I killed all the bosses but the door will not open. I have beaten this game four times and I have never had this problem before.  I do not know what to do, can someone help me?
In Dark Souls 2 there is a 'Shrine of Winter' near Sinners Rise. When you approach the shrine it will not let you open the door and a message pops up that says      Seek Mightier Souls   What does this mean and how do you open this shrine?
Why is a T distribution used for hypothesis testing a linear regression coefficient? In practice, using a standard T-test to check the significance of a linear regression coefficient is common practice. The mechanics of the calculation make sense to me.  Why is it that the T-distribution can be used to model the standard test statistic used in linear regression hypothesis testing? Standard test statistic I am referring to here:   $$ T_{0} = \frac{\widehat{\beta} - \beta_{0}}{SE(\widehat{\beta})} $$ Not sure if this option ever existed, but the new QGIS version hides a lot of things somewhere else and this seems so simple that I'm kind of convinced it was possible to do this before. I have quite a large data set with ship tracks. I now want to analyse individual ship tracks (made from point layer with  Plugin). However, it is so crowded that it is hard to look at individual tracks. I don't want to save each single track as a new layer, so I wonder if it is possible to select one track in the Attribute table and display only that one on the map/hide the others? Is there a printf converter to print in binary format? I can print with printf as a hex or octal number.  Is there a format tag to print as binary, or arbitrary base? I am running gcc. printf(&quot;%d %x %o\n&quot;, 10, 10, 10); //prints &quot;10 A 12\n&quot; print(&quot;%b\n&quot;, 10); // prints &quot;%b\n&quot; Why does just splitting an Ethernet cable not work? I thought the Ethernet is logically a one-line communication bus (for argument's sake, I am excluding hubs). All machines attached on the bus hears the same signals and the machines themselves try to avoid collisions by randomly backing off.    If so, why would splitting one Ethernet line from my home router into two and connecting two computers not work? Why do I have to add a switch to it?  What the Internet said would not work.                                                             +----------+                                                +----------&gt;|computer 1|                                                |           +----------+     +------------------+       +---------------+     |4 port home router+------&gt;|simple splitter|     +------------------+       +---------------+                                                |           +----------+                                                +----------&gt;|computer 2|                                                            +----------+   What the Internet said I should do                                                             +----------+                                                +----------&gt;|computer 1|                                                |           +----------+     +------------------+       +---------------+     |4 port home router+------&gt;|    switch     |     +------------------+       +---------------+                                                |           +----------+                                                +----------&gt;|computer 2|                                                            +----------+   Is this because of the signal degradation (reduced electric current)?    Thank you for all the answers! The reason why I did not just use the two ports of my home router is...  The 4-port gigabit router is in my room, and I had put a computer in another room (also my room, though). Since a wired network is far more reliable and secure, I had bought a long Ethernet cable and and connected the computer to the router. Now I was thinking about adding another computer to that room. I could buy another long Ethernet cable, but then there will be two cables between the rooms. The one line already is a minor annoyance, so I thought if I could share the one line between the two computers in that room. A switch would work, but it requires power and is a little bit pricey. That is why I wondered why it would not work to simply split the physical Ethernet cable.  Apparently I do not completely understand how Ethernet and a switch work. I just have some bit of knowledge I heard in my college class. Dismiss flags on the same posts with different validities Sometimes a post has both valid flags and invalid flags. As far as I can tell, all the flags on the same post have to be dismissed with the same validity.  