The Nineteenth Byte

OldSandboxPosts

01:00

Sandbox posts last active a week ago: Stronger and stronger radiation, Is it metric?, Challenge

l4m2

01:52

When a pair of cops-and-robbers worth one upvote but don't worth two, do you choose one or just a random one?

Aiden Chow

@l4m2 honestly it doesnt really matter, at the end of the day if u think smth is worth upvoting then u should upvote it

theres no set rules and what u r allowed and not allowed to upvote

lyxal

I upvote both :p

Sandbox Posts

02:36

0

A: Sandbox for Proposed Challenges

Comment Out the Comment I'm back! Lately I've been working on something, but finally I've finished it and can return to CGCC! I'll be continuing the Fast & Golfiest Series soon, and also try to start working on (given I have the time and energy) WumpusWars. For now, a new question as a gift for ...

Aiden Chow

hey guys so i just realized that some of the answers including link bubbler-4.github.io/piet have broken links, there are only 20 links to edit it seem like but i just want to say that cuz itll flood the homepage if i edit all at once now

lyxal

codegolf.meta.stackexchange.com/a/25664/78850 might be of interest on how to best handle it

Aiden Chow

it seem like i can just replace "bubbler-4.github.io/piet" with the new link "piet.bubbler.one/"

@lyxal its not really that big of edit, only 20 or so links

lyxal

I wasn't say it was big :p

Just to say that you might be interested in seeing what applies to your situation

> Mass edits should absolutely be split up into batches if you are expecting to do more than 10 edits. Roughly speaking, 10 edits should be the most you do at a time.

for example

Aiden Chow

so i do 10 rn, then wait till tmrw to do the other 10?

lyxal

02:44

> A good rule of thumb is for 10 unrelated challenges be moved above your edit batch before beginning the next batch.

for example

so maybe tomorrow, maybe today if there's enough answers

@AidenChow start with the 10 right now and see how things go from there

Aiden Chow

ok

Aiden Chow

02:57

i did the 10 edits lol, the homepage is flooded

New Posts

07:08

0

Q: Shift Braille down

Braille Patterns in Unicode contains 8 dots(⣿) and occupy area U+2800-U+28FF. Given a Braille Pattern character, where the bottom two dots are not used(aka. U+2800-U+283F), shift it down by one dot. Test cases: ⠀ => ⠀ (U+2800 -> U+2800) ⠅ => ⡂ ⠕ => ⡢ ⠿ => ⣶ Shortest code wins.

Sandbox Posts

07:52

0

A: Sandbox for Proposed Challenges

Braille Base64 Make a one-to-one map from the base64 character table(0-9a-zA-Z+/) to Braille(U+2800-U+283F). These characters are fixed map for compatibility. For other characters you can choose any map as long as there's no collision. ⠁ a ⠃ b ⠉ c ⠙ d ⠑ e ⠋ f ⠛ g ⠓ h ...

Simd

09:34

CMC Write code to output a string of random balanced parenthesese

The Thonnu

11:09

@Simd Python, 26 bytes: print({"()","(())"}.pop())

Outputs either () or (()) :p

lyxal

@TheThonnu oof that's nasty

Using a set like that is goofy golfing hours

Unrelated String

-2 :P

The Thonnu

Ooh nice

Simd

11:25

@TheThonnu excellent :)

Neil

12:35

surely the empty string is balanced for -4 bytes?

The Thonnu

13:04

I thought about that, but I wasn't sure

Simd

13:57

@Neil nice call

I have 5 different sorts of brackets and I want to iterate over all balanced bracketings. But the complication that I regard two bracketings as equivalent if I can swap one sorr of bracket for another

So ([]) is. equivalent to [()]

Unrelated String

14:34

you could just constrain the first opening bracket to be a parenthesis, you can't have a curly brace until there's been a square bracket, etc.

Bbrk24

@Simd So you want to iterate over all combinations (not permutations) of all subsets of the set with infinitely many of each of the five, to put it one way

The fact that it's infinite rules out some approaches

The fact that it's balanced means you only need one half of the string

Simd

14:57

@UnrelatedString what would I do if I have 5 different types of bracket?

@Bbrk24 let's say I only care about outputs of length at most 10

mousetail

CMC We all know the "Please excuse my please aunt Sally" mnemonic for the order of operations in basic arithmetic. Create a similar mnemonic for the full order of operations in C popularity-contest

Simd

@mousetail I had never heard the mnemonic before! Which country is it from?

