codegolf | Codegolf und verschiedene Programmier-Rätsel | Addon library

I have a vectors of column sums and row sums and I need to determine prior matrices that made these row sums and col sums. After looking through previous questions, I found this past page that gave me a good head start on this--

https://codegolf.stackexchange.com/questions/150012/enumerate-all-possible-grids-of-integers-with-constraints

However, my goal is a little different than the code presented above in that I have a 4x3 matrices instead of 3x3 matrices. I attempted to modify the r function to a 4x4 function--I figured I could add a 0 to my sums which would make it easily scalable. The function looks something like this--

So, Iwas reading through some Javascript Question answers on CodeGolf to try and improve my understanding of Javascript a lttle bit, & noticed this answer in the following question Generate Green Spraypaint Patterns Within the Answer I spotted a for loop that..[to me] is not the usual Syntax, & no matter how much I stare at it I can't figure out what this does. I can't find any referance to an operator ;; on Mozilla's Javascript Operators page & was wondering if anyone can explain What the double semi-colon does in the code below?

Is it possible to make a class A (can be either an ES 6 class or an old-style constructor function) such that class extends A {} instanceof A is true?

If it is possible, please give me an example. If it is not, please give me a proof.

Note that a class B such that new B instanceof B and new new B instanceof B are both true at the same time is possible by manipulating the prototype chain. It's possible to write a B such that new B satisfy the expected behaviour of a class extends B {}, but I would also like to investigate the possibility to make an A that class extends A {} satisfy the expected behaviour of new A as well.

This is related to my code golf which I have just posted, however in this question I would like to investigate the possibility of reverse - to make class extends A {} behaves like new A, while the code golf is to make new A behaves like class extends A {}.

I was looking at a code golf solution, and the code for it was(written by TFeld):

If I were to brute-force a PIN, I could iterate through all the possible PINs with this code:

This is a followup question to my previous finding a straight in a cribbage hand question and Counting Pairs in Cribbage Hand

Count the number of ways cards can be combined to a total of 15, then score 2 points for each pair. Ace worth 1, and J,Q,K are worth 10.

So my first poke at a solution required 26 different formulas. Basically I checked each possible way to combine cards to see if the total was 15. 1 way to add 5 cards, 5 ways to add 4 cards, 10 ways to add 3 cards, and 10 ways to add 2 cards. I thought I had this licked until I realized I was only looking at combinations, I had not considered the fact that I had to cap the value of cards 11, 12, and 13 to 10. I initially tried an array formula something along the lines of:

Inspired by this code-golf challenge, I thought it would be easy to achieve the end result with awk. The algorithm would be: convert everything to lowercase, then get only vowels to uppercase.

I thought this could work

I'm trying to solve a challenge on Codewars which requires you to reverse an array in JavaScript, in 16 characters or less. Using .reverse() is not an option.

The maximum number of characters allowed in your code is 28, which includes the function name weirdReverse, so that leaves you with just 16 characters to solve it in. The constraint -

Your code needs to be as short as possible, in fact not longer than 28 characters

Sample input and output -

Input: an array containing data of any types. Ex: [1,2,3,'a','b','c',[]]

Output: [[],'c','b','a',3,2,1]

The starter code given is -

What's the idiomatic way to implement longest common prefix in APL? In Aaron Hsu's thesis, footer page number 76, it is said:

the idiom +.= computes the length of the common prefix shared by two paths

This works in the case we are interested in the paper, since we are guaranteed that two paths, once they stop matching, will not have any more matches. However, this assumption does not hold in general. By way of examples: