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Minimal steps down to 1

From Rosetta Code
Task
Minimal steps down to 1
You are encouraged to solve this task according to the task description, using any language you may know.


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

C#[edit]

Works with: C sharp version 7
using System;
using System.Collections.Generic;
using System.Linq;
 
public static class MinimalSteps
{
public static void Main() {
var (divisors, subtractors) = (new int[] { 2, 3 }, new [] { 1 });
var lookup = CreateLookup(2_000, divisors, subtractors);
Console.WriteLine($"Divisors: [{divisors.Delimit()}], Subtractors: [{subtractors.Delimit()}]");
PrintRange(lookup, 10);
PrintMaxMins(lookup);
lookup = CreateLookup(20_000, divisors, subtractors);
PrintMaxMins(lookup);
Console.WriteLine();
 
subtractors = new [] { 2 };
lookup = CreateLookup(2_000, divisors, subtractors);
Console.WriteLine($"Divisors: [{divisors.Delimit()}], Subtractors: [{subtractors.Delimit()}]");
PrintRange(lookup, 10);
PrintMaxMins(lookup);
lookup = CreateLookup(20_000, divisors, subtractors);
PrintMaxMins(lookup);
}
 
private static void PrintRange((char op, int param, int steps)[] lookup, int limit) {
for (int goal = 1; goal <= limit; goal++) {
var x = lookup[goal];
if (x.param == 0) {
Console.WriteLine($"{goal} cannot be reached with these numbers.");
continue;
}
Console.Write($"{goal} takes {x.steps} {(x.steps == 1 ? "step" : "steps")}: ");
for (int n = goal; n > 1; ) {
Console.Write($"{n},{x.op}{x.param}=> ");
n = x.op == '/' ? n / x.param : n - x.param;
x = lookup[n];
}
Console.WriteLine("1");
}
}
 
private static void PrintMaxMins((char op, int param, int steps)[] lookup) {
var maxSteps = lookup.Max(x => x.steps);
var items = lookup.Select((x, i) => (i, x)).Where(t => t.x.steps == maxSteps).ToList();
Console.WriteLine(items.Count == 1
? $"There is one number below {lookup.Length-1} that requires {maxSteps} steps: {items[0].i}"
: $"There are {items.Count} numbers below {lookup.Length-1} that require {maxSteps} steps: {items.Select(t => t.i).Delimit()}"
);
}
 
private static (char op, int param, int steps)[] CreateLookup(int goal, int[] divisors, int[] subtractors)
{
var lookup = new (char op, int param, int steps)[goal+1];
lookup[1] = ('/', 1, 0);
for (int n = 1; n < lookup.Length; n++) {
var ln = lookup[n];
if (ln.param == 0) continue;
for (int d = 0; d < divisors.Length; d++) {
int target = n * divisors[d];
if (target > goal) break;
if (lookup[target].steps == 0 || lookup[target].steps > ln.steps) lookup[target] = ('/', divisors[d], ln.steps + 1);
}
for (int s = 0; s < subtractors.Length; s++) {
int target = n + subtractors[s];
if (target > goal) break;
if (lookup[target].steps == 0 || lookup[target].steps > ln.steps) lookup[target] = ('-', subtractors[s], ln.steps + 1);
}
}
return lookup;
}
 
private static string Delimit<T>(this IEnumerable<T> source) => string.Join(", ", source);
}
Output:
Divisors: [2, 3], Subtractors: [1]
1 takes 0 steps: 1
2 takes 1 step: 2,-1=> 1
3 takes 1 step: 3,/3=> 1
4 takes 2 steps: 4,-1=> 3,/3=> 1
5 takes 3 steps: 5,-1=> 4,-1=> 3,/3=> 1
6 takes 2 steps: 6,/2=> 3,/3=> 1
7 takes 3 steps: 7,-1=> 6,/2=> 3,/3=> 1
8 takes 3 steps: 8,/2=> 4,-1=> 3,/3=> 1
9 takes 2 steps: 9,/3=> 3,/3=> 1
10 takes 3 steps: 10,-1=> 9,/3=> 3,/3=> 1
There are 16 numbers below 2000 that require 14 steps: 863, 1079, 1295, 1439, 1511, 1583, 1607, 1619, 1691, 1727, 1823, 1871, 1895, 1907, 1919, 1943
There are 5 numbers below 20000 that require 20 steps: 12959, 15551, 17279, 18143, 19439

Divisors: [2, 3], Subtractors: [2]
1 takes 0 steps: 1
2 takes 1 step: 2,/2=> 1
3 takes 1 step: 3,-2=> 1
4 takes 2 steps: 4,-2=> 2,/2=> 1
5 takes 2 steps: 5,-2=> 3,-2=> 1
6 takes 2 steps: 6,/2=> 3,-2=> 1
7 takes 3 steps: 7,-2=> 5,-2=> 3,-2=> 1
8 takes 3 steps: 8,-2=> 6,/2=> 3,-2=> 1
9 takes 2 steps: 9,/3=> 3,-2=> 1
10 takes 3 steps: 10,/2=> 5,-2=> 3,-2=> 1
There is one number below 2000 that requires 17 steps: 1699
There is one number below 20000 that requires 24 steps: 19681

Go[edit]

package main
 
import (
"fmt"
"strings"
)
 
const limit = 50000
 
var (
divs, subs []int
mins [][]string
)
 
// Assumes the numbers are presented in order up to 'limit'.
func minsteps(n int) {
if n == 1 {
mins[1] = []string{}
return
}
min := limit
var p, q int
var op byte
for _, div := range divs {
if n%div == 0 {
d := n / div
steps := len(mins[d]) + 1
if steps < min {
min = steps
p, q, op = d, div, '/'
}
}
}
for _, sub := range subs {
if d := n - sub; d >= 1 {
steps := len(mins[d]) + 1
if steps < min {
min = steps
p, q, op = d, sub, '-'
}
}
}
mins[n] = append(mins[n], fmt.Sprintf("%c%d -> %d", op, q, p))
mins[n] = append(mins[n], mins[p]...)
}
 
func main() {
for r := 0; r < 2; r++ {
divs = []int{2, 3}
if r == 0 {
subs = []int{1}
} else {
subs = []int{2}
}
mins = make([][]string, limit+1)
fmt.Printf("With: Divisors: %v, Subtractors: %v =>\n", divs, subs)
fmt.Println(" Minimum number of steps to diminish the following numbers down to 1 is:")
for i := 1; i <= limit; i++ {
minsteps(i)
if i <= 10 {
steps := len(mins[i])
plural := "s"
if steps == 1 {
plural = " "
}
fmt.Printf("  %2d: %d step%s: %s\n", i, steps, plural, strings.Join(mins[i], ", "))
}
}
for _, lim := range []int{2000, 20000, 50000} {
max := 0
for _, min := range mins[0 : lim+1] {
m := len(min)
if m > max {
max = m
}
}
var maxs []int
for i, min := range mins[0 : lim+1] {
if len(min) == max {
maxs = append(maxs, i)
}
}
nums := len(maxs)
verb, verb2, plural := "are", "have", "s"
if nums == 1 {
verb, verb2, plural = "is", "has", ""
}
fmt.Printf(" There %s %d number%s in the range 1-%d ", verb, nums, plural, lim)
fmt.Printf("that %s maximum 'minimal steps' of %d:\n", verb2, max)
fmt.Println(" ", maxs)
}
fmt.Println()
}
}
Output:
With: Divisors: [2 3], Subtractors: [1] =>
  Minimum number of steps to diminish the following numbers down to 1 is:
     1: 0 steps: 
     2: 1 step : /2 -> 1
     3: 1 step : /3 -> 1
     4: 2 steps: /2 -> 2, /2 -> 1
     5: 3 steps: -1 -> 4, /2 -> 2, /2 -> 1
     6: 2 steps: /2 -> 3, /3 -> 1
     7: 3 steps: -1 -> 6, /2 -> 3, /3 -> 1
     8: 3 steps: /2 -> 4, /2 -> 2, /2 -> 1
     9: 2 steps: /3 -> 3, /3 -> 1
    10: 3 steps: -1 -> 9, /3 -> 3, /3 -> 1
  There are 16 numbers in the range 1-2000 that have maximum 'minimal steps' of 14:
    [863 1079 1295 1439 1511 1583 1607 1619 1691 1727 1823 1871 1895 1907 1919 1943]
  There are 5 numbers in the range 1-20000 that have maximum 'minimal steps' of 20:
    [12959 15551 17279 18143 19439]
  There are 3 numbers in the range 1-50000 that have maximum 'minimal steps' of 22:
    [25919 31103 38879]

With: Divisors: [2 3], Subtractors: [2] =>
  Minimum number of steps to diminish the following numbers down to 1 is:
     1: 0 steps: 
     2: 1 step : /2 -> 1
     3: 1 step : /3 -> 1
     4: 2 steps: /2 -> 2, /2 -> 1
     5: 2 steps: -2 -> 3, /3 -> 1
     6: 2 steps: /2 -> 3, /3 -> 1
     7: 3 steps: -2 -> 5, -2 -> 3, /3 -> 1
     8: 3 steps: /2 -> 4, /2 -> 2, /2 -> 1
     9: 2 steps: /3 -> 3, /3 -> 1
    10: 3 steps: /2 -> 5, -2 -> 3, /3 -> 1
  There is 1 number in the range 1-2000 that has maximum 'minimal steps' of 17:
    [1699]
  There is 1 number in the range 1-20000 that has maximum 'minimal steps' of 24:
    [19681]
  There is 1 number in the range 1-50000 that has maximum 'minimal steps' of 26:
    [45925]

Java[edit]

Algorithm works with any function that supports the Function interface in the code below.