The scenario I encountered (on ):   A low-quality question is closed. The asker improves the question and flags for reopening. A moderator reopens the question, but doesn't dismiss the flag yet. Someone else flags the question as “off topic”.   Since I agree with the reopening, and I don't consider the question off-topic, I would like to dismiss the first flag as valid and the second flag as invalid. But there's only one set of buttons for both flags.  And just now, two flags on a closed question: one that says “delete it”, the other one that says “reopen it”. The correct course of action is neither (the ideal outcome would be for the question to be edited then reopened). I should provide different feedback to the two flaggers, but I can't.  Unless I've missed something, please allow multiple flags on the same post to be dismissed separately. The new top bar is out on meta. Consider it a beta EDIT Dec 4 -  We're doing it live! Please post bugs or concerns as questions tagged [top-bar]  We have rolled out our 95% completed top bar for feedback.  Note:   We are still adding a few final features, so if you have a request for a feature that is not there you might hold off on that for a day or two Yes, the bar is black1. Also give that a day or two to sink in.   I also want to specifically address the meta/main/chat links.  I'm very aware that many of you like having the links at the top of the page and that this new version is different.  The thought process behind moving these links is that the "site switcher" should be an automatic place to go when you are on a site and want to go to another one. If you haven't already developed the muscle memory for these links, getting between them required a bit of a cognitive load: You must first decide that you want to switch sites and then you have to decide which area to go to to switch.  Also, we find that less familiar users aren't entirely clear what a standalone link that says "main" or "meta" actually does... Not to mention that it hurts my UX sensibilities that we are dynamically changing links in an element that should be fixed.  It is true that we have added a click in the process, but we have also removed a lot of clicks for other common actions that are much more frequent than switching between main and meta.  All I ask is that you give it a few days and see if the new "switch sites at the top left" paradigm grows on you.  What do you think?  1.  points out that technically it's not quite black. Is drinking water during or after a meal better? While we know that every finite set is compact, how to prove that every infinite subset of $\mathbf{R}$, the set of all real numbers, is also compact in the finite-complement topology (also called the co-finite topology)?   What is the situation in the case of an arbitrary infinite set $X$ with the finite-complement topology? Inspired by a , here's a list of the 'etymology' for the  hats. For pictures and a list of criteria, see  instead.  The list is a community wiki; feel free to edit it if you have more information about a hat. If you like to do some research, most hats have been featured in earlier editions, so you might be able to find more information in old Meta Stack Exchange questions. A locally constant sheaf on a locally connected space is a covering space; Proof? As part of my hobby i'm learning about sheaves from Mac Lane and Moerdijk. I have a problem with Ch 2 Q 5, to the extent that i don't believe the claim to be proven is actually true, currently. Here is the question repeated:     A sheaf on a locally connected space $X$ is locally constant if each point $x \in X$ has a basis of open neighborhoods  $\mathcal{N}_x$ such that whenever $U,V \in \mathcal{N}_x$ with $U \subset V$, the restriction $\mathcal{F}V \rightarrow \mathcal{F}U$ is a bijection. Prove that $\mathcal{F}$ is locally constant iff the associated etale space over $X$ is a covering.   I don't have a problem (i think) showing the "covering" $\rightarrow$ "locally constant" direction of the implication. My problem is in the reverse direction. In fact, in the process of trying to solve this problem i think i have come up with a very simple counterexample.  Here's my attempt at a counterexample:  Let $X = \{a,b,c\}$ be  finite space with open subsets $\{\{\},\{a\},\{a,b\},\{a,b,c\}\}$. I claim this is hyper-connected and locally connected.  