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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