Can I enchant a necklace with the equivalent of a healing potion? (DnD 5e)

I know nothing of magic items really, especially custom ones, so I’m just going to describe an ideal scenario and I’d love some help in how something like it could be achieved, please! Am looking for some way to enchant a necklace with the ability to give some HP to the wearer. In an ideal world this would be triggered if the wearer hits 0HP, and would be rechargeable in some way. My player is not a spell caster, so I would be looking for a merchant or someone in town who could do this for her, if you have any advice on how much this might cost, up front and for each recharge or something. (For flavour reasons, she already has a specific necklace she’d like to enchant) If anything like this is possible, I’d love some help! Thanks!

what is the equivalent of Lathander from the Dawn War Deities

A player getting his cleric ready for a campaign I am creating has told me he usually plays Forgotten Realms and his usual deity is Lathander, this is not a deity I am familiar with.

Largely I am basing my Pantheon for this campaign on the DawnWar Deities, with some tweaks being made. I am perfectly happy allowing my Cleric to play Lathander, and he is equally happy to worship an equivalent deity in my pantheon.

Who would be the closest equivalent to Lathander from the Dawn War Deities, either to be replaced by Lathander or to replace him as my players Deity?

Offered equivalent .com domain to purchase, what is going on, and how to proceed?

A quick bit of background: I run a small (single person) software company, and operate under a .com.au domain that I have had for around 8 years. There has always been a .com domain of the same name, operating in an entirely different sector and country, but with not much going on on their website. I have always wanted the .com equivalent of my current domain, just because over the 8 years not only has business improved a little, but I also left Australia and, of course, a .com is just a little more desirable. With that in mind, I had the .com on backorder with my Australian registrar for quite a few years – but it never came up.

In the last few weeks, I suddenly got an influx of offers to buy this .com domain, seemingly from a number of different entities. First thoughts – scam of some sort. After checking out the WHOIS, it does seem that the .com expired last September, but my backorder had lapsed (great!). Registrant details are hidden behind PERFECT PRIVACY, LLC in Florida, and the domain status is clientTransferProhibited and the domain servers are NS1.PENDINGRENEWALDELETION.COM.

I am really curious as to what is actually going on here. Is it just a scam, because people have scraped details of an expiring domain, or do these people actually have access? If so, how? Has it been purchased? I didn’t (and don’t) know much about the murky world of domain name transfers, but SnapNames came up in my search and I looked on there. I did find the .com domain there, with a status of "Closed". Does that mean it came up in an auction and was bought? Or is another possibility that it is coming up for sale, and multiple people are trying to work out if it is worth buying and flipping to me? If it has sold, is there any way of finding out how much it went for? That might give me a bit of leverage in any future negotiations.

It is not, I don’t think, a valuable name. One of the potential sellers has offered it to me for $ 1299, which I might consider paying, but another question would be how does one go about purchasing a domain safely from an internet unknown?

Is Exit (no square brackets) equivalent to Quit[] for refreshing the Kernel from within an Evaluation Notebook?

I prefer to use Exit as it conveniently requires fewer key presses over Quit[]. But before I use it regularly I need to know if there any subtle differences between Quit[] and Exit. The Wolfram documentation pages for Quit and Exit appear to be very similar and even call these two functions synonymous but I just need to be sure.

Thanks.

Transforming multi-tape Turing machines to equivalent single-tape Turing machines

We know that multi-tape Turing machines have the same computational power as single-tape ones. So every $ k$ -tape Turing machine has an equivalent single-tape Turing machine.

About the computability and complexity analysis of such a transformation:

Is there a computable function that receives as input an arbitrary multi-tape Turing machine and returns an equivalent single-tape Turing machine in polynomial time and polynomial space?

What is the AC equivalent of mirror image?

It is suggested (by the DMG) to account for defensive options by assuming increased AC, e.g. +2 AC for magic resistance.

I would like to give some of my monsters the spell mirror image as part of a spellcasting / innate spellcasting trait. There is precedence for this: the Alhoon, the Faerie Dragon, and the Lamia have it, for example.

What would be a rough estimate of the AC equivalent for mirror image for the purpose of determining CR / combat performance?

Tangentially related (compares mirror image to other spells for PCs concerning the defensive capabilities): What's my most efficient use of spell slots to help my AC?