Let $\mathcal{F}$ be the sheaf with $\mathcal{F}(\{a,b,c\}) = \{q\}$, $\mathcal{F}(\{a,b\}) = \{r,s\}$, and $\mathcal{F}(\{a\}) = \{t\}$, where $r = q|_{\{a,b\}}$, and $t = q|_{\{a\}} = r|_{\{a\}} = s|_{\{a\}}$. I claim this defines a locally constant sheaf. Indeed we can have $\mathcal{N}_{a} = \{\{a\}\}$, $\mathcal{N}_{b} = \{\{a,b\}\}$, and $\mathcal{N}_c = \{\{a,b,c\}\}$.  Now we can calculate the corresponding etale space $p : E \rightarrow X$ as follows: $E = \{\dot{q}(a), \dot{q}(b), \dot{q}(c), \dot{s}(b)\}$, and has these four distinct elements since $\dot{q}(a) = \dot{r}(a) = \dot{s}(a) = \dot{t}(a)$, and $\dot{q}(b) = \dot{r}(b)$. The action of $p$ follows by construction.  However, $p$ does not appear to me to be a covering space. In particular, the only neighborhoods of $b \in X$ are $\{a,b\}$and $\{a,b,c\}$ and we have $p^{-1}(\{a,b\}) = \{\dot{q}(a),\dot{q}(b),\dot{s}(b)\}$ and this cannot be partitioned into homeomorphic images of $\{a,b\}$ because $2 \nmid 3$. Similarly, $p^{-1}(\{a,b,c\}) = \{\dot{q}(a), \dot{q}(b), \dot{q}(c), \dot{s}(b)\}$ and this cannot be partitioned into homeomorphic images of $\{a,b,c\}$ since $3 \nmid 4$.  So i think i have a locally constant sheaf on a locally connected space whose corresponding etale space is not a covering. What have i done wrong? I have seen the claim mentioned elsewhere so i'm convinced i'm wrong somewhere. Perhaps someone knows where a correct proof is published otherwise? Iterating over array of objects javascript - odd behaviour? Nature of gravity: gravitons, curvature of space-time or both? General relativity tells us that what we perceive as gravity is curvature of space-time.   On the other hand (as I understand it) gravity can be understood as a force between objects which are exchanging (hypothetical) virtual particles called gravitons, similar to the way electromagnetic forces are due to objects exchanging virtual photons?  At least at first glance, the two concepts seem mutually exclusive. Is there a description of gravity which includes both, or is this contradiction one of the problems in combining GR with quantum mechanics? Chrome browser will not connect to intranet web addresses Motivation for Napier's Logarithms In the wikipedia article on logarithms, I am clueless about the approach and motivation for the following computations done by Napier (and the mysterious appearance of Euler's number) in this . Please help me understand:       By repeated subtractions Napier   calculated $10^7(1 − 10^{−7})^L$ for L   ranging from 1 to 100. The result for   $L=100$ is approximately $0.99999 = 1  − 10^{-5}$. Napier then calculated the   products of these numbers with $10^7(1  − 10^{−5})^L$ for L from 1 to 50, and   did similarly with $0.9995 ≈ (1 −  10^{−5})^{20}$ and $0.99 ≈ 0.995^{20}$.   These computations, which occupied 20   years, allowed him to give, for any   number ''N'' from 5 to 10 million, the   number ''L'' that solves the equation      $$N=10^7 {(1-10^{-7})}^L$$      In modern notation, the relation to   natural logarithms is:      $$L = \log_{(1-10^{-7})} \!\left(  \frac{N}{10^7} \right) \approx 10^7  \log_{ \frac{1}{e}} \!\left(  \frac{N}{10^7} \right) = -10^7 \log_e  \!\left( \frac{N}{10^7} \right)$$      where the very close approximation   corresponds to the observation that      $${(1-10^{-7})}^{10^7} \approx  \frac{1}{e}. $$   I understand the computations, but I don't know what's going on. What kind of ship was flying away from Rey in her flashback? The ship flying away from Rey in her flashback of being left on Jakku looks like one of the ships in the group that takes her away when she is captured.  If so, and it is a First Order ship that dropped her off, this is an interesting Easter egg. I'm trying to test a page that has an input but really its a drop down list. The user enters text and then a dropdown appears based on what the user inputted. They are groups that are in the database. If I look for these elements before typing anything they dont exist but once I start typing they do exist.  The problem is I cant just type the whole name and then press enter or return because an option has to be selected from the list.  I can press arrow down to get the focus on the item from the list but then if I say enter it doesnt select, it sends that enter to the original element and nothing gets selected.  I can find the select element:  var ddList = driver.FindElement(By.CssSelector(".ui-autocomplete-input")); ddList.SendKeys("Admins");   This is the first time the list appears in the html.  