The Nineteenth Byte

12:25 AM

My cmc got buried

yesterday, by Quintec

CMC: Given an integer n as input, output the kth prime such that the difference between the kth and (k +/- 1)th prime is at least n

ais523

that seems like the sort of challenge Husk would be really good at, but I don't know the language

Pavel

Also Prolog

ais523

it seems about average, it has the right primitives but fitting them into the syntax is hard

and evaluation order would be a problem

let me have a go in Brachylog, I know that pretty well

bleh, this is a pain, you can't easily express "two consecutive primes" or "nth prime"

you'd have to word it as "on some prefix of the sequence of primes…"

and then you're stringing together way too many primitives

H.PWiz

@ais523 Preliminary 10 bytes Husk: !V≥⁰Ẋ-İpİp. Annoying "nth prime" isn't short. I can probably golf it away though

Quintec

My solution in any language is far longer than I expected

H.PWiz

12:52 AM

@Quintec I think that is phrased is a confusing way. I'm assuming you want the first prime such that the difference between it and the next/prev prime is at least n

ais523

@H.PWiz my attempt after learning Husk produced the very similar !Vo≥⁰`-İpİp

which is one byte longer

H.PWiz

@ais523 I think you can use ≠ as absolute difference.

ais523

oh, then it's the same length (and a very similar program)

if only there was some easy way to get V to return the value itself, rather than its index…

H.PWiz

Lol, I found 8 bytes, now you said that.

ais523

1:12 AM

down to 9 using an entirely different approach: ←€mṗNR←⁰0

oh, duh, I can see an obvious byte saving

nope, doesn't work, because R can't figure out its argument order without some character to disambiguate

the two off-by-ones are annoying…

H.PWiz

I recall being of the opinion that the default argument order for R was wrong

ais523

also it doesn't work for very small inputs due to, effectively, treating -1 as prime

H.PWiz

(I'm still trying to understand it)

been a while since i've read husk

ais523

€mṗtNR←⁰0 works for input 2, but not lower inputs

(because it treats 0 as prime rather than -1)

the basic idea is to construct an infinite list of booleans specifying whether the number at that index is prime

then look for a sequence of sufficiently many zeroes (i.e. "composite") in a row

H.PWiz

Oh, cool solution. Mine is much easier

ais523

1:29 AM

here's a 9-byte solution that works for all inputs: →←ġo<⁰-İp

is that at all close to yours?

(form groups of primes that differ by less than the input, then take the last element of the first group)

H.PWiz

Sort, mine is closest to your 10 byte solution.

ais523

this question could probably make main

H.PWiz

(my solution is straight-forward but uses an obscure command)

ais523

1:44 AM

η, perhaps? I can't figure out what it does from the description

meanwhile, →V≥⁰Σm´RẊ-İp is a fun solution but way too long

H.PWiz

Mine has a levenshtein distance of 4 from your (first) 10 byte solution

ais523

S!Vo≥⁰≠İp gets rid of the İp repetition from that answer, but that's 9 bytes

Quintec

@ais523 Are you talking about mine? Can't read husk lol

ais523

yes, your question

finding a good language to solve it is interesting, golfing is also interesting

you made me learn a language to solve it, that's the sign of a good question

Quintec

Hm... okay, will do

Okay, posted

ais523

2:03 AM

here's an 8-byte solution that only works for even numbers as inputs: -⁰LU⁰mṗN

if you give it an odd number as input, it returns 1 less than the second solution for the input below

restricting the challenge to even input would probably increase the range of potential solutions, not sure if that's a good or a bad thing

New Main Posts

0

Q: Prime Difference

Given an integer n, output the smallest prime such that the difference between it and the next prime is at least n. For example, if n=5, you would output 23, since the next prime is 29, and 29-23>=5. More Input/Output Examples 1 -> 2 (2 - 1 >= 1) 2 -> 3 (5 - 3 >= 2) 3 -> 7 (11 - 7 >= 3) 4 -> 7...

Veskah

Slo-bot

ASCII-only

4:50 AM

@ais523 hmm. would using a match/switch not be more complex? especially if the language intends to have a more powerful switch statement (function)?

Pavel

@ais523 And then you get reference.wolfram.com/language/ref/For.html?q=For and it's weird

ASCII-only

5:12 AM

@Pavel weird?

Pavel

Actually Switch is way worse as a function

ASCII-only

@Pavel oi. excuse me

it's only because mathematica doesn't have postfix blocks in the argument list (like... groovy i think it is?)...

in fact, I don't think Mathematica has blocks at all... so there's the real reason they're bad

@Pavel how weird would it be if it looked exactly the same as normal loops (i.e. how you're used to)

Pavel

It kinda does, (statement; statement; statement)

That's not really a block, it's more like Perl's do

But Mathematica lets you keep args unevaluated so you can use that like a block

ASCII-only

but it still doesn't allow for args after the arglist. which is not the most readable

Pavel

@ASCII-only Try it online!

ASCII-only

5:21 AM

@Pavel hmm?

Pavel

@ASCII-only Arglist, then arg

Unless I don't understand what your point is

ASCII-only

not like that

If[foo](body)

Pavel

ah

Something like If[foo][body] would be possible I guess

Which I guess is Curry[If, {1, 2, 3}], or If[cond][true][false]

Jo King

5:47 AM

@Deadcode I've been trying out programming in EMCA regex and been banging my head against the wall, since I'm not that smart.

H.PWiz

Me too. Not gone any further than regexgolf (alf.nu) though

Jo King

I've been trying to do this question, but I'm stuck on how to check if one capture is larger than another >.<

ASCII-only

@JoKing is there even any way to do it other than ((.+).+)?

Jo King

?

@ASCII-only I've made two captures \1 (2^n) and \4 (k) and I can't figure out how to compare the lengths of the two using the original number

ASCII-only

@JoKing is it not possible to assume one is 2^n and the other is k, but capturing them as ((.+).+) wherever you capture the longer one? or does the longer one appear later in the regex/do i have no idea what you're doing

Jo King

5:54 AM

wait, I think I might have it...

nvm

@ASCII-only Here's what I've got so far. ^x(?=(((?=(x+)\3$)\3)*x)(x*)$)x*$, where 2^n and k are captured in the lookahead and the rest of the number is free (at least, I think that's how this is working). I just can't get around the backtracking

ASCII-only

oh btw have you tried using regex101 to see what the engine is doing

Jo King

yeah, but I think I'm still confused

i realise that i'm not capturing 2^n :(

ASCII-only

6:28 AM

@JoKing :| ok so it turns out this is very different from matching a polynomial

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