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7:14 PM
Japt magic, @cairdcoinheringaahing ;) The Í is the shortcut for n2, which gets passed to the î as 2 arguments, which only expects 1 argument. I'll add a full explanation in the morn'. — Shaggy 38 secs ago
I really want to know why Japt has a shortcut for n2??
Is n2 something special in JS?
 
7:25 PM
0
A: Sandbox for Proposed Challenges

Taco タコスPythagoras' Golfing Grid Recently, I created a binary word search that got me working with grids. It was fun, so I wanted to create some more similar content. Meet Pythagoras' Golfing grid: Input You'll be given a numeric value for T. Output You'll output the values for d through k in alphabetic...

 
When do we move challenges from the sandbox to a main question? I'm not sure I ever asked that lol
 
When it's ready to be posted :P (typically, wait a week and ask for feedback in chat a couple of times)
If it's especially complicated, then it should take longer to be ready. Otherwise, any time between 3 days (especially simple challenges) to a week (standard) works
 
Sounds good :) thanks!
 
@SandboxPosts @Taco It's a little hard to understand, as written. Do you mind if I do a bit of editing to try to make it clearer?
 
@cairdcoinheringaahing Feel free! Let me remove the runtime requirement first. I'll post here when done.
Done
 
7:39 PM
By "while taking into account the Pythagorean theorem", do you mean that \$a^2 + b^2\$ has to equal \$c^2\$?
 
7:50 PM
@Tacoタコス I think your example for 75 is wrong? Either that, or the "c = f = d * k" bit shouldn't include "= f", as f is 100 in your example, but d*k = 4*20 = 80
In fact, the "= f" is the only bit about it that doesn't work
Changing T to 55 and f to 80 would fix it
 
Does anyone understand what makes the difference between, say, a question with score 20 and one with score 200?
 
If I did, all my questions would have a score of 200 :P
 
Seems familiar, but I don't see any issues with it
 
8:02 PM
0
Q: Is it a row of Pascal's triangle?

emanresu APascal's triangle is a triangular diagram where the values of two numbers added together produce the one below them. This is the start of it: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 You can see that the outside is all 1s, and each number is the sum of the two above it. This continues forever. Yo...

 
Bye
 
@cairdcoinheringaahing Absolutely correct, sorry about that. I'm correcting that.
Fixed
 
What is this transformation called?
['123', '456', '789'] → ['147', '258', '369']
 
Why am I an idiot sometimes... Thanks a mil @cairdcoinheringaahing!!
 
8:17 PM
@Tacoタコス Updated, feel free to rollback/change anything you dislike/I got wrong :)
 
That looks perfect actually lol
Thanks for the assist!
 
I went 9 days without posting anything, and in the last 24 hours I've posted 1 challenge and 4 hours :P
 
> in the last 24 hours I've posted ... 4 hours
 
8:30 PM
But yes, I control time and have only posted 4 hours ;P
I'll get around to posting the other 20, but that's effort and people keep asking me to explain things when I post :P
 
8:41 PM
I was bored, so I decided to make Minesweeper in Vim. :P
Use 'x' to check a tile, and normal Vim controls to move the cursor around. It works best in Vim instead of TIO. I chose ▄ and ▀ for the mines and tiles because neither renders in my terminal, so they look the same.
 
Use homoglyphs of x :p
 
that would probably work too
You can make the grid as shown in the input whatever size you want, then paste the code in to set everything up.
I was too lazy to implement flagging tiles or a way to randomly generate the grid
I like doing stupid stuff in Vim :)
 
*looks at Vim GoL* really? :P
 
Hard to believe, right? :P
Turns out X homoglyphs actually render in my terminal, so I'm using that now: Try it online!
 
9:05 PM
@AaronMiller Insane
@BrowncatPrograms Why eat a quantum computer when you can eat a quantum banana?
 
Who said anything about eating a quantum computer wtf lol
 
You said you were going to grab your quantum computer
 
To use, not to eat :P
I've restarted work on Add++ v6 after about 3 months of not touching it, and I already forgot that I've added partial function application ಠ_ಠ
 
@cairdcoinheringaahing Wat's the difference?
 
