The Nineteenth Byte

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Λ̸̸

2:20 AM

Awesome! I cleared the latest questions queue.

If that's the case I'm posting my sandboxed question.

New Main Posts

Q: Is the input Bl lu ur rr ry?

This is based off this challenge. Given an input string, check whether the string is blurry. What's a blurry string? A blurry string is a string that's contained in the set of all blurrified pure-ASCII strings. Take a non-blurrified pure-ASCII string abc as an example. You repeat every charac...

Λ̸̸

2:59 AM

CGCC is playing alive mimicry...

Λ̸̸

3:10 AM

(It's a pun of death mimicry)

A: Is the input Bl lu ur rr ry?

Retina, 16 bytes ^.((.) \2)*.$|^$ Try it online!

6 hours later…

ComFreek

9:16 AM

In JS I got d=[] and then want to increment its values, e.g. d[3]++

But this doesn't work since d[3] is undefined, and incrementing that results in NaN

Hence, I now write d[j]=d[j]?d[j]+1:1. Any ideas to get that shorter?

Overall, I have c=[1,3,15,20,100,200];d=[];for(i=20;i--;)j=~~(Math.random()*6),d[j]=d[j]?d[j]+1:1;d.map((x,i,_,e=c[i])=>[e,e*x])

Ah, d[j=~~(Math.random()*6)]=d[j]?d[j]+1:1 is trivial optimization

ah d[j=~~(Math.random()*6)]=~~d[j]+1 does it, too!

John Dvorak

nice

|| has lower precedence than *

ComFreek

@JohnDvorak Where should I use it?

John Dvorak

instead of one of the ~~s

technically &&7 would also work in this case, I suppose

ComFreek

you mean j=Math.random()*6||?

Huh

Sorry, I don't get it

John Dvorak

Bitwise OR coerces to integers

ComFreek

9:29 AM

But it also changes the output, no?

E.g. 1.2|2 results in 3

John Dvorak

ah, sorry; |, not ||. That's even better.

ComFreek

e.g. Math.random()*6|1 never results in 0

John Dvorak

If you choose the right operand correctly, it will only coerce

ComFreek

ah just 0

dumb me

John Dvorak

That's it!

ComFreek

9:32 AM

Thank you

somehow I wonder if I can get rid of d=[];, too

ngn

@ComFreek which challenge are you solving?

ComFreek

None posted here

But here's the description: "Suppose you have coins with values 1, 3, 15, 20, 100, 200. Choose 20 such coins uniformly at random. Sort by coin and print the sums."

intuitition: e.g. output something like "coin with value 1 was chosen 8 times amounting to 1*8=8€, ..."

the program is allowed to output in any data structure as I do above

For reference, here's the current state:
`c=[1,3,15,20,100,200];d=[];for(i=20;i--;)d[j=Math.random()*6|0]=~~d[j]+1;d.map((x,i,_,e=c[i])=>[e,e*x])`

ngn

@ComFreek any data structure? :) what if you pack the bits

ComFreek

well, would that be allowed on this site?

ngn

depends on the definition of any

ComFreek

9:38 AM

The challenge is my own (to troll someone who asked "pls giv me teh code" on a forum)

so I'd be ready to accept any conventions this site has, too

ngn

i don't know if this has ever come up here

ComFreek

Which challenges would you recommend beginners?

The top questions often have many answers already

ngn

@ComFreek does this look correct? c=[1,3,15,20,100,200];b=0;for(i=20;i--;)b+=1<<5*(Math.random()*6|0);c.map((e,i)=>[e,e*(31&b>>5*i)])

ComFreek

Does << or * bind stronger?

ngn

@ComFreek *

ComFreek

9:52 AM

how I even parse the last subexpression?

31 & (b >> (5*i))?

ngn

@ComFreek b encodes what was previously d[0],d[1].. - bits 0 to 4 are reserved for d[0], bits 5 to 7 for d[1], etc. this expression extracts d[i] from b

ComFreek

why do you multiply by 5?

ngn

(bits are counted from the right of the binary representation)

@ComFreek because the bits for d[i] start from bit number i*5

ComFreek

but then I'd have done (Math.random()*6|0) << (5*i)

i.e. I don't see why they should be semantically equivalent

ngn

@ComFreek are you sure?

ComFreek

9:59 AM

I am not, I am rusty with bit fiddling

ngn

@ComFreek in that expression i is between 0 and 20, it's not an index in d

ComFreek

ngn

@ComFreek btw, you can inline c. it's used only once.

ComFreek

What does that mean?

ngn

@ComFreek b=0;for(i=20;i--;)b+=1<<5*(Math.random()*6|0);[1,3,15,20,100,200].map((e,i)=>[e,e*(31&b>>5*i)])

ComFreek

10:05 AM

oh right

ngn

@ComFreek i forgot to reply to this one.. yes - that's how it parses

Neil

10:17 AM

@ComFreek I think you could use -~d[j] instead of ~~d[j]+1.

ComFreek

wow

wait

shouldn't that be equivalent to -1?

Neil

CMC: Given a value and the exponents of g, cm and s, convert the value to SI units (kg, m, s). Examples: [10, 1, 1, -2] -> 1; [1, 1, -3, 0] -> 1000

@ComFreek see codegolf.stackexchange.com/a/7123/17602

ngn

10:36 AM

@Neil did you mean [10, 1, -1, -2] for the first example?

@ComFreek .map((e,i)=>[e,e*(31&b>>5*i)]) -> .map(e=>[e,e*(31&b>>5*++i)]) because we know i is -1 before the .map( )

ComFreek

nice

ngn

@Neil apl, 18 bytes: ⊃×.001 .01 1×.*1↓⊢

2 hours later…

Neil

12:11 PM

@ngn no, I meant [100000, 1, 1, -2], thanks for pointing out my mistake

5 hours later…

Adám

5:37 PM

@ngn Why not take the first element as left argument? ⊣×.001 .01 1×.*⊢ or even simply ×/⊣,.001 .01 1*⊢

Mats Granvik

6:01 PM

Hi! Is this a site where one can ask questions about how to write more efficient versions of a certain program? I have a code in Mathematica that computes 10000 terms of a sequence that appears that appears to be asymptotic to a square root, Sqrt(8*x). Already seeing a plot of the output would be nice. But I suspect that I need to go at least to x=10^6 to see any possible deviations.

Adám

@MatsGranvik It is on-topic, yes, even though the main focus is saving storage space for the program, not saving CPU resources.

Mats Granvik

ok. Thanks.

Adám

So in light of that, and greatly reduced traffic here, it might take a while to get an answer, if you get any at all. Have you considered posting on CodeReview.SE?

Mats Granvik

The name code review sounds familiar.

Adám

ngn

7:03 PM

@Adám if we can split it like that, sure, it's simpler. also: ×/⎕,⎕÷1e3 100 1

oops.. i misspoke. that's not multiplication.

4 hours later…

New Sandboxed Posts

11:23 PM

A: Sandbox for Proposed Challenges

Length of Longest Non-Repeating Substring code-golf string Your task is to take as input a single string (or list of characters, list of code points, etc.) and return the length of its longest non-repeating substring. Aside: This challenge is similar to Longest Non-Repeating Substring, but wit...