Does a Wizard copying a spell count as one of their two new spells for a level?

In D&D 5E Wizards get an additional two spells per level to add to their spellbook:

Each time you gain a wizard level, you can add two wizard spells of your choice to your spellbook.

If a Wizard has a handful of Scrolls and another Wizard’s spellbook they have found while adventuring, can they attempt to scribe them all (assuming they are of the necessary spell level and have the time and money to do so), or are they limited to 2 new spells per level?

Does a sprite familiar’s equipment count against its carrying capacity?

Under Lifting and Carrying (PHB 176) it says:

Your carrying capacity is your Strength score multiplied by 15. This is the weight (in pounds) that you can carry… You can push, drag, or lift a weight in pounds up to twice your carrying capacity (or 30 times your Strength score).

and also specifies that a tiny creature can carry half as much

A sprite familiar summoned through a warlock’s Pact of Chain feature has a strength score of 3 and so has a carrying capacity of 22.5 pounds and a push, drag, lift limit of 45 pounds. The sprite stat block also specifies, however, that they wear leather armor and carry a longsword and shortbow.

Supposedly these weigh 10 pounds, 3 pounds and 2 pounds respectively but considering they are smaller than those worn by a medium humanoid it is unlikely that they weigh as much.

Is there any official guidance as to how much the sprite familiar’s equipment counts against its carrying capacity?

Does mobile biometric authentication count as Knowledge, Possession, or Inherence authentication

Apple claims in this year’s WWDC that Face ID and Touch ID count for both Possession and Inherence identity factors, because they are using Biometrics (Inherence) to access the secure element on your phone (Possession) to retrieve a unique key. See here:

I think both claims are a stretch. For Inherence, yes, you have proved to iOS that the person who set up Face ID is again using the phone, and therefore given access to the secure key. So iOS can claim Inherence. But your app has no proof that the human possessing the phone is actually your user. Hence my app considers mobile local authentication merely a convenient Knowledge factor–a shortcut for your username and password that resolves common credential problems like human forgetfulness.

As for Possession, again, I think the claim is a stretch unless before writing the unique key to the phone’s secure element you somehow prove that the possessor of the phone is your actual intended user. I suppose if you enable Face ID login immediately after account creation you can have this proof–the brand-new user gets to declare this is their phone like they get to choose their username and password. But on any login beyond the first you would have to acquire proof of Possession using an existing factor before you could grant a new Possession factor. Else a fraudster who steals credentials can claim their phone is a Possession factor by enabling Face ID; a situation made extra problematic by Apple’s claim that Face ID also counts as Inherence!

Am I wrong in this assessment? Which of Knowledge, Possession, and Inherence should an app developer grant mobile local biometric authentication?

Airmon-ng/Airodump-ng – Low Beacon Count on certain networks

It’s been a few years since I’ve played around with this so I’m not sure if times have changed.

When using AR7921 chipset (Alfa AWUS036NHA) on Ubuntu 20.04 I can see many networks using the following commands

airodump-ng start wlx00c0ca84d0f8  airodump-ng mon0 

I’ve noticed all the VMxxxxx networks have a very low beacon count, perhaps 1 every 20 seconds (compared to e.g BT Broadband APs which seem to have a “normal” beacon count of several per second. The VMxxxx networks are Virgin Media home broadband networks- including the connection I’m legitimately using that is 6 feet away from me.

The rest of the networks have normal looking beacon counts. Has something changed/new technology in place that reduces the beacon count over the past few years or is the issue something else?

I can use the following command for half an hour on my VMxxxx network and find nothing connecting to it, even though I’ve got 4 different devices here using that access point (disconnected and reconnected them several times).

airodump-ng -d [APMACHERE] -c 6 mon0 

Sample output after 5 minutes

 CH  6 ][ Elapsed: 5 mins ][ 2020-06-13 16:53 ][ fixed channel mon0: -1                                            BSSID              PWR RXQ  Beacons    #Data, #/s  CH  MB   ENC  CIPHER AUTH ESSID   xx:xx:xx:xx:xx:xx    0   0        6        0    0   6  54e. WPA2 CCMP   PSK  VMxxxxxxx                                                           BSSID              STATION            PWR   Rate    Lost    Frames  Probe                                                                      

It is my understanding there should be a significant amount more beacons, especially from an AP in the same room.

Any idea what the problem might be?

