Settlers of Catan Cities and Knights: Can Longest Road Leave the Game? Background In C&amp;K, it is possible for your road length to decrease. If you have the longest road, and it gets shortened (by the Diplomat progress card or by an opponent's knight), what happens to the Longest Road card? The base game rules say the following:  The first player to build a continuous road (not counting forks) of at least 5 road segments receives the special card &quot;Longest Road.&quot; If another player succeeds in building a longer road than the one created by the current owner of the &quot;Longest Road&quot; card, they immediately take the special card (and its 2 victory points).  And the C&amp;K rules say:  Knights ... can be used to break another player's &quot;longest road...&quot; If the red player places his knight at 'B,' it will break blue's road, shortening his road for purposes of determining the longest road.  Consider the following situations Assume the rest of the other players' road lengths are too short to matter. In all cases, Player 1 starts with the longest road: a) Player 1's maximum road length is 7, Player 2's is 5. b) Player 1's maximum road length is 7, Player 2's is 6. c) Player 1's maximum road length is 7, Player 2's is 7. d) Player 1's maximum road length is 5, Player 2's is 5. e) Player 1's maximum road length is 5, Player 2's is 3. If in all scenarios Player 1's road is shortened by 1 length, who ends up with the Longest Road card? a) Player 1's road is still longer, so Player 1 keeps the Longest Road card. b) The two players are now tied, so Player 1 keeps the Longest Road card. c) Player 1's road is now shorter, so Player 2 takes the Longest Road card. d) Player 1's road is now shorter, so Player 2 takes the Longest Road card. NOTE: Player 1's road length fell below the 5-length threshold of earning the Longest Road card in the first place. But Player 2 still has a road of length 5, so there's no question: Player 2 takes the Longest Road card. e) What happens now? Does the Longest Road card get set aside, belonging to no player, or does Player 1 retain the card because no other player has (direct quote from the rules) &quot;succeed[ed] in building a longer road than the one created by the current owner of the 'Longest Road' card?&quot; Scenario e) According to what is written in the base rules, a player must take the Longest Road card for themselves in order for another player to lose the Longest Road card. There is no mechanism for the Longest Road card exiting the game because there is no way for road lengths to shorten, and thus no way for no player to have a maximum road length of less than five after any player has gained the Longest Road card. The C&amp;K rules do not explicitly specify whether or not the Longest Road card stays in play. They simply say that knights can affect the length of a road; the Diplomat card rules also do not mention anything of use here. The interpretation that my friends and I have settled on is that Player 1 retains the Longest Road card until another player takes it away by having a longer road, regardless of whether or not Player 1's road length is still above 5. However, Catan Universe seems to disagree with us on this one... Does anyone have an official ruling for this? Thanks!
Who would get the longest road? Imagine the following setup. There are 3 (or more) players with an equally long road. What would happen if the longest road was broken? How is it decided who obtains the longest road from there on?
Greyed-Out vs Invisible block all but a few ips with firewalld On a linux  networked machine,  i would like to restrict the set of addresses on the "public" zone  (firewalld concept), that are allowed to reach it. So the end result would be no other machine can access  any port or protocol, except those explicitly allowed, sort of a mix of     --add-rich-rule='rule family="ipv4" source not  address="192.168.56.120" drop'    --add-rich-rule='rule family="ipv4" source not  address="192.168.56.105" drop'   The problem above is that  this is not a real list, it will block everything since if its one address its blocked by not being the same as the other, generating an accidental "drop all" effect,  how would i  "unblock" a specific non contiguous set? does source accept a list of addresses? i have not see anything in my look at the docs or google result so far.     