After resuming I have a graphics issue with textures (not sure, see screenshots).  Windows frame:    Viber app:    Also some other apps are affected.  Some video info:  $ lspci | grep 'VGA\|3D' 00:02.0 VGA compatible controller: Intel Corporation 4th Gen Core Processor Integrated Graphics Controller (rev 06) 01:00.0 3D controller: NVIDIA Corporation GM204M [GeForce GTX 970M] (rev a1)  $ glxinfo | grep OpenGL OpenGL vendor string: NVIDIA Corporation OpenGL renderer string: GeForce GTX 970M/PCIe/SSE2 OpenGL core profile version string: 4.5.0 NVIDIA 375.39 OpenGL core profile shading language version string: 4.50 NVIDIA OpenGL core profile context flags: (none) OpenGL core profile profile mask: core profile OpenGL core profile extensions: OpenGL version string: 4.5.0 NVIDIA 375.39 OpenGL shading language version string: 4.50 NVIDIA OpenGL context flags: (none) OpenGL profile mask: (none) OpenGL extensions: OpenGL ES profile version string: OpenGL ES 3.2 NVIDIA 375.39 OpenGL ES profile shading language version string: OpenGL ES GLSL ES 3.20 OpenGL ES profile extensions:   Unity reload (Alt+F2 → unity) helps, but it crashes the most of open apps.  Any suggestions?  UPD: seems like it is bug of NVidia card/driver (reproduced with nvidia-367, nvidia-370, nvidia-375, nvidia-378).
After some gorgeous updates I witness this:
I found the following proof arguing for the irrationality of $\sqrt{2}$.     Suppose for the sake of contradiction that $\sqrt{2}$ is rational, and choose the least integer $q &gt; 0$ such that $(\sqrt{2} - 1)q$ is a nonnegative integer. Let $q'::= (\sqrt{2} - 1)q$. Clearly $0 &lt; q' &lt; q$. But an easy computation shows that $(\sqrt{2} - 1)q'$ is a nonnegative integer, contradicting the minimality of $q$.   Clearly, this is a vague proof in that it expects the reader to work out the math involved on his own. I took that as an exercise and worked out the following:  Assuming $\sqrt{2}$ to be rational, let $$\sqrt{2} = \frac{p}{q}$$ where $p$ and $q$ are least possible integers.  Clearly then, $\sqrt{2} - 1$ would also be rational, $$\sqrt{2} - 1 = \frac{p-q}{q}$$ where both $(p-q)$ and $q$ are also integers.  Multiplying both sides by $q$, we get $$(\sqrt{2} - 1)q = (p-q) = q'$$ where $q'$ is an integer. Using, $2&gt;\sqrt{2} &gt; 1$ , we get $$1&gt;\sqrt{2}-1&gt;0$$ proving that $q'$ is a non-negative integer.  As stated in the proof, computing $(\sqrt{2}-1)q'$ we get, $$(\sqrt{2}-1)q' = \frac{(p-q)^2}{q}$$which is not necessarily an integer (in fact necessarily not an integer). However, the proof claims it to be.  Am I missing something or something is wrong with how I proceeded?  Here is a link to the textbook where I found this proof: , (p 32, problem 1.14).  P.S. This is my first post here and my first time with TeX. I'd also like to know if there is anything wrong with the way my question is presented. How can I effectively use the Coppersmith? I've been playing Dominion with the base + Intrigue sets for a while now, and one of the cards that I have trouble wrapping my mind around is the Coppersmith.  In theory, it seems like a great card - it allows you to make all of your Coppers into Silvers, and in some cases - even Gold. However, I have yet to build a deck during a game that manages to use it effectively. Usually, in order to pull one or two of those with a nice amount of coppers, I need to create a deck that relies heavily on +cards/+actions chains, and the Coppersmith only begins to affect the game too late for it to be of any use.  So... how can this card be effectively used? I set up Windows 10 and have been using it for months on a spare laptop.  When installing, I did not use a Microsoft account to enable all of the app connectivity because I don't need that functionality.  I do not want these features active on this device, and I used a simple user name during the install.  Everything was going fine until I read about the Linux/bash features that are available in the anniversary update.  I decided to try them out and entered my Microsoft Insider credentials to enable the features in the developer options of the Control Panel.  I didn't realize the primary account on this device would now become linked to that Microsoft account.  Is there a simple way to disassociate the Microsoft account from the local account, or are they now linked forever in an unholy union?  I can set up a new local account, but I'd rather not lose all the settings and folder preferences prior to my gaffe. Deploy a project using Git push Countable union of countable sets(ZF) How do I disable client side decoration globally in GNOME? I'm using Ubuntu GNOME 17.04. One encounters client side decoration (CSD) as most of the official GNOME applications (Nautilus, gedit, Evince etc.) use  (a combined title bar and toolbar, see linked screenshots) whereas other applications don't. This feature is disabled in Unity, all the applications use traditional separate title bars.  