary as the two forces balanced?
Answer: The frictional force acting on the object when at rest is $\le \mu R$ where $\mu$ is the coefficient of friction between the two surfaces and R is the normal or contact force experienced by the object. Note that $\le$ means that friction is not always equal to $\mu R$ rather this is the maximum value that friction can take. Friction will oppose the applied force up until the applied force becomes greater that $\mu R$ at which point it will be unable to increase and there will be a resultant force causing the object to accelerate (start moving).
Edit: Once in motion the object will experience a constant resistive friction; however it will not in fact be equal to $\mu R$. When moving the object experiences kinetic friction as opposed to static friction when it is stationary - this kinetic friction is $< \mu R$ and constant. A graph of the frictional force against the aplied force would look like this:

TLDR; Yes the applied force must be greater than $\mu R$ for the object to move.
Comment: Bro, when the force overcomes the force of static friction, the kinetic friction comes into play which is smaller than static one, then a net external force creates which moves the block. Then it must accelerate. But can it not move in uniform Velocity? If it moves then how? That's my problem.
Comment: @PradipDebnath Im not sure I understand you. Yes there is a net force producing acceleration and as a reult the velociy will not be constant. It may seem constant (e.g say you are pushing a very heavy box) but this is becuse the force you apply is not much greater than the kinetic friction so the acceleration will be small.
Comment: @PradipDebnath This answer says nothing about the velocity being constant.
Comment: @PradipDebnath After the block has **accelerated** to a certain velocity, you can reduce the applied force from being $\mu_sN$ to $\mu_kN$. This will help you achieve a constant velocity.
Comment: @PradipDebnath Regarding your first comment, see my answer. After you overcome the max static friction force and sliding occurs the object will accelerate, because your force will be greater than the kinetic friction force. But you can then reduce your force to equal the kinetic friction force and the object will continue on at constant velocity. If you reduce your force below the kinetic friction force the object will stop and now you will have to increase your force to more than the max static friction force to get it moving again. This is because of the difference between two forces.
Answer: It overcomes the force of static friction. Which is why it starts moving.
Soon after that, it is set equal to the kinetic friction. Which is why is isn't accelerating. Remember that motion doesn't require force. Only changes in motion do. An object "moving uniformly due to a force applied" has a net force of zero.
Comment: Steeven, since kinetic friction is less than static, the body will accelerate if the applied force that overcame static friction is not reduced to equal the static friction force. Don't you agree?
Comment: @BobD Agreed, yes.
Comment: Oops just realized I meant to say "is not reduced to equal the KINETIC friction force". But I guess you figured that.
Comment: @BobD I did. Still agree. But I do not feel such explanation fitting in my answer here, though. But thanks for the heads-up.
Comment: You're right Steeven. You did say after it starts moving its set to kinetic friction. That's why I'm up voting it.
Answer: You have two types of friction, static and kinetic (a.k.a sliding) friction. When you begin to apply a force on an object of mass $m$ on a table it is the static friction force that resists your effort to move it. As you keep increasing your force, the static friction force opposing you continues to increase, but only up to a point where your force equals the maximum possible static friction which is $μ_{s}mg$ where $μ_s$ is the coefficient of static friction.
When the maximum static friction force is reached the object starts sliding. Now the friction force that opposes your force is kinetic friction and is $μ_{k}mg$ where $μ_k$ is the coefficient of kinetic friction.
Now here's the important point. Usually the coefficient of kinetic friction is less than static friction. So the kinetic friction force will be less than the force you applied to overcome static friction. Therefore, if you maintain that same level of force the object will accelerate based on the difference between your force and the opposing kinetic friction force.
On the other hand, if you reduce your force so as to equal the kinetic friction force, the net force will be zero and now the object will continue at constant velocity, the  velocity it had just before you reduced your force.
The diagram in @Cal answer below shows how the friction force drops from static to kinetic. It is based on a diagram on the Hyperphysics web site.
Hope this helps.
Answer: It's just that the scenario "when a body kept on a table starts moving uniformly due to a force applied, it overcomes force of friction" is a bit nastier than the simple sentence would lead us to believe. It makes us think of an object which (i) immediately starts moving, and (ii) keeps a constant speed. Well, it would have to accelerate first. So the event that most closely resembles the scenario expressed in the sentence, but yet physically sound, is the one @harshit54 alluded to: The force would have to be somewhat larger than the maximum static friction to initiate motion. Then the object will accelerate to the speed in question (during which time the force may be reduced, since the force of friction will reduce due to typical kinetic friction coefficients being smaller, but that's not crucial, I think). Then once the speed is at that value, the force would need to equal the force of kinetic friction from then on.