EDIT: I  just  created this:  # firewall-cmd  --zone=encrypt --list-all encrypt (active)   interfaces: eth1   sources: 192.168.56.120   services: ssh   ports: 6000/tcp   masquerade: no   forward-ports:    icmp-blocks:    rich rules:    But i can still reach port 6000 from  .123   my intention was that   if a source is not listed, it should not be able to reach any service or port I noticed something strange recently when trying to use rounded corners in pgfplots: it totally ruins the plot. See below without rounded corners:    and then with rounded corners:    Here is a minimal example:  \documentclass{standalone} \usepackage{pgfplots} \pgfplotsset{compat=newest}  \begin{document}  \begin{tikzpicture}[scale=0.8,trim axis left, trim axis right] \tikzset{every mark/.append style={scale=0.6}} \pgfplotsset{legend style={font=\footnotesize}}  \begin{axis}[   grid=both,             rounded corners,             ]     \addplot[samples=100,color=blue,very thick,smooth,domain=-10:10] {cos(x^(3))}; \end{axis} \end{tikzpicture}  \end{document}   Thanks in advance for your ideas ! systemctl --user not available for www-data user systemctl --user seems to be working fine for the desktop user:  dev@dev-VirtualBox:~$ systemctl --user &gt; /dev/null dev@dev-VirtualBox:~$ echo $? 0   But when running the same command under the www-data user I get an unexpected response  dev@dev-VirtualBox:~$ sudo su www-data -s /bin/bash www-data@dev-VirtualBox:~$ systemctl --user &gt; /dev/null  Failed to connect to bus: No such file or directory  www-data@dev-VirtualBox:~$ echo $?  1   How to enable systemctl --user here?  Running Ubuntu 16.04 Z-axis issue with the floating review info panel, permits various elements to flow over it n tasks assigned to n computers, what is the EX value of a computer getting 5 or more tasks? Say a central server assigns each of n tasks uniformly and independently at random to n computers connected to it on a network. Say a computer is ‘overloaded’ if it receives 5 or more tasks.  Q: Calculate the expected number of overloaded computers.  I thought of doing [1 - Pr(a computer is not overloaded)] but that leads me to a complicated expression of: $$1 - PR(NotOver) = 1 - \sum_{i=0}^4  \left( \frac{1}{n} \right)^{i} { \left( \frac{n-1}{n} \right)}^{n-i}$$ multiplying this by n would(hopefully) give the Expected value. But the answer seems not very elegant atall, is there something I'm missing or an easier way to tackle this? Thanks! Why a mono-atomic crystal layer (2D) can't be stable? According to Peierls and Landau, 2D crystals were thermodynamically unstable. They can't exist! Of course, this theory was disapproved in 2004 (example: graphene).  What is the general definition of stability of a general system?  What is the thermodynamics' stability? How does one find which files are taking up 80% of the space on a Linux webserver? The drive is constantly filling up. You've hunted down all the spare and random files you can. grep'd for coredump files and even removed some of the un-needed backups...  What would be your next move.  The actual server in question has 10GB of website files and the OS shouldn't take any more than 10GB so how do you track down what's filling a 50GB (virtual) drive? To apply for a Schengen business visa, you need a letter from the company in the Member state inviting you. My company is based in the UK, owns properties in Schengen countries, and I have to inspect their condition.  What letter do I get? Is there any freely available pointcloud data?   I'm especially interested in dense clouds of small areas(5m to 100m radius) but anything else would be fine too.   I've found that these pages have very nice and easy to access point cloud data. Creating a node fitting the horizontal width of two other nodes I have the two PGF nodes foo and bar positioned in a row.  \documentclass{article} \usepackage{tikz} \usetikzlibrary{positioning} \begin{document}   \begin{tikzpicture}[node distance=0.2cm,mynode/.style={rectangle,draw}]     \node[mynode] (foo)                {foo};     \node[mynode] (bar) [right=of foo] {bar};   \end{tikzpicture} \end{document}   How can I create a third node below those two nodes with the width of foo, bar and the node distance between them? Is there a way to do this with the fit library or is this the wrong approach? Adding values to a picklist using apex Find all natural numbers $n$ and $m$ for which $n^m=(n-1)!+1$ If someone could claim me as a dependent, but they will not, am I required to check the "Someone can claim me..." box on my 1040? Let's say that an adult is living with his parents.  He meets all of the  of his parents.  Now let's say that he wants to file a tax return and claim the , as he did not receive any of the stimulus payments in 2020.  His parents decide not to claim him as a dependent on their taxes.  Can he leave the &quot;Someone can claim me as a dependent&quot; box on his  unchecked and claim the Recovery Rebate Credit?  This does not apply to me, but it could apply to many people this year, with the new Recovery Rebate Credit putting people in the situation where it could be advantageous to not be claimed as a dependent. Surprisingly, the  do not provide any guidance on checking this box in any way, that I can find.  I have seen it asserted many times in answers and comments here on Money.SE that because the form says, &quot;Someone CAN claim...,&quot; the box must be checked whether or not anyone actually will claim them.  But I have not yet found this explicitly stated in any official IRS website or publication.  