This is a working solution. Not pretty, but it gets the job done.
onesList =: (0<{.)`($:@}. + ((10 ^ <:@#) * 1<{.) + ((1 + 10 #. }.) *
1={.) + <:@# * {. * 10 ^ 2 -~ #)@.(1<#)
ones =: 10 onesList@:(#.^:_1)"0 ]
NB. tests
ones 30 521
13 213
That was easy. Now for the “inverse:”
dlog =: 10&^.
flog =: <.&.dlog
NB. no $: inside of DDs
dekasect =: {{
0 dekasect y;0;0;(* flog@<.@dlog) flog y
:
'N offset start step' =: y
if. step < 1 do. offset return. end.
stips =. start + steps =. step * i. 11
stops =. (x * steps) + ones stips
totals =. stops + ones offset - start
idx =. stops I. N
if. idx < # steps do.
if. N = idx { totals do. idx { stips return. end.
end.
(x + idx = 2) dekasect N; ((;~offset&+) (<:idx) { steps), < step % 10
}}
NB. test
dekasect 213
530
It doesn’t return the smallest/largest one but the earliest one found.
That’s been fun.
Am 17.01.22 um 20:18 schrieb Skip Cave:
If I want to find the number of '1' digits in an integer, I can design a
verb c1, (count ones) to do that:
*c1=.{{+/1=10#.^:_1]y}}"0*
Test it:
* c1 10*
*1*
* c1 11*
*2*
* c1 103416*
*2*
* c1 56161764121*
*4*
I can also find the total number of '1' digits in a list of sequential
integers from 1 to n:
* to=:[+i.@:>:@-~ *
* +/c1 1 to n=.30*
*13*
* +/c1 1 to n=.521*
*213*
My question is: Given the total number of 1s in a sequence from 1 to n,
what is n?
A verb F(x) should return the final integer n, in the sequential list 1 to
n, that contains x ones.
Using the previous results as an example:
* F 213 *NB. A sequence 1 to n has 213 ones. What is n?
*521 *NB. The sequence 1 to 521 has 213 ones
What are some options to design the verb F?
1) Explicit
2) Implicit
3) Iteration
4) Recursion
Skip Cave
Cave Consulting LLC
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