Principle of friction force


Correct order of horizontal and vertical strokes

Question: Could someone explain to me why with kanji like 生 after the top horizontal stroke I have to go on with the vertical one, while instead with kanji like 金 I have to draw the two horizontal strokes and then the vertical one? Is there a simple rule to be applied?
Comment: to add on your confusion, take a look to the stroke order of 全 compared to金. There is no hard rule for stroke order, only guidelines.
Answer: The primary reason the stroke orders are what they are is that in Japanese, lots of characters are optimised for writing efficiency and/or cursive script. While they seem illogical at first, if you follow these stroke orders, you will eventually start writing more efficiently with less tired hands for longer pieces of writing (while simultaneously maintaining legibility).

This is「生」written in one or two strokes, which can only be achieved if you follow the hand motions of the Japanese stroke order.

This is「金」written in three or four strokes. Again, it follows the hand motions of the stroke order that you're taught.

「全」is no exception!

The secondary reason is that if you don't follow these stroke orders, you risk getting marked wrong in your tests.

All images from cidianwang.com and jisho.org.
Answer: This site here may help:
https://www.tofugu.com/japanese/kanji-stroke-order/
Summary:
Kanji Stroke orders are:

start left-to-right, top-to-bottom;
then horizontal lines coming before vertical lines;
then symmetrical lines on either side of a vertical line;
then symmetrical strokes not on a line;
then inside-to-outside (unless it's a box kanji, then box it in before writing the rest);
then right-to-left diagonals;
then line that cross a lot of other lines;
then dashes at the end unless the dash is the top-left line, in which case it comes first.

Knowing these stroke orders can help you write faster for longer. The only other place I can think this is useful in is school if they make you write them step-by-step.


How do I get the extras?

Question: How do I get any of the Social Club extras, like the Annis Elegy RH8? Yes, I am a member in the social club website, but the only problem is that I don't know how to get the stuff that comes with it.
Answer: You bring up your phone by pressing up on the directional pad. Then go to internet. There you should find the cars on the car sites accordingly. Clothes will be in stores to buy. And finally, weapons will be on the far right hand wall of all Ammunation stores.

How do I get the extras?


Dual-Boot with Vista?

Question: I want to know how to dual boot my computer with ubuntu 12.04 and windows vista(step by step would be best) and also wanted to know if it would slow down my pc even more than it already is?
Comment: Installing Ubuntu alongside Windows Vista won't affect the performance of Vista in any way.
Answer:
You first install Vista, as you have done it until now.
You then free up some space with your partition manager. (This is what I use, it is free.)
You create a bootable Ubuntu USB or CD (details on how to do that on Ubuntu page).
You install Ubuntu with the option "Install along side Windows."
Comment: As an alternative to "Partition Tool," you can use Vista's built-in Disk Management utility (`diskmgmt.msc`), which supports resizing NTFS partitions (and will schedule dynamic resizing operations on mounted partitions for shutdown, so they can be performed safely). Or use GParted on the Ubuntu CD. Or just run Ubuntu's installer on the CD, and tell it to install Ubuntu **alongside** other operating systems (it will dynamically shrink the Windows partitions for you, if no unpartitoned space currently exists). That is, most of the time, *all* you have to do is steps 3 and 4.

