Is there any official (or semi-official) clarification on how often Psychic Virtuoso allows a character to use occult skill unlocks?

The description of the Psychic Virtuoso feat from Occult Adventures states "You can use all of your occult skill unlocks more often and you are more talented at using them." However, the Benefit section only lists a bonus for using them, nothing about using them more often. I cannot find any errata or clarification for this feat indicating either it doesn’t actually let you use occult skill unlocks more often or, if it does, how often it allows a character to use them.

Gestalt rules clarification

I am playing in a D&D 3.5 game. We are level 5 gestalt with 2 free LA (as in if you have a race or template with LA, you can get up to 2 total of them for free. Any Racial HD are paid as normal).

I chose a water orc (0 LA) with the natural lycanthrope template, Weretiger version (3LA normally, 1 with the rules. and with 3RHD). On one side of my Gestalt, I put 4 levels of Warblade and 1 level of Warshaper. On the other I put the LA of the weretiger, and 4 levels of the Weretiger class.

We had down how our characters were going to progress and for the next 2 levels, I put 2 more levels of Warshaper and 2 more levels of weretiger and the other side had 2 more levels of Weretiger.

This is where I need help because at this point, I started to progress Warblade and Warshaper at the same time, which I was told I could not do, as you can not move the side a gestalt class is on. I cannot find any such restriction on the d20SRD page, and I can also not find the source book for gestalt rules. I looked in PHB1&2 as well as the DMG1&2.

Is this a rule in the gestalt ruleset, or is it a houserule? Also, what specific book or books were gestalt rules introduced in?

If my explanation didn’t make sense, I put down a visual aid below this:

lvl 1: WeretigerLA |||| Warblade 1

lvl 2: Weretiger 1 |||| Warblade 2

lvl 3: Weretiger 2 |||| Warblade 3

lvl 4: Weretiger 3 |||| Warblade 4

lvl 5: Weretiger 4 |||| Warshaper 1

lvl 6: Weretiger 5 |||| Warshaper 2

lvl 7: Weretiger 6 |||| Warshaper 3

lvl 8: Warshaper 4 |||| Warblade 6

lvl 9: Warshaper 5 |||| Warblade 7

Echo Knight Legion of One movement clarification

Upon reaching 18th level as an Echo Knight Fighter you gain the ability Legion of One.

You can use a bonus action to create two echoes with your Manifest Echo feature, and these echoes can coexist. If you try to create a third echo, the previous two echoes are destroyed. Anything you can do from one echo’s position can be done from the other’s instead.

Can you move both echos 30′ each or is it just a shared 30′ of movement between the two?

Freerunning talent clarification

The Fantasy Flight Games SWRPG talent "Freerunning" found in the Cyphers and Masks and Endless Vigil sourcebooks states:

Once per round, before performing a Move maneuver, the character may suffer one strain. If he does so, he may use his Move maneuver to move to any location within short range (even straight up) as long as there is some sort of object to move across or a path to move along.

Can someone explain what exactly this talent does, specifically clarifying:

any location within short range (even straight up) as long as there is some sort of object to move across of a path to move along.

Clarification of the proof involving the regularity condition in Master Theorem

I was going the text Introduction to Algorithms by Cormen Et. al. Where I came across the following statement in the proof of the third case of the Master’s Theorem.

(Master theorem) Let $ a \geqslant 1$ and $ b > 1$ be constants, let $ f(n)$ be a function, and let $ T (n)$ be defined on the nonnegative integers by the recurrence( the recursion divides a problem of size $ n$ into $ a$ problems of size $ n/b$ each and takes $ f(n)$ for the divide and combine)

$ T(n) = aT(n/b)+ f (n)$ ;

where we interpret $ n/b$ to mean either $ \lceil b/n \rceil$ or $ \lfloor b/n \rfloor$ . Then $ T(n)$ has the following asymptotic bounds:

  1. If $ f(n)=O (n^{log_ba – \epsilon})$ for some constant $ \epsilon > 0$ , then $ T(n)=\Theta (n^{log_ba})$ .

  2. If $ f(n)=\Theta (n^{log_ba})$ , then $ T(n)=\Theta (n^{log_ba}lg n)$

  3. If $ f(n)=\Omega (n^{log_ba + \epsilon})$ for some constant $ \epsilon > 0$ , and if $ af(n/b) \leqslant cf(n)$ for some constant $ c < 1$ and all sufficiently large n, then $ T(n)=\Theta (f(n))$ .

