The Nineteenth Byte

trichoplax

13:00

Do I seem human to you? I just spent 5 minutes identifying endless rivers and street numbers and shop fronts before being allowed to post an SE answer.

Kenny Lau

Shit happens.

zyabin101

@trichoplax Daaaaarn, this is an advanced ICCS (Image CAPTCHA Challenge Syndrome).

Did you try a sane browser? If so, which?

trichoplax

I use Firefox, which has allowed me to submit questions and answers unhindered up until now. Maybe my history of questions and answers don't seem sufficiently human...

Kenny Lau

Chat mini-challenge: Returns if x is power of y

x>=1, y>=2

both integers

zyabin101

>_<

Kenny Lau

13:07

@zyabin101 Don't spoil please, others haven't seen it

zyabin101

Challenge idea: reduce an unary fraction in Retina.

Kenny Lau

@zyabin101 32 bytes

zyabin101

\(o_O)/

Kenny Lau

@zyabin101 How many bytes did you use?

zyabin101

@KennyLau I didn't do the challenge.

Another challenge: detect if a fraction cannot be reduced in Retina.

Kenny Lau

13:17

@zyabin101 22 bytes

still 22 bytes

zyabin101

Even more \(o_O)/

Kenny Lau

Do you understand it?

zyabin101

@KennyLau shrug

Kenny Lau

try to

it's useful

zyabin101

Okay.

Mars Ultor

13:21

Fractions are represented as <unary anything but ;> ; <unary anything but ;> in the input

Kenny Lau

@MarsUltor Thanks, I used x in the fractions

Mars Ultor

Capturing group 1 captures a number, 2 optionally repeats it, and 3 repeats it. If all are repeats, it passes and returns 0, otherwise it returns 1

@KennyLau Oh wait my bad

Kenny Lau

18 bytes, I forgot to golf it

Mars Ultor

@KennyLau Maybe you should use 0 instead of x so it's more like unary?

Kenny Lau

@MarsUltor We use 1 instead?

Mars Ultor

13:27

Huh, TIL

Why though?

Kenny Lau

Well, so that the value is actually 1

the standard base conversion algorithm would work

Mars Ultor

Oh

Kenny Lau

for example, 1001_2 is really 1*2**3 + 1*2**0

I've added GCD here

Mars Ultor

@KennyLau Not this?

^(1+)\1*;\1+$
$.1

Kenny Lau

Well, the output should be in unary also

Mars Ultor

13:38

\d+
$*
^(1+)\1*/\1+$
$.1

Wow, it's only 24 bytes to return GCD for integer numerator+denominator proper/improper fractions

is sure that's shorter than Java

Wait, is Retina optimized for arithmetic?

Apparently not

Kenny Lau

Depends on what you mean by "optimized"

Sherlock9

Any golfing suggestions are welcome: codegolf.stackexchange.com/a/79916/47581

Mars Ultor

Maybe Retina should have arithmetic mode? i.e, long repeats are stored as repeats, not massive strogns

Kenny Lau

what does that mean? (sorry for bad English)

@Sherlock9 Would it be faster to use reverse and base conversion?

What is your native language?

(programming langauge)

Sherlock9

My native programming language is Python, and it doesn't have anything for converting to another base, just from

Kenny Lau

13:52

->n{s=(1..n).inject(:*).to_s(10);s.length-s.reverse.to_i(10).to_s(10).length}

@Sherlock9 I see

oops, that's 77 bytes

Sherlock9

->n{s=(1..n).reduce(:*).to_s;s.size-s.reverse.to_i.to_s.size}

Is much shorter at 61 bytes

Kenny Lau

alright.

->n{s=(1..n).reduce(:*).to_s(5).reverse;s.map.with_index{|x,i|x*(5**i-1)/4}}

oops, very long

mınxomaτ

I just noticed what Google is using instead if the traditional Lorem Ipsum

9

quartata

hahaha

El'endia Starman

> shank kevin

Good thing I don't know anyone named Kevin.

Sherlock9

14:13

Kevin not Kenny Lau?

