So I'm checking out the 5e DMG on encounter building, and it seems that XP is awarded based on monsters defeated (ie; defeating 3 Goblins worth 50xp each is 150xp), but encounter difficulty is based on an additional multiplier for the number of creatures.  This means that fighting a single CR 1 Bugbear earn a 4 player party 200xp and would be considered a Medium difficulty encounter, yet fighting a group of 3 CR 1/4 Goblins would earn them 150xp (3x50xp) even though according to the encounter difficulty calculator this would actually be a Hard encounter.  Is there any reason given anywhere why supposedly harder encounters are worth less experience than easier ones, just because they happen to involve more monsters?
Based on the table Encounter XP Multipliers from the Basic Rules' DM section, Page 57, when you're building encounters in 5E, you include an XP modifier depending on the number of creatures your PCs will be fighting: 1.5x for a pair, 2x for 3-6 and so on. However, the PCs only get the base XP as a reward.     This doesn't change the actual xp reward ... just your calculations of how difficult the encounter is.   Why is this? After acknowledging that battles vs multiple opponents are more difficult, why not reward them as such?
I inherited a project using SQL Server 2005.  Recently there were some performance problems and I started looking at the indexes, and found  _dta_index_survey_25_135059617__K1_K10_K19:     id ASC, sent_date ASC, group_id ASC   and  _dta_index_survey_21_364632442__K18_K1_K2_K9:     group_id ASC, id ASC, campaign_id ASC, sent_date ASC   which appears to be generated by some kind of automatic performance tuning tool.  I think SQL Server comes with something like that but I'm not familiar with it.  Anyway, id is the primary key, and what I'm trying to understand here is how it could possibly be beneficial to include the primary key within a secondary index... because you have to have the leftmost pieces of information to use an index and if you know id then you've got the row anyway.  Can someone explain why a tool would have generated these indexes and if there's any point in keeping them?  Just because the indexes are getting used doesn't mean they're a good design.  SQL Server is using the wrong indexes lately, and I'm trying to determine why. More specifically, queries against this table that ran in milliseconds a month ago are now taking multiple seconds.  The table is huge (4M rows), and growing steadily, but at the same pace it's been going at for years.  Almost every query joins across group_id or campaign_id.  Often both.  The application almost never specifically selects or sorts by the PK, but I could see how ID could be useful when comparing or merging lists of IDs within the database engine.  ... But id is also an identity column, and an index holds lists of row ids, so it still seems redundant to have the ID in the key when it's also in the index "buckets".  And the additional fields could still be covered by the index without being part of the key.  I discovered that (id, sent_date, group_id) hardly ever gets read.  Meanwhile (group_id, id, campaign_id, sent_date) is one of the hottest indexes.  Next I found that an index which I thought was pretty reasonable (campaign_id, sent_date, group_id, id) also has a index_columns_include of every other column in the table, and occupies more space than the primary clustered index on id!  (but it's also pretty hot...)  Surely I can replace it with a simple (campaign_id, sent_date, group_id)? What is 'Currying'? I've seen references to curried functions in several articles and blogs but I can't find a good explanation (or at least one that makes sense!) Cohen's d for 2x2 interaction I have a significant 2x2 interaction in a between-subjects design.  I would normally report omega-squared as my effect-size measure, but I've been asked to provide Cohen's d instead?  Can Cohen's d be computed for an interaction?  If so, how? What does this cryptic Bash command mean? According to the Wikipedia article on  the sample SD   $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \overline{x})^2}$$  is a biased estimator of the SD of the population. It states that $E(\sqrt{s^2}) \neq \sqrt{E(s^2)}$.   NB. Random variables are independent and each $x_{i} \sim N(\mu,\sigma^{2})$   My question is two-fold:   What is the proof of the biasedness? How does one compute the expectation of the sample standard deviation    My knowledge of maths/stats is only intermediate. Private IP address in public DNS We have an SMTP only mail server behind a firewall which will have a public A record of mail..  The only way to access this mail server is from another server behind the same firewall.  We do not run our own private DNS server.  