Now in the proof of Master’s Theorem with $ n$ as exact power of $ b$ the expression for $ T(n)$ reduces to :

$ $ T(n)=\Theta(n^{log_ba})+\sum_{j=0}^{log_bn -1} a^jf(n/b^j)$ $

Let us assume,

$ $ g(n)=\sum_{j=0}^{log_bn -1} a^jf(n/b^j)$ $

Then for the proof of the 3rd case of the Master’s Theorem the authors prove that,

If $ a.f(n/b)\leqslant c.f(n)$ for some constant $ c<1$ and for all $ n\geqslant b$ then $ g(n)=\Theta(f(n))$

They say that as, $ a.f(n/b)\leqslant c.f(n) \implies f(n/b)\leqslant (c/a).f(n)$ then interating $ j$ times yeilds, $ f(n/b^j)\leqslant (c/a)^j.f(n)$

I could not quite get the mathematics used behind iterating $ j$ times.

Moreover I could not quite get the logic behind the assumption of $ n\geqslant b$ for the situation that $ n$ should be sufficiently large.(As the third case of the Master’s Theorem says).

Moreover in the similar proof for the third case of the general master theorem( not assuming $ n$ as exact powers of $ b$ ) there again the book assumes that $ n\geqslant b+b/(b-1)$ to satisfy the situation of sufficiently large $ n$ .

I do not quite understand what the specific value has to do and why such is assumed as sufficiently large $ n$

(I did not give the details of the second situation as I feel that it shall be something similar to the first situation)

Trivial clarification with the analysis of the Dijkstra Algorithm as dealt with in Keneth Rosen’s “Discrete Mathematics and its Application”

I was going through the text, “Discrete Mathematics and its Application” by Kenneth Rosen where I came across the analysis of the Dijkstra Algorithm and felt that the values at some places of the analysis are not quite appropriate. The main motive of my question is not the analysis of the Dijkstra Algorithm in general( a better version and more clearer version exists in the CLRS text) but my main motive is analysis of the algorithm acurately as far as the mathematics is concerned, considering the below algorithm as just an unknown algorithm whose analysis is required to be done. I just want to check my progress by the fact that whether the thing which I pointed out as being weird, is actually weird or not.

Lets move on to the question. Below is the algorithm in the text.

ALGORITHM: Dijkstra’s Algorithm.

procedure Dijkstra(G: weighted connected simple graph, with all weights positive)        {G has vertices a = v[1], ... ,v[n] = z and weights w(v[j], v[j])      where w(v[j], v[j]) = ∞ if {v[i],v[j]) is not an edge in G}      for i: = 1 to n         L(v[i]) := ∞      L(a) := 0      S:=∅      {the labels are now initialized so that the label of a is 0 and all          other labels are ∞, and S is the empty set}       while z ∉ S          u := a vertex not in S with L(u) minimal          S:= S ∪ {u}          for all vertices v not in S              if L(u) + w(u, v) < L(v) then                  L(v) := L(u) + w(u, v)              {this adds a vertex to S with minimal label and updates the labels of vertices not in S}            return L(z)  {L(z) = length of a shortest path from a to z} 

The following is the analysis which they used:

We can now estimate the computational complexity of Dijkstra’s algorithm (in terms of additions and comparisons). The algorithm uses no more than $ n − 1$ iterations where $ n$ is the number of vertices in the graph, because one vertex is added to the distinguished set at each iteration. We are done if we can estimate the number of operations used for each iteration. We can identify the vertex not in S in the $ k$ th iteration with the smallest label using no more than $ n − 1$ comparisons. Then we use an addition and a comparison to update the label of each vertex not in S in the $ k$ th iteration . It follows that no more than $ 2(n − 1)$ operations are used at each iteration, because there are no more than $ n − 1$ labels to update at each iteration.