Kenny Lau

I requested the name change... he really followed xd

Sherlock9

I was just telling El that he probably does know that Kevin

Also Kenny. would you mind I post that 61-byte version of your code?

Kenny Lau

what do you mean?

oh

wait

Sherlock9

Since you came up with the idea, and I just golfed it, I figured you might want to post it

Kenny Lau

->n{(n-n.to_s(5).split("").map{|x|x.to_i}.reduce{:+})/4}

Try this

Agawa001

14:20

why ppcg isnt getting its design ?

@KennyLau sorry

i unstarred it now

Kenny Lau

lol

zyabin101

@Agawa001 Because it is in the depths of the Stack Exchange Design Department's TODO list. :P

Agawa001

who will guarantee it lasts on the top

Kenny Lau

@Sherlock9 I don't know why it doesn't work in the online compiler

Agawa001

@zyabin101 ofcourse, this site deserves too much time to ameliorate

Sherlock9

14:22

You used {:+} in the reduce when it should be (:+)

Kenny Lau

oh ok

Sherlock9

->n{(n-n.to_s(5).chars.map{|x|x.to_i}.reduce(:+))/4}

For 52 bytes

Kenny Lau

> 1234.to_s(5).split("")
=> #<NoMethodError: undefined method `t' for main:Object>

Sherlock9

I want someone to post all three of these answers somewhere because they're great algorithms

Kenny Lau

What happened?

Sherlock9

14:23

That is really weird

Try retyping it

Kenny Lau

A Nerd - I

This answer is invalid, right?

Kenny Lau

@Sherlock9 That is black magic. It worked.

Adnan

@ANerd-I It's abusing a standard loophole

Kenny Lau

@Sherlock9

Agawa001

14:24

@ANerd-I yes it is

Sherlock9

@KennyLau Sometimes, copy-pasting throws is random characters for no good reason as I found out today

A Nerd - I

@Adnan Oh wow you ninja'd me even there.

Sherlock9

Alrighty then

Adnan

Hahaha

Kenny Lau

@Sherlock9 It's true. It inserted U+00AD between spli and t

1234.to_s(5).split("")

Agawa001

14:25

@ANerd-I languages not fitting the contest-conditions are alll loopholes

Kenny Lau

for some reason, chars also doesn't work online

Agawa001

oups, ninja'd

Sherlock9

@KennyLau Retyping didn't work this time?

Kenny Lau

@Sherlock9 Nope. Different version, I guess.

mınxomaτ

Also, more Google shenanigans: They named their natural language parser "Parsey McParseface" (source):

Sherlock9

14:30

@KennyLau I don't quite understand why this works

Kenny Lau

Black magic.

Agawa001

@Fatalize the middler one .

Kenny Lau

Challenge: Retina this

@Sherlock9

lambda n:reduce(lambda a,b:a+b, [d*(5**i-1)/4 for i,d in enumerate(str(n,5)[::-1])])

(5**n-1)/4 == reduce(lambda a,b:a+b, 5**i for i in range(n))

that's basically how it works

Sherlock9

but why n - sum of digits of n in base 5 and then divide by 4?

That's the part I don't get

Kenny Lau

well, it's basically the same

Sherlock9

14:38

I do remember that 1+5+25+...+5**n=5**(n+1)/4

Kenny Lau

consider 2413-2-4-1-3 as (2000-2) + (400-4) + (10-1) + (3-3)

all in base5, of course

then consider it as 2*(5**0+5**1+5**2) + 4*(5**0+5**1) + 1*(5**0) + 3*()

then convert it back to the famous repeated floor-division algorithm

Sherlock9

OOOH

Now how do I explain all that in the post?

Kenny Lau

f=lambda n:n and n/5+f(n/5)

link them to oeis xd

oh, that's already posted xd

mınxomaτ

I managed to discover a several versions-old bug in FF and now we can't even work on it, since Bugzilla is down >.>

flawr

3

@AlexA. Still better than I.