Screenshots (from ):     Nautilus with GtkHeaderBar   Nautilus without GtkHeaderBar     For consistency I want to disable GtkHeaderBars globally (in all the applications using it) and use separate title bars. Is there a way to achieve that?  I have heard of gtk3-nocsd package, but couldn't make it work with GNOME for all the applications. Implement MicroController Power Switch and re-use the button for other features This is similar to other questions, but I feel it is different because of the reuse of the "power" button.   I want a "latching" Momentary Power button.  Behavior:   User Pushes momentary button, holds down as long as necessary and the device powers up. Between Power On and Off the same button will be re-used for other functions  When it is held down for 5 +/- seconds the MicroController will turn itself off.    Design Constraints:   Battery provides between 3.7V to 2.9V during it's usable life The Voltage regulator requires Vin +/- 0.3V on the Enable Pin, 3.1V output. MicroController operates at 3.1V MicroController Output Pin High state is 2.7V Max   Debouncing: In this case I'm not concerned about debouncing the switch, if the user doesn't hold it down long enough for the MicroController to set it's Digital Output pin  High the device doesn't power on.  I've worked up the following schematic, which kind of works.    It assumes 3.7V from the battery. Uses an Analog Input for the Power button, which then uses the variation in voltage to determine if the button has been pressed. In the current design the difference is 0.12 V (from 3.4 to 3.52), but the MicroController has a 12 bit ADC so that shouldn't be a problem, in addition I can adjust the sensitivity range of the ADC.   Questions:  What is the difference between reality and the simulator?  Is there a better way?  How can I get a greater voltage difference on the button input? I've tried many different combinations but they increase the voltage range above the 3.6V Input High Maximum specified in the datasheet.  I'm not excited about the "Leakage current" when the uC Controlled Power Pin is Low, any suggestions?   Thank you for any ideas/suggestions/answers. 8 bits representing the number 7 look like this:  00000111   Three bits are set.     What are algorithms to determine the number of set bits in a 32-bit integer? Let's disallow moderator nominations from people who've been suspended in the past year The locus of the intersection point of two perpendicular tangents to a given ellipse Effects:  +25 attack damage +15% life steal +40% attack speed Passive  Unique: Your attacks deal 5% of the target's current health in physical damage (60 max vs minions and monsters). Active   Unique: Drains target champion, dealing 15% of the champion's maximum Health in physical damage and healing you by the same amount. Additionally you steal 30% of their movement speed for 4 seconds - 60 second cooldown. (500 range)  Cost: 3200g VS.   Effects:  +70 attack damage +12% life steal Passive: Gain +1 attack damage and +0.2% life steal per kill (maximum +30 attack damage and +6% life steal). Half of the bonuses are lost upon death.  Cost: 3200g What is better for what situations, and when should you build BT instead of BotRK and Vice Versa. Which provides more lifesteal, the only variable being the stats of the weapon? I find myself constantly questioning my decision on which weapon to get and would really love some in depth answers. Would be great if any examples given were given from the ADC side of the house. just after the Big Bang and before inflation the currently observable universe was the size of a coin. One millionth of a second later the universe was the size of the Solar System, which is an expansion much much faster than speed of light. Can space expand with unlimited speed? Batch export Photoshop layers to individual PNG files I am a web dev and competent in Fireworks, but not so much in Photoshop.  I have just received a layered PSD file to turn into a web page.  Can anyone tell me the easiest way to export all layers to individual png files?  There are lots of layers and doing this manually seems wrong.  I have seen  but it does seem there should be native functionality for this in PS.  I have access to Photoshop CS4. Any pointers appreciated. Any software for windows Writing custom (intelligent) microsoft windows firewall extension (C#/C++/C) Writing custom (intelligent) firewall for windows Installing nginx etc. on the same computer and use it as proxy with rules.   "You could not solve all Ddos problems with software" is not answer for this question.I want to reduce Ddos vulnarebility of my server.(It will stop low-level Ddos attacks)  Hosted solutions or cloud services are not related with this question.  Closing some of the ports and protocols via Microsot Windows Firewall is not related with this question.  Any idea?  