Dual-Boot with Vista?


Can you figure out if your friend is sharing their location from their iPhone or MacBook?

Question: Can you figure if your friend is sharing their location from their iPhone or  MacBook ?
Comment: As bmike correctly said, no–however MacBook's aren't even a valid source–only iPhones, iPads, iPod touches, and Apple Watches can be used for this.
Comment: They could be as "Find my device" lists all of these devices, including Macs
Answer: No. Apple lets people choose which device feeds the location for them. Apple lets them choose to share that location with you but not the name or type of system that feeds the geo location.
You get to see that location and not which device they chose to feed the location data. I'm not aware of any Mac client to input from iCloud find my friends, so you can be pretty sure they're not sharing from their Mac due to system design and not due to Apple letting you know what your friend chose. They could use a watch, iPad, iPod or iPhone currently in iCloud Find My Friends app.


WS2812B current draw

Question: I want to put my led strip in my PC and control it with Arduino or ESP. Since led strip has 60 leds it's going to draw ~3.5 amps. I'm asking if anything would happen if I connect led strip to 20 amps molex cable of my PSU? I am already writing my code, and this is the only thing that I'm interested in. Thanks!
Comment: There is no generic LED strip, no generic molex cable and no generic power supply that would allow anyone to make a concise answer.
Comment: Sorry, I don't understand what are you saying
Comment: If you mean the 4-pin PC peripheral cable, sometimes referred to as the Molex cable, then you need to match the voltage, and be sure that you don't overload your PC. None of these are typically good for 20A, so you need to do more research, or post more details so we can help.
Comment: So, molex cable has 2 GND, 12V and 5V. Led strip works with 5 volts, and so every led draws 60mA so I need ~3.5A for whole led strip. I'm just interested if led strip would blow up or something if I connect 5V 20A to it. Voltage is good but I'm not sure about the current. PSU won't be overloaded.
Answer: This question has been asked here so many times it is getting boring. A properly designed load will only draw the current from the power source that it needs. The power source, if it has greater current capacity than the load requires, is not going to somehow force the load to take five or six times the current that it needs. Assuming that everything stays at the proper voltage of course.
Think about your cell phone charger that you plug into the AC mains receptacle in your house. That receptacle can likely supply 15A at the mains voltage and yet the charger only needs maybe 0.1A or so. By plugging in your cell phone charger it does not explode due to the 15A capability of the receptacle.
Comment: So, led strip won't die or anything?
Comment: @Kule - If you understand what I wrote then you can derive you own answer to your question. If you do not understand then I suggest you do more reading and research till you do before connecting up any circuitry.
Comment: So, you're saying led strip will only draw the amount of current it needs? If it's like a so then I'm safe?
Comment: If the combined existing load and additional load does not exceed PSU rating then you should be OK.
Comment: I know nothing will happen to the PSU but will anything happen to led strip?
Answer: Both the power supply and the LED strip are fine with that, as long as

you don't draw that from the standby power pin
you don't give a signal to the LED strip while it's unpowered (WS2812 are notoriously unhappy with that)
you supply the LED strip in multiple places (because the copper is too thin to transport 3.5A all from one end)
Comment: Thank you, I won't power it from multiple supplies, I'm just going to power it from molex cable, and standby wire is already hooked up to motherboard and it can only supply 2A. Thank you sir!

WS2812B current draw


Programs to draw pictures of maths problems

Question: I studied mathematics and I was taught to draw pictures if problem is too hard to solve. What kind of tools/programming languages there are aimed to draw pictures for math problems? Should one learn Tikz or some programming language?
Answer: The range of software in this category is quite vast. You will need to do some research to find items that satisfy your precise needs.
I have used https://www.desmos.com/ to make drawings in two-space. You can use https://alternativeto.net/software/desmos/ to investigate related possibilities. And this site may help you to find something like Tikz.
I think it's reasonable to say that, most open source mathematics systems have some form of graphics system for displaying various mathematical entities. Look for pages like https://en.wikipedia.org/wiki/List_of_open-source_software_for_mathematics for possibilities.
To answer your question directly, yes, you could learn Tikz. Alternatively, you could learn Python or Ruby. I am familiar with the former language and know that many products are available for it for manipulating mathematical objects. It's not difficult to learn the rudiments of it.