The ideal answer here will cite an official source. points out that certain key and block lengths were a requirement for the AES submissions:     The candidate algorithm shall be capable of supporting key-block combinations with sizes of 128-128, 192-128, and 256-128 bits.   My question is this; how can having a key length that is greater than the block size enhance security?  If we look at ECB mode for simplicity, when the key length is equal to the block length we have a permutation relationship between the set of all plaintexts and the set of all ciphertexts.  Once you increase the key length beyond that, you start to introduce collisions. In fact, you only introduce collisions from then on don't you? Collisions in the sense that a key will create the same ciphertext-plantext relationship as another key.  This would lead me to believe that only part of the key-space is unique, and only that part of key-space needs to be brute-forced?  I know that I am clearly incorrect, I just don't understand why. Does proving (second countable) $\Rightarrow$ (Lindelöf) require the axiom of choice? I get double question marks when referring to labels inside figures I have some charts embedded in figures, and I want to be able to refer to them easily using the "Figure x" notation. I'm doing it like this (in this order):  \documentclass[12pt,letterpaper]{article}  ...  \section*{Results} ...we performed a linear regression to obtain a standard curve  (Figure~\ref{0:stdcurve})....  \pagebreak \section*{Results (Graphs)} \subsection*{Construction of Standard Graph for PNP} \begin{figure}[h]   % Figure 1     \centering     \includegraphics[scale=1]{kinematics-res/graph0-1}     \caption{The standard curve for PNP.}     \label{0:stdcurve} \end{figure}   However, instead of getting the output "...to obtain a standard curve (Figure 1)...", I just get "...to obtain a standard curve (Figure ??)..."  I also noticed that the PDFLaTeX compiler is generating warnings saying that it couldn't find the reference of the name 0:stdcurve. Why am I getting these warnings? I have to prove by induction that $$\sum_{r=0}^{n}\begin{pmatrix}n \\ r \end{pmatrix}=2^k$$, using the identity $\begin{pmatrix}n \\ r-1 \end{pmatrix}+\begin{pmatrix}n \\ r \end{pmatrix}=\begin{pmatrix}n+1 \\ r \end{pmatrix}$. I'm fine with everything except the inductive step, where I only managed to get as far as $$\sum_{r=0}^{k+1}\begin{pmatrix}n \\ r \end{pmatrix}= \sum_{r=0}^{k}\begin{pmatrix}n \\ r \end{pmatrix}+\begin{pmatrix}n \\ k+1 \end{pmatrix}$$ I tried to then use the identity and substitute $\begin{pmatrix}n \\ r \end{pmatrix}=\begin{pmatrix}n+1 \\ r \end{pmatrix}-\begin{pmatrix}n \\ r-1 \end{pmatrix}$ but all this did was just take me around in a circle. I also couldn't substitute $2^k$ for $\sum_{r=0}^{k}\begin{pmatrix}n \\ r \end{pmatrix}$ since this didn't properly simplify to give $2^{k+1}$. However, I think that I have to obtain the expression $\sum_{r=0}^{k}\begin{pmatrix}n \\ r \end{pmatrix}$ since it simplifies to $2^{k+1}$. But I've not been able to obtain this in anyway so I'm wondering if anyone can guide in what I need to or what direction I need to take to complete the inductive step. Clearing Recently used coordinate reference systems using QGIS? After using some non usual SRC for demonstration purpose, I wish to clear my  "Recently used coordinate reference systems" list for the sake of praticity.  Is it possible?   I am using QGIS 2.10.1. Including +/- signs to battery in Circuitikz I'm attempting to draw a battery (using the battery1 component) that includes positive and negative signs at the appropriate ends. A  that was answered here (using version 2.4) shows exactly what I mean:    Only, of course, I want to be using a battery1 symbol instead. So, I have my code as follows:  \usepackage[americanvoltages,fulldiodes,siunitx]{circuitikz} \begin{document} \begin{circuitikz}     \draw (0,0) to [battery1=\SI{9}{V}] (3,0)                 to [pC,v=$v_C$] (6,0); \end{circuitikz} \end{document}   And I end up with this:    Notice the lack of +/- signs (I know how circuits work so I don't need to be actually shown these signs, but for the sake of copying my class notes diagrams I want these signs to be included in my latex diagram).  In either case, I think it may have something to do with the version of Circuitikz I'm using because I'm using the latest version (I believe, 0.3.0) but in the previous question linked above, they were using 2.4 and it worked out for them.  I would really like to find a way to add +/- without having to manually add it with Tikz or something, i.e. just have it naturally part of the component without having to add nodes or extra Tikz code etc. If that's not possible, then I'll have to settle for that as a last result.   Thank you for your help!  EDIT: I should also mention that I've tried using battery1, v=$9V$ instead of battery1=\SI{9}{V} and to no avail (the +/- signs still don't show up).