EDIT After Questions  Low level: sample scenerio: 500 zombie clients and each of them sending 100 Custom Get/Post requests per second)   Web Application is on IIS. I want to stop attacks before they reach to IIS.  EDIT After Helpful Comment  How can the DDOS attack achieve its aim?   Exhausting bandwidth ( Not related with my question) Exhausting network hardware of server (NIC) ( Not related with my question) Exhausting server resources (RAM,CPU)  (Yep, I mean this)   If you stop attacks late, then you lost more resources! (Is this false?)   You can stop attacks on Web Application Level (in Asp.net MVc Code or in Handler) You can stop attacks on IIS (Dynamic Ip Restrictions)  You can stop attacks on lower levels (Firewall ..)   EXTRA INFO ABOUT THE LAST ATTACK (7 Hours ago)  When I connected server during the attack IIS Cpu Utilization was %92-%99. And when I try to connect homepage I got this error:  A connection was successfully established with the server, but then an error occurred during the pre-login handshake. (provider: SSL Provider, error: 0 - The wait operation timed out.) Root filling up although it should have some space This has been perplexing me for some time. I'm on Ubuntu 11.04. My root partition is filling up and I can't figure out what's causing it. Even stranger, when I run df -h, this is the output:  Filesystem            Size  Used Avail Use% Mounted on /dev/sda2             9.2G  8.8G     0 100% /   As you see, there should be about 400 MB free, but Avail shows 0. What could be causing this? I've even uninstalled some programs to free the space and it fills up again. Suggestions on how to find out which, if any, files are filling up the space are also welcome. Prove that given $A\subseteq\mathbb{R}$, if has a maximum $a_0$, then $a_0$ must be the supremum.   In order to prove this, I have written the following:  Let $A\subseteq \mathbb{R}$, nonempty set because $a_{0}$ $\in A$ for hypotesis. Then, since $a_{0}$ $\in A$ is the largest element of the set $A$   (i.e. for each element $A$ of A we have that $ a\leq a_{0}$) $A$ is bounded above. Then, A has a supremum. Let $s$ be the least upper bound of $A,$ so we have that for each element $a$ of $A$, $a\leq s$, for being $s$ an upper bound. In particular, $a_0\leq s$, which implies that $s=a_0$.   Then, we have that $\sup(A)=\max(A)=a_0$, which is exactly what we wanted to prove.  Observation: this proof is true if $s \in A$   I'm not sure if my proof is correct. I will be thankful if someone tells me. Catalina not allowing to change read/write permissions despite having SIP disabled Currently Running: macOS Catalina 10.15.1 Issue: Ever since I upgraded to Catalina, macOS is not allowing me to change read and write permissions on system protected files, hence, I am not able to tweak the system as per my likes and dislikes. Steps Tried To Solve The Issue: SIP, always remains . So, no question arises regarding any conflict with system integrity protection. I have also tried to force change the permissions on the system files I want to edit using certain brute force commands, which I can't recall the name of, right now. But, I have tried. As of now, I am still unable to change read and write permissions on system protected files and resources. Wired connection not working Ubuntu 12.04 Power of a generator generates the group iff this power is coprime to the group order. Let $\langle g\rangle$ be a cyclic group of order $n$. Suppose $1\leq q \leq n-1$, I want to show that $g^q$ generates $\langle g\rangle$ if and only if $\gcd(n,q)=1$.  Suppose $g^q$ generates $\langle g\rangle$, then   $$ 1,g^q,g^{2q},...,g^{(n-1)q} $$  are all distinct and $g^{nq}=1$, since $n$ is the order of the group. Note that this is the lowest multiple of $q$ for which this is the case. Therefore $\operatorname{lcm}(q,n)=qn$ and it follows that $q$ and $n$ are coprime.  On the other hand, if $\gcd(q,n)=1$, then $\operatorname{lcm}(q,n)=qn$, we have $g^{nq}=1$ and hence all of  $$ 1,g^q,g^{2q},...,g^{(n-1)q} $$  are distinct again and $g^q$ is a generator.  Is my proof correct at all and is there maybe a more elegant argument? How many solutions are there to $x_1 + x_2 + ... + x_5 = 21$? How many solutions are there to the equation      $x_1 + x_2 + x_3 + x_4 + x_5 = 21$    where $x_1, i = 1,2,3,4,5$ is a nonnegative integer such that $0 ≤ x_1 ≤ 3, 1 ≤ x_2 &lt; 4$, and $x_3 ≥ 15$?  I have correctly completed the previous parts of the question but am having trouble with this particular restriction.  The total number of solutions to the equation is $C(25,21) = 12650$.  I have tried to do each restriction individually:  For $0 ≤ x_1 ≤ 3$, I solved this by subtracting the number of solutions when $x_1 ≥ 4$ from the total number of solutions. This is $C(25,21) - C(21,17) = 5985$.  For $1 ≤ x_2 &lt; 4$, I have $C(24,20) - C(21,17) = 5985$.  For $x_3 ≥ 15$, I have $C(10,6) = 210$.  However, the restriction is when these are all together. I am not sure how to proceed. The answer is $106$.