Is it a good idea to use the private IP address as an A record in a public DNS server - or is it best to keep these server records in each servers local hosts file? Drush hangs when trying to bootstrap Drupal ArrayList object sorting confusion Why shouldn't I mix tabs and spaces? The cartesian product $\mathbb{N} \times \mathbb{N}$ is countable I'm examining a proof I have read that claims to show that the Cartesian product $\mathbb{N} \times \mathbb{N}$ is countable, and as part of this proof, I am looking to show that the given map is surjective (indeed bijective), but I'm afraid that I can't see why this is the case. I wonder whether you might be able to point me in the right direction?  Indeed, the proof begins like this:  "For each $n \in \mathbb{N}$, let $k_n, l_n$ be such that $n = 2^{k_n - 1} \left(2l_n - 1 \right)$; that is, $k_n - 1$ is the power of $2$ in the prime factorisation of $n$, and $2 l_n - 1$ is the (necessarily odd) number $\frac{n}{2^{k_n - 1}}$."  It then states that $n \mapsto \left(k_n , l_n \right)$ is a surjection from $\mathbb{N}$ to $\mathbb{N} \times \mathbb{N}$, and so ends the proof.  I can intuitively see why this should be a bijection, I think, but I'm not sure how to make these feelings rigorous?   I suppose I'd say that the map is surjective since given any $\left(k_n , l_n \right) \in \mathbb{N} \times \mathbb{N}$ we can simply take $n$ indeed to be equal to $2^{k_n - 1} \left(2l_n - 1 \right)$ and note that $k_n - 1 \geq 0$ and thus $2^{k_n - 1}$  is both greater or equal to one so is a natural number (making the obvious inductive argument, noting that multiplication on $\mathbb{N}$ is closed), and similarly that $2 l_n - 1 \geq 2\cdot 1 - 1 = 1$ and is also a natural number, and thus the product of these two, $n$ must also be a natural number. Is it just as simple as this?  I suppose my gut feeling in the proving that the map is injective would be to assume that $2^{k_n - 1} \left(2 l_n - 1 \right) = 2^{k_m - 1} \left(2 l_m - 1 \right)$ and then use the Fundamental Theorem of Arithmetic to conclude that $n = m$. Is this going along the right lines? The 'implicit' definition of the mapping has me a little confused about the approach.    On a related, but separate note, I am indeed aware that if $K$ and $L$ are any countable sets, then so is $K \times L$, so trivially, taking the identity mapping we see trivially that this map is bijective and therefore that $\mathbb{N}$ is certainly countable (!), and thus so is $\mathbb{N} \times \mathbb{N}$. Hence, it's not really the statement that I'm interested in, but rather the exciting excursion into number theory that the above alternative proof provides. How do I connect a PS3 Controller with Xperia P? How do I install drivers for the RTL8821CE Network Controller on Ubuntu 20.04? My Laptop (HP Laptop 14-ck1xxx) came with the RTL8821CE Network Controller. When I run lspci | grep Network I get: 03:00.0 Network controller: Realtek Semiconductor Co., Ltd. RTL8821CE 802.11ac PCIe Wireless Network Adapter  I‘m unable to get the Wi-Fi to work. What I did so far:  Download the repo at . Unzip the file and rename the folder from rtl8821ce-master to rtl8821ce. Run the following commands in the terminal:  cd Downloads/rtl8821ce/ sudo apt install bc module-assistant build-essential dkms sudo m-a prepare sudo ./dkms-install.sh    Everything seemed to install and execute successfully. Then, I rebooted my laptop and made sure secure boot is disabled. However, Ubuntu still displays: No Wi-Fi Adapter Found - make sure you have a Wi-Fi adapter plugged in and turned on  The command uname -r returns: 5.4.0-7634-generic  The command sudo lshw -C network returns: *-network                         description: Ethernet interface        product: RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller        vendor: Realtek Semiconductor Co., Ltd.        physical id: 0        bus info: pci@0000:02:00.0        logical name: eno1        version: 15        serial: c4:65:16:08:52:bf        size: 1Gbit/s        capacity: 1Gbit/s        width: 64 bits        clock: 33MHz        capabilities: pm msi pciexpress msix bus_master cap_list ethernet physical tp mii 10bt 10bt-fd 100bt 100bt-fd 1000bt-fd autonegotiation        configuration: autonegotiation=on broadcast=yes driver=r8169 duplex=full firmware=rtl8168h-2_0.0.2 02/26/15 ip=192.168.178.37 latency=0 link=yes multicast=yes port=MII speed=1Gbit/s        resources: irq:16 ioport:4000(size=256) memory:a1204000-a1204fff memory:a1200000-a1203fff   *-network UNCLAIMED        description: Network controller        product: RTL8821CE 802.11ac PCIe Wireless Network Adapter        vendor: Realtek Semiconductor Co., Ltd.        physical id: 0        bus info: pci@0000:03:00.0        version: 00        width: 64 bits        clock: 33MHz        capabilities: pm msi pciexpress cap_list        configuration: latency=0        resources: ioport:3000(size=256) memory:a1100000-a110ffff  What am I doing wrong? How can this be solved? This applies to all d20 systems, but with an example from the from the Pathfinder  (emphasis mine):  Stonecunning: Dwarves receive a +2 bonus on Perception checks to potentially notice unusual stonework, such as traps and hidden doors located in stone walls or floors. They receive a check to notice such features whenever they pass within 10 feet of them, whether or not they are actively looking.  