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QUESTION
ANSWER
Answered 2021-Apr-26 at 11:13In the end I came up with this, which is a little unwieldy but does the job:
QUESTION
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.
ANSWER
Answered 2021-Mar-20 at 10:57to show good faith, this is the program rewritten.
QUESTION
Pretty new to Rust, decided to brush up using the Advent of Code 2020 Day 1 Puzzle. I'm using the following function:
...ANSWER
Answered 2021-Jan-18 at 21:19You can omit the return
keyword only if the returned value is the last expression in the function block, otherwise you need to explicitly use return
. Adding return
to your example fixes that particular error but a bunch of new ones come up in its place. This is how I'd write the function:
QUESTION
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:
- For any even integer, keep the same relative position in the output list.
- 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:
...ANSWER
Answered 2020-Dec-27 at 02:23This is much easier to do without recursion than with it! Since you only care about the order of evens relative to other evens, and likewise for odds, you can start by splitting the list in two. Then, map the right function over each, and simply concatenate the results.
QUESTION
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
...ANSWER
Answered 2020-Dec-22 at 19:40Your nested loop condition is bad - your algorithm will never count more than the policy’s max value
QUESTION
For Advent of Code Day 17 (problem statement isn't super important), what I want to do is have a type:
...ANSWER
Answered 2020-Dec-17 at 20:49As mentioned in the comments, the issue is that you cannot return a reference to something while subsequently wanting to mutate it. Because that something is still being "borrowed" and referenced elsewhere.
If that was possible, then suppose you collect()
ed everything into a Vec<&Cubes>
, then if that resulted in a Vec<&Cubes>
with 3 items. Then given that it's 3 references to the same instance, then all 3 items would have the same state as the last item.
In short you'd not end up with x
, f(x)
, f(f(x))
as you want, but instead f(f(x))
, f(f(x))
, f(f(x))
Since you want less cloning, then it sounds more like you just want an fn next_state(&mut self)
method. Then you could iterate and call cubes.next_state()
which would update cubes
to its next state, while no needless cloning would occur.
QUESTION
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":
...ANSWER
Answered 2020-Dec-16 at 14:12I do not understand the code and not indent to, but the loop is strange. Anyway:
"-Wall -Wextra -pedantic" don't even issue a warning or something.
And there are also other ways to detect UB! Compiling your code with some more -fsanitize=*
options results with:
QUESTION
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?
...ANSWER
Answered 2020-Dec-13 at 13:07As a self response and with the purpose of make a wiki to find this solution for those who in the future need a CRT implementation in javascript/typescript:
First think is to implement Modular Multiplicative Inverse, for this task what we try to find is an x such that:
a*x % modulus = 1
QUESTION
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:
ANSWER
Answered 2020-Dec-12 at 19:48forEach
does not return a value. If you want to sum all the results, you can use Array#reduce
. In addition, you can set the accumulator's initial value to 0
by passing in a second argument, so you can remove the check for the array's length being 0
.
QUESTION
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,
...ANSWER
Answered 2020-Dec-12 at 00:36Each node with the same values could be evaluated many times, as there is no memoization. For example, I see that the "e OR f" is run over a million times.
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