Frequency of cantrips outside of battle

Despite being “at will”, how often can a cantrip be cast outside of battle? For example, how long must a player wait before shooting 2 fire bolts, and is there a limit to how many could be cast or can they just go at it continuously for a day without incurring penalties?

Some Background

I have recently started DM’ing and there’s a few concepts that still elude me, or at least I am finding some trouble reconciling my interpretation of the rules and practical balance.

As I understand it, cantrips can be cast “at-will”. That’s fine inside of battle, but outside of it, things begin to get extremely broken. I have a player whose solution to most challenges is to “fire bolt it”. Is there a reinforced door? A stone wall? A shielded altar? He will just fire bolt it for 8 consecutive hours… Now, on most occasions I have been able to come up with counter-measures, like, bouncing the fire bolts back at him when they hit the altar’s shield, but still, it sounds ridiculous to me that you could just cast continuously for 2/3 of a day if you wished it so, without incurring any penalties… it’s simply absurd on so many levels! And yes, I know you technically could say “I hit it with my club for 8 hours”, and it would be just as ridiculous in my mind that there would be no fatigue, no weapon deterioration, etc…

I have told my player he can cast once every 6 seconds (since that’s the duration of a turn in battle), but I have no idea if that is correct, and it still feels unbalanced.

Given node and value find frequency of given value from node to root in a tree

A tree with N vertices and N-1 edges is present.

A value x can be inserted in any of the nodes.

A single node may contain multiple values.

How do I answer queries of the given type?

Given a node n and value v find frequency of v in the path from node n to the root of the tree.

I require an algorithm that answers such queries in log(N) time.

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Information Theory: Comparing surprisal of words with varying count frequency

This is a very broad question, I’m not sure if cstheory is the better place.

How can I compare the conditional surprisal of words that vary in frequency?

$ S(w|context)=−log(p(w|context))=−log(\frac{count(w,context)}{count(context)})$

The $ count(w,context)$ depends on the frequency of the word w because this can be further broken down to $ p(w|context)count(w)$ . This means a word that is more frequent will have a lower surprisal.

Is there a way to compare the surprisal of words with varying count/frequency, i.e control for frequency? Do I just divide $ count(w,context)$ by $ count(w)$ to normalize by count?

Disable CPU frequency scaling?

I need to run ATLAS on my ubuntu laptop, but the software won’t run unless you dissable CPU frequency scaling. I’ve tried the instructions from here but they aren’t working on my Toshiba laptop. With freq-scaling on the software becomes useless.

Can anyone help?

Frequency analysis based anagram checker

I uploaded code solutions for some problems of the book Cracking the Coding Interview, 6th Edition to GitHub, I would like to know your rating and potential improvement of the code I wrote.

Here is the first problem of chapter 1 (anagrams):

#include <stdio.h> #include <string.h>  /*check if one string is an anagram of another, it uses an int array  * called alphabet to store frequencies of chars in both strings, add  * 1 for s1 and subtract 1 for s2*/ int are_anagrams(const char *s1, const char *s2) {     int alphabet[26] = { 0 };     int index1, index2;     size_t l1 = strlen(s1), l2 = strlen(s2), i;      /*if the strings have different lengths are not anagrams */     if (l1 != l2) return 0;      /* count the frequencies of characters */     for (i = 0; i < l1; ++i) {         index1 = s1[i] - 'a';         index2 = s2[i] - 'a';         ++alphabet[index1];         --alphabet[index2];     }     /* all the alphabet letters should be 0, otherwise the strings are not      * anagrams */     for (i = 0; i < 26; ++i)         if (alphabet[i] != 0) return 0;      return 1; }  int main() {     char s1[] = "aaabbbccc";     char s2[] = "aabbaccbc";     printf("%d\n", are_anagrams(s1, s2));     return 0; } 

Any suggestion or advice is welcome, thanks for your time.

Building a frequency dictionary from a pandas dataframe without looping

I need to make a frequency dictionary from a pandas series (from the ‘amino_acid’ column in dataframe below) that also adds an adjacent row for each entry in the dictionary (from ‘templates’ column).

    templates   amino_acid 0   118         CAWSVGQYSNQPQHF 1   635         CASSLRGNQPQHF 2   468         CASSHGTAYEQYF 3   239         CASSLDRLSSGEQYF 4   51          CSVEDGPRGTQYF 

My current approach of iterating through the dataframe seems to be inefficient and even an anti-pattern according to this post. How can I improve the efficiency/use best practice for doing this?

My current approach:

sequence_counts = {} seqs= list(zip(df.amino_acid, df.templates))  for seq in seqs:     if seq[0] not in sequence_counts:         sequence_counts[seq[0]] = 0     sequence_counts[seq[0]] += seq[1] 

I’ve seen people the below way, but can’t figure out how to adjust it to add each respective ‘templates’ entry:

sequence_counts = df['amino_acid'].value_counts().to_dict() 

Any help/feedback would be greatly appreciated! 🙂

Is frequency vector a good choice for minimizing chain code derivative vector?

I’m trying to implement a feedforward neural network that recognize a type of cable based on chain codes derivative vector. The size of my chain codes derivative vectors is variable, and I would like them to converge to the same length. As a solution I’m thinking about a frequency vector that will contain each direction(8 in total). So from a variable length, all the vectors will converge to vectors of length 8. Is this approach ok?