3D printing of alumina

Question: I want to print a structure of alumina, and the 3D printer that I am using is an extrusion-based printer. So I need your help in knowing that what composition (binders and deflocculant) of alumina will be suitable to print the 3D structure in such a manner that it holds its structure (as it is printed)? And to what extent should rheological properties be maintained so that structure doesn't fail?
Comment: How are you going to turn the slurry into a solid? Existing processes for similar things (eg Saffil fibres) require rapidly heating the slurry in a furnace.
Answer: Use methylcellulose(MC) as a binder and Ammonium Polyacrylate (NHPA) as your dispersant or deflocculant. The concentration of methylcellulose can be set according to your requirement as its non-ionic (if you want liquid slurry use MC in large amount and if you don't use less amount). For NHPA you have to make different samples of fixed alumina concentration and NHPA of different concentrations. Check viscosity of each sample and the one which gives you least viscosity means the best-dispersed sample and hence your operating concentration of NHPA.

3D printing of alumina


Is there any tool or method to detect spoofed CCs?

Question: One of our phishing scenarios during a pentest went like this: we send an email saying something along the lines of "As per the manager (CCd) please download the latest software included". Then we attach the manager as CCd in the headers but never actually send the CC email using our server, making it look like the manager is aware and approves, while the real manager has never actually received the email. However what we're finding an issue with is how do we actually fix this problem? Does a solution exist which checks the CCd users and whether they've actually received the emails? Is there any other way to detect this?
Answer: There is no way to detect this and there is no tool for this. You would need to actually check with the claimed recipient or at least with the log files of the mail server - the latter is not something a normal employee could do.
I think you need to adjust the processes instead, i.e. employees should not trust anything written in the body, any attachments, the subject or the Cc of the mails since these could all be arbitrary made up - similar to snail mail. From is a bit more reliable if DMARC together with DKIM/SPF is used but with only a bit of social tricks From can be made look trustable too even if DMARC is enabled.
Comment: Are you aware of any SMTP server that does those checks? That is a server which checks the mail logs automatically for any CCd users. I suppose that we could script it but oh boy would that be annoying. But yea educating the users is definitely a priority
Comment: @trallgorm: since the CC'ed users might belong to a completely different domain where the server has no access to one cannot create a mail server which does such checks.


How have DNS recursion attacks happened?

Question: There are a few resources describing how to carry out DNS recursion attacks. Is there any good explanation to help me understand how these attacks work and if systems on my network are still vulnerable?
Comment: Wait, do you mean DNS cache poisoning?
Comment: What do you mean by "DNS recursion attacks"?  This is not a standard phrase that I've heard used in the field, so I suspect you've either heard it somewhere or made it up yourself.  (There are a variety of DNS attacks that might make use of recursion in some capacity, though I've never heard them called a "recursion attack".)  I can make some plausible guesses about what you might mean by that phrase, but I'd rather not guess.  What do *you* mean by "DNS recursion attacks"?  What makes you focus on "DNS recursion attacks"?
Answer: The basic "Mother of all Recursion Attacks" was published by Dan Kaminsky.
The basic summary is that you ask for a site and provide an answer with the appropriate query idea for the recursive server before that server gets the real answer.
The best resource I've got for you is An Illustrated Guide to the Kaminsky DNS Vulnerability which describes how the recursive system worked and why the flaw was possible.