How to make jquery count down timer function manually editable

I have a wordpress theme and I would like to display a certain promo start time in front page. It uses jquery.countdown.min.js and has following default timer settings

var siteCountDown = function() {          if ( $  ('#date-countdown').length > 0 ) {             $  ('#date-countdown').countdown('2020/10/10', function(event) {               var $  this = $  (this).html(event.strftime(''                 + '<span class="countdown-block"><span class="label">%w</span> weeks </span>'                 + '<span class="countdown-block"><span class="label">%d</span> days </span>'                 + '<span class="countdown-block"><span class="label">%H</span> hr </span>'                 + '<span class="countdown-block"><span class="label">%M</span> min </span>'                 + '<span class="countdown-block"><span class="label">%S</span> sec</span>'));             });         }      };     siteCountDown(); 

It displays that promo will start in four months later after now. But I would like to set timer manually via wordpress admin dashboard. Is it possible to insert some php codes in .countdown(‘2020/10/10’ something like

<?php if ($  year = get_option('of_year') && $  month = get_option('of_month') &&  $  day = get_option('of_day') ) { ?> .countdown('<?php echo $  year; ?>/<?php echo $  month; ?>/<?php echo $  day; ?> <?php }  ?> 

so that I can edit promo start time manually

My options are:

$  options[] = array( "name" => "Year",                     "desc" => "promo start year",                     "id" => $  shortname."_promo_year",                     "std" => "",                     "type" => "tex");   $  options[] = array( "name" => "Mount",                     "desc" => "promo start month",                     "id" => $  shortname."_promo_month",                     "std" => "",                     "type" => "tex");  $  options[] = array( "name" => "Day",                     "desc" => "promo start day",                     "id" => $  shortname."_promo_day",                     "std" => "",                     "type" => "tex"); 

Would this algorithm fail to count solutions $>$ $1$ for Exact-3-cover?

Decision Problem: Given a set $ S$ , is there at least a given $ N$ $ >$ $ 1$ amount of solutions, for an $ Exact~Cover~by~3-sets$ for $ C%$ ?

$ s$ = $ 1,2,3,4,5,6$

$ c$ = $ [[1,2,3],[4,3,2],[4,5,6],[5,1,6],[5,6,3]]$


$ [1,2,3],[4,5,6]$

$ [4,3,2],[5,1,6]$

$ N$ = $ 2$

Yes, there are $ N$ solutions.


  1. Remove sets that have repeating elements

    (eg. [1,1,2] is deleted from $ C$ )

  2. Remove sets that have elements that don’t exist in $ S$

    (eg. [9,5,6] is deleted because $ 9$ not in $ S$ )

  3. Make sure all elements in $ S$ exist in $ C$ .

$ for$ a $ in$ $ range(0, length(s)):$

$ ~~~~~~~~$ $ IF$ $ s[a]$ $ not$ in $ c$ :

$ ~~~~~~~~~~~~~$ OUTPUT NO

Convert $ C$ into a complete list

$ WHILE$ $ c[i]$ has [brackets]:

$ ~~~~~~~~~~$ DELETE [BRACKETS] FROM $ C$

now $ c$ = $ [1, 2, 3, 4, 3, 2, 4, 5, 6, 5, 1, 6, 5, 6, 3]$

Finally, Decide

$ n$ = $ (‘Enter~for~N:~’))$

$ yes$ = $ 0$

$ for$ a $ in$ $ range$ (0, $ length(c)):$

$ ~~~~~~$ $ if$ $ c$ .count($ c$ [a]) >= $ n$ :

$ ~~~~~~~~~~$ $ yes$ = $ 1$

$ ~~~~~~$ else:

$ ~~~~~~~~~~$ OUTPUT NO

$ ~~~~~~~~~~$ HALT

$ if$ $ yes$ == $ 1$ :

$ ~~~~$ OUTPUT YES

Edit: The above should do the same below.

yes = 0 for a in range(0, length(s)):     if c.count(s[a]) >= n:         yes = 1     else:         OUTPUT('No')         break  if yes == 1:     OUTPUT('yes') 

Facts to consider

  1. There cannot be any sets with elements that don’t exist in $ S$ .

  2. There cannot be any sets with repeating elements.

  3. All elements in $ S$ must exist in $ C$ . Else, a $ no$ is given.
  4. $ N$ must be > $ 1$
  5. If any element in $ C$ occurs < $ N$ times then the output must be $ No$ , because there wouldn’t be at least $ N$ solutions.


Will this algorithm always work if the input is > $ 1$ , and if no how would it fail?

Count how many posts have a specified tag AND category

I know how to count how many posts has a certain tag, or category

For example:

$  term_slug = 'some-post-tag'; $  term = get_term_by('slug', $  term_slug, $  post_tag); echo $  term->count; 

BUT! Is there anyway to count how many posts that have a tag AND a specified category?

I want to count how many posts that have the tag(slug) “cats” and the category slug “allow-list”

Is this even possible?

If I multiclass as a warlock and another class, and I get spell slots from that other class, do they count as my highest-level slot or not? [duplicate]

Since warlocks have so few spell slots, they always cast with the highest level spell slot possible. So, if I got spell slots by multiclassing as a sorcerer (or wizard) and I cast a warlock spell using one of those sorcerer (or wizard) slots, would I cast it as the highest level I have (level 3), or would I only be able to cast a spell that slot level, but it would act as if I had used a level 3 spell slot or would I only be able to cast the level the slot is and that’s all? I know this is probably a stupid question but it might just work.