Can a character take 10 or 20 on a Spellcraft check to identify magical items?

This came up in this question: What's an efficient way to handle magical item identification?. Many people suggested taking 10 on Spellcraft checks to identify magical items. I certainly see the appeal—but is it possible in all cases?

I thought that taking 10 or 20 was only possible if many attempts could be made and failure was not harmful (or if it was, the character accepted the harm). Since some magical items could be cursed or booby-trapped, it seems that trying to identify the item could be enough to trigger its harmful effects. By trying to identify such items, a character may be in "immediate danger" without knowing it. Does this mean that one cannot take 10 or 20?

For convenience, here’s what Paizo says about it (link):

Taking 10 and Taking 20

A skill check represents an attempt to accomplish some goal, usually while under some sort of time pressure or distraction. Sometimes, though, a character can use a skill under more favorable conditions, increasing the odds of success.

Taking 10:

When your character is not in immediate danger or distracted, you may choose to take 10. Instead of rolling 1d20 for the skill check, calculate your result as if you had rolled a 10. For many routine tasks, taking 10 makes them automatically successful. Distractions or threats (such as combat) make it impossible for a character to take 10. In most cases, taking 10 is purely a safety measure—you know (or expect) that an average roll will succeed but fear that a poor roll might fail, so you elect to settle for the average roll (a 10). Taking 10 is especially useful in situations where a particularly high roll wouldn’t help.

Taking 20:

When you have plenty of time, you are faced with no threats or distractions, and the skill being attempted carries no penalties for failure, you can take 20. In other words, if you roll a d20 enough times, eventually you will get a 20. Instead of rolling 1d20 for the skill check, just calculate your result as if you had rolled a 20.

Taking 20 means you are trying until you get it right, and it assumes that you fail many times before succeeding. Taking 20 takes 20 times as long as making a single check would take (usually 2 minutes for a skill that takes 1 round or less to perform).

Since taking 20 assumes that your character will fail many times before succeeding, your character would automatically incur any penalties for failure before he or she could complete the task (hence why it is generally not allowed with skills that carry such penalties). Common "take 20" skills include Disable Device (when used to open locks), Escape Artist, and Perception (when attempting to find traps).

What is the highest minimum and maximum roll possible on a (Dex) sleight of hand/pickpocket check? [closed]


FOR THE KLEPTOMANIACS OUT THERE:

Let’s say we’re attempting to pickpocket something from someone and would like to reduce our chances of failure as much as possible. What is the highest minimum roll and maximum roll reliably possible in the game? In order to make this easier to replicate for others, I’m going to offer some restrictions:

  • Things that are not largely guaranteed in most games, such as: Boons, Artifacts, and Manuals.
  • This can be any level character up to 20th level, all that matters is that it still remains the highest min and max roll for the check.
  • Must be acting alone and receive assistance only from themselves, their items, feats, racial abilities, class features, etc.

These restrictions should ensure that the check can be performed reliably any time it is made by anyone with the proposed build in nearly any campaign with the average amount of magical items and quest rewards, etc.

I will post my own findings as an answer below. Feel free to use it as a frame for further answers, or offer one all your own.

Check if coroutine crashed

I can check if coroutine still running by setting boolean variable outside the coroutine and then letting the coroutine itself set the value to true when it runs and set it to false when it decide to stop .

But this method of detecting whether coroutine still active or not doesn’t work when the coroutine crashed due to error (invalid input, null reference , ect) because the boolean variable remain true while the coroutine itself has stopped due to error.

How do i check if coroutine still running or not in current frame ? i can’t spend my code checking for next frame .

Preferably C# , but answer in javascript also accepted.

Does Improved Abjuration affects spells that only offer an option of having an ability check?

Consider the Abjuration Wizard’s 10th level feature:

Improved Abjuration

Beginning at 10th level, when you cast an abjuration spell that requires you to make an ability check as a part of casting that spell (as in Counterspell and Dispel Magic), you add your proficiency bonus to that ability check.

Some spells, such as Telekinesis, include cases where an ability check is required, either as soon as the spell is casted, or later one.

If the object is worn or carried by a creature, you must make an ability check with your spellcasting ability contested by that creature’s Strength check.

  1. Does an Abjuration Wizard ever adds his proficiency bonus when casting Telekinesis?
  2. If so, does that only apply to the case where a check is made on the same turn as the casting?

How to check effeciently that some changes preserve the topological ordering?

I have an acyclic directed graph with $ k$ connected components. I also have a topological ordering $ L$ (a sequence of vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering).

So, let’s say for example $ L = i_0,i_1,..,i_n$ .

I have the following operations:

  1. Exchange(i,j): it simply exchange the positions of i and j in the list. Meaning that $ L = ..,i,..,j,..$ becomes $ L’=..,j,..,i..$

  2. Inserting(i,p): it simply remove $ i$ from its position and inserts it in position $ p$ . Meaning that we have $ L=a,b,c,i,d,e,f$ then inserting(i,6) gives $ L”=a,b,c,d,e,i,f$ .

Now how to check that $ L’$ and $ L”$ are also valid topological sorting? Apart from doing a graph traversal of course because I want to exploit the two facts: i) the graph comes in $ k$ connected components and ii) some parts of $ L$ remain unchanged in $ L’$ and $ L”$ so its not useful to check them. Please consider indicating the complexity of your approaches.