The Nineteenth Byte

Cat Wizard

18:00

That's what I thought but it doesn't work

Is that supposed to work?

DJMcMayhem

Which synonym are you trying to vote on?

Cat Wizard

Well at this point any one I haven't proposed.

Initially I just wanted to vote on integer

The entire synonym interface is absurdly obtuse.

OH, you have to go to the master tag not the synonym

Man that's dumb

This is so awful.

Pavel

Posted HW in Q# because I guess I forgot to earlier.

Cat Wizard

When you suggest a synonym a vote box shows up on that page, which implies that you vote on the synonym page, however if you navigate away and back it's not there because you need to go to the master page instead.

DLosc

@Adám If the falsey output is allowed to be different for different input lists, then Pip, 3 bytes: MNg (gives nil instead of 0 when the list is empty). Otherwise, MNg|0

Pavel

18:16

@Adám C#: l=>!l.Contains(false)

Adám

@Pavel What does that return for an empty list?

Pavel

Oh nvm

@Adám F#: let rec f=function|[b]->b|true::t->f t|_->false

Probably not optimally golfed, at all.

AKA: If the input list has one element, return that element. If the first element is true, check if the rest of the list is true recursively. Otherwise (empty list or first element is false), return false.

Could be a bit shorter if you substitute true/false for 1/0.

Theo

@Adám Python: lambda l:all(l) and bool(l)

Probably not optimal but I can't think of a way to do it in just one check

Pavel

@Adám F#, better golfed: fun i->(Seq.all id i)&not<|Seq.isEmpty i

DJMcMayhem

@Theo s/bool/len saves you 1 byte

DLosc

18:29

If you don't mind getting 0 for false, that is.

Erik the Outgolfer

actually, it should be lambda l:len(l)and all(l)

Theo

@EriktheOutgolfer Yeah ofc no idea what I was thinking

Erik the Outgolfer

or even lambda l:len(l)*all(l)>0

Theo

Ooh that one's good

DLosc

@EriktheOutgolfer Ooh, nice. That looks like a QBasic trick right there. ;)

Erik the Outgolfer

18:32

or: lambda l:min(l+[0])

(might return 0 instead of False)

DLosc

Won't that give 0 all the time?

Erik the Outgolfer

oh right

DLosc

(I already thought of that myself. :P)

DJMcMayhem

~~you want max, right?~~ I'm dumb

Erik the Outgolfer

lambda l:min(l or[0])

Theo

18:33

Yeah

Erik the Outgolfer

no I don't want max

Theo

Then it would give True if there was any true

Erik the Outgolfer

max(0,0,1,0,1)==1

nindja

DLosc

This one's longer, but I like it anyway: lambda l:sum(l)==len(l)>0

Erik the Outgolfer

and invalid too

sum([])==len([])

DLosc

18:35

Which is why the >0 is there.

Erik the Outgolfer

oh hm

Theo

I don;'t think operators right associate, do they?

Erik the Outgolfer

Python's comparison operators are confusing >_>

DLosc

s/confusing/awesome

Theo

Oh. Guess it does

Erik the Outgolfer

18:36

a<w>e<s>o>m>e

DJMcMayhem

I have a 34 byte brain-flak answer, anyone wanna try to beat me?

(inb4 cat-wizard)

Theo

I love posting my wildly sub-par answers here and just watching literally everyone destroy me

DLosc

@Theo Python's comparison operators work just like they do in math: a > b > c means "a is greater than b, which is greater than c." So sum(l)==len(l)>0 means "sum of l is equal to length of l, which is greater than 0."

Erik the Outgolfer

just like they do in math?

(sum(l)==len(l))>0

I don't think so

Theo

Yeah

It's evaluating the len(l)>0 first, no?

DLosc

18:39

@EriktheOutgolfer That's not math, that's C.

Cat Wizard

@DJMcMayhem What's the CMC?

Theo

a > b > c compares the b > c first I'm pretty sure

Erik the Outgolfer

beat drop

DJMcMayhem

9 hours ago, by Adám

CMC: Given a Boolean list, return whether it is all true AND length>0. E.g. [true,false] → false. [true,true] → true. [true] → true. [] → false.

Cat Wizard

Are we assuming that true is 1?

or non-zero

DJMcMayhem

18:41

I have no idea

I didn't assume either

Erik the Outgolfer

usually it's whatever it is for your language

(note: that's subjective)

Cat Wizard

@DJMcMayhem What does that mean?

