Discrepancy regarding AoE point of origin between English and German PHB

I own both the English and German version of the D&D 5e Player’s Handbook. On pages 204 and 205 it talks about area of effect spells and whether or not the point of origin is included in it.

The English PHB states:

A [cone,cube,line]’s point of origin is not included in the [cone,cube,line]’s area of effect, unless you decide otherwise.


A [cylinder,sphere]’s point of origin is included in the [cylinder,sphere]’s area of effect.

While the English version differentiated between two types of point of origin, the German version states the same sentence for all 5 types. Which is a translation of the sentence used in the English version for the cone, cube and line.

Der Ursprungspunkt [des/der] [Kegels,Linie,Sphäre,Würfels,Zylinders] ist nicht Teil [seines/ihres] Flächeneffekts, es sei denn, du möchtest, dass er es ist.

Both the Basic Rules PDF and the System Reference Document use the same wording as the English PHB.

My English PHB says its version is:

Tenth Printing: October 2018.

While my German one says:

  1. überarbeitete Auflage, 2019

Is this a mistake in the translated version? If so, who can I report this too?

Discrepancy between Google Ads clicks and Google Analytics sessions

After a month of ads with Google Ads Shopping campaigns and continuously identifying, day after day, a significant discrepancy of clicks on Google Ads and Sessions in Google Analytics (as well as another analytics tool that brings numbers very close to analytics’s numbers), I contacted Google Support. For more than 3 weeks, after numerous interactions and different/contradictory suggestions/explanations/solutions, it finally seems that my case has taken a direction towards the solution, but the result that has been passed to me, despite having convinced me that it is the which may actually be the cause of the problem, was not satisfactory to me. I was really disappointed with the procedure Google Support provided me, and would like more explanations and information from the community.

What I got last through Google’s support was this:

“There are two methods for logging out of a session: a session can be opened on the same day or on several days, weeks or months. Expiration by time: After 30 minutes of inactivity At midnight Campaign Change: If a user enters through a campaign, he leaves and then comes back through another campaign. Why is there a discrepancy between clicks and sessions? I have separated this link from the help center that explains specifically what we said in connection If you’ve noticed one of the following discrepancies between clicks and sessions, use the troubleshooter for clicks × sessions to identify and resolve issues. As well as reporting the discrepancy between clicks and sessions up to 20% is considerable. “

Among the Google help links passed by the support, it has not been clear to me whether a 20% less discrepancy in the number of sessions in relation to clicks is normal or if only when the session number is higher (the phrase “A single user can open multiple sessions” suggests that it is only normal when the number of sessions is higher).

But if it’s really normal that the number of sessions is 20% less than the number of clicks, I’m led to think of the possibilities:

1 – Analytics fails as a tool to count 100% the number of sessions (accessed by the same person or by different people). Once when a click occurs, the user is taken to the site and must count as a session. If it is normal for analytics to compute 20% fewer sessions than clicks, soon the analytics fail to compute all the sessions.

2 – Google uses this explanation to be able to charge up to 20% more clicks than clicks that actually occurred from its users.

I do not like to believe in either hypothesis, first to rely on Google Analytics as a tool to identify sessions and because their numbers are very similar to that of a third analysis tool used (built by my ecommerce platform), and also for not believing that a company like Google would lend itself to deliberately charging 20% ​​more clicks from its customers. So I turn to the community so maybe someone will bring me a third alternative that I have not been able to identify and/or a totally different explanation from the one I received from the support (as they have done so many times over the course of those weeks).

Graph ordering with smallest max vertex “discrepancy”

Consider an undirected graph $ G=(V,E)$ and a bijective function $ f:V \rightarrow [|V|]$ which orders the vertices by mapping them onto the first $ |V|$ natural numbers.

Define the cost of an ordering of a graph to be:

$ $ \text{cost}(G,f) = \max_{(u,v) \in E} f(u)-f(v)$ $

(I refer to $ f(u) – f(v)$ in the title as “discrepancy” to avoid confusion with terms such as length or distance)

Intuitively, if we were to construct $ G$ by adding on one vertex at a time and adding the appropriate edges to the already-existing vertices, the cost is how far back in our list of vertices we’d have to go and add edges to.

Question: How difficult is the problem of minimizing this cost? Is there an efficient algorithm for finding an optimal ordering? Also is there a name for this cost that i’m not aware of?


A BFS ordering can do arbitrarily bad on this problem: Consider a complete binary tree — the optimal cost is $ O(\log N)$ but a BFS ordering can cost up to $ O(N)$ .

A DFS ordering can do arbitrarily bad on this problem: Consider a path graph modified such that each vertex is connected to an additional leaf vertex — the optimal cost is $ 2$ but a DFS ordering can cost up to $ O(N)$ .

RAM usage discrepancy? Memory leak? Ubuntu 18.10 with Cinnamon

So I’m fairly new to using Ubuntu, but what I’ve been finding is that I need to reboot regularly because eventually (over the course of a day or two) I seem to run out of usable RAM and stop being able to play the games I play because they lock up and start disk thrashing and if I’m lucky I’ll be able to wait it out and force-close the game, and if I’m not I’ll need to just hard-reboot. The weirdest thing about it, though is this: https://i.imgur.com/mD8wVRE.png

This shows that just about half my RAM is in use. But then I tab over to the processes page and: https://i.imgur.com/cZ4sZqm.png

As you can see, this is without even running things like a browser or anything. So where has the RAM gone?

Transaction discrepancy between Google Analytics and Woocommerce

Hey guys!

I am currently experiencing a 25% discrepancy between the numbers reported in Google Analytics and those reported in Woocommerce backend and don't know the cause of it. I have read many articles (ie: Google Analytics and Woocommerce transaction mismatch, etc) but so far I haven't been able to fix my problem and Google Analytics is still reporting 25%…

Transaction discrepancy between Google Analytics and Woocommerce

How should the DM manage the discrepancy between the player’s memory and their PC’s memory?

It may happen that, during a session, players don’t remember the name of a NPC that they met (or, more generally, information about something that happened) during the previous session. Obviously, their PC remembers that information. Conversely, a player may have taken notes about a not so important event that happened several years ago (in game). In this case, it is possible that the PC does not remember it.

How should the DM manage the discrepancy between the player’s memory and their PC’s memory? In the case of 5e, should he have the PCs make Intelligence checks?

Lower bound of disjointness by discrepancy?

I need to show that $ Disc_\mu(Disj) \geq \frac{1}{2n+1}$ for any distribution $ \mu$ . Disjointness is defined as

$ Disj(X,Y)=\left\{ \begin{array}[ll]+1 & \text{if $ X \cap Y = \emptyset$ } \ 0 & \text{otherwise}\end{array}\right.$

The discrepancy for a function is defined as

$ Disc_\mu(f) = max_R \{Disc_\mu(R,f)\}$ for any rectangle $ R$ and further

$ Disc_\mu(R,f) = |\Pr_\mu [(x,y) \in R \wedge f(x,y)=1] – \Pr_\mu[(x,y) \in R \wedge f(x,y) =0] |$ .

I don’t understand where the $ \frac{1}{2n+1}$ comes from and how to define a proper rectangle $ R$ .