ADVENT | Adversarial Entropy Minimization for Domain Adaptation | Machine Learning library

I am new to go and I am trying to learn some basic use of signal functions in goroutines. I have an infinite for loop in go. Through this for loop, I pass values to a goroutine through a channel. I also have a threshold value after which I will like to stop sending values indefinitely to the goroutine (i.e. close the channel). When the threshold value is reached, I will like to break the for loop. Following is what I have tried so far.

In this particular example, thresholdValue = 10 and I would like to print values from 0 , ..., 9 and then stop.

I followed this post on medium and this post on stackoverflow. I picked elements from these posts which I could use.

This is what I have done at the present. In the main function of my code, I purposefully make the for loop an infinite loop. My main intention is to learn how to have the goroutine readValues() take the threshold value and then stop transmission of values indefinitely in the channel.

Pretty new to Rust, decided to brush up using the Advent of Code 2020 Day 1 Puzzle. I'm using the following function:

Is there any way to process a changing list using higher-order functions in Clojure and not using explicit recursion? For example, consider the following problem (that I made up to illustrate what I have in mind):

Problem: Given a list of unique integers of unknown order. Write a that produces an output list as follows:

1. For any even integer, keep the same relative position in the output list.
2. For any odd integer, multiply by ten, and put the new number at a new place: at the back of the original list.

So for example, from original vector [1 2 3 4 5], we get: [2 4 10 30 50]

I know how to solve this using explicit recursion. For example:

I'm trying to solve a puzzle of 'Advent of code', which entails finding out how many passwords are valid of a list of passwords and their policicy.

But my code returns always 627 instead 474, which is the official solution to my puzzle input.

Can someone explain what could be wrong?

The puzzle: AdventOfCode_2020_Day2

My puzzle input: GoogleDrive_PuzzleInput

For Advent of Code Day 17 (problem statement isn't super important), what I want to do is have a type:

I wrote this C code to solve Advent of Code 13 2020. I know, it's probably not viable trying to solve it via brute force, but the program gives the correct answer for the example input.

If I try to let gcc optimize the code, it gives the correct result with -O1, but creates an endless loop with -O2. After all the research my conclusion is that there is undefined behavior in my code, I guess it has to do with the probability that "found" may never be higher than 0 and so "time" would overflow.

Here is the question: Does anybody know how to patch that undefined behavior?

"-Wall -Wextra -pedantic" don't even issue a warning or something.

I just can't find a solution. If I, for example, change the head of the while loop to (!found && time < 10000000000), so that no overflow can occur, it just breaks the loop right away with a time value of 10000000003 when compiled with -O2, but still gives the right result with -O1.

Here is the code, the correct result would be "1068781":

I have been trying to solve Advent of Code 2020 day 13 part 2 task. I found a lot of hints talking about something called Chinese Remainder Theorem. I have tried some implementations following npm's nodejs-chinesse-remainders but this implementation since to be quite old (2014) and also requires extra libraries for Big Int cases.

How could I implement the modular multiplicative inverse ? How could I refactor the CRT algorithm define in the npm module for which I provided a link?

I am working through day 7 of the Advent of Code and am stuck on why my code is returning undefined when the base case in findNestedBags hits the first condition (data[key].includes(color)). It returns 0 when the second case is true so I am confused why one works and the other doesn't.

The objective is to figure out how many different bags will lead to a bag that contains a 'shiny gold' bag. If a certain colored bag contains a shiny gold bag as an option (found in the value of key/value pair), it's an option. Alternatively, if one of the bags in list then has shiny gold as an option in its key/value pair, that can count as well...

This is a sample of the data structure as found in the variable data, just with many more rows:

I'm solving the 2015 version of Advent of Code and got, at day 7, an unexpected behavior that perhaps someone can help me understand.

The problem asks to compute the values (unsigned 16-bit) that a network of wires produce, like,