Erik the Outgolfer

for example, all non-zero values are truthy in Brain-Flak (that is, unless negative ones are falsy and I've forgot stuff)

DJMcMayhem

@CatWizard Sorry, I mean I didn't think about it, but mine will work for non-zero

Cat Wizard

Ok

DLosc

18:42

@Theo Nope. Python docs: "Unlike C, expressions like a < b < c have the interpretation that is conventional in mathematics... Comparisons can be chained arbitrarily, e.g., x < y <= z is equivalent to x < y and y <= z"

Erik the Outgolfer

btw the bottom of the stack is an infinite pool of zeroes in Brain-Flak heh

quartata

I don't know why other languages don't have chained comparisons

DJMcMayhem

Yes, but there is also a stack-height operator, so there are explicit and implicit 0's

Theo

Oh so essentially you're doing sum(l) == len(l) and len(l) > 0

quartata

So annoying

Theo

18:43

I'm just dumb

Erik the Outgolfer

@DJMcMayhem I mean, below an empty stack are zeroes

New Main Posts

2

Q: Eating fish, growing in size

You are fish in a pond that needs to survive by eating other fish. You can only eat fish that are the same size or smaller than yourself. You must create a program that takes a shoal of fish as an input string. From this you must work out how many fish you can eat and ultimately the size you will...

DJMcMayhem

Yeah. I'm just pointing out that those zeroes are different than real zeroes on the bottom of the stack

DLosc

@quartata I agree! They're a little harder to implement, but not THAT hard. Pip has them.

Cat Wizard

@DJMcMayhem 30

Erik the Outgolfer

18:45

it's really just chained-comparison ::= expression rel-operator [expression | chained-comparison]

DJMcMayhem

@CatWizard Hmm. That's almost the exact same solution as mine. Why does yours have <{}> where the wiki has (<{}>)?

Cat Wizard

@DJMcMayhem 22

@DJMcMayhem The stack only has one thing so when I pop, there is guarenteed to be a zero.

Zacharý

I got it to 11

DJMcMayhem

@CatWizard ([]<{}>[[]]<>) is really clever

Cat Wizard

20

3

Theo

18:48

@Zacharý Think we're gonna get past 200 on this one too?

Cat Wizard

Don't need to zero because it can only ever be zero.

Erik the Outgolfer

@DJMcMayhem wait...

DJMcMayhem

Yes?

Erik the Outgolfer

I mean, wait until he, uh, finishes first...

DJMcMayhem

It's still clever

dzaima

18:49

@Adám JS, 17

Cat Wizard

Ok I don't think 20 can be beat.

Zacharý

@Theo Maybe: but we don't have a Stasoid/chance war there.

Erik the Outgolfer

now do it in Brain-Flak Classic :P (jk no way one can do that)

Theo

@Zacharý I say we forbid them from answering until there's an accepted answer, then let them have at it

Adám

Clever. That should be possible in a few languages. Edit: Actually, nvm, 'cause JS.

Erik the Outgolfer

18:50

btw the 30 has got a star but there's also a 20

DJMcMayhem

Fixed :P

Theo

@EriktheOutgolfer Sorry I'm slow on my stars

Erik the Outgolfer

wow, that does remind me of Contact

Zacharý

@Theo No. Let's just try not to summon them until we're stuck.

Theo

@Zacharý Sounds like a good plan

Cat Wizard

18:55

CMC (That might be moved to main): Write the longest program such that the program errors, however removing any continuous substring of the program causes it to run without errors.

Pavel

you wat D:

Zacharý

@Theo He may have saw it, since he's still going on the behemoth

Theo

@CatWizard That eliminates any languages where any sort of 1-character program will crash

Erik the Outgolfer

so?

Theo

@Zacharý 2 answers in like an hour, on a roll

Cat Wizard

18:57

@Theo Not certainly, as long as the first and last characters of your program run fine it may be possible.

Theo

True. Sorry I was just kinda thinking out loud

Zacharý

HOLY HASKELL.

Cat Wizard

No problem.

I spent the last half an hour trying to come up with a python program of length > 1. And failed.

Theo

Of length > 1?

Cat Wizard

yes?

Erik the Outgolfer

19:00

print'Hello, World!'

Theo

Doesn't pretty much every valid python program have length > 1?

Cat Wizard

Sorry a python answer to the CMC of length > 1

Theo

Ah. Gotcha

Zacharý

@Theo Digit, space(?), newline, comment, tab(?).

