## Does Elemental Wild Shape work like regular Wild Shape?

A Druid of the Circle of the Moon gets a feature called Elemental Wild Shape which allows them to transform into elementals. I couldn’t find anywhere that said one way or the other, do the usual restrictions of Wild Shape apply to Elemental Wild Shape? For example, can a Druid cast spells while shapeshifted into an elemental?

If not, the Beast Spells feature allows a Druid to cast spells while shapeshifted, but specifically mentions beast forms. Can a Druid with the Beast Spells feature cast spells while in elemental form?

## How to add a goal in Google analytics with regular expression such as

I want to add Goal in Google Analytics with the same URL twice with a different regular expression such as

www.example.com/checkout#shipping www.example.com/checkout#payment 

these are on the same page with different tabs. So how I track user go which tab and other go where because google analytics just track my checkout page, not #shiiping and #payment

## Why metamagic feats for spell-like abilities that Warlocks use are so much stronger than regular metamagic feats at early levels?

Warlock is able to stack maximize and empower spell like abilities as early as level 6 without increasing effective caster level, which is not true for maximize/empower spell feat (for Sorcerers and Wizards)

Warlock is able to maximize and empower their magic items with that boost their eldritch blast such as gloves of eldritch admixture, which adds 4d6 to their total damage, but it doesn’t hold true for regular metamagic feats.

How come there is such a power gap between metamagic feats for spell vs spell like abilities?

## Automatic-Dodge vs. Regular Dodge Bonus

I’ve been consulting a few RIFTS and Palladium system forums online to try and figure out exactly how the Automatic Dodge system in the games work and have been rather unsuccessful. So I’m turning to the RPG stack to try and come up with a legitimate explanation for the ability and its mechanics.

Automatic dodge is an ability conferred by certain Hand to hand skills, classes such as Juicer, super powers such as Extraordinary Physical Prowess, Super speed, etc. that confers the ability to dodge without using up one of your melee actions for the round.

There are other dodge bonuses in the game systems that confer similar bonuses to dodging. It doesn’t state anywhere in any book I’ve read in the Palladium system whether or not the dodge bonuses from hand to hand, or other skills can be added to automatic dodge bonuses to determine if an attack is dodged.

So my question is thus:

Can Automatic-Dodge Bonuses and Dodge Bonuses added together to determine whether or not an attack is evaded without using up a melee action, and where if any can I find the rules to support said answer?

## A regular language derived from another

This is similar to a previous question I asked, but doesn’t seem aminable to the same technique. Given a regular language $$A$$, show the following language is regular: $$\{x|\exists y \; |y| = 2^{|x|} and \; xy \in A\}$$

I’m aware of the notion of regularity preserving functions, and that it would suffice to show that $$f(x) = 2^x$$ satisfies the property that for an ultimately periodic set $$U$$, $$f^{-1}(U) = \{m|f(m) \in U\}$$ is ultimately periodic. I’m struggling to $$f$$ has this property, but the book from which this comes implies a solution not using this is possible. It appears to be looking for a construction.

I can see that by repeated application of the idea behind the Pumping Lemma, if $$A$$ has DFL with $$k$$ states, that for any $$x$$ with $$|x| \geq k$$ then $$\exists y \; |y| = 2^{|x|} and \; xy \in A\ \implies \exists y \; |y| \leq k \; and \; xy \in A\$$

But this doesn’t give anything going in the opposite direction, that shows that some suitably short $$y$$ guarantees the existence of a $$y$$ of the required length.

Any help in solving this, or hint at how to progress would be very helpful.

## Is every unambiguous grammar regular?

While searching for an answer to this question I found out that there is an unambiguous grammar for every regular language. But is there a regular language for every unambiguous grammar? How can I prove that this is/isn’t true?

## Can this language be called regular?

Recently, I was facing some problems in effectively proving the following :

Consider the alphabet Σ ={0,1,2,…,9,#}, and the language of strings of the form x#y#z, where x,y and z are strings of digit such that when viewed as numbers, satisfy the mathematical equation x+y=z.

Is this language regular and why ?

I was trying to apply the Pumping Lemma, but am unsure of how to complete the proof. Could anyone please help ?

## How to apply Arden’s theory to determine a regular expression

If P=(ab) and Q=(a)* How do I use Arden’s theorem to determine the regular expression R. I’m not sure if I am supposed to just substitute the values of P and Q in the equation R= Q + RP. Also how would I use that to check that R satisfies Arden’s equation

## Proof that L^2 is regular => L is regular

I’m trying to show $$L^2 \in \mathsf{REG} \implies L \in \mathsf{REG}$$ with $$L^2 = \{w = w_1w_2 \mid w_1, w_2 \in L\}$$ but I cant seem to find a proof that feels right.

I first tryed to show $$L \in \mathsf{REG} \implies L^2 \in \mathsf{REG}$$, by constructing an machine $$M$$ that consists of two machines $$A=A’$$ with $$A$$ recognizing $$L$$. $$M$$ has the same start states as $$A$$ but the final states of $$A$$ are put together with the starting states of $$A’$$. Further $$M$$ uses the same accepting states as $$A’$$. Hope that makes sens so far 😀

Now to show $$L^2 \in \mathsf{REG} \implies L \in \mathsf{REG}$$ I’d argue the same way, but:

The machine $$M’$$ that accepts $$L^2$$ has to recognize $$w_i \in L$$ in some way, and because $$L^2$$ is regular, $$M’$$ has to be a NFA/DFA. So the machine has to check if $$w_i \in L$$ and this cant be done by using something else than a NFA/DFA.

This feels wrong and not very mathematical, so maybe somebody knows how to do this?

## Problem using ‘Regular expression’ in order to split characters of a column when there is no delimiter between them

I have a table with below structure:

create table TBL_TEST (   col_id   NUMBER,   col_name VARCHAR2(500) ) 

Some example data :

col_id | col_name    -----------------   1    | aetnap           2    | elppa          3    | ananab        

What I need to do is to split characters of column col_name for each col_id for example for col_id=1 we must have :

col_id | col_name    -----------------   1    | a   1    | e   1    | t   1    | n   1    | a   1    | p 

this query is fine when there is only one record in the table :

SELECT col_id, REGEXP_SUBSTR(col_name, '[a-z]{1}', 1, LEVEL) AS VAL   FROM tbl_test t CONNECT BY REGEXP_SUBSTR(col_name, '[a-z]{1}', 1, LEVEL) is not null 

but as soon as I insert another record in the table (say col_id=2 and col_id=3) I can not have the desired result. I want to know two things:

1. Why is this query works fine for one record and it does not for more ?
2. what is the best way to split the characters when there is no delimiter between them?