I should have removed from 'seno' the line that says:
assert. hi<1000
The purpose of that line was to save me from having to use jbrk if I
hit an infinite loop from getting my conditions backwards. It's not a
useful mechanism, and should have been removed before I posted the
implementation.
FYI,
--
Raul
On Tue, Jan 18, 2022 at 2:27 AM Hauke Rehr <hauke.r...@uni-jena.de> wrote:
>
> I didn’t do it much differently, conceptually.
> But instead of binary search (bisection)
> I split it up in ten buckets (dekasect).
> Apologies for the bad naming:
> another greek-latin hybrid, oh my …
>
> I split that way because I wanted to evaluate
> ones more often on small inputs than large.
> Didn’t measure if it atcually helps, though.
> And maybe I should have added an M. in order
> to memoize?
>
> But seno definitively reads better.
>
> Am 18.01.22 um 02:54 schrieb Raul Miller:
> > Your 'ones' is a bit more efficient than the routine I had come up
> > with, so I'll stick with your implementation here.
> >
> > Meanwhile, it's perhaps worth noting
> >
> > I.213=ones i.1000
> > 521 522 523 524 525 526 527 528 529 530
> >
> > So, building the inverse... it's typical for there to be ten numbers
> > who had n "one digits in the sequence". But, for example, there's no
> > number which has exactly 3 "one digits in the sequence". Ten has 2,
> > eleven has four. But we can find the minimum value which had at least
> > n preceding one digits.
> >
> > Personally, I'd use a binary search here:
> >
> > seno=:{{
> > if.y=ones y do.y return.end.hi=.2*lo=.y
> > while.y>ones hi do. hi=.2*lo=.hi end.
> > while.hi>lo do.
> > assert. hi<1000
> > select.*y-ones mid=.<.-:lo+hi+1
> > case._1 do.if.hi=mid do.lo=.hi break.end.hi=.mid
> > case.0 do.lo=.hi=.mid break.
> > case.1 do.lo=.mid
> > end.
> > end.
> > while.y <: ones lo do.lo=.<:hi=.lo end. NB. maybe too crude...
> > hi
> > }}
> >
> > seno 213
> > 521
> > ones 521
> > 213
> >
> > FYI,
> >
>
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