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Does the verbal component for spellcasting have to be words?

To make a long story short: for my first D&D campaign, I want to create a warlock that had to give up her voice as a part of her deal with her Patron.

As such, she is incapable of speaking, but she can still produce sounds with her mouth. I was wondering if that would incapacitate her from casting spells with a Verbal component.

Do spells needs a specific phrase to be cast, or does gibberish work?

How to remove any words containing two adjacent characters with different cases?

I have a list of permutations of ABCabc and I want to remove any permutations with two adjacent characters with different cases (uppercase and lowercase).

For example,

  • ABCcab is kept.
  • ABCacb must be removed because Ca contains two adjacent characters with different cases.
  • AbBcaC must be removed as well.


Here is my attempt but without filtering.

Select[StringJoin /@ Permutations[Characters@"ABCabc"],....] 

Would a two word domain name with a shared letter between the words be treated the same for SEO as the fuller name?

I found a domain name (2 words) that is not available. Then I found almost the same domain name with a slight difference, the last letter of the first word is the same of the first letter of the second word.

Ex : (not available) (available) 

In terms of SEO, is it seen as the same to Google? Would search engines interpret them the same?

If we want to map abbreviations of full-English words (e.g. map “Jan” to “January”), how can we identify abbreviations which map to multiple words?

Short Version:

How can we construct such a trie which maps abbreviations of names-of-the-month to full-month (we map the abbreviation "mar" to "march")?

  • The set of all abbreviations is formed by:
    • keeping the first letter of the month name. (all abbreviations of "january" begin with "j")
    • deleting 1 or more characters ("jan" deletes "uary" from "january")

The Looonnnnggg Version:

How can we construct such a trie?
What algorithm will build the appropriate trie from the container of verbose strings.

Consider the English names for months of the year:

  • January
  • February
  • March
  • April
  • [… truncated …]
  • October
  • November
  • December

We find it useful say that the English names for months of the year are "verbose" strings.
For any "verbose" string $ v$ , and any string $ a$ we say that $ a$ is an "abbreviation" of $ v$ if and only if all of the following conditions are met:

  • $ a$ non-empty. $ |a| \geq 1$
  • $ a$ can formed by deleting 1 or more characters from "verbose" string $ v$
  • $ a(1) = v(1)$ . Assume that string indexing begins at $ 1$ , and not $ 0$ .

For example, "jan" is an abbreviation of "january."

Suppose you want to write an algorithm which:

  • accepts a list of verbose strings as inputs.
  • the algorithm outputs a "trie" data-structure (information retrieval tree) $ T$ such that:
    • The trie $ T$ accepts any ASCII string as input.
    • An output (leaf node) of the trie should be set of strings $ S$ such that:
      • every string in $ S$ is a verbose string
      • the string fed as input into trie $ T$ is an abbreviation of every verbose string in container $ S$

Some examples of input to the trie and output of the trie are shown below:

  • Example 1

    • Input: "Ma"
    • Output: $ \{$ "March", "May"$ \}$
  • Example 2

    • Input: "Mar"
    • Output: $ \{$ "March"$ \}$
  • Example 3

    • Input: "Decuary"
    • Output: $ \{$ " "$ \}$ ….. the empty-set

The output from the trie should be one of:

  • the empty set
  • a set of one item
  • a set of two or more items

For months of the year, we might write javascript so that an end-user can type in any half-way reasonable date-format, instead getting an error message when they put slashes instead of dashes, etc….

If you do not like the months of the year application, a different use-case would be to write write your own Linux Shell (similar to BASH). Maybe any half-way reasonable abbreviation of "make directory" will map to "mkdir" In that case, we could have many-to-one mapping from high-level shell-commands to low-level Linux commands.

The question is:

How can we construct such a trie?
What algorithm will build the appropriate trie from the container of verbose strings.

Also, can we avoid brute-force generating a list of all aberrations before-hand? The set of all strings form-able by deleting 1 or more characters from the verbose strings is quite large. We would like to avoid combinatorial explosion, if we can.

The programming language (Java, python, C+ + ) does not matter for answering this question.

Number of words of length n for special language

Let $ \Sigma$ be an alphabet and let $ L$ be a language over it with the following properties:

  1. if $ w\in L$ then there exists $ v\in \Sigma^*$ such that $ wv \in L$ and for every $ s\in \Sigma$ the word $ wvs$ does not lie in $ L$
  2. $ wv\in L$ then $ vw \in L$
  3. It is prefix-closed, i.e. prefix of any word is still in the language.

Note that by the definition, it is not cyclic language. I’m trying to compute its growth function, by that I mean $ \gamma_n:= |\{w\in L \mid |w| = n\}|$ . I know about my specific case that it is not regular and my hypothesis is that function $ \Gamma(x) = \sum_{n=1}^\infty \gamma_nx^n$ is not rational. However, I couldn’t find any information about these functions for non-regular languages. Maybe, there’s a formula that connects entropy of language, i.e. $ e(L):= \limsup\limits_{n\to\infty} \frac{\log\gamma_n}{n}$ and the $ \Gamma$ function. Or for such a language there’s a way to describe its growth throughout the growth of the language $ \operatorname{End}(L) = \{ w\in L \mid \forall s\in \Sigma \,ws \text{ is not in } L \}$ .

Is my recursive algorithm for Equivalent Words correct?

Here is my problem.

Problem Given two words and a dictionary, find out whether the words are equivalent.

Input: The dictionary, D (a set of words), and two words v and w from the dictionary.

Output: A transformation of v into w by substitutions such that all intermediate words belong to D. If no transformation is possible, output “v and w are not equivalent.”

I need to write both recursive and dynamic programming algorithm. As for recursion, I came up with this algorithm. Is it correct?

 EquivalentWordsProblem(v, w, D)  1.m <- len (v)  2.n <-  len (w)  3.substitutions = [] #array to save substitutions   4.if m != n:  5.  return "v and w are not equivalent"  6.else  7.for i <- m to 1 <-1 do  8.  for j <- n to j <- 1 do  9.    if v[i] != w[j]:  10.           substituted_word <- v[1…i-1]+v[j] #we substitute v[i] for w[j]  11.              if substituted_word in D:  12.                substitutions.append(substituted_word)  13.                   return EquivalentWordsProblem(v[1…m-i], w, D) #recur on the string of length m - i  14.   else:          return EquivalentWordsProblem(v[1…m-1], w, D) #recur on the string decreasing its length by 1  15.if len(substitutions) != 0:  16.  return substitutions   17.else  18.   return (“v and w are not equivalent”)