Five page essay outline

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Five page essay outline

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by: dreamgoodlife
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Category: Link Building
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by: spotifystreamko
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How does the ‘five minute adventuring day’ affect class balance?

In the basic rules the The Adventuring Day is described as follows:

Assuming typical adventuring conditions and average luck, most adventuring parties can handle about six to eight medium or hard encounters in a day. If the adventure has more easy encounters, the adventurers can get through more. If it has more deadly encounters, they can handle fewer.

In reality many home games, mine included, often don’t have as many as six to eight encounters, covering a range of difficulties, between long rests.

The ‘five minute adventuring day’ takes this to the extreme and describes a situation where PCs are allowed to long rest between pretty much every encounter, and thus don’t have to worry about the ongoing effects of resource depletion from one encounter to another.

It is fairly well established that this is bad for game balance and makes the PCs much stronger (see Are there rules for shortening the adventuring day and reducing the number encounters without unbalancing them? amongst others). However, this questions is not about game balance (i.e. PCs v DM game balance).

I’d like to know how the five minute adventuring day affects intra-party class balance specifically.

For example:

  1. It’s well established that Paladins have extremely powerful burst damage but which other classes benefit most from a shorter adventuring day?
  2. Which classes are most adversely affected?
  3. Are all non-spellcasting classes going to be outdamaged by all spellcasting classes under these conditions?
  4. Does the answer to this question vary much from low to high level play?

For the purposes of this question please assume two hard / deadly encounters, lasting three – five rounds each, and no short rests in an adventuring day – a not uncommon occurrance in my home games. Consider basic class features but don’t worry about archetypes.

Python program to find a word ladder transforming “four” to “five”

I saw this Puzzling problem and thought I would try to write a Python program to solve it. The task is to transform “four” to “five”, forming a new four-letter word at each step, replacing one letter at each step, in as few steps as possible.

But turns out I don’t know how to optimize recursion, so I’m posting here for help. I’m mostly just confused on why the code to change the past needs to be at the top of the function, but I would also like advice on how to speed this up in general. Right now it takes about 10x as long for each step up max_depth gets on my computer.

There won’t be any matches until you change max_depth – I didn’t want anyone copy-pasting and lagging out. There should be a solution at depth 5, according to Puzzling. However, my words file doesn’t have the Foud or the word Fous, which that answer uses. Bumping up to max_depth six will take my computer ~10 minutes, which I don’t want to try yet.

def hamming(string1, string2):     assert len(string1) == len(string2)     return sum(char1 != char2 for char1, char2 in zip(string1, string2))  max_depth = 3 start_word = "five" end_word = "four" all_words = open("/usr/share/dict/words", "r").read().lower().splitlines() all_words = list(filter(lambda word: word.isalpha(), all_words)) all_words = list(filter(lambda word: len(word) == len(start_word), all_words))  sequences = []  def search(current_word, past = []):     # Needs to be first to be fast for some reason     past = past[:]     past.append(current_word)      if len(past) > max_depth:         sequences.append(past)         return      for word in all_words:         if hamming(word, current_word) == 1 and word not in past:             search(word, past)  search(start_word)  sequences = [sequence[:sequence.index(end_word) + 1] for sequence in sequences if end_word in sequence]  if len(sequences) == 0:     print("No matches") else:     print(min(sequences, key=len))  

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by: spotifytech
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Category: Audio & Music
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rxJs retryWhen just five times

I’d like to retry a request no more than 5 times.

What happens: if the first request fails by timeout but the second request succeeds, it still issues the other 3 requests.

I also suspect that a memory leak is lurking there.

retryOnUnknownError: (errors: Observable<any>) => Observable<any> = (errors: Observable<any>) => {      const maxRetrials = 2;     const waitDelayBeforeRetrial = 500;      return errors.pipe(         concatMap((err, count) => {             console.log('Errored for the ', count, 'th time', err);              if (count >= maxRetrials)             {                 console.log('Max retrials limit reached. Not retrying');                 return throwError(err);             }              if (err instanceof HttpErrorResponse && err.status !== 403) {                 return throwError(err);             }              if (err instanceof TimeoutError) {                  console.log('Got timeout error. Retrying');               }              // try again             return of(err).pipe(delay(waitDelayBeforeRetrial));         })     ); };   this.http.get('bla').pipe(retryWhen(retryOnUnknownError));