 
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
 
public class MinimalStepsDownToOne {
 
public static void main(String[] args) {
runTasks(getFunctions1());
runTasks(getFunctions2());
runTasks(getFunctions3());
}
 
private static void runTasks(List<Function> functions) {
Map<Integer,List<String>> minPath = getInitialMap(functions, 5);
 
// Task 1
int max = 10;
populateMap(minPath, functions, max);
System.out.printf("%nWith functions:  %s%n", functions);
System.out.printf(" Minimum steps to 1:%n");
for ( int n = 2 ; n <= max ; n++ ) {
int steps = minPath.get(n).size();
System.out.printf("  %2d: %d step%1s: %s%n", n, steps, steps == 1 ? "" : "s", minPath.get(n));
}
 
// Task 2
displayMaxMin(minPath, functions, 2000);
 
// Task 2a
displayMaxMin(minPath, functions, 20000);
 
// Task 2a +
displayMaxMin(minPath, functions, 100000);
}
 
private static void displayMaxMin(Map<Integer,List<String>> minPath, List<Function> functions, int max) {
populateMap(minPath, functions, max);
List<Integer> maxIntegers = getMaxMin(minPath, max);
int maxSteps = maxIntegers.remove(0);
int numCount = maxIntegers.size();
System.out.printf(" There %s %d number%s in the range 1-%d that have maximum 'minimal steps' of %d:%n  %s%n", numCount == 1 ? "is" : "are", numCount, numCount == 1 ? "" : "s", max, maxSteps, maxIntegers);
 
}
 
private static List<Integer> getMaxMin(Map<Integer,List<String>> minPath, int max) {
int maxSteps = Integer.MIN_VALUE;
List<Integer> maxIntegers = new ArrayList<Integer>();
for ( int n = 2 ; n <= max ; n++ ) {
int len = minPath.get(n).size();
if ( len > maxSteps ) {
maxSteps = len;
maxIntegers.clear();
maxIntegers.add(n);
}
else if ( len == maxSteps ) {
maxIntegers.add(n);
}
}
maxIntegers.add(0, maxSteps);
return maxIntegers;
}
 
private static void populateMap(Map<Integer,List<String>> minPath, List<Function> functions, int max) {
for ( int n = 2 ; n <= max ; n++ ) {
if ( minPath.containsKey(n) ) {
continue;
}
Function minFunction = null;
int minSteps = Integer.MAX_VALUE;
for ( Function f : functions ) {
if ( f.actionOk(n) ) {
int result = f.action(n);
int steps = 1 + minPath.get(result).size();
if ( steps < minSteps ) {
minFunction = f;
minSteps = steps;
}
}
}
int result = minFunction.action(n);
List<String> path = new ArrayList<String>();
path.add(minFunction.toString(n));
path.addAll(minPath.get(result));
minPath.put(n, path);
}
 
}
 
private static Map<Integer,List<String>> getInitialMap(List<Function> functions, int max) {
Map<Integer,List<String>> minPath = new HashMap<>();
for ( int i = 2 ; i <= max ; i++ ) {
for ( Function f : functions ) {
if ( f.actionOk(i) ) {
int result = f.action(i);
if ( result == 1 ) {
List<String> path = new ArrayList<String>();
path.add(f.toString(i));
minPath.put(i, path);
}
}
}
}
return minPath;
}
 
private static List<Function> getFunctions3() {
List<Function> functions = new ArrayList<>();
functions.add(new Divide2Function());
functions.add(new Divide3Function());
functions.add(new Subtract2Function());
functions.add(new Subtract1Function());
return functions;
}
 
private static List<Function> getFunctions2() {
List<Function> functions = new ArrayList<>();
functions.add(new Divide3Function());
functions.add(new Divide2Function());
functions.add(new Subtract2Function());
return functions;
}
 
private static List<Function> getFunctions1() {
List<Function> functions = new ArrayList<>();
functions.add(new Divide3Function());
functions.add(new Divide2Function());
functions.add(new Subtract1Function());
return functions;
}
 
public abstract static class Function {
abstract public int action(int n);
abstract public boolean actionOk(int n);
abstract public String toString(int n);
}
 
public static class Divide2Function extends Function {
@Override public int action(int n) {
return n/2;
}
 
@Override public boolean actionOk(int n) {
return n % 2 == 0;
}
 
@Override public String toString(int n) {
return "/2 -> " + n/2;
}
 
@Override public String toString() {
return "Divisor 2";
}
 
}
 
public static class Divide3Function extends Function {
@Override public int action(int n) {
return n/3;
}
 
@Override public boolean actionOk(int n) {
return n % 3 == 0;
}
 
@Override public String toString(int n) {
return "/3 -> " + n/3;
}
 
@Override public String toString() {
return "Divisor 3";
}
 
}
 
public static class Subtract1Function extends Function {
@Override public int action(int n) {
return n-1;
}
 
@Override public boolean actionOk(int n) {
return true;
}
 
@Override public String toString(int n) {
return "-1 -> " + (n-1);
}
 
@Override public String toString() {
return "Subtractor 1";
}
 
}
 
public static class Subtract2Function extends Function {
@Override public int action(int n) {
return n-2;
}
 
@Override public boolean actionOk(int n) {
return n > 2;
}
 
@Override public String toString(int n) {
return "-2 -> " + (n-2);
}
 
@Override public String toString() {
return "Subtractor 2";
}
 
}
 
}
 
Output:
With functions:  [Divisor 3, Divisor 2, Subtractor 1]
  Minimum steps to 1:
     2: 1 step : [-1 -> 1]
     3: 1 step : [/3 -> 1]
     4: 2 steps: [/2 -> 2, -1 -> 1]
     5: 3 steps: [-1 -> 4, /2 -> 2, -1 -> 1]
     6: 2 steps: [/3 -> 2, -1 -> 1]
     7: 3 steps: [-1 -> 6, /3 -> 2, -1 -> 1]
     8: 3 steps: [/2 -> 4, /2 -> 2, -1 -> 1]
     9: 2 steps: [/3 -> 3, /3 -> 1]
    10: 3 steps: [-1 -> 9, /3 -> 3, /3 -> 1]
  There are 16 numbers in the range 1-2000 that have maximum 'minimal steps' of 14:
    [863, 1079, 1295, 1439, 1511, 1583, 1607, 1619, 1691, 1727, 1823, 1871, 1895, 1907, 1919, 1943]
  There are 5 numbers in the range 1-20000 that have maximum 'minimal steps' of 20:
    [12959, 15551, 17279, 18143, 19439]
  There is 1 number in the range 1-100000 that have maximum 'minimal steps' of 24:
    [77759]

With functions:  [Divisor 3, Divisor 2, Subtractor 2]
  Minimum steps to 1:
     2: 1 step : [/2 -> 1]
     3: 1 step : [-2 -> 1]
     4: 2 steps: [/2 -> 2, /2 -> 1]
     5: 2 steps: [-2 -> 3, -2 -> 1]
     6: 2 steps: [/3 -> 2, /2 -> 1]
     7: 3 steps: [-2 -> 5, -2 -> 3, -2 -> 1]
     8: 3 steps: [/2 -> 4, /2 -> 2, /2 -> 1]
     9: 2 steps: [/3 -> 3, -2 -> 1]
    10: 3 steps: [/2 -> 5, -2 -> 3, -2 -> 1]
  There is 1 number in the range 1-2000 that have maximum 'minimal steps' of 17:
    [1699]
  There is 1 number in the range 1-20000 that have maximum 'minimal steps' of 24:
    [19681]
  There is 1 number in the range 1-100000 that have maximum 'minimal steps' of 28:
    [98413]

With functions:  [Divisor 2, Divisor 3, Subtractor 2, Subtractor 1]
  Minimum steps to 1:
     2: 1 step : [-1 -> 1]
     3: 1 step : [-2 -> 1]
     4: 2 steps: [/2 -> 2, -1 -> 1]
     5: 2 steps: [-2 -> 3, -2 -> 1]
     6: 2 steps: [/2 -> 3, -2 -> 1]
     7: 3 steps: [-2 -> 5, -2 -> 3, -2 -> 1]
     8: 3 steps: [/2 -> 4, /2 -> 2, -1 -> 1]
     9: 2 steps: [/3 -> 3, -2 -> 1]
    10: 3 steps: [/2 -> 5, -2 -> 3, -2 -> 1]
  There is 1 number in the range 1-2000 that have maximum 'minimal steps' of 13:
    [1943]
  There are 2 numbers in the range 1-20000 that have maximum 'minimal steps' of 17:
    [17495, 19439]
  There are 22 numbers in the range 1-100000 that have maximum 'minimal steps' of 19:
    [58319, 69983, 76463, 77759, 78623, 87479, 89423, 90071, 90287, 90359, 90383, 90395, 91259, 91355, 91367, 93311, 95255, 95471, 95903, 96119, 96191, 98171]

Julia[edit]

This is a non-recursive solution that is memoized for the count-only portions of the task, but not memoized when an example is to be listed. Interestingly, for the specified tasks only the starting points of 2 and 3 are processed without use of memoization.