Kenny Lau

14:50

@Sherlock9 You know, f=lambda{|n|n<1?n/5+f(n/5):n}

Sherlock9

@KennyLau Yeah, i saw that too :D

Kenny Lau

So have you updated your solution?

(For some reason recursion doesn't work online)

zyabin101

@flawr @AlexA. Cute billiard dog alert! >.<

Kenny Lau

@Sherlock9 Wait... there is a ruby answer using this

2

A: Zeroes at the end of a factorial

Ruby, 22 bytes One of the few times where the Ruby 0 being truthy is a problem for byte count. f=->n{n>0?f[n/=5]+n:0}

Sherlock9

@KennyLau Not yet. Still working on the explanation

Did update with the second algorithm though

Suever

14:53

Question for you guys. When a challenge says input can be in any format, does that mean that anything is fair game?

Kenny Lau

@Suever Link?

Suever

Specifically this challenge

8

Q: Do these squares overlap?

Given the coordinates of the upper left corners of two squares and their side lengths, determine whether the squares overlap. A square includes the top and left lines, but not the bottom and right lines. That is, a point (a,b) is inside a square with side length k that starts at (x,y) if and only...

Could I have two inputs 1) [k1, k2] 2) [x1, y1; x2, y2]

Kenny Lau

I would say yes.

Downgoat

@mınxomaτ what's the bug?

Kenny Lau

@Sherlock9 You just achieved "longer-than-java":

0

A: Zeroes at the end of a factorial

Java, 38 bytes int z(int n){return n>0?n/5+z(n/5):0;} Full program, with ungolfed method: import java.util.Scanner; public class Q79762{ int zero_ungolfed(int number){ if(number == 0){ return 0; } return number/5 + zero_ungolfed(number/5); } ...

Sherlock9

14:59

What the heck XD

mınxomaτ

@Downgoat The encoding routine FF uses for all it's hash digests is essentially broken for anything but pure ASCII.

Downgoat

@mınxomaτ oh... That seems bad (and possibly quite problematic)

Kenny Lau

@Sherlock9 Have you seen my challenge?

mınxomaτ

@Downgoat That's not a problem per se. It just rejects valid hashes for no reason, marking valid websites as untrusted.

Kenny Lau

2 hours ago, by Kenny Lau

Chat mini-challenge: Returns if x is power of y

2 hours ago, by Kenny Lau

x>=1, y>=2

2 hours ago, by Kenny Lau

both integers

Downgoat

15:01

@mınxomaτ Oh. I thought you meant its ssl would break on non-ASCII

Sherlock9

Hm, I did but I haven't considered it yet

I just finished editing the post, I was going to try to make a regex that did the challenge since I don't know Retina

mınxomaτ

@Downgoat No. That would be an OpenSSL problem. The bug is in the encoder that transfers files from memory to OpenSSL. They get mangled somehow in the process.

Kenny Lau

@Sherlock9 Who and what are you replying to?

Downgoat

@Quill how did you get coveralls working ._.

zyabin101

@Downgoat Coverage tool, Coveralls plugin for Node.js and Travis. :P

Eᴀsᴛᴇʀʟʏ Iʀᴋ

15:05

@Fatalize Never had wine.

Downgoat

Plugin? Oh.......

Eᴀsᴛᴇʀʟʏ Iʀᴋ

Cheese, yes.

@Agawa001 ?

Downgoat

>_> how did i not know about this

Fatalize

@EᴀsᴛᴇʀʟʏIʀᴋ good

Sherlock9

@KennyLau Did I see your challenge? Yes. But I will work on the following challenge first

33 mins ago, by Kenny Lau

Challenge: Retina this

But with regex since I am unfamiliar with Retina

Kenny Lau

15:06

well okay.

zyabin101

@Sherlock9 Retina is basically regex.

Sherlock9

Er, hang on. How do you convert from decimal to base 5 in regex?

zyabin101

With a special notation.

Sherlock9

@zyabin101 Sure, basically, but every regex system has its own quirks

See ^^

Kenny Lau

@Sherlock9 By black magic.

mınxomaτ

15:10

@Sherlock9 Yeah, but everything should just be using PCRE.