centre tap transformer 240V – 24V @ 250mA

Question: I have a centre tap transformer 240V – 24V @ 250mA. I know I will get 12V between the centre tap and either of the other connections (I think one will be -12V), but will I have two 12V supplies at 250mA or 125mA?   And will the 24V supply be 250mA?
Answer: Each half can still supply 250mA. So you 2 12V outputs @250mA.
A common way to use a center tap transformer is to provide postive and negative supplies with 4 diodes.
Be careful about using a bridge rectifier from each half as I see in some circuits, that won't work as some of the diodes short out others.
The total power output depends upon how the transformer was rated but it probably is good for 250mA DC output from each of the positive and negative supplies.
Comment: If I ignore the centre tap and just use the two other connections will I get 24V and will it be 250mA?
Comment: You would get 24V between the +12 and -12 outputs in Kevin's answer, but with the regulators, that circuit still needs the centertap common internally.  What you cannot do is draw 250ma from say the 12V output in Kevin White's circuit, and simultaneously draw 24mA from the overall 24V.
Comment: You should be able to draw close to 250mA from each supply.  It is the same as drawing 250mA from 24v using a bridge rectifier.  You only use each winding half of the time for each polarity output.
Comment: thanks for the help kevin, are all the capacitors in the above circuit electrolytic?
Comment: The big values marked with polarity could be aluminum electrolytics. The small values (0.01uF) should be ceramic.


Fundamental group of $\mathbb{R}^n\backslash \{0\}$

Question: I am wondering about what the fundamental group of $\mathbb{R}^n \backslash \{0\}$ or more generally $\mathbb{R}^n \backslash U$ where $U$ is a subset of $\mathbb{R}^n$ for $n>1$. For $n=1$ I understand that the fundamental group is not trivial since the resulting space is not simply-connected. In all other case though the space is simply connected. Thus, logic says that the fundamental group is trivial. Therefore how do we classify spaces that we subtract various points out of them. E.g. how we can tell the difference between $\mathbb{R}^n \backslash U$ and $\mathbb{R}^n \backslash V$ for different sets $U,V$ (with different number of elements as well)? Is this the reason we need homology?
Answer: For $n = 1$, the space $\mathbb{R} \setminus \{0\}$ is the disjoint union of two intervals, so both connected components are contractible and have trivial fundamental group.
When $n = 2$, you can see that $\mathbb{R}^2 \setminus \{0\}$ deformation retracts onto $$S^1 = \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 = 1 \},$$
the retraction being $z \mapsto \frac{z}{\|z\|}$. This is the standard circle, in particular its fundamental group is $\mathbb{Z}$ as usual. This space is not simply connected as you claimed.
More generally, the space $\mathbb{R}^n \setminus \{0\}$ deformation retracts onto
$$S^{n-1} = \left\{ (x_1, \dots, x_n) \in \mathbb{R}^n : \sum x_i^2 = 1 \right\}.$$
This is the standard $n$-sphere, and it's a classical exercise that this space is simply connected when $n > 2$.
Now when $U \subset \mathbb{R}^n$ is an arbitrary set, even open, you cannot say anything at all about the fundamental group of $\mathbb{R}^n \setminus U$, and not even its homology, without knowing anything about $U$. Indeed it's a theorem of Whitehead that any closed manifold embeds into $\mathbb{R}^n$ for some $n$; if you take $U$ to be the complement of the image of the embedding, you can see that $\mathbb{R}^n \setminus U$ can have pretty much any fundamental group or homology you can imagine (there are some restrictions for closed manifolds of course, but you can imagine even more exotic spaces).