Bbrk24

"Please excuse my dear aunt Sally" is what I was taught in middle school

user

US?

mousetail

According to Wikipedia US and France

Simd

15:11

@user ah ok. That explains it

Funny lot :)

@mousetail surely in France they have something in french

mousetail

I assume the french version happens to start with the same letters

Simd

@Bbrk24 we are just taught pemdas or bodmas

Bbrk24

Barentheses????

user

Brackets

mousetail

But how would you add bitshifts and ternaries?

Simd

15:15

@mousetail yes. It starts parenthèses, exposant...

user

@mousetail There's a ton of operators there, some of them with the same precedence. Do you only want one operator from each group?

mousetail

It's a pop con, put it in whatever form makes them easiest to remember

Unrelated String

16:11

@Simd the "etc." is there for a reason :P

Simd

16:23

@UnrelatedString :)

@UnrelatedString I have a new formulation which might help

Unrelated String

but yeah you could do it by either only considering brackets you've already used + the next one as you generate, rerolling strings that don't obey the constraint, or replacing the brackets after generating

Simd

@UnrelatedString I wonder how hard that would be to code

Unrelated String

i know there's a challenge from a while back that's essentially for checking the constraint

Simd

@UnrelatedString oh do you know which one?

Unrelated String

nop

e

Simd

16:28

☹️

RubenVerg

18:31

@Simd sure

mark each pair with its grade of the first element, then sort those indices by the grade of each float

(which is what APL (Dyalog APL), 34 bytes: f←{(⍋1/⍵)[⌈2÷⍨⍋,⍵]} does, the previous solution is just the same but worse)

The Empty String Photographer

18:46

I have a sandbox post that has 2 upvotes but hasn’t been interacted with lately so I don’t think it would be a good idea to post it yet:

2

A: Sandbox for Proposed Challenges

Hashers and Crashers What a secure hash! Related, but hashes numbers instead of strings and uses a different scoring system. Definitions A hash collision is when two different strings produce the same hash. Summary In this challenge, the cops will make a hash function, with a collision they know...

Simd

@RubenVerg that is cool

The Empty String Photographer

@Simd print("") # Chosen by fair random balanced bracket string generator. Guaranteed to be random.

Simd

@TheEmptyStringPhotographer :)

Consider all arrays of 20 digits were each digit occurs exactly twice. I can move from one array to another by choosing two digits and swapping them

So far so good

But I regard two arrays as equivalent if there is a bijection from [0..9] to [0..9] that makes them the same

So 1 2 2 1 is equivalent to 2 1 1 2

How I can move around the space of non equivalent arrays?

The Empty String Photographer

@Simd then it’s a set, not an array.

@Simd is 1 2 3 4 equivalent to 5 6 7 8?

Simd

@TheEmptyStringPhotographer yes. Except every digit must occur twice

@TheEmptyStringPhotographer the problem is that every digit must occur twice

The order matters

The Empty String Photographer

19:00

Oh

Simd

You can think of them as open and close brackets of different sorts

Each array has 10! equivalents I think

RubenVerg

19:24

@Simd I think always swapping one first occurrence and one second occurrence might work

not sure though

Simd

19:39

@RubenVerg take 3 2 1 3 2 1 as an example

What would you swap?

RubenVerg

how is that a valid string?

Simd

@RubenVerg it wasn't :)

RubenVerg

any of 0/4, 0/5, 1/3, 1/5, 2/3, 2/4 generates a different-up-to-bijection string, afaict

Simd

Yes

RubenVerg

actually doesn't any swap of two different things also do?

Simd

19:42

Is there any swap of different values which fails?

@RubenVerg yes

So I want to be able to visit all non equivalent arrays from there by swapping

RubenVerg

abcabc -> BbcaAc is equivalent to abcabc -> aAcbBc, up to bijection

@Simd as in, you want a procedural way to start from one such array and eventually get through all of them?

Simd

@RubenVerg yes, exactly

@RubenVerg good point

RubenVerg

20:15

@Simd thought about it for a while, I think it might be way easier to detect whether a given permutation is a bijection of one that appears earlier in the sequence

would an alg that just goes through all possible maps and then skips the ones that it doesn't want be fine?

Simd

@RubenVerg it's much better than nothing :). But would it have to store a massive cache?

RubenVerg

don't think it would

you can replace each entry in the string with its unique index

if they are the same, that's the "canonical" up-to-bijection-equivalent unique string

I think that's equivalent to checking that the unique elements are equal to their grade

Simd

That would be cool

RubenVerg

okay no the last thing I said is clearly wrong

you just want [1..len(unique(x))] to be the same as unique(x)

Simd

20:32

@RubenVerg ah yes

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