Kenny Lau

@Sherlock9 Regex isn't turing-complete without repeat.

Sherlock9

@KennyLau Check with 35 xs. I think you're missing a zero, there.

Kenny Lau

@Sherlock9 Oops.

Sherlock9

Alright, so the next steps are to sum the digits, subtract from the original and divide it by 4

Where was that page on unary arithmetic?

Actually, it would be simpler just to repeatedly divide by 5 and sum. If I knew how to sum in the first place

Kenny Lau

Here.

sum is a no-op

or just remove the delimiter

I will leave this part to you

Repeatedly is + in Retina

Sherlock9

15:26

After fiddling with Retina for a bit, I should really be studying for my exams

As it might be more productive than trying to relearn regex one token at a time

Kenny Lau

lol

Downgoat

Chat mini-challenge: Write a program which outputs a version of this file, but fixed (supports empty arrays), in as few bytes as possible

Eᴀsᴛᴇʀʟʏ Iʀᴋ

...

Agawa001

@EᴀsᴛᴇʀʟʏIʀᴋ ??

Sherlock9

->x,y{a=x.to_s(y).chars;a==[?1]+[?0]*~-a.size}

@KennyLau This was much faster

Eᴀsᴛᴇʀʟʏ Iʀᴋ

15:31

@Downgoat Cheddar, 22 bytes: getSource()

@Agawa001 ??

Kenny Lau

@Sherlock9 How does that work?

Agawa001

27 mins ago, by Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Agawa001 ?

???

Downgoat

@EᴀsᴛᴇʀʟʏIʀᴋ :2965374 doesn,t woek

Sherlock9

@KennyLau If x in base y is "1" plus (len(a)-1) "0"s then x is a power of y

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Downgoat does on y machine

try updating to the latest release

Downgoat

15:32

@EᴀsᴛᴇʀʟʏIʀᴋ ...

Eᴀsᴛᴇʀʟʏ Iʀᴋ

link

Downgoat

@EᴀsᴛᴇʀʟʏIʀᴋ i am

Eᴀsᴛᴇʀʟʏ Iʀᴋ

try again?

Kenny Lau

@Sherlock9 There is a much shorter solution

Sherlock9

Sorry for the multiple pings

Oh?

Eᴀsᴛᴇʀʟʏ Iʀᴋ

15:33

lol

Downgoat

@EᴀsᴛᴇʀʟʏIʀᴋ Can you share screetshot of your output?

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Sherlock9 chat-minichallenge output

@Downgoat no, has PII

Agawa001

i was annoyed by the decimal remainder 1 of my previous rep so i downvoted, it is divisible by 10 now

Sherlock9

@EᴀsᴛᴇʀʟʏIʀᴋ I was replying to Kenny saying there is a much shorter solution

Downgoat

@EᴀsᴛᴇʀʟʏIʀᴋ then redact it

Kenny Lau

15:34

@Sherlock9 25 bytes

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Downgoat no

Kenny Lau

excluding ->x,y{}

Eᴀsᴛᴇʀʟʏ Iʀᴋ

the sequence of [redacted]s are PII

Downgoat

How can i know your solution actually works then?

Eᴀsᴛᴇʀʟʏ Iʀᴋ

... u no trust innocent cat?

Sherlock9

15:36

@KennyLau I found this ->x,y{a=x.to_s(y);a==?1+?0*~-a.size}

Kenny Lau

@Sherlock9 29 bytes

Downgoat

@EᴀsᴛᴇʀʟʏIʀᴋ i trust you but i want to know the output

Eᴀsᴛᴇʀʟʏ Iʀᴋ

>_> it wiped cheddar completely

i need to reinstall cheddar for that

sorry, not today

Adnan

Sure, but brainbool's native integer size is 1 bit. The fact that it is turing complete is not related to the concept of native sizes. The fact that the language is turing complete means we can compute arbitrary precision numbers with it, however. But then, we can say the same about every language. Would you complain about a C solution limited to 64 bits numbers? This is not a serious answer, however, maybe you should state a minimal integer size, like 8 or 16 (or even 32) bits, to avoid these kinds of problems ;) — Nax 37 mins ago

I don't even...

