On my laptop I configure my Virtual WiFi in an elevated cmd with  netsh wlan set hostednetwork mode=allow ssid=VWiFi key=xxx keyUsage=persistent netsh wlan start hostednetwork   and can share my internet connection (UMTS connection) via WLAN to another device. This works fine, but after I connected my laptop once via WLAN to the internet I'm not able to start my Virtual WiFi anymore. When I use my UMTS internet connection again and try to share it with Virtual WiFi I get the error message that the hosted network could not be started because a system device is not working (no more specific information is given).  After I reinstall the WLAN adapter driver the Virtual WiFi will work again.  As I'd like to switch between UMTS and WLAN quite often this is a pain. Is there any command I can use to restart the WLAN driver or something similar to avoid the reinstalltion of the WLAN adapter driver? A simple disable and enable doesn't help.
I try to share my internet connection from my laptop (win7) via virtual wifi to my android phone (Samsung Gio). In an elevated cmd I enter  netsh wlan set hostednetwork mode=allow ssid=VWiFi key=xxx keyUsage=persistent netsh wlan start hostednetwork   The android phone connects to VWifi, but the internet connection is hardly usable on the phone while the internet connection on my laptop is very good at the same time (some Mbps)  What works: I can start WhatsApp, type and receive text messages, so I guess there's no general problem, i.e. config can't be completely wrong - but when I try to send pics or audio or try to receive pics or audio messages then I see from the network monitor on my laptop that the connection doesn't even try to send or receive anything (0 kbps up/down).  What sometimes works: connection to some websites with acceptable speed - but sometimes I don't get a single byte. Also this is not reproducible - the same website sometimes shows up, sometimes android tells my the site was not available.  What never worked so far: connecting to Google Play Store. It always tells me to check my connection and try again.  What I tried so far: I read on problems with IPv6, but no matter if I enable or disable IPv6 in the settings of the virtual Wifi, it doesn't help.  I also read about problems related to static/dynamic IPs, my netsh interface ip show config tells my that dchp isn't enabled on my virtual Wifi connection. If I try to enable it by netsh interface ip set address "myWificonnection" dhcp the android phone won't even connect anymore to the virtual Wifi at all. So there may be a solution here if android would connect with dhcp enabled.  I also read about problems with MAC filters, but this is no router, there's no MAC filter in the virtual Wifi.  When I directly connect to a real WLAN router, everything is fine and working. I have the problems only when trying to get into the internet via the virtual Wifi.  Any ideas what may be configured wrong? Either on laptop or on android? Thanks.
Ruby Style: How to check whether a nested hash element exists How to select the architecture to download? I've had my Trek 3900 for about a year now, and I replaced the chain about 3 months ago, but otherwise it's running with the same parts it came with (not counting tubes and tires).  I had to replace my chain because it had gotten really rusty from riding in rain/mud/water on trails, and had started skipping.    My problem is, whenever I ride, if I get above or below a very light pressure when pedaling, the chain skips. I was wondering if this could be caused by my chain, which has gotten a little rusty but not as stretched as the other one was, or if I might need a new cassette?    I just readjusted my derailer, but it still skips.   Does anybody know what could be causing this? What pins on a nRF51822 do I connect to a STM32 programmer? How do I add VBA in MS Office? Connected set in $\mathbb{S}^1$ Let $A$ be a connected set in $\mathbb{S}^1$. Show that $A^c$ is also connected.  My idea: suppose $A^c$ is not connected. Then, there exist $U,V \subset \mathbb{S}^1$ open and disjoint such that $U \cup V =A^c  $. Therefore $\mathbb{S}^1 = A \cup U \cup V$. If I had that $A\cup U$ was open I could arrive at  contradiction, since $\mathbb{S}^1$ is connected, but I can't say that... What is the simplest way to convert char[] to/from tchar[] in C/C++(ms)? This seems like a pretty softball question, but I always have a hard time looking up this function because there seem there are so many variations regarding the referencing of char and tchar. A general explanation of rim, tire and tube specifications I have a dozen or so image files on my iMac's desktop.  Most of the time, after booting up my computer, it displays a thumbnail of each image (as it should).  But sometimes, the thumbnails won't show up (see screenshot below).    It appears to happen randomly.  I estimate that it happens about 20% of the time.  Why is this, and how can I prevent it from happening?  I am running OS X 10.9.4.  