The Nineteenth Byte

Mars Ultor

5:04 AM

What should I put in a personal website?

Alex A.

Personal things

Quill

Personal or profile

Vihan

@MarsUltor show off some of your work?

El'endia Starman

Vihan

Photoshop is much more difficult than it seems

Mars Ultor

5:15 AM

@Vihan Not much point. Also, I don't think I really have anything to show off.

@Vihan No, you're just using the wrong tools

Clone stamp is your friend

Vihan

@MarsUltor I tried using that, lemme see if I can find an example

Mars Ultor

I need help on adding 'alternative character' key shortcuts

Quill

The failures usually don't hurt anything, and if it installs several versions, it increases the chance that one of them is right. (Note: The 'yes' command and '2>/dev/null' are recommended additions.)

Vihan

aha...

clone stamp still made the G look messed up...

El'endia Starman

@Quill ah-hahaha

Mars Ultor

5:22 AM

@Vihan Whoever did that broke the text shadow

Vihan

@MarsUltor The reason it's so messed up is I couldn't find the font so I had to crop it in from another book. I couldn't get the shadow because photoshop is weird

Mama Fun Roll

0

A: Generate an aperiodic integer sequence

RETURN, 9 bytes (noncompetitive) 0[' ,1+$¦{\3}§.1!]! Try it here. Make sure to run using "timed run," because this is an infinite stream. Also replace \3 with character U+0003. Explanation Basically just outputs each number reversed. I couldn't find any periodicity, but please tell me if yo...

Is this aperiodic?

@Vihan waitwat? no it's not.

Vihan

@MamaFunRoll see how messed up the above image looks. I tried for like an hour to get that looking right...

Mars Ultor

I tried for like an hour to get my profile pic right (it's also copied btw)

Mama Fun Roll

Psh, I never try.

I just am.

Vihan

5:27 AM

@MamaFunRoll you should change your profile picture to a Mama Fun Roll

though I'm not sure I want to know what a "mama fun roll" is :/

Mama Fun Roll

google.com/…

My pic comes up first.

Vihan

@MamaFunRoll haha, perfect

Mama Fun Roll

Oh god...

Alex A.

Say I have two arrays, x and e. x can contain duplicates. What I want to do is reduce x and e together, so that x contains only distinct values and e is the sum of the values in e for each duplicated value in x. What's the best way to approach this?

Mama Fun Roll

@AlexA. Use Javascript

Vihan

5:28 AM

@AlexA. you should use jquery, it's great and does all sorts of things

Alex A.

I don't know what I expected asking that in here

Mama Fun Roll

You should know that you shouldn't know what to expect

Alex A.

@El'endiaStarman Any ideas? ^^^^^

El'endia Starman

@AlexA. haha, indeed... :P

Okay, asking for Python, or just a general algorithm?

Vihan

@MamaFunRoll ^

Alex A.

5:30 AM

@El'endiaStarman In general, but Python would be an acceptable way to represent it.

El'endia Starman

Clarify what you mean by "e is the sum of the values in e for each duplicated value in x"?

Alex A.

x: 1 1 2 3 4 4 4 5 6
e: 0 1 1 0 1 1 1 0 0

Say that's what we have

then I want

x: 1 2 3 4 5 6
e: 1 1 0 3 0 0

El'endia Starman

ahhh, okay

Mama Fun Roll

@Vihan That is not a mama.

Vihan

@MamaFunRoll photoshop?

Mama Fun Roll

5:33 AM

Hold up, how do you know I'm Asian?

Vihan

googles how do you photoshop something into a mama

@MamaFunRoll it was an image result when I googled "fun roll"

El'endia Starman

@AlexA. Do you want this done in-place or can new lists be created?

Mama Fun Roll

@Vihan Oh god...

Alex A.

@El'endiaStarman New lists are fine, whatever is easiest.

Mama Fun Roll

At least there's a mama (I think)

Vihan

5:35 AM

@MamaFunRoll you can never know for sure until you enter the world of Photoshop...

Mama Fun Roll

@AlexA. Oh I might be able to do this one. Gimme some time.

El'endia Starman

def foo(x,e):
    y=[];f=[]
    for i,j in zip(x,e):
        if y[-1] == i:
            f[-1] += j
        else:
            y.append(i)
            f.append(j)
    return y,f

Alex A.

-1 is gets the last element?

El'endia Starman

correct

Mars Ultor

@MamaFunRoll Your Twitter/Github have your real name.

