What exactly does the mindless trait entail

So I was skimming through posts about Touhou and found a 5e homebrew race called the satori (located here) and that made me wonder how accurate that is for pathfinder. So I looked into it a bit and found this post though was more about someone not liking a gm ruling rather than someone wanting to know about how mindless creatures actually work and respond causing replies like “your gm is wrong”. It did, however, give examples of how mindless is described in the creature types and subtypes section of the rules in the spoiler section as well as the paragraph that followed.

So what I want to know is what exactly is a mindless creature and how does it respond to its environment.

Cast Sleep spell and the hit points are exactly tied, which creature falls asleep?

so playing DnD 5e, what happens if I cast the first level Sleep spell, get 27 for my roll, but there are two 15 hit point creatures in the area of effect?

I did not roll high enough to put both creatures to sleep. Only one can fall asleep (27-15=12). How do you determine which of the two creatures falls asleep?

Caster’s choice? Random chance? Higher constitution modifier?

Sleep spell text:

This spell sends creatures into a magical slumber. Roll 5d8; the total is how many hit points of creatures this spell can affect. Creatures within 20 feet of a point you choose within range are affected in ascending order of their current hit points (ignoring unconscious creatures). Starting with the creature that has the lowest current hit points, each creature affected by this spell falls unconscious until the spell ends, the sleeper takes damage, or someone uses an action to shake or slap the sleeper awake. Subtract each creature’s hit points from the total before moving on to the creature with the next lowest hit points. A creature’s hit points must be equal to or less than the remaining total for that creature to be affected. Undead and creatures immune to being charmed aren’t affected by this spell.

Thanks for your time!

Is it possible to update exactly 1 byte in RAM

For example I have a static C++ array {'d', 'o', 'c', 's'}. And I have x86 architecture, with 32-bits length words.

I want to replace letter c with g. As far as I understand, when we make a read operation from a byte addressable RAM, CPU always reads 1 word from it. So it will read the whole array.

After CPU will read the array, I will replace letter c with g and write it back to memory.

Will CPU write 1 bytes or 4 byte in RAM?

If it writes 4 bytes, what’s the point of using byte addressable memory instead of word addressable?

Graph with exactly 2 Minimum Spanning Trees


Say that a graph, $ G = (V, E)$ has 2 minimum spanning trees (MSTs). Given this condition stipulated, prove that any cycle formed by all the edges in both the MSTs (i.e., the union of the edges in of the 2 MSTs) that at minimum, 2 of the edges in the set which is the union of the edges have equal weight. Also show that either this edge is the largest weight in the cycle, or not the largest weight in the cycle.

Overall am pretty stuck on this question.

My initial thoughts are the following: In any graph with more than 1 MST, clearly this means that the edge weights can’t be distinct, otherwise there wouldn’t be multiple MSTs. Also the graph $ G$ must contain cycles, otherwise, it wouldn’t have multiple MSTs.

My idea for proving that any cycle formed by the union of the edges of the two MSTs would be that in $ MST_1$ there is some edge, $ e$ that is not in $ MST_2$ and there is also some edge $ f$ that is not in $ MST_1$ . Using the cut property if $ e$ was not placed in $ MST_2$ and $ f$ was not placed in $ MST_1$ then then we have that the weight of $ f$ , and $ e$ , $ w(f) = w(e)$ .

Having trouble formalizing this though, and wondering if its actually a correct deduction. I feel that it makes sense given some examples and drawing, but not quite certain that’s actually true. Then from there I felt that there had to be some node, $ z$ such that $ z$ had 2 edges with the same weights, and when we combine the edges from $ MST_1$ and $ MST_2$ we end up with both the edges from $ z$ forming a cycle, and the edges are the same weights, so we know at least 2 of the edges form a cycle… Or the union of the edges could form a cycle graph itself which would then show that the 2 edges with the same weights are part of a cycle, I think? Is this somewhat on the right track? Is there some sort of condition for a graph, $ G$ , in order for it to have exactly 2 MSTs? Or is there some property I’m missing?

If someone could please provide a bit of guidance in the right direction, it would be extremely appreciated. Thanks.

Which damage dice exactly does the Great Weapon Fighting fighting style allow you to reroll?

This has been a point of some controversy within our local AL community, with different DMs giving different rulings. From what I could see, this site has only touched on the Maneuver/Smite issue, with, according to some of my DMs, an out of date answer.

This also leaves the problem of extra dice on magical weapons such as an Ild rune or Hazirawn, maneuver damage dice given by another player through Commander’s Strike, damage dice given by a College of Valor bard, Sneak Attack dice, and likely others which currently escape my mind.