Compared to simply the fundamental group, homology also allows you to detect higher dimensional structure of your space: $\pi_1$ doesn't distinguish the $n$-spheres $S^n$ for $n \ge 2$, but these spaces are intrinsically very different. But the higher homotopy groups $\pi_n$ also detect higher dimensional structure, and do distinguish the $n$-spheres.
From my perspective, the main reason you want to use homology instead of (or in addition to) homotopy groups is because of the Mayer–Vietoris exact sequence, an analog of the Seifert–van Kampen theorem for homology. This allows you to break up your space in smaller pieces. See this question and its excellent answer by Ronnie Brown.
Comment: @Marion Well, it's possible that $\mathbb{R}^n \setminus U$ is bounded for example, take $U = \{ \|x\| > 1 \}$ for example. Then you can't start "very far away" from $0$. Another thing to consider is that even if locally you can trivialize the fundamental group, globally there might still be nontrivial loops. Think about a solid torus (a "donut"): locally it's contractible, but in globality there's a loop that cannot be trivialized. As for classification, as I explained it's at least as hard as classifying all manifolds: something that's simply impossible in high dimensions.
Comment: @Marion Yes, a space is simply connected iff it is path-connected and its fundamental group is trivial. It is true that $\mathbb{R}^n \setminus \{0\}$ is simply connected for $n \ge 3$, but not for a general set $U$! For example, take the complement of a circle embedded in $\mathbb{R}^3$...
Comment: @Marion It's almost that. Homology uses singular [simplices](https://en.wikipedia.org/wiki/Simplex) to "detect" higher dimension holes. It produces powerful invariants, but unfortunately that's not enough to completely classify spaces. For example there exist spaces $X$ such that $H_k(X) = H_k(S^n)$ for all $k$, but $X$ and $S^n$ are not homotopy equivalent: the so-called [homology spheres](https://en.wikipedia.org/wiki/Homology_sphere). If you then consider $X \setminus \{x\}$, that space has $H_k = 0$ for all $k$ but is not contractible! Homology is still an extremely useful tool, though.
Comment: I am confused. Imagine I start a path at $x$ far away from $0$ at $\mathbb{R}^n$. Then locally the fundamental group at that point is trivial. But on the other hand, you re right in the sense that if I try to take a loop near $0$ indeed I get that the fundamental group is $\mathbb{Z}$. Also, how do you classify such spaces in principle (imagine you know what these sets $U,V$ are)?
Comment: I see what you mean "globally" and I will agree with you for $\mathbb{R}^2$. I "see" it but, surly for $\mathbb{R}^n$ with $n \geq 3$ the fundamental group is trivial. Simple-connectedness requires/implies that the fundamental group is trivial. Correct me if mistaken. Also thanks for your wonderful answer.
Comment: yes yes, I understand it now quite well. If I understand it well, homology throws spheres and higher dimensional spheres to classify spaces instead of loops, right? Could not we somehow classify spaces then by counting the number of such spheres?
Answer: You are right, in higher dimensions ($n>2$) the (first) fundamental group of $\mathbb{R}^n$ minus a point is trivial. This is, however, not true for the removal of an arbitrary subset. Think of removing a tubular shape or just a circle from $\mathbb{R}^3$.
Other means of classification are so called higher homotopy groups, where you look at homotopy classes of maps from spheres of any dimension into the space you want to classify. See (almost) any book on algebraic topology.


Requesting info from Service de publicité foncière - France

Question: Is "demande d'extrait de matrice cadastrale" (https://www.service-public.fr/particuliers/vosdroits/R18819) a right form to request info on property ownership in France? If not - what is? If it is, will it work if only either a reference cadastrale/addresse or personne inscrite, but not both is known? What kind of a mandate is required from an applicant?
Comment: Also see related answer [here](https://expatriates.stackexchange.com/a/13882/15178), on how to obtain info on property ownership in France.
Answer: The document you are mentioning in your question will only tell you who is paying taxes on a given property, so it might either be outdated or not reflecting the real owner(s).
The way to be sure of the owner of a property is to go to the service de publicité foncière. This can be accomplished remotely or by going to the appropriate office. There's usually a fee associated with these requests. It is unfortunately not clear which of the proposed forms will be required for the purpose of identifying the owner.
You can identify address, cadastral section and number from the cadastre website or geoportail.
See related question here.
Comment: Thanks.  It seems that these forms won't work when only owner's name and commune where property is located are known, but property address and cadastral reference are unknown.  Or am I reading it incorrectly?  Is there a way to search for properties based on owner's name?  Thanks again.