The algorithm uses no more than $ n − 1$ iterations where $ n$ is the number of vertices in the graph, because one vertex is added to the distinguished set at each iteration., What I feel is that it shall be $ n$ iterations and not $ n$ as in the very first iteration the vertex $ a$ is included in the set $ S$ and the process continues till $ z$ is inserted into the set $ S$ and $ z$ may be the last vertex in the ordering i.e.$ v_n$ .

The rest statements are fine I hope.

Clarification on cooperative actions

Numenera Discovery, page 118 states:

Helping: If you use your action to help someone with a task, you ease the task. If you have an inability in a task, your help has no effect. If you use your action to help someone with a task that you are trained or specialized in, the task is eased by two steps. Help is considered an asset, and someone receiving help usually can’t gain more than two assets on a single task if that help is provided by another character.

Am I understanding this correctly? Does this mean players can always help each other? Would this mean that in every non-combat situation when players are together as a group, the players are effectively guaranteed to have two levels of assets (from either a trained character, or multiple characters helping)?

Rise of the Separatists sourcebook talent clarification

In the Fantasy Flight Games Star Wars RPG, in the sourcebook Rise of the Separatists, the talent called “Consider Our Options” states:

The character may take the Consider Our Options action, making a Hard Negotiation check. If the check succeeds, until the start of the character’s next turn, any enemy that attacks the character suffers 2 strain, plus 1 strain per additional 2 successes on the character’s check, before resolving the attack. (If this strain causes an adversary to become incapacitated, the GM may determine that adversary simply bows out of the fight.) The effect ends if the character inflicts damage on an enemy.

In the case that a Minion or Rival, as opposed to a Nemesis, attacks the character, taking strain as damage per the rules of Minions and Rivals, count as inflicting damage and thus ending the effect according to “The effect ends if the character inflicts damage on an enemy.” which is stated at the end of the talent?

If it might help clarify, the “Improved” version of the talent states:

When the character takes the Consider Our Options action, the character may choose to have the effects also apply to attacks that target the character’s allies who are within short range. The effect ends for all characters if the character or an affected ally inflict damage on an enemy.

clarification of importing sites

on the individual projects i have :-

options>
tick box (yes) use urls from global site lists if enabled – identified & verified
(so it imports from global identified lists & shares the verified lists

on global options
“build site lists that can be used globally by each project”
identified & verified ticked

i then right clicked on project > show urls> show stats about remaining urls
& project said 0 urls left

i then imported sites from a .txt list
right click on project>import target urls>from file

let gsa ser go through all the urls
i then right clicked on project > show urls> show stats about remaining urls
& project said 0 urls left

then right click on project>import target urls>from identified list
& it says 57630

so could you please explain what gsa does in “global site lists” on the global options & the individual projects.

ultimately i want to import urls (whichever way is best / fastest), let gsa go through them & then gsa will share the verified urls
between each project without doing any other importing.

Memory Addressing – Alignment Clarification

I’m reading “Computer Architecture A Quantitative Approach” (5th edition) and I’m having a hard time understing this table:

enter image description here

I understand how Misalignment happens, i.e., some byte, half-word, word, or double word it’s not a multiple of the address where the element is being stored or accessed. The part that I don’t understand is that it says in the caption: “that the byte offsets that label the columns specify the low-order 3 bits of the address”

My question is why are the low-order 3 bits of the address important or relevant in this situation? I’m assuming that the low order bits are the lowest binary values in the address. My intuition is that the 3 low order bits can tell us if the address is a multiple(2,4, or 8) of whichever data is being accessed, but I’m not sure.