Implemented as a generic solution for any functions acting on an integer and taking any range of second arguments, with the goal solution also specifiable. To do so generically, it is also necessary to specify a failure condition, which in the example is falling below 1.

import Base.print
 
struct Action{T}
f::Function
i::T
end
 
struct ActionOutcome{T}
act::Action{T}
out::T
end
 
Base.print(io::IO, ao::ActionOutcome) = print(io, "$(ao.act.f) $(ao.act.i) yields $(ao.out)")
 
memoized = Dict{Int, Int}()
 
function findshortest(start, goal, fails, actions, verbose=true, maxsteps=100000)
solutions, numsteps = Vector{Vector{ActionOutcome}}(), 0
seqs = [ActionOutcome[ActionOutcome(Action(div, 0), start)]]
if start == goal
verbose && println("For start of $start, no steps needed.\n")
return 0
end
while numsteps < maxsteps && isempty(solutions)
newsequences = Vector{Vector{ActionOutcome}}()
numsteps += 1
for seq in seqs
for (act, arr) in actions, x in arr
result = act(seq[end].out, x)
if !fails(result)
newactionseq = vcat(seq, ActionOutcome(Action(act, x), result))
numsteps == 1 && popfirst!(newactionseq)
if result == goal
push!(solutions, newactionseq)
else
push!(newsequences, newactionseq)
end
end
end
if !verbose && isempty(solutions) &&
all(x -> haskey(memoized, x[end].out), newsequences)
minresult = minimum(x -> memoized[x[end].out], newsequences) + numsteps
memoized[start] = minresult
return minresult
end
end
seqs = newsequences
end
if verbose
l = length(solutions)
print("There ", l > 1 ? "are $l solutions" : "is $l solution",
" for path of length ", numsteps, " from $start to $goal.\nExample: ")
for step in solutions[1]
print(step, step.out == 1 ? "\n\n" : ", ")
end
end
memoized[start] = numsteps
return numsteps
end
 
failed(n) = n < 1
 
const divisors = [2, 3]
divide(n, x) = begin q, r = divrem(n, x); r == 0 ? q : -1 end
 
const subtractors1, subtractors2 = [1], [2]
subtract(n, x) = n - x
 
actions1 = Dict(divide => divisors, subtract => subtractors1)
actions2 = Dict(divide => divisors, subtract => subtractors2)
 
function findmaxshortest(g, fails, acts, maxn)
stepcounts = [findshortest(n, g, fails, acts, false) for n in 1:maxn]
maxs = maximum(stepcounts)
maxstepnums = findall(x -> x == maxs, stepcounts)
println("There are $(length(maxstepnums)) with $maxs steps for start between 1 and $maxn: ", maxstepnums)
end
 
function teststeps(g, fails, acts, maxes)
println("\nWith goal $g, divisors $(acts[divide]), subtractors $(acts[subtract]):")
for n in 1:10
findshortest(n, g, fails, acts)
end
for maxn in maxes
findmaxshortest(g, fails, acts, maxn)
end
end
 
teststeps(1, failed, actions1, [2000, 20000, 50000])
empty!(memoized)
teststeps(1, failed, actions2, [2000, 20000, 50000])
 
Output:
With goal 1, divisors [2, 3], subtractors [1]:
For start of 1, no steps needed.

There are 2 solutions for path of length 1 from 2 to 1.
Example: divide 2 yields 1

There is 1 solution for path of length 1 from 3 to 1.
Example: divide 3 yields 1

There are 3 solutions for path of length 2 from 4 to 1.
Example: divide 2 yields 2, divide 2 yields 1

There are 3 solutions for path of length 3 from 5 to 1.
Example: subtract 1 yields 4, divide 2 yields 2, divide 2 yields 1

There are 3 solutions for path of length 2 from 6 to 1.
Example: divide 2 yields 3, divide 3 yields 1

There are 3 solutions for path of length 3 from 7 to 1.
Example: subtract 1 yields 6, divide 2 yields 3, divide 3 yields 1

There are 3 solutions for path of length 3 from 8 to 1.
Example: divide 2 yields 4, divide 2 yields 2, divide 2 yields 1

There is 1 solution for path of length 2 from 9 to 1.
Example: divide 3 yields 3, divide 3 yields 1

There is 1 solution for path of length 3 from 10 to 1.
Example: subtract 1 yields 9, divide 3 yields 3, divide 3 yields 1

There are 16 with 14 steps for start between 1 and 2000: [863, 1079, 1295, 1439, 1511, 1583, 1607, 1619, 1691, 1727, 1823, 1871, 1895, 1907, 1919, 1943]
There are 5 with 20 steps for start between 1 and 20000: [12959, 15551, 17279, 18143, 19439]
There are 3 with 22 steps for start between 1 and 50000: [25919, 31103, 38879]

With goal 1, divisors [2, 3], subtractors [2]:
For start of 1, no steps needed.

There is 1 solution for path of length 1 from 2 to 1.
Example: divide 2 yields 1

There are 2 solutions for path of length 1 from 3 to 1.
Example: divide 3 yields 1

There are 2 solutions for path of length 2 from 4 to 1.
Example: divide 2 yields 2, divide 2 yields 1

There are 2 solutions for path of length 2 from 5 to 1.
Example: subtract 2 yields 3, divide 3 yields 1

There are 3 solutions for path of length 2 from 6 to 1.
Example: divide 2 yields 3, divide 3 yields 1

There are 2 solutions for path of length 3 from 7 to 1.
Example: subtract 2 yields 5, subtract 2 yields 3, divide 3 yields 1

There are 5 solutions for path of length 3 from 8 to 1.
Example: divide 2 yields 4, divide 2 yields 2, divide 2 yields 1

There are 2 solutions for path of length 2 from 9 to 1.
Example: divide 3 yields 3, divide 3 yields 1

There are 2 solutions for path of length 3 from 10 to 1.
Example: divide 2 yields 5, subtract 2 yields 3, divide 3 yields 1

There are 1 with 17 steps for start between 1 and 2000: [1699]
There are 1 with 24 steps for start between 1 and 20000: [19681]
There are 1 with 26 steps for start between 1 and 50000: [45925]

Mathematica/Wolfram Language[edit]

$RecursionLimit = 3000;
ClearAll[MinimalStepToOne, MinimalStepToOneHelper]
MinimalStepToOne[n_Integer] := Module[{res},
res = Reap[MinimalStepToOneHelper[{n}]][[-1, 1]];
SortBy[res, Length]
]
MinimalStepToOneHelper[steps_List] := Module[{n, out},
n = Last[steps];
If[n == 1,
Sow[steps];
,
If[Divisible[n, 3],
out = steps~Join~{" /3=> ", n/3};
MinimalStepToOneHelper[out]
];
If[Divisible[n, 2],
out = steps~Join~{" /2=> ", n/2};
MinimalStepToOneHelper[out]
];
If[n > 1,
out = steps~Join~{" -1=> ", n - 1};
MinimalStepToOneHelper[out]
];
]
]
Do[
sel = First[MinimalStepToOne[i]];
Print[First[sel],
": (" <> ToString[(Length[sel] - 1)/2] <> " steps) ", sel // Row]
,
{i, 1, 10}
]
 
$RecursionLimit = 3000;
ClearAll[MinimalStepToOne2, MinimalStepToOneHelper2]
MinimalStepToOne2[nn_Integer] :=
Module[{res, done, solution, maxsteps},
done = False;
solution = {};
MinimalStepToOneHelper2[steps_List] := Module[{n, out},
n = Last[steps];
If[n == 1,
solution = steps;
,
If[solution === {},
If[Divisible[n, 3],
out = steps~Join~{" /3=> ", n/3};
MinimalStepToOneHelper2[out]
];
If[Divisible[n, 2],
out = steps~Join~{" /2=> ", n/2};
MinimalStepToOneHelper2[out]
];
If[n > 1,
out = steps~Join~{" -1=> ", n - 1};
MinimalStepToOneHelper2[out]
];
];
];
];
MinimalStepToOneHelper2[{nn}];
maxsteps = (Length[solution] - 1)/2;
maxsteps
]
 
$RecursionLimit = 3000;
ClearAll[MinimalStepToOneMaxSteps, MinimalStepToOneMaxStepsHelper]
MinimalStepToOneMaxSteps[n_Integer, maxsteps_Integer] :=
Module[{res},
res = Reap[MinimalStepToOneMaxStepsHelper[{n}, maxsteps]][[-1, 1]];
(Min[Length /@ res] - 1)/2
]
MinimalStepToOneMaxStepsHelper[steps_List, maxsteps_Integer] :=
Module[{n, out},
n = Last[steps];
If[n == 1,
Sow[steps];
,
If[maxsteps > 0,
If[Divisible[n, 3],
out = steps~Join~{" /3=> ", n/3};
MinimalStepToOneMaxStepsHelper[out, maxsteps - 1]
];
If[Divisible[n, 2],
out = steps~Join~{" /2=> ", n/2};
MinimalStepToOneMaxStepsHelper[out, maxsteps - 1]
];
If[n > 1,
out = steps~Join~{" -1=> ", n - 1};
MinimalStepToOneMaxStepsHelper[out, maxsteps - 1]
];
];
];
]
 
allsols = Table[
max = MinimalStepToOne2[i];
{i, MinimalStepToOneMaxSteps[i, max]}
,
{i, 1, 2000}
];
 
a = Last[SortBy[GatherBy[allsols, Last], First /* Last]];
{a[[1, 2]], a[[All, 1]]}
 