Downgoat

@EᴀsᴛᴇʀʟʏIʀᴋ :(

Sherlock9

15:39

@KennyLau I found this for 31 bytes ->x,y{/^10*$/.match(x.to_s(y))}

Kenny Lau

@Sherlock9 In fact I found a 24-byte

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Adnan To be fair, the answer hits a loophole.

Adnan

@EᴀsᴛᴇʀʟʏIʀᴋ Yes, but it seems that he ignores that fact

quartata

@EᴀsᴛᴇʀʟʏIʀᴋ Yeah, one that's a standard loophole that was never funny

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Adnan yeah... idk what he means?

Downgoat

15:41

;_;

Eᴀsᴛᴇʀʟʏ Iʀᴋ

Q_Q

zyabin101

@EᴀsᴛᴇʀʟʏIʀᴋ curl http://install.cheddar.vihan.org/ | bash? >_>

Eᴀsᴛᴇʀʟʏ Iʀᴋ

Q_Q

Downgoat

;_;

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Downgoat @zyabin101 has a good idea. install.cheddar.vihan.org should show the most recent cheddar downloads.

@BassdropCumberwubwubwub I love how your name spreads across the whole list of users when you come in... :P

Sherlock9

15:44

@KennyLau Welp, I don't know. What do you have?

Downgoat

@EᴀsᴛᴇʀʟʏIʀᴋ It should but im too lazy to make a website...

Bassdrop Cumberwubwubwub

@EᴀsᴛᴇʀʟʏIʀᴋ Did you just call me fat?

Kenny Lau

@Sherlock9 x.to_s(y).reverse.to_i<2

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Downgoat I'll make it?

@BassdropCumberwubwubwub no.

zyabin101

@EᴀsᴛᴇʀʟʏIʀᴋ @Downgoat install.cheddar.vihan.org should be the install script for Cheddar and CPM.

Bassdrop Cumberwubwubwub

15:44

@EᴀsᴛᴇʀʟʏIʀᴋ :P kidding mate

Downgoat

@EᴀsᴛᴇʀʟʏIʀᴋ you'd want to? :D

Eᴀsᴛᴇʀʟʏ Iʀᴋ

I called your name a type of spread, like onion dip.

@Downgoat Yes. I need more stuff that other people think is cool so I can skip learning java next year...

@zyabin101 yeah

or at least a link with a download?

Sherlock9

@KennyLau O_O Damn, that's brilliant and I am pissed that I didn't think of it XD

->x,y{x.to_s(y).reverse.to_i<2} is 32 bytes though?

And my regex is 31 bytes

Kenny Lau

well, I said excluding ->x,y{}

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Downgoat i gtg, talk about wubsite later.

Sherlock9

15:47

Well, then this is also 24 bytes /^10*$/.match(x.to_s(y))

zyabin101

@EᴀsᴛᴇʀʟʏIʀᴋ BAI

Kenny Lau

@Sherlock9 oh, alright

that's brilliant

Sherlock9

YAY! :D

Kenny Lau

@Sherlock9 spoiler: retina solution

Sherlock9

It's not much of a spoiler if I barely understand what's going on XD

Ah, wait. I think I get the first stage

Downgoat

15:49

@EᴀsᴛᴇʀʟʏIʀᴋ bai

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@zyabin101 BAI

@Downgoat bai

Bassdrop Cumberwubwubwub

@Sherlock9 Retina, the new malbolge

Sherlock9

Until \b. Hang on, looking at docs

Kenny Lau

@Sherlock9 \b is regex meaning word-boundary

orlp

if you look closely enough in her eyes you can see photoshop staring back at you

2

Bálint

15:59

So, we post cats again?

Sherlock9

@KennyLau I see. Is this because $ was not available for end of string?

Bálint

3

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