There are pen and paper approaches, but assuming a brute force, and function
that answers basic first part of counting 1s in a number
c1 =: 10&(1 +/@:= #.inv)"0
and if you almost always want the sum from a list of arguments then
c2 =: +/@:c1
a brute force tacit iterative way of answering your final question is by
tracking 2 numbers (next value to count 1s, cummulative sum so far)
5 ( (0 >:@{ ]) , (1 { ]) + 0 c1@>:@{ ]) 10 2
11 4
Note the function is ambivalent while ignoring the right parameter. It returns
the pair of "last" number in presumptive sequence along with the cummulative
sum of 1s in that sequence.
to iterate until the sum will match a provided right argument:
203 ( (0 >:@{ ]) , (1 { ]) + 0 c1@>:@{ ])^:([ > 1 {:: ])(^:_) 1 1
511 204
keeps the pair number. That might be useful to know that there is no sequence
that has an exact 203 cummulative sum, but otherwise a function can lopp off
that extra information:
203 ( (0 >:@{ ]) , (1 { ]) + 0 c1@>:@{ ])^:([ > 1 {:: ])(^:_)({.@) 1 1
511
On Monday, January 17, 2022, 02:19:38 p.m. EST, Skip Cave
<s...@caveconsulting.com> wrote:
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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