

Given:

 A starting, positive integer (greater than one), N.
 A selection of possible integer perfect divisors, D.
 And a selection of possible subtractors, S.
The goal is find the minimum number of steps necessary to reduce N down to one.

At any step, the number may be:

 Divided by any member of D if it is perfectly divided by D, (remainder zero).
 OR have one of S subtracted from it, if N is greater than the member of S.


There may be many ways to reduce the initial N down to 1. Your program needs to:

 Find the minimum number of steps to reach 1.
 Show one way of getting fron N to 1 in those minimum steps.


Examples

No divisors, D. a single subtractor of 1.

Obviousely N will take N-1 subtractions of 1 to reach 1
Single divisor of 2; single subtractor of 1:

N = 7 Takes 4 steps N -1=> 6, /2=> 3, -1=> 2, /2=> 1
N = 23 Takes 7 steps N -1=>22, /2=>11, -1=>10, /2=> 5, -1=> 4, /2=> 2, /2=> 1
Divisors 2 and 3; subtractor 1:

N = 11 Takes 4 steps N -1=>10, -1=> 9, /3=> 3, /3=> 1
Task

Using the possible divisors D, of 2 and 3; together with a possible subtractor S, of 1:

1. Show the number of steps and possible way of diminishing the numbers 1 to 10 down to 1.
2. Show a count of, and the numbers that: have the maximum minimal_steps_to_1, in the range 1 to 2,000.
Using the possible divisors D, of 2 and 3; together with a possible subtractor S, of 2:

3. Show the number of steps and possible way of diminishing the numbers 1 to 10 down to 1.
4. Show a count of, and the numbers that: have the maximum minimal_steps_to_1, in the range 1 to 2,000.


Optional stretch goal

2a, and 4a: As in 2 and 4 above, but for N in the range 1 to 20_000


Reference

 Learn Dynamic Programming (Memoization & Tabulation) Video of similar task.
step=: {{
  ~.((#~ 1<:]),y-/m),(#~ (=<.)),y%/n
}}
 
steps=: {{
  m step n^:(1 - 1 e. ])^:a:
}}
 
show=: {{
  paths=.,:,:0 0 1   NB. operator, operand, net value
  m=.,m [ n=.,n      NB. m: subtractors, n: divisors
  for_ok.}.|.m steps n y do.  NB. ok: valid net values
    last=.{."2 paths  
    subs=. (1,.m,.0)+"2]0 0 1*"1 last+"1 0/m
    divs=. (2,.n,.0)+"2]0 0 1*"1 last*"1 0/n
    prev=. subs,"2 divs   NB. we are working backwards from 1
    paths=. (,({:"1 prev)e.ok)#,/prev,"1 2/"2 paths
  end.
  ;@((<":y),,)"2((' -/'{~{.);":@{:)"1}:"2}:"1 paths
}}
 