Theo

space and tab both work

Cat Wizard

19:01

semicolon

Oh no it doesn't

interesting.

Theo

Nope

Otherwise something like ;0 mighta worked

DLosc

@CatWizard QBasic, 41 bytes: 9999999999999999999999999999999999999999

Cat Wizard

I thought there might be something like that.

@DLosc Can you add a byte by using 1s instead?

dzaima

@CatWizard APL, ⎕←

DLosc

No, I tried. It seems valid line numbers can have up to 40 digits.

Cat Wizard

19:03

Integeresting.

DLosc

Same length restriction applies to labels, incidentally.

And variable names.

Cat Wizard

@dzaima How does that work? I don't know APL.

dzaima

@CatWizard ⎕ takes no args. By itself it returns an input line evaluated. ← is set a variable. ⎕← has no right arg for it, but by itself (←) it evaluates to the function of itself

Zacharý

←← should also work

Adám

@CatWizard Dyalog APL, 6146 bytes: Try it online!

DLosc

19:10

Hahaha... QB64, 42 bytes: 99999999999999999999999999999999999999999

Cat Wizard

It seems like there is a prevailing strategy

ngn

@Adám ngn/k: {7!&/x} (or in some future version: 7!&/), note that maxlong mod 7 is 0

DLosc

@CatWizard It makes sense. The best way to make sure that deleting any character removes the error, is to make all the characters contribute to the same error.

Cat Wizard

Yeah. I'm thinking of requiring there be no reapeating characters

But that challenge would be brutally difficult. I've yet to see a score better than 2.

DLosc

Then all the answers would be 1234567890 :P

Oh, I guess that wouldn't error.

Nvm

Theo

19:15

would something like this 1<==0 in python spark any ideas for anyone?

Cat Wizard

more likely 9876543210

Theo

It works if any of the operators are removed, but not either of the numbers

Cat Wizard

@Theo Ewauls are hard, because you basically can never start or end a program with = unless there is a comment.

DJMcMayhem

CMC: What's the longest English word of length N, where you can make a chain of N English words where each word can be formed by removing one letter from the previous word?

For example, fine works because fine -> fin -> in -> I are all words, but the does not work because the -> he -> ?? has length 2 instead of 3

Cat Wizard

@CatWizard But then you need to find a language where that is out of boudns

DLosc

19:19

@DJMcMayhem Hm. Puzzling.SE has done stuff like this, but I think they allowed anagramming at each step. The max length was something like 18. Without anagrams, it'll be rather shorter, of course.

Cat Wizard

I think we've done this too in chat before.

DLosc

Is the -> he invalid, or is it just a chain of length 2?

DJMcMayhem

It's a chain of length 2, but the goal is to find a word length n with chain length n.

So it's invalid for the CMC because it's too short (even though it's a valid chain)

DLosc

7 letters: PLANETS -> PLANET -> PLANE -> PANE -> PAN -> AN -> A

DJMcMayhem

@CatWizard I've done a very similar CMC, but that was for words where every prefix is also a word

Cat Wizard

19:21

spinned -> pinned -> pined -> pine -> pin -> in -> I

sliders -> slider -> slide -> slid -> lid -> id -> I

Theo

spineda -> spined -> spine -> sine -> sin -> in -> I

Cat Wizard

What is spineda? it's not in my dictionary

Theo

Oh looks like it's a place. Probably invalid. It showed up on thefreedictionary.com which has a useful tool for this

Cat Wizard

supines -> supine -> spine -> pine -> pin -> in -> I

Theo

Looks like 7 is the one to beat

satangs -> stangs -> tangs -> tang -> tan -> an -> a

Anush

19:33

hi all

DJMcMayhem

@Theo Apparently satsang is a word, so satsangs -> satsang -> satang -> stang -> tang -> tan -> an -> a

New record :)

Theo

Wow can't believe I missed that

Well done

Wonder how hard this would be to code up

Anush

how about this slightly tricky question. Given two sets of integers X and Y of sizes m and n and m<n, find a bipartite matching (missing out n-m integers in Y of course) that minimizes the sum of the absolute differences of the pairs of integers. The tricky part is to make it run in O(m^2 + n) time