In this case, does the GM roll the check in secret or ask the player roll the check? If the GM asks the player to roll for the check, what stops situations like:  GM: (interrupting) &quot;Gimli, Roll a perception check.&quot; Dwarf player: &quot;Oh, it's a natural one, I guess I don't see anything.&quot; Human player: &quot;Okay, I'll roll a perception check for anything unusual in this room. Nice! 19. How about 19, does 19 work for you?&quot; GM: &quot;Gimli, everything looks kosher to you. Aragon, you notice a rock coloured tarp on the ground here, covering what looks like a pit trap.&quot;  In this case, by telling the dwarf to roll his &quot;free&quot; perception check, he's telegraphed to the players that there is something to notice. How can this be fixed, without bluffing checks for nothing in particular to throw off the signal to noise ratio of similar checks? Last night I was updating my Ubuntu 10.04 to either 11.04 or 12.04 (I'm not entirely sure which one it was, but I think it was 11.04) and I left it overnight to finish. When I came down in the morning, I noticed that the computer hadn't barely begun installing the updates! The update thing said that my computer was on the installation step, but it was still preparing the libc bin (or whatever its called) . It's like the first part of the step installing the upgrades. So, I stupidly restarted the computer, and when it rebooted, I got five options on a purple screen:  Ubuntu, with Linux 3.0.0-28-generic Ubuntu, with Linux 3.0.0-28-generic (recovery mode) Previous Linux versions Memory test (memtest86+) Memory test (memtest86+, serial console 115200)   Long story short, not matter which way I enter Ubuntu to try and reinstall the updates using the root thing (it's like terminal) after I enter sudo apt-get update it basically said it failed to fetch every url link, and it then says:  W: Some index files have failed to download. They have been ignored, or old ones used instead. W: Not using locking for read only lock file /var/lib/dpkg/lock E: dpkg was interrupted, you must manually run 'sudo dpkg --configure -a' to correct the problem.   So, I entered sudo dpkg --configure -a, and it says:  dpkg: error: unable to access dpkg status area: Read-only file system"   I'm left like wtf, and I really just some guidance here. I just got this laptop, and I can not afford a new one if I broke this one. Can I register a domain name that contains a company name or trademark? I wonder is it legal to register a domain name that contains a company name. For example cococola.cc (considering that this domain name is available). Can I buy and use any domain names that contain a name of a trademark? Should there be a comma in 'a 30-day, money-back guarantee'? How much will a poor first semester affect my grad school chances? How does Cantor's diagonal argument work? I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the  article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):  \begin{align} s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.  My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:  \begin{align} s_0 &amp;= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\ s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ s_4 &amp;= (0,1,1,\mathbf{1},\dots)\\ s_4 &amp;= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\ s_5 &amp;= (1,0,0,0,\dots)\\ s_6 &amp;= (1,0,0,1,\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:     For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,      $$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$      ...snip...      Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.      Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.   I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:  $$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 &amp; \mathrm{if\ } s_{m,m} = 1\\1 &amp; \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$  This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because...   Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.  Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".  What am I missing? Getting full JS autocompletion under Sublime Text Use early 2009 20" iMac as external display for 2015 rMBP