$RecursionLimit = 3000;
ClearAll[MinimalStepToOne, MinimalStepToOneHelper]
MinimalStepToOne[n_Integer] := Module[{res},
res = Reap[MinimalStepToOneHelper[{n}]][[-1, 1]];
SortBy[res, Length]
]
MinimalStepToOneHelper[steps_List] := Module[{n, out},
n = Last[steps];
If[n == 1,
Sow[steps];
,
If[Divisible[n, 3],
out = steps~Join~{" /3=> ", n/3};
MinimalStepToOneHelper[out]
];
If[Divisible[n, 2],
out = steps~Join~{" /2=> ", n/2};
MinimalStepToOneHelper[out]
];
If[n > 2,
out = steps~Join~{" -2=> ", n - 2};
MinimalStepToOneHelper[out]
];
]
]
Do[
sel = First[MinimalStepToOne[i]];
Print[First[sel],
": (" <> ToString[(Length[sel] - 1)/2] <> " steps) ", sel // Row]
,
{i, 1, 10}
]
 
$RecursionLimit = 3000;
ClearAll[MinimalStepToOne2, MinimalStepToOneHelper2]
MinimalStepToOne2[nn_Integer] :=
Module[{res, done, solution, maxsteps},
done = False;
solution = {};
MinimalStepToOneHelper2[steps_List] := Module[{n, out},
n = Last[steps];
If[n == 1,
solution = steps;
,
If[solution === {},
If[Divisible[n, 3],
out = steps~Join~{" /3=> ", n/3};
MinimalStepToOneHelper2[out]
];
If[Divisible[n, 2],
out = steps~Join~{" /2=> ", n/2};
MinimalStepToOneHelper2[out]
];
If[n > 2,
out = steps~Join~{" -2=> ", n - 2};
MinimalStepToOneHelper2[out]
];
];
];
];
MinimalStepToOneHelper2[{nn}];
maxsteps = (Length[solution] - 1)/2;
maxsteps
]
 
$RecursionLimit = 3000;
ClearAll[MinimalStepToOneMaxSteps, MinimalStepToOneMaxStepsHelper]
MinimalStepToOneMaxSteps[n_Integer, maxsteps_Integer] :=
Module[{res},
res = Reap[MinimalStepToOneMaxStepsHelper[{n}, maxsteps]][[-1, 1]];
(Min[Length /@ res] - 1)/2
]
MinimalStepToOneMaxStepsHelper[steps_List, maxsteps_Integer] :=
Module[{n, out},
n = Last[steps];
If[n == 1,
Sow[steps];
,
If[maxsteps > 0,
If[Divisible[n, 3],
out = steps~Join~{" /3=> ", n/3};
MinimalStepToOneMaxStepsHelper[out, maxsteps - 1]
];
If[Divisible[n, 2],
out = steps~Join~{" /2=> ", n/2};
MinimalStepToOneMaxStepsHelper[out, maxsteps - 1]
];
If[n > 2,
out = steps~Join~{" -2=> ", n - 2};
MinimalStepToOneMaxStepsHelper[out, maxsteps - 1]
];
];
];
]
 
allsols = Table[
max = MinimalStepToOne2[i];
{i, MinimalStepToOneMaxSteps[i, max]}
,
{i, 1, 2000}
];
 
a = Last[SortBy[GatherBy[allsols, Last], First /* Last]];
{a[[1, 2]], a[[All, 1]]}
Output:
1: (0 steps) 1
2: (1 steps) 2 -1=> 1
3: (1 steps) 3 /3=> 1
4: (2 steps) 4 -1=> 3 /3=> 1
5: (3 steps) 5 -1=> 4 -1=> 3 /3=> 1
6: (2 steps) 6 /2=> 3 /3=> 1
7: (3 steps) 7 -1=> 6 /2=> 3 /3=> 1
8: (3 steps) 8 /2=> 4 -1=> 3 /3=> 1
9: (2 steps) 9 /3=> 3 /3=> 1
10: (3 steps) 10 -1=> 9 /3=> 3 /3=> 1
{14, {863, 1079, 1295, 1439, 1511, 1583, 1607, 1619, 1691, 1727, 1823, 1871, 1895, 1907, 1919, 1943}}

1: (0 steps) 1
2: (1 steps) 2 /2=> 1
3: (1 steps) 3 -2=> 1
4: (2 steps) 4 -2=> 2 /2=> 1
5: (2 steps) 5 -2=> 3 -2=> 1
6: (2 steps) 6 /2=> 3 -2=> 1
7: (3 steps) 7 -2=> 5 -2=> 3 -2=> 1
8: (3 steps) 8 -2=> 6 /2=> 3 -2=> 1
9: (2 steps) 9 /3=> 3 -2=> 1
10: (3 steps) 10 /2=> 5 -2=> 3 -2=> 1
{17, {1699}}

Nim[edit]

We use two recursive functions, the first one to find the minimal length sequences, the second one to find the minimal number of steps. For both, we use memoization. The program takes about 10 ms to execute.

import strformat, strutils, tables
 
type
 
Sequence = seq[Natural]
 
Context = object
divisors: seq[int]
subtractors: seq[int]
seqCache: Table[int, Sequence] # Mapping number -> sequence to reach 1.
stepCache: Table[int, int] # Mapping number -> number of steps to reach 1.
 
 
proc initContext(divisors, subtractors: openArray[int]): Context =
## Initialize a context.
for d in divisors: doAssert d > 1, "divisors must be greater than 1."
for s in subtractors: doAssert s > 0, "substractors must be greater than 0."
result.divisors = @divisors
result.subtractors = @subtractors
result.seqCache[1] = @[Natural 1]
result.stepCache[1] = 0
 
 
proc minStepsDown(context: var Context; n: Natural): Sequence =
# Return a minimal sequence to reach the value 1.
 
assert n > 0, "“n” must be positive."
if n in context.seqCache: return context.seqCache[n]
 
var
minSteps = int.high
minPath: Sequence
 
for val in context.divisors:
if n mod val == 0:
var path = context.minStepsDown(n div val)
if path.len < minSteps:
minSteps = path.len
minPath = move(path)
 
for val in context.subtractors:
if n - val > 0:
var path = context.minStepsDown(n - val)
if path.len < minSteps:
minSteps = path.len
minPath = move(path)
 
result = n & minPath
context.seqCache[n] = result
 
 
proc minStepsDownCount(context: var Context; n: Natural): int =
## Compute the mininum number of steps without keeping the sequence.
 
assert n > 0, "“n” must be positive."
if n in context.stepCache: return context.stepCache[n]
 
result = int.high
 
for val in context.divisors:
if n mod val == 0:
let steps = context.minStepsDownCount(n div val)
if steps < result: result = steps
 
for val in context.subtractors:
if n - val > 0:
let steps = context.minStepsDownCount(n - val)
if steps < result: result = steps
 
inc result
context.stepCache[n] = result
 
 
template plural(n: int): string =
if n > 1: "s" else: ""
 
proc printMinStepsDown(context: var Context; n: Natural) =
## Search and print the sequence to reach one.
 
let sol = context.minStepsDown(n)
stdout.write &"{n} takes {sol.len - 1} step{plural(sol.len - 1)}: "
var prev = 0
for val in sol:
if prev == 0:
stdout.write val
elif prev - val in context.subtractors:
stdout.write " - ", prev - val, " → ", val
else:
stdout.write " / ", prev div val, " → ", val
prev = val
stdout.write '\n'
 
 
proc maxMinStepsCount(context: var Context; nmax: Positive): tuple[steps: int; list: seq[int]] =
## Return the maximal number of steps needed for numbers between 1 and "nmax"
## and the list of numbers needing this number of steps.
 
for n in 2..nmax:
let nsteps = context.minStepsDownCount(n)
if nsteps == result.steps:
result.list.add n
elif nsteps > result.steps:
result.steps = nsteps
result.list = @[n]
 
 
proc run(divisors, subtractors: openArray[int]) =
## Run the search for given divisors and subtractors.
 