Are these two methods of handling Elven Accuracy “Double Advantage” mathematically equivalent?

For context, part of the Elven Accuracy feat (Xanathar’s Guide to Everything, p. 74) states:

Whenever you have advantage on an attack roll using Dexterity, Intelligence, Wisdom, or Charisma, you can reroll one of the dice once.

So the obvious way to handle this, mechanically, is to roll two dice, pick the lowest, and roll it again. But as an effort to save time, I’ve proposed instead simply rolling three dice simultaneously, and picking the highest rolled value.

The problem is that I’m not certain that this is mathematically correct.

I created a code simulation that was intended to model the probability curve of both methods, and it suggests that the two methods are mathematically equivalent, but the simulation only performs direct sampling of random numbers and their results; it has unavoidable error in the results, and it doesn’t attempt to solve the underlying mathematical principles involved.

//Roll 3, pick highest ResultSet: Double Advantage      Average: 15.48246                                                                                                      Variance: 14.94721234837884                                                                                            Std. Deviation: 3.8661624834425727                                                                                     95% range: [6, 20]                                                                                                     Mode: 20                                                                                                               Median: 16         //Roll 2, reroll lowest, pick highest ResultSet: Alternate Double Advantage                                                                                      Average: 15.488486                                                                                                     Variance: 14.944649427739675                                                                                           Std. Deviation: 3.8658310138623073                                                                                     95% range: [6, 20]                                                                                                     Mode: 20                                                                                                               Median: 16             

Is it correct to say that these two dice-rolling methods are equivalent, or should I stick to the RAW interpretation of how these dice should be rolled?


For full context, I’m planning out a build for a Shadow Sorcerer that fights only in melee combat, and if this character has the ability to nearly-permanently shroud themselves in Darkness (which is one of their class features), it’ll give them nearly permanent Advantage against creatures that don’t have Devil’s Sight or Truesight (or a reliable, spammable Counterspell/Dispel Magic). So simplifying this roll can matter in terms of time saved.

Is my recursive algorithm for Equivalent Words correct?

Here is my problem.

Problem Given two words and a dictionary, find out whether the words are equivalent.

Input: The dictionary, D (a set of words), and two words v and w from the dictionary.

Output: A transformation of v into w by substitutions such that all intermediate words belong to D. If no transformation is possible, output “v and w are not equivalent.”

I need to write both recursive and dynamic programming algorithm. As for recursion, I came up with this algorithm. Is it correct?

 EquivalentWordsProblem(v, w, D)  1.m <- len (v)  2.n <-  len (w)  3.substitutions = [] #array to save substitutions   4.if m != n:  5.  return "v and w are not equivalent"  6.else  7.for i <- m to 1 <-1 do  8.  for j <- n to j <- 1 do  9.    if v[i] != w[j]:  10.           substituted_word <- v[1…i-1]+v[j] #we substitute v[i] for w[j]  11.              if substituted_word in D:  12.                substitutions.append(substituted_word)  13.                   return EquivalentWordsProblem(v[1…m-i], w, D) #recur on the string of length m - i  14.   else:          return EquivalentWordsProblem(v[1…m-1], w, D) #recur on the string decreasing its length by 1  15.if len(substitutions) != 0:  16.  return substitutions   17.else  18.   return (“v and w are not equivalent”)  

Is checking if regular languages are equivalent decidable?

Is this problem algorithmically decidable?

L1 and L2 are both regular languages with alphabet $ \Sigma$ . Does L1 = L2?

I think that it is decidable because you can write regular expressions for each language and see if they are the same. But I’m not sure how to prove it since I see that you prove something is decidable by showing a Turing Machine

Are Online Problems always harder than the Offline equivalent?

I am currently studying Online-Algorithms, and I just asked myself if online Problems are always harder than the offline equivalent.

The most probable answer ist yes, but I can’t figure the reason out why.

Actually I have a second more specific question. When an offline Problem has some integrality gap ($ IG\in[1,\infty) $ ) we know in an offline setting, that there is generally no randomized rounding algorithm which achieves a ratio $ C\geq IG$ .

Can this just be adapted to the online problem? If some fractional algorithm has competitive ratio $ c_{frac}$ can some randomized rounding scheme only reach competitive ratio as good as $ \frac{c_{frac}}{IG}$ ?

Thanks in advance.