## 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?

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## 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.

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## 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?

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## 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

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## 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?

## L = { w x (w^r)* ∣ w , x ∈ ( a + b ) * } is it regular language? [duplicate]

L = { w x (w^r)* ∣ w , x ∈ ( a + b ) * } is it regular language? If so, why?

The notation W^r means the reverse string of W.

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## When you cast Flock of Familiars, is your regular familiar’s telepathy improved to match the others?

My problem is with the D&D 5e paradigm ‘spells do only as they say they do’ which if I am reading flock of familiars correct means that already possessing a familiar weakens this spell to the point where you’d have been better off not having that original familiar.

The spell description of the flock of familiars spell states (emphasis mine):

If you already have a familiar conjured by the Find Familiar spell or similar means, then one fewer familiars are conjured by this spell.

Familiars summoned by this spell can telepathically communicate with you and share their visual or auditory senses while they are within 1 mile of you.

Whereas the spell description of the spell find familiar states (emphasis mine):

While your familiar is within 100 feet of you, you can communicate with it telepathically. Additionally, as an action, you can see through your familiar’s eyes and hear what it hears until the start of your next turn, gaining the benefits of any special senses that the familiar has. During this time, you are deaf and blind with regard to your own senses.

This makes it seem as if the familiars summoned by flock of familiars are much, much stronger than those summoned by the regular find familiar because (1) their range is increased from 100 foot to 1 mile and (2) the flock of familiars does not state that one would be deaf and blind to their own senses. However, possessing a regular familiar does cause the Flock of Familiars spell to summon one familiar less effectively turning three strong familiars into two strong familiars and one weak familiar.

I very strongly doubt that this is the RAI and if this comes up on my table – which it might since I plan to take this spell when my wizard levels to lvl 3 – I, for one, will be arguing to give the regular familiar the same benefits for as long as flock of familiars lasts, but is this correct RAW or is it just wishful thinking of a player planning to use the spell? And is the not deaf and blind part supposed to be a benefit of flock of familiars (allowing you to see through all three’s senses at once) or is it an ommision on Wizard of the Coast’s part?

EDIT: Before anyone makes any wrong assumptions, I am not trying to rules lawyer my DM. If she says I cannot, I will not. However, we are on very good footing and I dare believe that if I ask her for those benefits that she will instantly say yes. I just don’t want to ask for those benefits unless they are actually RAI and if they are even RAW, all the better.

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## Set difference of recursively enumerable language and regular language

Consider a recursively enumerable language L1, and a regular language L2.

Is L1\L2 regular, context-free, or recursively enumerable?

Is L2\L1 regular, context-free, or recursively enumerable?

My thought process is that L1\L2 is recursively enumerable and L2\L1 is regular. Is this correct?

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