var context = initContext(divisors, subtractors)
echo &"Using divisors: {divisors} and substractors: {subtractors}"
for n in 1..10: context.printMinStepsDown(n)
for nmax in [2_000, 20_000]:
let (steps, list) = context.maxMinStepsCount(nmax)
stdout.write if list.len == 1: &"There is 1 number " else: &"There are {list.len} numbers "
echo &"below {nmax} that require {steps} steps: ", list.join(", ")
 
 
run(divisors = [2, 3], subtractors = [1])
echo ""
run(divisors = [2, 3], subtractors = [2])
Output:
Using divisors: [2, 3] and substractors: [1]
1 takes 0 step: 1
2 takes 1 step: 2 - 1 → 1
3 takes 1 step: 3 / 3 → 1
4 takes 2 steps: 4 / 2 → 2 - 1 → 1
5 takes 3 steps: 5 - 1 → 4 / 2 → 2 - 1 → 1
6 takes 2 steps: 6 / 2 → 3 / 3 → 1
7 takes 3 steps: 7 - 1 → 6 / 2 → 3 / 3 → 1
8 takes 3 steps: 8 / 2 → 4 / 2 → 2 - 1 → 1
9 takes 2 steps: 9 / 3 → 3 / 3 → 1
10 takes 3 steps: 10 - 1 → 9 / 3 → 3 / 3 → 1
There are 16 numbers below 2000 that require 14 steps: 863, 1079, 1295, 1439, 1511, 1583, 1607, 1619, 1691, 1727, 1823, 1871, 1895, 1907, 1919, 1943
There are 5 numbers below 20000 that require 20 steps: 12959, 15551, 17279, 18143, 19439

Using divisors: [2, 3] and substractors: [2]
1 takes 0 step: 1
2 takes 1 step: 2 / 2 → 1
3 takes 1 step: 3 - 2 → 1
4 takes 2 steps: 4 - 2 → 2 / 2 → 1
5 takes 2 steps: 5 - 2 → 3 - 2 → 1
6 takes 2 steps: 6 / 2 → 3 - 2 → 1
7 takes 3 steps: 7 - 2 → 5 - 2 → 3 - 2 → 1
8 takes 3 steps: 8 / 2 → 4 - 2 → 2 / 2 → 1
9 takes 2 steps: 9 / 3 → 3 - 2 → 1
10 takes 3 steps: 10 / 2 → 5 - 2 → 3 - 2 → 1
There is 1 number below 2000 that require 17 steps: 1699
There is 1 number below 20000 that require 24 steps: 19681

Perl[edit]

#!/usr/bin/perl
 
use strict; # https://rosettacode.org/wiki/Minimal_steps_down_to_1
use warnings;
no warnings 'recursion';
use List::Util qw( first );
use Data::Dump 'dd';
 
for ( [ 2000, [2, 3], [1] ], [ 2000, [2, 3], [2] ] )
{
my ( $n, $div, $sub ) = @$_;
print "\n", '-' x 40, " divisors @$div subtractors @$sub\n";
my ($solve, $max) = minimal( @$_ );
printf "%4d takes %s step(s): %s\n",
$_, $solve->[$_] =~ tr/ // - 1, $solve->[$_] for 1 .. 10;
print "\n";
printf "%d number(s) below %d that take $#$max steps: %s\n",
$max->[-1] =~ tr/ //, $n, $max->[-1];
($solve, $max) = minimal( 20000, $div, $sub );
printf "%d number(s) below %d that take $#$max steps: %s\n",
$max->[-1] =~ tr/ //, 20000, $max->[-1];
}
 
sub minimal
{
my ($top, $div, $sub) = @_;
my @solve = (0, ' ');
my @maximal;
for my $n ( 2 .. $top )
{
my @pick;
for my $d ( @$div )
{
$n % $d and next;
my $ans = "/$d $solve[$n / $d]";
$pick[$ans =~ tr/ //] //= $ans;
}
for my $s ( @$sub )
{
$n > $s or next;
my $ans = "-$s $solve[$n - $s]";
$pick[$ans =~ tr/ //] //= $ans;
}
$solve[$n] = first { defined } @pick;
$maximal[$solve[$n] =~ tr/ // - 1] .= " $n";
}
return \@solve, \@maximal;
}
Output:
---------------------------------------- divisors 2 3 subtractors 1
   1 takes 0 step(s):  
   2 takes 1 step(s): /2  
   3 takes 1 step(s): /3  
   4 takes 2 step(s): /2 /2  
   5 takes 3 step(s): -1 /2 /2  
   6 takes 2 step(s): /2 /3  
   7 takes 3 step(s): -1 /2 /3  
   8 takes 3 step(s): /2 /2 /2  
   9 takes 2 step(s): /3 /3  
  10 takes 3 step(s): -1 /3 /3  

16 number(s) below 2000 that take 14 steps:  863 1079 1295 1439 1511 1583 1607 1619 1691 1727 1823 1871 1895 1907 1919 1943
5 number(s) below 20000 that take 20 steps:  12959 15551 17279 18143 19439

---------------------------------------- divisors 2 3 subtractors 2
   1 takes 0 step(s):  
   2 takes 1 step(s): /2  
   3 takes 1 step(s): /3  
   4 takes 2 step(s): /2 /2  
   5 takes 2 step(s): -2 /3  
   6 takes 2 step(s): /2 /3  
   7 takes 3 step(s): -2 -2 /3  
   8 takes 3 step(s): /2 /2 /2  
   9 takes 2 step(s): /3 /3  
  10 takes 3 step(s): /2 -2 /3  

1 number(s) below 2000 that take 17 steps:  1699
1 number(s) below 20000 that take 24 steps:  19681

Phix[edit]

Using simple iterative (vs recursive) memoisation. To make things a bit more interesting it maintains separate caches for any&all {d,s} sets.

sequence cache = {},
ckeys = {}
 
function ms21(integer n, sequence ds)
integer cdx = find(ds,ckeys)
if cdx=0 then
ckeys = append(ckeys,ds)
cache = append(cache,{{}})
cdx = length(cache)
end if
for i=length(cache[cdx])+1 to n do
integer ms = n+2
sequence steps = {}
for j=1 to length(ds) do -- (d then s)
for k=1 to length(ds[j]) do
integer dsk = ds[j][k],
ok = iff(j=1?remainder(i,dsk)=0:i>dsk)
if ok then
integer m = iff(j=1?i/dsk:i-dsk),
l = length(cache[cdx][m])+1
if ms>l then
ms = l
steps = prepend(cache[cdx][m],sprintf("%s%d -> %d",{"/-"[j],dsk,m}))
end if
end if
end for
end for
if steps = {} then ?9/0 end if
cache[cdx] = append(cache[cdx],steps)
end for
return cache[cdx][n]
end function
 
procedure show10(sequence ds)
printf(1,"\nWith divisors %v and subtractors %v:\n",ds)
for n=1 to 10 do
sequence steps = ms21(n,ds)
integer ns = length(steps)
string ps = iff(ns=1?"":"s"),
eg = iff(ns=0?"":", eg "&join(steps,","))
printf(1,"%d takes %d step%s%s\n",{n,ns,ps,eg})
end for
end procedure
 
procedure maxsteps(sequence ds, integer lim)
integer ms = -1, ls
sequence mc = {}, steps, args
for n=1 to lim do
steps = ms21(n,ds)
ls = length(steps)
if ls>ms then
ms = ls
mc = {n}
elsif ls=ms then
mc &= n
end if
end for
integer lm = length(mc)
string {are,ps,ns} = iff(lm=1?{"is","","s"}:{"are","s",""})
args = { are,lm, ps, lim, ns,ms, mc}
printf(1,"There %s %d number%s below %d that require%s %d steps: %v\n",args)
end procedure
 
show10({{2,3},{1}})
maxsteps({{2,3},{1}},2000)
maxsteps({{2,3},{1}},20000)
maxsteps({{2,3},{1}},50000)
 
show10({{2,3},{2}})
maxsteps({{2,3},{2}},2000)
maxsteps({{2,3},{2}},20000)
maxsteps({{2,3},{2}},50000)
Output:
With divisors {2,3} and subtractors {1}:
1 takes 0 steps
2 takes 1 step, eg /2 -> 1
3 takes 1 step, eg /3 -> 1
4 takes 2 steps, eg /2 -> 2,/2 -> 1
5 takes 3 steps, eg -1 -> 4,/2 -> 2,/2 -> 1
6 takes 2 steps, eg /2 -> 3,/3 -> 1
7 takes 3 steps, eg -1 -> 6,/2 -> 3,/3 -> 1
8 takes 3 steps, eg /2 -> 4,/2 -> 2,/2 -> 1
9 takes 2 steps, eg /3 -> 3,/3 -> 1
10 takes 3 steps, eg -1 -> 9,/3 -> 3,/3 -> 1
There are 16 numbers below 2000 that require 14 steps: {863,1079,1295,1439,1511,1583,1607,1619,1691,1727,1823,1871,1895,1907,1919,1943}
There are 5 numbers below 20000 that require 20 steps: {12959,15551,17279,18143,19439}
There are 3 numbers below 50000 that require 22 steps: {25919,31103,38879}

With divisors {2,3} and subtractors {2}:
1 takes 0 steps
2 takes 1 step, eg /2 -> 1
3 takes 1 step, eg /3 -> 1
4 takes 2 steps, eg /2 -> 2,/2 -> 1
5 takes 2 steps, eg -2 -> 3,/3 -> 1
6 takes 2 steps, eg /2 -> 3,/3 -> 1
7 takes 3 steps, eg -2 -> 5,-2 -> 3,/3 -> 1
8 takes 3 steps, eg /2 -> 4,/2 -> 2,/2 -> 1
9 takes 2 steps, eg /3 -> 3,/3 -> 1
10 takes 3 steps, eg /2 -> 5,-2 -> 3,/3 -> 1
There is 1 number below 2000 that requires 17 steps: {1699}
There is 1 number below 20000 that requires 24 steps: {19681}
There is 1 number below 50000 that requires 26 steps: {45925}