Answering how long I plan to stay

Question: I'm visiting the US for 2 weeks, for example 10th of May until the 24th of May, when the officer asks how long are you staying in the US do I say 2 weeks or 15 days?
Answer: It doesn't matter. Both honestly describe your plan. The same goes for saying one week when planning staying for 8 days or one month when planning to stay for 25 days.
If you are still anxious about it, just say "I'm leaving on the 24th" or "my return ticket is on 24th" (if you do have a return ticket).
Unlike some countries or what you might have thought, U.S. generally allows you to stay for six months (if you are Canadian or Bermudian, or have a B1/B2 visa) or 90 days (if you have an ESTA) regardless of your stated plan, though you should always honestly declare your travel plans. In the rare case where they limit the period of your stay, you will know that and more inquiries on your return date are likely to follow.
Comment: Reminds me of old - pre 2001 - times. I'd go to USA for work a couple of times every year. "When was your last vist to the US?" -- Me: Uhm, about 2 weeks ago, there's a stamp in my passport somewhere - I had quite a few of those stamps.

Answering how long I plan to stay


Have causative verb get causative verb

Question: I am in confusion  of using causative verb 'have'
If i write
"I wanted to have you sit on the chair".
" i wanted to get you to sit on the chair".
"Why did you get him to come with you?"
Are those grammatical and idiomatic.
Comment: To me, the _have_ causitive is not idiomatic with a full verb (eg _sit_), though it is with a bare participle (eg _Have a house built_). The construction you use (_have someone sit_) sounds American to me.
Answer:
"I wanted to have you sit on the chair"
and
"I wanted to get you to sit on the chair"

are for all intents and purposes identical in meaning, and both are grammatically correct. Both are in common use, although they're slightly less formal than using other, more specific verbs (like "I wanted you to sit").

Have causative verb get causative verb


Cat behaving strangely

Question: My cat is starting to act strange sometimes. His eyes will get large, and he'll start meowing loudly and trying to bite at me, if I try to touch him he just meows louder. I usually just let my arm go limp when we'd play and he'd get nippy, but now if I do he latches on and starts to pull like he wants to drag me somewhere. This comes and goes in spells, I thought he might be hurt somewhere and thats why he was avoiding touch but when he's not in a mood and I pet him he's fine. He's only 3 so I don't think he has brain problems. And he's fixed too so I really don't know whats going on.
Comment: Any case of a change in behavior should always result in a visit to your vet.
Answer: Interesting that he is neutered and doing this! On that note, I think males are a little more crazy than females.
Our cat gets like this too. I usually refer to him as amped or being a spaz. He goes totally crazy. He gets huge eyes like he is on the hunt. He will pounce on everything. He loves to run and bite my feet and legs. If I'm sitting, he will bite my hands and arms. His back fur will twitch, and he will crouch to the ground. He may also run from room to room, jump on furniture, and attack marks on the wall.
I would just find ways to calm the cat down. Don't know if these work at all, but I've seen online that there is aroma therapy for pets or cbd oil for pets. What I normally do is throw him under a blanket. Don't know why, but maybe the weight of the blanket or the darkness or something soothes him and gets him to calm down.
To stop him from biting me, I try to avoid it. I pick him up (even if he doesn't want it) and move him to a different spot or room. I will also close doors to separate us.
Another thing I do is go and get toys and play rough with him. I have a feather wand and just whip it all over the place and it helps get his energy out. Or, I will chuck balls throughout the house and let him go chase them.
Lastly, maybe you could train him? I know people recommend positive reinforcement (via treats). This wasn't good enough for our cat, so we use a spray bottle and say down. He is pretty responsive to that.
Since our cat got neutered last week, the amount of these "amped up" episodes has drastically gone down!
Good luck!

Cat behaving strangely