9:23 PM
Dangit lol I can't use sum of row n = 2^n to solve that pascal's triangle one
That was the shortest way in C# IMO lol
 
10:02 PM
@Tacoタコス Still works, but you also have to check if the input is a palindrome
 
@cairdcoinheringaahing Fair point :D
 
I'll circle back lol
 
[3,1,1,3] would pass that check, but isn't a row
 
:/ lol
I'll still circle back to see if I can come up with some swift solution. Perhaps testing coefficients?
 
10:04 PM
Use a golfing language with a binomial coefficient builtin :P
 
If I understood golfing languages lol
I'm still learning those, so I'm using C# since I know it :D
 
Try Scala! It's somewhat similar to C# and won't help you learn golflangs but it will make you cool
@LuisMendo Here to promote MATL's builtins? :P
 
@Tacoタコス ...that's just made my find an incredibly stupid, but 5 byte solution in Jelly :P
 
If it's short, it ain't stupid :P
 
@user It wasn't my intention, but that's always an option :-)
 
10:08 PM
> Your video game enemies are like me except less good looking.
 
Wait, that's not a stupid idea :P Thinking about why it works, it's kinda brilliant (not to pat my own back :P)
 
Brilliant! :P
@user From this answer. Looking at the answerer's pfp, it seems like those video game enemies must be really ugly
 
ಠ_ಠ
You're kidding me
Mine fails for single element lists
Heartbreaking
 
Just omit that test case and delete this message :P
@cairdcoinheringaahing Oh no, you won't be able to sell your heart now! :P
 
For those who were interested, it used the fact that the \$n\$th row of Pascal's triangle gives the coefficients of \$(x+1)^n\$. If we then get the roots of the polynomial represented by that list of coeffs, it'll be a list of all -1s (unless the list is a single element, as that has no roots)
 
10:21 PM
CMC: Given a list of numbers (integers, positive integers, floats, whatever you want), verify that the list is a palindrome and that it is increasing (not strictly) up to the center (and, of course, that means it must be decreasing from the center to the end)
 
@cairdcoinheringaahing Maybe convolve with 1,1 and then find the roots?
 
Convolving with 1,1 already takes 4 bytes, then +2 for finding the roots and you're at 6 bytes (my current score)
 
@user e.g. [1, 2, 3, 4] -> F (not palindrome, not decreasing after the center), [1, 2, 3, 2, 1] -> T, [1, 3, 4, 2, 1] -> F (not palindrome), [3, 1, 0, 1, 3] -> F (not increasing until the center and decreasing afterwards)
 
Lol, and then you still need to make sure the leading coefficient is also 1
 
Yep, fails for [2,2] without that ಠ_ಠ
¯\_(ツ)_/¯ my current way is fairly nice, although that is bugging me
 
10:29 PM
Nice, Scala's risen to 1049 from 816 in August and 772 in 2016. I can't tell if someone's been secretly posting really short Scala answers or if other answerers are getting worse, but either way, I'm happy :)
 
I think ELO score rises as new answers (or games) are posted, so long as you aren't the loser all the time
 
There haven't been too many Scala answers, though, because I'm the only one using it afaik and I haven't been too active recently
 
11:16 PM
Funny how K manages to beat 05AB1E in this challenge
 
@user JavaScript, 4 bytes: _=>1 (uses the Z_1 set of numbers)
 
nice abuse of "whatever you want" :P
Also works with the number set being an empty set
 
@tjjfvi There are now 4 rules :P
 
@Bubbler I don't think the empty set is really a valid set of numbers
 
Why not?
 
11:31 PM
It's not even a group
 
With an empty set, both _=>0 and _=>1 would work, I guess
 
No, [] is a valid list with an empty set, so only _=>1 works
 
@tjjfvi Doesn't matter for this challenge :P
@Bubbler Ah, true
The only valid list, really
 
Yes
 
0
Q: Game of Life, but on a 4-8-8 tiling

BubblerBackground The 4-8-8 tiling looks like this: For the purpose of this challenge, we take the orientation of the tiling as exactly shown above. In plain English words, we take the tiling so that it can be seen as a rectangular grid of alternating octagons and squares, the upper left corner being a...

 
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