Python[edit]

Python: Recursive, with memoization[edit]

Although the stretch goal could be achieved by changing the recursion limit, it does point out a possible issue with this type of solution. But then again, this solution may be more natural to some.

 
from functools import lru_cache
 
 
#%%
 
DIVS = {2, 3}
SUBS = {1}
 
class Minrec():
"Recursive, memoised minimised steps to 1"
 
def __init__(self, divs=DIVS, subs=SUBS):
self.divs, self.subs = divs, subs
 
@lru_cache(maxsize=None)
def _minrec(self, n):
"Recursive, memoised"
if n == 1:
return 0, ['=1']
possibles = {}
for d in self.divs:
if n % d == 0:
possibles[f'/{d}=>{n // d:2}'] = self._minrec(n // d)
for s in self.subs:
if n > s:
possibles[f'-{s}=>{n - s:2}'] = self._minrec(n - s)
thiskind, (count, otherkinds) = min(possibles.items(), key=lambda x: x[1])
ret = 1 + count, [thiskind] + otherkinds
return ret
 
def __call__(self, n):
"Recursive, memoised"
ans = self._minrec(n)[1][:-1]
return len(ans), ans
 
 
if __name__ == '__main__':
for DIVS, SUBS in [({2, 3}, {1}), ({2, 3}, {2})]:
minrec = Minrec(DIVS, SUBS)
print('\nMINIMUM STEPS TO 1: Recursive algorithm')
print(' Possible divisors: ', DIVS)
print(' Possible decrements:', SUBS)
for n in range(1, 11):
steps, how = minrec(n)
print(f' minrec({n:2}) in {steps:2} by: ', ', '.join(how))
 
upto = 2000
print(f'\n Those numbers up to {upto} that take the maximum, "minimal steps down to 1":')
stepn = sorted((minrec(n)[0], n) for n in range(upto, 0, -1))
mx = stepn[-1][0]
ans = [x[1] for x in stepn if x[0] == mx]
print(' Taking', mx, f'steps is/are the {len(ans)} numbers:',
', '.join(str(n) for n in sorted(ans)))
#print(minrec._minrec.cache_info())
print()
Output:
MINIMUM STEPS TO 1: Recursive algorithm
  Possible divisors:   {2, 3}
  Possible decrements: {1}
    minrec( 1) in  0 by:  
    minrec( 2) in  1 by:  /2=> 1
    minrec( 3) in  1 by:  /3=> 1
    minrec( 4) in  2 by:  /2=> 2, /2=> 1
    minrec( 5) in  3 by:  -1=> 4, /2=> 2, /2=> 1
    minrec( 6) in  2 by:  /3=> 2, /2=> 1
    minrec( 7) in  3 by:  -1=> 6, /3=> 2, /2=> 1
    minrec( 8) in  3 by:  /2=> 4, /2=> 2, /2=> 1
    minrec( 9) in  2 by:  /3=> 3, /3=> 1
    minrec(10) in  3 by:  -1=> 9, /3=> 3, /3=> 1

    Those numbers up to 2000 that take the maximum, "minimal steps down to 1":
      Taking 14 steps is/are the 16 numbers: 863, 1079, 1295, 1439, 1511, 1583, 1607, 1619, 1691, 1727, 1823, 1871, 1895, 1907, 1919, 1943


MINIMUM STEPS TO 1: Recursive algorithm
  Possible divisors:   {2, 3}
  Possible decrements: {2}
    minrec( 1) in  0 by:  
    minrec( 2) in  1 by:  /2=> 1
    minrec( 3) in  1 by:  /3=> 1
    minrec( 4) in  2 by:  /2=> 2, /2=> 1
    minrec( 5) in  2 by:  -2=> 3, /3=> 1
    minrec( 6) in  2 by:  /3=> 2, /2=> 1
    minrec( 7) in  3 by:  -2=> 5, -2=> 3, /3=> 1
    minrec( 8) in  3 by:  /2=> 4, /2=> 2, /2=> 1
    minrec( 9) in  2 by:  /3=> 3, /3=> 1
    minrec(10) in  3 by:  /2=> 5, -2=> 3, /3=> 1

    Those numbers up to 2000 that take the maximum, "minimal steps down to 1":
      Taking 17 steps is/are the 1 numbers: 1699

Python: Tabulated[edit]

The tabulated algorithm solves the allied problem of "Find the minimum steps in going from 1 to N, where at each step a member of D can be a multiplier or a member of S can be added".

The stretch goal is attempted.
The table to solve for N contains all the results from 1 up to N. This is used in the solution.

class Mintab():
"Tabulation, memoised minimised steps to 1"
 
def __init__(self, divs=DIVS, subs=SUBS):
self.divs, self.subs = divs, subs
self.table = None # Last tabulated table
self.hows = None # Last tabulated sample steps
 
def _mintab(self, n):
"Tabulation, memoised minimised steps to 1"
divs, subs = self.divs, self.subs
 
table = [n + 2] * (n + 1) # sentinels
table[1] = 0 # zero steps to 1 from 1
how = [[''] for _ in range(n + 2)] # What steps are taken
how[1] = ['=']
for t in range(1, n):
thisplus1 = table[t] + 1
for d in divs:
dt = d * t
if dt <= n and thisplus1 < table[dt]:
table[dt] = thisplus1
how[dt] = how[t] + [f'/{d}=>{t:2}']
for s in subs:
st = s + t
if st <= n and thisplus1 < table[st]:
table[st] = thisplus1
how[st] = how[t] + [f'-{s}=>{t:2}']
self.table = table
self.hows = [h[::-1][:-1] for h in how] # Order and trim
return self.table, self.hows
 
def __call__(self, n):
"Tabulation"
table, hows = self._mintab(n)
return table[n], hows[n]
 
 
if __name__ == '__main__':
for DIVS, SUBS in [({2, 3}, {1}), ({2, 3}, {2})]:
print('\nMINIMUM STEPS TO 1: Tabulation algorithm')
print(' Possible divisors: ', DIVS)
print(' Possible decrements:', SUBS)
mintab = Mintab(DIVS, SUBS)
mintab(10)
table, hows = mintab.table, mintab.hows
for n in range(1, 11):
steps, how = table[n], hows[n]
print(f' mintab({n:2}) in {steps:2} by: ', ', '.join(how))
 
for upto in [2000, 50_000]:
mintab(upto)
table = mintab.table
print(f'\n Those numbers up to {upto} that take the maximum, "minimal steps down to 1":')
mx = max(table[1:])
ans = [n for n, steps in enumerate(table) if steps == mx]
print(' Taking', mx, f'steps is/are the {len(ans)} numbers:',
', '.join(str(n) for n in ans))
Output:
MINIMUM STEPS TO 1: Tabulation algorithm
  Possible divisors:   {2, 3}
  Possible decrements: {1}
    mintab( 1) in  0 by:  
    mintab( 2) in  1 by:  /2=> 1
    mintab( 3) in  1 by:  /3=> 1
    mintab( 4) in  2 by:  /2=> 2, /2=> 1
    mintab( 5) in  3 by:  -1=> 4, /2=> 2, /2=> 1
    mintab( 6) in  2 by:  /3=> 2, /2=> 1
    mintab( 7) in  3 by:  -1=> 6, /3=> 2, /2=> 1
    mintab( 8) in  3 by:  /2=> 4, /2=> 2, /2=> 1
    mintab( 9) in  2 by:  /3=> 3, /3=> 1
    mintab(10) in  3 by:  -1=> 9, /3=> 3, /3=> 1

    Those numbers up to 2000 that take the maximum, "minimal steps down to 1":
      Taking 14 steps is/are the 16 numbers: 863, 1079, 1295, 1439, 1511, 1583, 1607, 1619, 1691, 1727, 1823, 1871, 1895, 1907, 1919, 1943

    Those numbers up to 50000 that take the maximum, "minimal steps down to 1":
      Taking 22 steps is/are the 3 numbers: 25919, 31103, 38879

MINIMUM STEPS TO 1: Tabulation algorithm
  Possible divisors:   {2, 3}
  Possible decrements: {2}
    mintab( 1) in  0 by:  
    mintab( 2) in  1 by:  /2=> 1
    mintab( 3) in  1 by:  /3=> 1
    mintab( 4) in  2 by:  /2=> 2, /2=> 1
    mintab( 5) in  2 by:  -2=> 3, /3=> 1
    mintab( 6) in  2 by:  /3=> 2, /2=> 1
    mintab( 7) in  3 by:  -2=> 5, -2=> 3, /3=> 1
    mintab( 8) in  3 by:  /2=> 4, /2=> 2, /2=> 1
    mintab( 9) in  2 by:  /3=> 3, /3=> 1
    mintab(10) in  3 by:  /2=> 5, -2=> 3, /3=> 1

    Those numbers up to 2000 that take the maximum, "minimal steps down to 1":
      Taking 17 steps is/are the 1 numbers: 1699