Mama Fun Roll

5:40 AM

@MarsUltor True true.

Quill

var e = {};
var newX = [];
x.forEach(function(item){
    if (newX.includes(item)){
        e[item]++;
    } else {
        newX.push(item);
        e[item] = 0;
    }
});

Vihan

@Quill +2 for JavaScript -1 for not golfed

El'endia Starman

@Vihan Alex wanted "easiest". :P

Quill

jokes on you, I still get +1

El'endia Starman

@Quill If I'm reading this right, it actually wouldn't work.

Quill

5:43 AM

works fine for me

Mars Ultor

@El'endiaStarman How?

Quill

Alex A.

@El'endiaStarman This appears to work! Any idea if this is an efficient algorithm?

El'endia Starman

Yep, thought so.

11 mins ago, by Alex A.

x: 1 2 3 4 5 6
e: 1 1 0 3 0 0

Mars Ultor

@Quill e[item] = 1

El'endia Starman

5:44 AM

@AlexA. Well, it's O(n), and it depends on x being sorted.

Quill

why does x=2 make e=1 in @Alex's example

El'endia Starman

@MarsUltor That wouldn't work either.

Vihan

(x,e={})=>x.reduce((l,i)=>~l.indexOf(i)?e[i]=-~e[i],[])

Mars Ultor

@El'endiaStarman Why not?

Alex A.

13 mins ago, by Alex A.

x: 1 1 2 3 4 4 4 5 6
e: 0 1 1 0 1 1 1 0 0

sum(e) when x == 2 is 1

Mars Ultor

5:46 AM

Oh wait

Quill

so why does 3 = 1

im confuse

Mars Ultor

@Quill You're supposed to add the e values, not 1

El'endia Starman

@AlexA. What I'm doing is going through each element of x and asking, "is this in y?", taking a little shortcut by only looking at the last element of y because I'm assuming x is sorted. If it is, I add the corresponding item of e to the last element of f, because that corresponds with y. Otherwise, I append to both lists.

Alex A.

Makes sense!

Thanks :)

El'endia Starman

Welcome. :)

Alex A.

5:49 AM

So if this is O(n), that's pretty good, right?

El'endia Starman

It's a bit more involved if x is not sorted.

O(n) is very good.

Quill

oh. it's two lists. not just an arbitrary one. right

Alex A.

@El'endiaStarman Hm, if x isn't sorted going into it, we'd have to sort e such that the elements still match up with their corresponding elements of x after sorting.

That seems like it'd be difficult

New Main Posts

0

Q: Word Search Helper

I've been doing word searches recently, and I thought it would be so much easier if all of the words read left-to-right. But rewriting all the lines takes a lot of effort! So I'm enlisting code golfers to help. (Disclaimer: The above story may or may not be remotely accurate.) Your code will ta...

El'endia Starman

Alright, lemme write some Python for the unsorted case. It's not a big change.

def foo(x,e):
    y=[];f=[]
    for i,j in zip(x,e):
        if i in y:
            f[y.index(i)] += j
        else:
            y.append(i)
            f.append(j)
    return y,f

I think this is O(n^2) in the worst case, which is not terrible.

If it is possible for you to sort pairs, then do that instead, since sorting is O(n log n).

Alex A.

5:54 AM

What's y.index(i)? Is that the index in y that contains element i?

El'endia Starman

Yep.

Vihan

Barfy gif

Chatgoat can finally wat. :^)

El'endia Starman

And by "sort pairs", I mean, sort [ [2,0], [1,1], [1,0] ] to get [ [1,0], [1,1], [2,0] ]. Python does this automatically.

Alex A.

@El'endiaStarman So if I can sort pairs, the complexity is O(n log n) for the sort and O(n) for the aggregation? And that's better than O(n^2), right?

El'endia Starman

Yes, because overall, the complexity is O(n log n).

Alex A.

5:58 AM

Because the bigger term dominates, yeah?

El'endia Starman

yep

Alex A.

I have only a very basic understanding of complexity so I appreciate your patience. :)

El'endia Starman

:)

If you really want to be precise, it's basically O(n (log n + 1) ), and the constant 1 just doesn't matter as much as n increases.

The algorithm that uses indexing is O(n^2) because you have to look through the whole list every time, which is n(n-1)/2 in the worst case (think triangular numbers), which is O(n^2) because constants don't matter.

Alex A.