Neil

snowing -> sowing -> swing -> sing or wing -> sin or win -> in -> I

Riker

complexity challenges are no fun imo

4

Cat Wizard

19:36

atheists -> atheist -> theist -> heist -> hist -> hit -> hi -> I

Anush

@Riker :(

Riker

I mean I just do'nt find htem interesting

@CatWizard nice

Anush

@Riker understood.. this one may be a bad challenge

but some are fun

Pavel

but y tho

DJMcMayhem

def isFine(word):
    if not isWord(word):
        return False

    if len(word) == 1:
        return True

    return any(isFine(word[:i] + word[i+1:]) for i in range(len(word)))

@Theo ^

Just need to hook up isWord

Riker

19:39

are you on mobile still

Theo

@DJMcMayhem I'll do that real quick, won't be super efficient tho

Neil

@DJMcMayhem startling -> starting -> staring -> string -> sting -> sing -> sin -> in -> I

2

Theo

@Neil Well done

DJMcMayhem

@Theo No, but I don't think it get's exponentially slow until 15+ letter words or so

Theo

@DJMcMayhem Yeah, but it won't be great for actually finding them

DJMcMayhem

19:43

@Theo True... How about this?

fineWords = {}

def isFine(word):
    global fineWords

    if not isWord(word):
        return False

    if len(word) == 1:
        return True

    return any((word[:i] + word[i+1:]) in fineWords for i in range(len(word)))

for word in sorted(wordList):
    if isFine(word):
        fineWords.add(word)

Trades time inefficiency for memory inefficiency :P

Theo

That's my kind of tradeoff

I was thinking it might be more efficient to work from the bottom up, considering that you can for example only start with i or a, then only go to an, at, as, ah etc.

DJMcMayhem

This sorta does that

Theo

Except that you're essentially checking for each word

There's a lot of words in the english language to check

Cat Wizard

strimmers -> strimmer -> trimmer -> trimer -> timer -> time -> tie -> ti -> I

Pavel

You missed a step between time and ti

Cat Wizard

19:47

Fixed, thanks

Theo

Oh shit that's 9

Cat Wizard

I believe there already was 9

Riker

no only 8

Cat Wizard

startling is 9

Pavel

.oO(What's a strimmer)

Cat Wizard

19:48

It's a variety of lawnmower. It might be a trademark.

Theo

Turns out not only can I not code, I also can't count

Riker

@Pavel weed eater

quartata

string trimmer

Pavel

Someone should write a script that operates on /usr/share/dict/words

Riker

already running DJ's on it

at 'r' so far, gimme another 60 sec

quartata

19:50

You could do this really efficiently with a suffix automata I think

Hmm

DJMcMayhem

@Theo New approach:

tio.run/##bVA7D4IwEN77K26E@IjEjcTVidHEgTCUcEhjPUipafz12PJoIXpbv/…

Riker

10: branchiate

Rod

[wheatley -> wheatly -> wheaty -> wheat -> heat -> hat -> at -> a] wizard

Riker

according ot the program

Theo

What's the breakdown?

Riker

19:53

not sure checking

Cat Wizard

brachiate is a word is branchiate?

Theo

I think so

DJMcMayhem

@CatWizard It means having gills

Cat Wizard

Well brachiate is probably the next step then

Riker

hold up I think I did it wrong

Pavel

19:55

D:

The Mattbat999

Hello Code Golf people!

Riker

I can't tell yet I accidentally crashed my text editor

ngn

@Rod why not add an -s? if Wheatley is a surname, the Wheatleys are the family bearing it

DJMcMayhem

@TheMattbat999 Hello!

Riker

hey @TheMattbat999

The Mattbat999

19:56

@DJMcMayhem hello. I have a very simple question. What is Code Golf?

DJMcMayhem

It's a competition to write the shortest code to achieve a particular task

Rod

is programming, but ~~useless~~ fun

Cat Wizard

codegolf.stackexchange.com/tags/code-golf/info might also be helpful.

Rod

@Riker nice pic

The Mattbat999

Oh ok. I was confused. I seen the site, and read a few questions, but had no idea what the purpose was. Thanks!

Pavel

19:58

Useless until someone writes the fastest implementation of the Pi function ever: codegolf.stackexchange.com/questions/74269/…

Riker

@Rod my laser cat? huh ok

thanks lol

I think wheat made it actually

Cat Wizard

Man the starboard has kindof been ruined.

But I guess that's ok.

Riker

yep

Rod

@Riker that makes more sense now lol

DJMcMayhem

@CatWizard Hopefully that's a bit better now

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