It looks like this:  &lt;li class="ui-menu-item" role="presentation"&gt; For $G$ group and $H$ subgroup of finite index, prove that $N \subset H$ normal subgroup of $G$ of finite index exists A toplogical space $X$ is said to be second countable   if there exists a countable basis for the topology.      $X$ is separable if there exist a countable dense subset.   Show that a second countable space always is separable.      Give an example of a space that is separable but not    second countable   The first part was easy and i managed to solve it on my own  Assume that $X$ is second countable and let $B_1, B_2, B_3...$ be a countable basis. Construct the set $A$ by taking one element $x_n$ out of every $B_n$. Then $A$ becomes dense because every open set $U$ contain a $B_n$ which in turn contain the element $x_n$ in $A$.    Again, like in the question i posted yesterday, i had trouble to come up with an example but our professor gave an example during the last lecture but i dont fully understand it:  "For an example of a separable but not second countable space let $X$ be the real line provided with the cofinite topology, i.e. where a set is open if its either empty or if its complement is finite.   Then every infinite set is dense and hence $X$ is separable.  But for an arbitrary countable family of open sets $U_1, U_2...$ write $U_i = X$\ $A_i$, where $A_i$ is finite, there is an element $x \in X$\ $(\cup_i Ai)$, because $A_i$ is countable,  such that the open set $X$\ $\{x\}$ is then not able to contain any $U_i$. Consequently $X$ is not second countable."  We asked him to explain it in more detail but his english is poor and nobody understood anything...:)       Def: A subset $A$ of a topological space $X$ is dense in $X$ if for any point $x$ in $X$, any neighborhood of $x$ contains at least one point from $A$.   $1.$ "Then every infinite set is dense"  Take $A$ to be one of these infinite sets. If $x \in A \subset X$ then it trivially fulfills the definition. But if we take a $x \not\in A$ why is it not possible to find a neighborhood in these finite sets that does not contain an element in $A$?  edit: its obvious if it is only open neighborhoods, open sets in the topology, but that is not specified in the definition.  $2.$ I am not fully understanding this either  "But for an arbitrary countable family of open sets $U_1, U_2...$ write $U_i = X$\ $A_i$, where $A_i$ is finite, there is an element $x \in X$\ $(\cup_i Ai)$, because $A_i$ is countable,  such that the open set $X$\ $\{x\}$ is then not able to contain any $U_i$. Consequently $X$ is not second countable."  Is it possible to express this in a simpler way? Vertex cover in a graph with maximum degree of 3 and average degree of 2. Ubuntu 20.04 Network Performance Extremely Slow I just upgraded from Ubuntu 18.04 to 20.04 on one of the  machines, and the network performance is abhorrent. Internet speed is very fast from a google speed test (>300Mb/s for both upload and download).     However, whenever I try to browse a website or even run sudo apt update the connection will often time out. My internet is very fast on all other devices and I really don't know how to debug this problem.  sudo lshw -class network -short &amp;&amp; nmcli device status    shows  H/W path           Device           Class          Description ============================================================== /0/100/1c.6/0      wlp2s0           network        Wi-Fi 6 AX200 /3                 br-96f5b790e29e  network        Ethernet interface /4                 br-15e76fb81c05  network        Ethernet interface /5                 docker0          network        Ethernet interface  DEVICE           TYPE      STATE         CONNECTION       wlp2s0           wifi      connected     CSA HQ           br-15e76fb81c05  bridge    connected     br-15e76fb81c05  br-96f5b790e29e  bridge    connected     br-96f5b790e29e  docker0          bridge    connected     docker0          p2p-dev-wlp2s0   wifi-p2p  disconnected  --               lo               loopback  unmanaged     --               gpd0             tun       unmanaged     --                Solution:  It seems like this is a pretty common problem, and one without a definitive cause. For me, it was because Ubuntu resets the TLP configuration, and the power management went back to "on". I disabled Wi-Fi power management directly by editing:  sudo nano /etc/NetworkManager/conf.d/default-wifi-powersave-on.conf   Then changed the value from 3 to 2, saved changes and rebooted.  I actually got this solution from Reddit and this worked for my case specifically but it seems like it could be caused by a bunch of other things. Read all the proposed solutions below!