I implemented your algorithm in Julia and it seems to work absolutely perfectly. Thanks so much!

Quick too

El'endia Starman

You're welcome. :)

I think it'd be hard to do worse than O(n^2) unless it's code-trolling. :P

Alex A.

6:13 AM

Hahaha

I honestly didn't have any idea of how to approach this going into it, so I really do appreciate your help.

El'endia Starman

:)

What did you need it for, anyway?

Alex A.

I'm working on some code for survival analysis and this is part of computing the Kaplan-Meier estimator of the survivor function.

El'endia Starman

ahh, neat

I have no idea what that is... :P

Your turn to educate? :D

Mars Ultor

@El'endiaStarman Here

Helka Homba

Whats the oldest thing about the internet that you can remember that no longer exists?

I can remember when Amazon only sold books.

Mars Ultor

6:16 AM

Presumably @AlexA. is using those to find the number of steps down at each value of x

Alex A.

Basically, the premise of survival analysis is trying to make inferences about the amount of time it takes for a group to die.

Optimizer

so... Civil War trailer anyone?

El'endia Starman

@HelkaHomba Dial up? Oh wait, that still exists. :P

@Optimizer Saw that like an hour ago.

Optimizer

saw that 7 hrs ago, but OMG

my inner marvel fangirl is crying out loud inside of me

Mars Ultor

What is this even used for

El'endia Starman

6:17 AM

I'm pretty excited for it too.

Optimizer

its going to be rad

Helka Homba

Marvel movies are still coming out? urg

El'endia Starman

Of course.

They're tapping into a gold mine of comic book stories.

New Main Posts

1

Q: Decompose binary into alternating subsequences

This was inspired by Problem 13 - Non-Repeating Binary of HP CodeWars' recent competition. Let's take a random decimal number, say 727429805944311 and look at its binary representation: 10100101011001011111110011001011101010110111110111 Now split that binary representation into subsequence...

code-golf sequence subsequence

Optimizer

@HelkaHomba oh really?

Vihan

6:25 AM

@Optimizer I do not understand why they called it Civil War when the trailer makes it looks like Avengers 3

Mars Ultor

Marvel Cinematic Universe

The Marvel Cinematic Universe (MCU) is an American media franchise and shared fictional universe that is centered on a series of superhero films, independently produced by Marvel Studios and based on characters that appear in publications by Marvel Comics. The franchise has expanded to include comic books, short films, and television series. The shared universe, much like the original Marvel Universe in comic books, was established by crossing over common plot elements, settings, cast, and characters. Clark Gregg has appeared the most in the franchise, portraying Phil Coulson, a character original...

Optimizer

@Vihan coz avengers 3 is bigger?

El'endia Starman

@Vihan I doubt they'd be fighting each other in Avengers 3.

Optimizer

also, avengers are supposed to be fighting alongside, not with each other

avengers 3 , 1 and 2 are everyone against thanos

El'endia Starman

Also, what's more interesting is that it's a Captain America movie.

Vihan

6:26 AM

Avengers: Civil War

Optimizer

so you can imagine the exponential growth

Vihan

when does the movie come out?

Optimizer

everyone = every one from every marvel studios movie

Vihan

@MarsUltor this article would of been more helpful

Mars Ultor

@Vihan But ^^^^^^^^^^ has a list of the rest of the movies in phase 3 as well (I think)

Vihan

6:31 AM

what is all this "Phase One", "Phase Two" stuff mean?

Helka Homba

@El'endiaStarman I'd just respect them more if they came up with some new characters or stories. The Avengers are 50 years old.

Mars Ultor

There were so many movies they split them into phases. (basically, splitting trilogies/duologies into different phases)
Also, I'm pretty sure each phase represents a certain period of time.

Optimizer

and a general direction and theme of the movies in the phase

i still miss iron man movies

those are the only ones where the suits get real upgrades

Marvel's Captain America: Civil War - Trailer 2 - REACTION

►

El'endia Starman

6:58 AM

@HelkaHomba Not cinematically!

Optimizer

@HelkaHomba dude, marvel in general is coming up with new stuff all the time.

New Sandboxed Posts

7:15 AM

0

A: Sandbox for Proposed Challenges

How fast can I Flood-It? [WIP] code-golf grid sequence There's a one-player game called Flood-It, in which the player must flood the whole board with one color within a certain number of moves. The board is a two-dimensional square grid of size n. At the start, each square in the grid is one ...

Transcript for