    Those numbers up to 50000 that take the maximum, "minimal steps down to 1":
      Taking 26 steps is/are the 1 numbers: 45925

Raku[edit]

(formerly Perl 6)

Works with: Rakudo version 2019.11
use Lingua::EN::Numbers;
 
for [2,3], 1, 2000,
[2,3], 1, 50000,
[2,3], 2, 2000,
[2,3], 2, 50000
-> @div, $sub, $limit {
my %min = 1 => {:op(''), :v(1), :s(0)};
(2..$limit).map( -> $n {
my @ops;
@ops.push: ($n / $_, "/$_") if $n %% $_ for @div;
@ops.push: ($n - $sub, "
-$sub") if $n > $sub;
my $op = @ops.min( {%min{.[0]}<s>} );
 %min{$n} = {:op($op[1]), :v($op[0]), :s(1 + %min{$op[0]}<s>)};
});
 
my $max = %min.max( {.value<s>} ).value<s>;
my @max = %min.grep( {.value.<s> == $max} )».key.sort(+*);
 
if $limit == 2000 {
say "
\nDivisors: {@div.perl}, subtract: $sub";
steps(1..10);
}
say "
\nUp to {comma $limit} found {[email protected]max} number{[email protected]max == 1 ?? '' !! 's'} " ~
"
that require{[email protected]max == 1 ?? 's' !! ''} at least $max steps.";
steps(@max);
 
sub steps (*@list) {
for @list -> $m {
my @op;
my $n = $m;
while %min{$n}<s> {
@op.push: "
{%min{$n}<op>}=>{%min{$n}<v>}";
$n = %min{$n}<v>;
}
say "
($m) {%min{$m}<s>} steps: ", @op.join(', ');
}
}
}
Output:
Divisors: [2, 3], subtract: 1
(1) 0 steps: 
(2) 1 steps: /2=>1
(3) 1 steps: /3=>1
(4) 2 steps: /2=>2, /2=>1
(5) 3 steps: -1=>4, /2=>2, /2=>1
(6) 2 steps: /2=>3, /3=>1
(7) 3 steps: -1=>6, /2=>3, /3=>1
(8) 3 steps: /2=>4, /2=>2, /2=>1
(9) 2 steps: /3=>3, /3=>1
(10) 3 steps: -1=>9, /3=>3, /3=>1

Up to 2,000 found 16 numbers that require at least 14 steps.
(863) 14 steps: -1=>862, -1=>861, /3=>287, -1=>286, -1=>285, /3=>95, -1=>94, -1=>93, /3=>31, -1=>30, /3=>10, -1=>9, /3=>3, /3=>1
(1079) 14 steps: -1=>1078, /2=>539, -1=>538, /2=>269, -1=>268, /2=>134, /2=>67, -1=>66, /2=>33, /3=>11, -1=>10, -1=>9, /3=>3, /3=>1
(1295) 14 steps: -1=>1294, /2=>647, -1=>646, /2=>323, -1=>322, /2=>161, -1=>160, /2=>80, /2=>40, /2=>20, /2=>10, -1=>9, /3=>3, /3=>1
(1439) 14 steps: -1=>1438, -1=>1437, /3=>479, -1=>478, -1=>477, /3=>159, /3=>53, -1=>52, /2=>26, /2=>13, -1=>12, /2=>6, /2=>3, /3=>1
(1511) 14 steps: -1=>1510, /2=>755, -1=>754, /2=>377, -1=>376, /2=>188, /2=>94, -1=>93, /3=>31, -1=>30, /3=>10, -1=>9, /3=>3, /3=>1
(1583) 14 steps: -1=>1582, /2=>791, -1=>790, -1=>789, /3=>263, -1=>262, -1=>261, /3=>87, /3=>29, -1=>28, -1=>27, /3=>9, /3=>3, /3=>1
(1607) 14 steps: -1=>1606, /2=>803, -1=>802, /2=>401, -1=>400, /2=>200, /2=>100, /2=>50, /2=>25, -1=>24, /2=>12, /2=>6, /2=>3, /3=>1
(1619) 14 steps: -1=>1618, /2=>809, -1=>808, /2=>404, /2=>202, /2=>101, -1=>100, /2=>50, /2=>25, -1=>24, /2=>12, /2=>6, /2=>3, /3=>1
(1691) 14 steps: -1=>1690, /2=>845, -1=>844, -1=>843, /3=>281, -1=>280, -1=>279, /3=>93, /3=>31, -1=>30, /3=>10, -1=>9, /3=>3, /3=>1
(1727) 14 steps: -1=>1726, -1=>1725, /3=>575, -1=>574, -1=>573, /3=>191, -1=>190, -1=>189, /3=>63, /3=>21, /3=>7, -1=>6, /2=>3, /3=>1
(1823) 14 steps: -1=>1822, /2=>911, -1=>910, -1=>909, /3=>303, /3=>101, -1=>100, /2=>50, /2=>25, -1=>24, /2=>12, /2=>6, /2=>3, /3=>1
(1871) 14 steps: -1=>1870, -1=>1869, /3=>623, -1=>622, -1=>621, /3=>207, /3=>69, /3=>23, -1=>22, /2=>11, -1=>10, -1=>9, /3=>3, /3=>1
(1895) 14 steps: -1=>1894, /2=>947, -1=>946, /2=>473, -1=>472, /2=>236, /2=>118, -1=>117, /3=>39, /3=>13, -1=>12, /2=>6, /2=>3, /3=>1
(1907) 14 steps: -1=>1906, /2=>953, -1=>952, /2=>476, /2=>238, /2=>119, -1=>118, -1=>117, /3=>39, /3=>13, -1=>12, /2=>6, /2=>3, /3=>1
(1919) 14 steps: -1=>1918, -1=>1917, /3=>639, /3=>213, /3=>71, -1=>70, /2=>35, -1=>34, /2=>17, -1=>16, /2=>8, /2=>4, /2=>2, /2=>1
(1943) 14 steps: -1=>1942, /2=>971, -1=>970, /2=>485, -1=>484, /2=>242, /2=>121, -1=>120, /2=>60, /2=>30, /3=>10, -1=>9, /3=>3, /3=>1

Up to 50,000 found 3 numbers that require at least 22 steps.
(25919) 22 steps: -1=>25918, /2=>12959, -1=>12958, /2=>6479, -1=>6478, /2=>3239, -1=>3238, /2=>1619, -1=>1618, /2=>809, -1=>808, /2=>404, /2=>202, /2=>101, -1=>100, /2=>50, /2=>25, -1=>24, /2=>12, /2=>6, /2=>3, /3=>1
(31103) 22 steps: -1=>31102, /2=>15551, -1=>15550, /2=>7775, -1=>7774, /2=>3887, -1=>3886, /2=>1943, -1=>1942, /2=>971, -1=>970, /2=>485, -1=>484, /2=>242, /2=>121, -1=>120, /2=>60, /2=>30, /3=>10, -1=>9, /3=>3, /3=>1
(38879) 22 steps: -1=>38878, /2=>19439, -1=>19438, /2=>9719, -1=>9718, /2=>4859, -1=>4858, /2=>2429, -1=>2428, /2=>1214, /2=>607, -1=>606, /2=>303, /3=>101, -1=>100, /2=>50, /2=>25, -1=>24, /2=>12, /2=>6, /2=>3, /3=>1

Divisors: [2, 3], subtract: 2
(1) 0 steps: 
(2) 1 steps: /2=>1
(3) 1 steps: /3=>1
(4) 2 steps: /2=>2, /2=>1
(5) 2 steps: -2=>3, /3=>1
(6) 2 steps: /2=>3, /3=>1
(7) 3 steps: -2=>5, -2=>3, /3=>1
(8) 3 steps: /2=>4, /2=>2, /2=>1
(9) 2 steps: /3=>3, /3=>1
(10) 3 steps: /2=>5, -2=>3, /3=>1

Up to 2,000 found 1 number that requires at least 17 steps.
(1699) 17 steps: -2=>1697, -2=>1695, /3=>565, -2=>563, -2=>561, /3=>187, -2=>185, -2=>183, /3=>61, -2=>59, -2=>57, /3=>19, -2=>17, -2=>15, /3=>5, -2=>3, /3=>1

Up to 50,000 found 1 number that requires at least 26 steps.
(45925) 26 steps: -2=>45923, -2=>45921, /3=>15307, -2=>15305, -2=>15303, /3=>5101, -2=>5099, -2=>5097, /3=>1699, -2=>1697, -2=>1695, /3=>565, -2=>563, -2=>561, /3=>187, -2=>185, -2=>183, /3=>61, -2=>59, -2=>57, /3=>19, -2=>17, -2=>15, /3=>5, -2=>3, /3=>1

Wren[edit]

Translation of: Go
Library: Wren-fmt
import "/fmt" for Fmt
 
var limit = 50000
var divs = []
var subs = []
var mins = []
 