Thanks. Yet another "inconsistent reputation history" How to prove floor identities? passthrough graphics card to virtualbox So I'm trying to get my laptops NVIDIA fx 880m to pass to a virtual box running window seven on a linux mint 17 install.  So far everything seems to be (maybe) heading in the right direction:  the device passed through happily, when I booted the virtual box it installed a bunch of new drivers, but when I try to install the NVIDIA driver on the guest it can't find the card. I looked under the device manager and there's no listing for the nvidia card under the pci bus, my guess would be that I need to disable it in the host so that it can be passed through to the guest (the virtual box manual said they can't be shared) but I'm not sure how to do so. Can anybody help with this? This question does not seem to me to be a duplicate of , because here the given solution was that it was not possible with a windows host, however in this case the host is linux How to remove session entries from LightDM? Constructing a HoTT proof term of 1≠0 As an exercise in HoTT basics, I am trying to construct a term that has the type $Id_{Nat}(S(O),O)\to\bot$. If this were a Coq proof, I'd be done after a single inversion on the premise, as the impossible identity would leave zero cases to consider. I guess I could do something similar here, but I'm not sure how. What would be an HoTT term that has the type? I wrote a simple backup script which back ups my stuff on a remote server, everything is working great but I'd like to have some sort of report like "backup successful or not".  this is my script   # backup CHECK date1=`date +"%d.%m.%y - %H.%M"` host=`hostname` twdate=`date +"%d.%m"` performtw=`ls -la|grep "tw"|grep "$twdate" &gt; tmp1.txt` echo $performtw if cat tmp1.txt|grep "tw" then echo backup successfull printf "tw backup success!" | mail -s "tw backup check $date1 repor$ rm tmp1.txt else echo backup failure printf "sitename backup failure!" | mail -s "site backup check $date1 repor$ rm tmp1.txt fi exit   But this isn't working really well and I ask you if there's some simpler and more powerful way to do it? Basically it just needs to check if file exists with the name starting xyz and was created at date xyz. Irreducibility check for polynomials not satisfying Eisenstein Criterion. My Question is to check Irreducibility for polynomials not satisfying Eisenstein Criterion.  As an Illustration, to check whether $x^{p-1}+.....+x+1$ for p a prime is irreducible or not, we replaced $x$ by $x+1$ and by using Eisenstein's Criterion for the resulting polynomial we conclude that resulting polynomial is irreducible and so is the original polynomial.  In general, for a given irreducible polynomial $f(x)$ with coefficients in a known U.F.D, is there some element $a$ such that we can apply Eisenstein's criterion to $f(x+a)$?  I am sure there would be no general structure for this but I expect there to be at least some special cases.   Any Reference/suggestion would be appreciated.  Thank You. I just proved that any finite group of order $p^2$ for $p$ a prime is abelian. The author now asks to show that there are only two such groups up to isomorphism. The first group I can think of is $G=\Bbb Z/p\Bbb Z\oplus \Bbb Z/p\Bbb Z$. This is abelian and has order $p^2$. I think the other is $\Bbb Z/p^2 \Bbb Z$.   Now, it should follow from the fact that there is only one cyclic group of order $n$ up to isomorphism that these two are unique up to isomorphism. All I need to show is these two are in fact not isomorphic. It suffices to show that $G$ as before is not cyclic. But this is easy to see, since we cannot generate any $(x,y)$ with $x\neq y$ by repeated addition of some $(z,z)$.  Now, it suffices to show that any other group of order $p^2$ is isomorphic to either one of these two groups. If the group is cyclic, we're done, so assume it is not cyclic. One can see that $G=\langle (1,0) ,(0,1)\rangle$. How can I move on? Regular Expressions: Is there an AND operator? Debt condition abuse? The rules states that you cannot have two of the same conditions.  You can take a debt condition to gain two additional successes on your influence test to gain assets from the reserve. Thus if you already have a debt, does that mean that you automatically gain two additional successes for each gain asset actions? Why is "our today's meeting" wrong? One of the answers to  question states that &quot;We shall discuss it in our today's meeting&quot;  is grammatically correct. To me, that sentence is clearly wrong. While in today's meeting is fine and in our today meeting is OKish (though at the very least clumsy), there's something about the possessive there (our today's) that makes it wrong for me. I would read that sentence as in our today's (as opposed to your today's) meeting. Similar to in our car's trunk where the our clearly modifies car and not trunk or car trunk, the our in our today's seems to be modifying today's and not meeting. So, my questions are i) is it actually grammatically wrong to say in our today's or is it just a question of usage? and ii) if it is indeed wrong, how can we explain its wrongness?