dnd 4e – Do I have to hold the Power Jewel to use it?

Power Jewel (Advendurers Vault, p176):

Power ♦ Daily (Minor Action)
This power allows you to regain the use of an encounter power of 1st or 3rd level.

Do I have to get it from my backpack before using it?

What about items that are normally1 used, like a Whistle of Warning2?

  1. Normally meaning their mundane versions have some applications, as opposed to a Power Jewel
  2. Adventurer’s Vault 2, page 78: Encounter(Minor) Each ally within 10 squares of you who can hear the whistle can shift 1 square and draw a weapon or retrieve an implement as a free action.

user expectation – Terminology: Long press or Touch & hold?

I would suggest “Touch and hold”. Many users tend not to understand the term “long” in longpress. Therefore “Press and (then) hold” is a better description of the action.


OK, due to the comments below, here is a more descriptive answer 🙂

“Longpress” means “press for a long time”.The word “long” comes first but that is the second action the user has to commit. We also have to tell the user how much time (s)he shall spend during that “long” time.

The word “longpress” is not (yet) a term that common people understand.

So, every term beginning with the word “long…” aught to be avoided.

“Press and hold” on the other hand tells the user to 1: press on the item and then 2: keep on doing so until something else happens, something that will inform the user (s)he can release his/her finger from the device.

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c# – List to hold copies of a prefab for later deletion

A List<string> is all well and good if you just want names. But you don’t want to delete the name, you want to delete the game object.

Also, if you always want to delete from the start of the collection (First-In, First-Out), and you don’t need to traverse or randomly access items in the collection, then the structure you want is a Queue.

So, replace:

List<string> obstacleList = new List<string>(10);


Queue<GameObject> obstacles = new Queue<GameObject>(10);

Next, capture the newly-created instance when you make it:

var instance = Instantiate(Obstacle, new Vector3(nextObstacle + randH, randV, 0), Quaternion.identity);

And add it to your queue:


Your removal can then look like this (put it before you add so you don’t exceed your promised capacity):

if (obstacles.Count >= 10)
    var toBeDestroyed = obstacles.Dequeue();

Can a Ledger Nano X hold multiple BitCoin Wallets?

If I recall, you can access different wallets by using a different passphrase, which is added to the seed phrase as a “25th word”, resulting in a completely different entropy seed and different key-pairs at every index in every path.


So you could have one passphrase with your “actual” money in it, and a second passphrase with just a couple of bucks in and out. If you are held at gun point and forced to unlock your ledger, you can enter the second password and show the attackers how broke you are :-]

Disclaimer: hypothetical dramatization only

Warning Newbie: How does a Ledger hold multiple currencies?

So you startup your ledger, copy down your 24 words, and you now have a public and private key. (I think). Now you add the BitCoin App to your Ledger and you can add BTC to your “hardware wallet” using those keys. (Though phrasing it that way is a bit of a misnomer since the only thing your Ledger ever really stores is your private/public keys.)

So far so good for one currency. However, a Ledger will hold multiple currencies. That I can find there is no asking for another 24 words. You just add the second app and you now have a wallet in that new currency. So I assume that means that if you purchase Ethereum coins it is going to use the same public/private keys that you used on BitCoin?

That would also imply that you cannot have two BitCoin wallets on the same Ledger because attempting to do so wouldn’t make sense without a second set of keys which the Ledger doesn’t have.

Is what I am saying correct?


linear algebra – Does the eigenvalue equality hold for my expression?


$g(boldsymbol{theta},boldsymbol{theta_0}) = trace (
boldsymbol{Omega{(boldsymbol{theta})}}^{-1} boldsymbol{Omega{(boldsymbol{theta_0})}})-ln(det(boldsymbol{Omega{(boldsymbol{theta_0})}})/det(boldsymbol{Omega{(boldsymbol{theta})}}))-N $

where $boldsymbol{theta} in boldsymbol{Theta}$ with $boldsymbol{Theta}$ a compact subset of $R^{n}$, $n$ and $N$ are fixed numbers, and $boldsymbol{theta_0}$ belongs to the interior of $boldsymbol{Theta}$.

Denote the eigenvalues of the symmetric matrices $boldsymbol{Omega{(boldsymbol{theta_0})}}$ and $boldsymbol{Omega{(boldsymbol{theta})}}$ by $lambda_{0s}$ and $lambda_s$ $(s=1,2,…,N)$ respectively, where $lambda_{0s}>0$ and $lambda_s>0$ for all $s$.

Given the above, does the following hold or is a further condition required, and if so which one?

$g(boldsymbol{theta},boldsymbol{theta_0}) = sum_{s=1}^N ((lambda_{0s}/lambda_{s})-ln(lambda_{0s}/lambda_{s})-1) ?$

Can a hold configuration of a Turing machine have the same status as another configuration?

At first, I thought that since the state that a hold configuration is in will be a hold state, the TM will stop when a configuration goes into that state.
Therefore, there should not be two configurations that are displayed at different times with the same current status.

However, after reading some introductions of the one-state Turing machine, I am confused about my initial conclusion, since I understand that a one-state Turing machine has its starting state as an acceptance state
Can anyone explain to me? Thank you so much!