The Nineteenth Byte

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!

nice

|| has lower precedence than *

@JohnDvorak Where should I use it?

instead of one of the ~~s

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

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

Huh

Sorry, I don't get it

Bitwise OR coerces to integers

9:29 AM

But it also changes the output, no?

E.g. 1.2|2 results in 3

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

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

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

ah just 0

dumb me

That's it!

9:32 AM

Thank you

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

@ComFreek which challenge are you solving?

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])`

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

well, would that be allowed on this site?

depends on the definition of any

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

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

Which challenges would you recommend beginners?

The top questions often have many answers already

@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)])

Does << or * bind stronger?

@ComFreek *

9:52 AM

how I even parse the last subexpression?

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

@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

why do you multiply by 5?

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

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

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

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

@ComFreek are you sure?

9:59 AM

I am not, I am rusty with bit fiddling

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

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

What does that mean?

@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)])

10:05 AM

oh right

@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.

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

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( )

nice

@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…

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.

@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.

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.