Which damage dice exactly does the Great Weapon Fighting fighting style allow you to reroll?

What exactly determines which entities/servers “on the Internet” get to read my e-mails when I send them from X to Y?

Let’s say, for example, that I have a Gmail account. I compose and send an e-mail to info@somecompany.com. They aren’t using Google services, just so that we can exclude “special cases” where they just keep the entire thing internal to their own network.

On a technical level, doesn’t the e-mail client/software just look up somecompany.com’s MX records and then connect directly to that IP address on the “e-mail port” and, assuming it is online and accepts the “handshake”, just transfers the e-mail directly to it?

Why have I heard all my life that e-mails just get “flung out in cyberspace randomly” and bounced around the entire world, allowing everyone and their grandmother to read it before it finally gets routed to its final destination?

Is this a total misconception? Was it something that was done in the 1960s because they couldn’t afford to have computers on at all times, so they had to do it like this? Is it by design in order to allow spying?

I feel ashamed for still not having a good grasp on this after all these years. I’ve probably tried to ask about this dozens of times over the years, but never got what I considered a clear and conclusive answer.

How exactly does Windows Defender in Windows 10 determine when to upload your local files to Microsoft?

Every time I install Windows 10, I painstakingly go through every setting that can be found in any GUI setting for the OS, disabling everything that sounds creepy.

One of the most disturbing things I’ve found is what I believe is called “automatic sample submission“, which means that the built-in anti-virus tool in Windows 10 can, by default, decide to upload any file it deems “potentially risky” to Microsoft, “for further analysis”. It also mentions that it doesn’t do this for files which “may contain personal data”.

But how can it know that? Does it:

  1. Simply look at the file extension and only upload .EXE and other “obvious binaries”?
  2. Does it ignore the file extension and instead look inside the file to check if it contains executable code?
  3. A combination of both?

What happens if I have a word processing document full of private information, but which also has a malicious macro or something accidentally baked (embedded) into it?

What happens if I have an EXE which actually has had all data files baked into it while I’m developing a game as to be a single file? (This is an actual situation I’ve been in in the past.)

Does it deem the data files for my local PostgreSQL database full of ultra-private information as “potentially dangerous” and upload those?

I can think of numerous situations where even the smartest code in the world would not be able to determine what contains private data or not. And, frankly, I have virtually zero confidence left in Microsoft’s judgment at this point, having wasted a huge amount of my life fighting the OS to be able to use it at all. I’ve found numerous typos in their “stable” releases, making me extremely scared of how much data has been uploaded in spite of all the care I’ve tried to take to avoid it.

I also remember that it eagerly wanted to re-enable this feature, even harassing me about it. I can imagine that the vast majority of users have no idea about this, let alone have gone through the trouble of force-disabling it.

Can someone explain what the “Proficiency Bonus” is in D&D 5e/Next exactly?

I am reading the 5e rules, and I keep seeing how you can add proficiency bonus to attack rolls checks with weapons you are proficient in and skill checks in skills you are proficient in, but I can’t find anything on what this bonus actually is. How do you determine what the numerical actual bonus you add to your roll is? And at what times is it used?

How exactly does a creature with an alignment subtype suffer alignment-based effects?

For example, say you “redeem” a demon (an outsider with the chaotic subtype), so its alignment is now true neutral. Let’s cast order’s wrath at caster level 10th on our demon. The spell deals 10d6 damage to chaotic outsiders, or half damage to creatures who are neither chaotic nor lawful (aka neutral).

The chaotic subtype says:

Any effect that depends on alignment affects a creature with this subtype as if the creature has a chaotic alignment, no matter what its alignment actually is. The creature also suffers effects according to its actual alignment.

I don’t know if this is relevant to this situation, but the ordered chaos feat from Fiendish Codex I seems to imply that being affected “as if you were chaotic, as well as your actual alignment” means that you take the least harmful version of the two effects: in this case, half damage (or, if the demon were lawful neutral, no effect)

But the wording of the SRD’s chaotic subtype seems to imply something even worse than the opposite: the demon would take the effects of the spell as a chaotic outsider, 10d6, and also as a neutral creature/outsider, which is half of 10d6 again (or half of 5d8, depending on when you consider its outsider type). In other words, having the subtype means that the creature takes more damage than a truly chaotic demon (so, instead being chaotic neutral or lawful neutral would cause it to only take 10d6. Though this would raise the question, are chaotic evil demons affected 3 times by the spell? Once for their alignment, once for the chaotic subtype, and once for the evil subtype? At least, an interpretation that goes to that extent seems silly, but it could still be a lesser version of that?)

Then I suppose it’s possible that the spell simply deals the worse of its two effects, 10d6.

Which is correct?