Can a creature kill itself by dashing during a chase?

DMG p.252

During the chase, a participant can freely use the Dash action a number of times equal to 3 + its Constitution modifier. Each additional Dash action it takes during the chase requires the creature to succeed on a DC 10 Constitution check at the end of its turn or gain one level of exhaustion.

Suppose a creature was in a chase and already at Level 4 of exhaustion. It chose to continue dashing and at the end of its turn failed its constitution check.

DMG p. 252

A participant drops out of the chase if its exhaustion reaches level 5, since its speed becomes 0.

Now suppose a creature that was [allowed to Dash twice on its turn] started its turn at exhaustion level 4 and chose to Dash twice. At the end of its turn, it would be required to make two Con checks. If it failed the first check and got to Exhaustion Level 5 would this immediately remove it from the chase and thus remove the need to check again, or would a second check still be required with the possibility of death if it failed?

Note that exhaustion from chases has a different recovery mechanic than other exhaustion, in that (DMG p. 252)

A creature can remove the levels of exhaustion it gained during the chase by finishing a short or long rest.

whereas exhaustion gained by other means requires a long rest, food, and drink. Thus there is some evidence that while chases are temporarily exhausting, the exhaustion gained from them is not as serious or as lasting. Might this include the chase-induced ‘death’ as well?

What is the theoretical result of flattening a list containing only itself?

Consider the following python code

X = [None] X[0] = X 

This has many fun properties such as

X[0] == X 

and

X[0][0][0][0][0][0][0][0] == X 

I understand how this works from a practical standpoint, as it’s all just the same pointer.

Now, when flattening a list, we tend to convert things from

[[1, 2, 3], [[4, 5], [6, 7]], 8, 9] 

to

[1, 2, 3, 4, 5, 6, 7, 8, 9] 

In this case, I am considering flattening to be reducing a multi-dimensional list down to a single list of only non-list elements.

In practice, flattening this list would be impossible, as it would create an infinite loop.

This may be more mathematical in nature, but I’m unsure how to put it in mathematical terms. Suppose we could flatten this list, would the result simply be the empty list?

For context, I initially got this idea by considering the list

X = [[], []] X[0] = X 

It is clear to see that at each pass of flattening, the empty list that is the second element simply disappears. This lead me to think that the overall result, may be the empty list.

Could it be possible that flattening this list would theoretically produce an infinitely long list of the list itself, as in

X == [X, X, X, X, X, X, ..., X] 

This is purely a fun thought exercise. Any insight and discussion on this would be appreciated.

P.S. Although I’m looking for an answer in plain terms, if anyone is more mathematically inclined, I would be interested to see how this problem could be formulated in some sort of set notation. Please feel free to point me to a relevant math exchange thread as well.

P.P.S. I would also be interested in a solid proof (not formal) to go along with the answer.

Storing the password to a vault in the vault itself?

Is there an additional risk associated with storing the master password to a vault inside the vault itself?

I would assume not, since in order to decrypt the vault you must already have that password. But maybe I’m missing something?

And without reuse concerns, anything that can steal the password from the unlocked vault can also just steal the vault itself, so no additional information is being exposed that way.

As to why, besides academic curiosity, I’ve also noticed that sometimes the web version of the vault does not automatically log me in, even if the native app is unlocked. So adding the vault password would simplify that process.

How does flexible casting interact with itself?

As I find builds about the Coffeelock (for peoples who doesn’t know, it’s a build that hollow you to have an infinite amount of 5th level spells slot RAW, involving a sorcerer multiclassing in warlock), I turned out asking myself a question:

How does flexible casting interact with itself?

To be clear, I want to know if you can turn created spell slots into sorcery points RAW (which could mean you have an infinite source of sorcery points if you need to do metamagic with your infinites spells).

How would Google treat a GitHub-Pages site, that is based on the same content as the GitHub repository itself?

I’ve just created a GitHub Pages "site", out of one of my repositories on Github. Basically it’s just a landing page, that pretty-prints my readme.md file(that is found in the repo). GitHub allows you to create such a site automatically.

I don’t know too much about SEO, but i’ve heard that google takes into account "duplicate" content, as a "devaluation" factor.

So i was wondering: If google finds a Github repo page, that outputs some readme.md file, and then finds a Github.io(GitHub Pages) page, that has almost the identical content- would it treat it as some duplicate, and apply some kind of penalty to it?

Can I make a creature target itself?

Hypnotism (Heroes of Fallen Lands p205):

Hit: […] The target uses a free action to make a melee basic attack against a creature of your choice[…]

Friendly Fire (Dragon 384):

Effect: The target repeats the attack as a free action against a creature you choose within 2 squares of the target of its original attack. The new target must still be legal for the attack.

Can I make the target of Hypnotism or Friendly Fire target itself with the MBA?

Read More excerpt link on a post is not linking to the post itself

My Read More link on several pages (showing an excerpt of those posts) is not linking to the posts themselves when you click on Read More. It looks like it is a link and should do so, but it just update the same page and keeps you on that same page…

I don’t know how to code so very very basic layman’s terms is greatly appreciated. I am using Astra Theme if that makes a difference at all.

Thank you in advance for your assistance.

Suggest how to allocate and deallocate storage for elements within the hash table itself by linking all unused slots into a free list

Suggest how to allocate and deallocate storage for elements within the hash table itself by linking all unused slots into a free list. Assume that one slot can store a flag and either one element plus a pointer or two pointers. All dictionary and free-list operations should run in O(1) expected time. Does the free list need to be doubly linked, or does a singly linked free list suffice?

Please tell me if my understanding is correct: each slot in T[0,1,2….m-1] is storing (1) a flag (2) an element (3) 2 pointers

So, instead of the slot T(h(k)) being a pointer to a list, it is able to save one element and then that element points to a list

There is also a “free list”

There is only one flag for each slot T[0,1,2….m-1]

To insert an element ‘x’ we will first check T[h(x.key)] to see if it is free. If it is free then we insert T[h(x.key)] and change the flag = 1
If it was not free, then there is a list, so we insert it at the beginning of the list
To delete we will check x.prev and x.next to see if they are NULL. So when we delete the element ‘x’ we will be left with an empty list at that slot. So we insert T[h(x.key)] into the free list and update the flag to 0 and then delete the element x from the list its stores in

What I don’t understand is the functionality of maintaining a free list?
Is there a single free list for the whole hash table or does each slot have its own free list?
Also why insert the element to be deleted into the free list and then delete it? Why don’t just delete it without using the intermediate free list?

I am unable to visualize the use of the free list as mentioned in the question.