// Assumes the numbers are presented in order up to 'limit'.
var minsteps = Fn.new { |n|
if (n == 1) {
mins[1] = []
return
}
var min = limit
var p = 0
var q = 0
var op = ""
for (div in divs) {
if (n%div == 0) {
var d = (n/div).floor
var steps = mins[d].count + 1
if (steps < min) {
min = steps
p = d
q = div
op = "/"
}
}
}
for (sub in subs) {
var d = n - sub
if (d >= 1) {
var steps = mins[d].count + 1
if (steps < min) {
min = steps
p = d
q = sub
op = "-"
}
}
}
mins[n].add("%(op)%(q) -> %(p)")
mins[n].addAll(mins[p])
}
 
for (r in 0..1) {
divs = [2, 3]
subs = (r == 0) ? [1] : [2]
mins = List.filled(limit+1, null)
for (i in 0..limit) mins[i] = []
Fmt.print("With: Divisors: $n, Subtractors: $n =>", divs, subs)
System.print(" Minimum number of steps to diminish the following numbers down to 1 is:")
for (i in 1..limit) {
minsteps.call(i)
if (i <= 10) {
var steps = mins[i].count
var plural = (steps == 1) ? " " : "s"
var mi = Fmt.v("s", 0, mins[i], 0, ", ", "")
Fmt.print(" $2d: $d step$s: $s", i, steps, plural, mi)
}
}
for (lim in [2000, 20000, 50000]) {
var max = 0
for (min in mins[0..lim]) {
var m = min.count
if (m > max) max = m
}
var maxs = []
var i = 0
for (min in mins[0..lim]) {
if (min.count == max) maxs.add(i)
i = i + 1
}
var nums = maxs.count
var verb = (nums == 1) ? "is" : "are"
var verb2 = (nums == 1) ? "has" : "have"
var plural = (nums == 1) ? "": "s"
Fmt.write(" There $s $d number$s in the range 1-$d ", verb, nums, plural, lim)
Fmt.print("that $s maximum 'minimal steps' of $d:", verb2, max)
System.print("  %(maxs)")
}
System.print()
}
Output:
With: Divisors: [2, 3], Subtractors: [1] =>
  Minimum number of steps to diminish the following numbers down to 1 is:
     1: 0 steps:  
     2: 1 step : /2 -> 1
     3: 1 step : /3 -> 1
     4: 2 steps: /2 -> 2, /2 -> 1
     5: 3 steps: -1 -> 4, /2 -> 2, /2 -> 1
     6: 2 steps: /2 -> 3, /3 -> 1
     7: 3 steps: -1 -> 6, /2 -> 3, /3 -> 1
     8: 3 steps: /2 -> 4, /2 -> 2, /2 -> 1
     9: 2 steps: /3 -> 3, /3 -> 1
    10: 3 steps: -1 -> 9, /3 -> 3, /3 -> 1
  There are 16 numbers in the range 1-2000 that have maximum 'minimal steps' of 14:
   [863, 1079, 1295, 1439, 1511, 1583, 1607, 1619, 1691, 1727, 1823, 1871, 1895, 1907, 1919, 1943]
  There are 5 numbers in the range 1-20000 that have maximum 'minimal steps' of 20:
   [12959, 15551, 17279, 18143, 19439]
  There are 3 numbers in the range 1-50000 that have maximum 'minimal steps' of 22:
   [25919, 31103, 38879]

With: Divisors: [2, 3], Subtractors: [2] =>
  Minimum number of steps to diminish the following numbers down to 1 is:
     1: 0 steps:  
     2: 1 step : /2 -> 1
     3: 1 step : /3 -> 1
     4: 2 steps: /2 -> 2, /2 -> 1
     5: 2 steps: -2 -> 3, /3 -> 1
     6: 2 steps: /2 -> 3, /3 -> 1
     7: 3 steps: -2 -> 5, -2 -> 3, /3 -> 1
     8: 3 steps: /2 -> 4, /2 -> 2, /2 -> 1
     9: 2 steps: /3 -> 3, /3 -> 1
    10: 3 steps: /2 -> 5, -2 -> 3, /3 -> 1
  There is 1 number  in the range 1-2000 that has maximum 'minimal steps' of 17:
   [1699]
  There is 1 number  in the range 1-20000 that has maximum 'minimal steps' of 24:
   [19681]
  There is 1 number  in the range 1-50000 that has maximum 'minimal steps' of 26:
   [45925]

XPL0[edit]

int     MinSteps,       \minimal number of steps to get to 1
Subtractor; \1 or 2
char Ns(20000), Ops(20000), MinNs(20000), MinOps(20000);
 
proc Reduce(N, Step); \Reduce N to 1, recording minimum steps
int N, Step, I;
[if N = 1 then
[if Step < MinSteps then
[for I:= 0 to Step-1 do
[MinOps(I):= Ops(I);
MinNs(I):= Ns(I);
];
MinSteps:= Step;
];
];
if Step >= MinSteps then return; \don't search further
if rem(N/3) = 0 then
[Ops(Step):= 3; Ns(Step):= N/3; Reduce(N/3, Step+1)];
if rem(N/2) = 0 then
[Ops(Step):= 2; Ns(Step):= N/2; Reduce(N/2, Step+1)];
Ops(Step):= -Subtractor; Ns(Step):= N-Subtractor; Reduce(N-Subtractor, Step+1);
]; \Reduce
 
proc ShowSteps(N); \Show minimal steps and how N steps to 1
int N, I;
[MinSteps:= $7FFF_FFFF;
Reduce(N, 0);
Text(0, "N = "); IntOut(0, N);
Text(0, " takes "); IntOut(0, MinSteps); Text(0, " steps: N ");
for I:= 0 to MinSteps-1 do
[Text(0, if extend(MinOps(I)) < 0 then "-" else "/");
IntOut(0, abs(extend(MinOps(I))));
Text(0, "=>"); IntOut(0, MinNs(I)); Text(0, " ");
];
CrLf(0);
]; \ShowSteps
 
proc ShowCount(Range); \Show count of maximum minimal steps and their Ns
int Range;
int N, MaxSteps;
[MaxSteps:= 0; \find maximum number of minimum steps
for N:= 1 to Range do
[MinSteps:= $7FFF_FFFF;
Reduce(N, 0);
if MinSteps > MaxSteps then
MaxSteps:= MinSteps;
];
Text(0, "Maximum steps: "); IntOut(0, MaxSteps); Text(0, " for N = ");
for N:= 1 to Range do \show numbers (Ns) for Maximum steps
[MinSteps:= $7FFF_FFFF;
Reduce(N, 0);
if MinSteps = MaxSteps then
[IntOut(0, N); Text(0, " ");
];
];
CrLf(0);
]; \ShowCount
 
int N;
[Subtractor:= 1; \1.
for N:= 1 to 10 do ShowSteps(N);
ShowCount(2000); \2.
ShowCount(20_000); \2a.
Subtractor:= 2; \3.
for N:= 1 to 10 do ShowSteps(N);
ShowCount(2000); \4.
ShowCount(20_000); \4a.
]
Output:
N = 1 takes 0 steps: N 
N = 2 takes 1 steps: N /2=>1 
N = 3 takes 1 steps: N /3=>1 
N = 4 takes 2 steps: N /2=>2 /2=>1 
N = 5 takes 3 steps: N -1=>4 /2=>2 /2=>1 
N = 6 takes 2 steps: N /3=>2 /2=>1 
N = 7 takes 3 steps: N -1=>6 /3=>2 /2=>1 
N = 8 takes 3 steps: N /2=>4 /2=>2 /2=>1 
N = 9 takes 2 steps: N /3=>3 /3=>1 
N = 10 takes 3 steps: N -1=>9 /3=>3 /3=>1 
Maximum steps: 14 for N = 863 1079 1295 1439 1511 1583 1607 1619 1691 1727 1823 1871 1895 1907 1919 1943 
Maximum steps: 20 for N = 12959 15551 17279 18143 19439 
N = 1 takes 0 steps: N 
N = 2 takes 1 steps: N /2=>1 
N = 3 takes 1 steps: N /3=>1 
N = 4 takes 2 steps: N /2=>2 /2=>1 
N = 5 takes 2 steps: N -2=>3 /3=>1 
N = 6 takes 2 steps: N /3=>2 /2=>1 
N = 7 takes 3 steps: N -2=>5 -2=>3 /3=>1 
N = 8 takes 3 steps: N /2=>4 /2=>2 /2=>1 
N = 9 takes 2 steps: N /3=>3 /3=>1 
N = 10 takes 3 steps: N /2=>5 -2=>3 /3=>1 
Maximum steps: 17 for N = 1699 
Maximum steps: 24 for N = 19681 

zkl[edit]

var minCache;  // (val:(newVal,op,steps))
fcn buildCache(N,D,S){
minCache=Dictionary(1,T(1,"",0));
foreach n in ([2..N]){
ops:=List();
foreach d in (D){ if(n%d==0) ops.append(T(n/d, String("/",d))) }
foreach s in (S){ if(n>s) ops.append(T(n - s,String("-",s))) }
mcv:=fcn(op){ minCache[op[0]][2] }; // !ACK!, dig out steps
v,op := ops.reduce( // find min steps to get to op
'wrap(vo1,vo2){ if(mcv(vo1)<mcv(vo2)) vo1 else vo2 });
minCache[n]=T(v, op, 1 + minCache[v][2]) // this many steps to get to n
}
}
fcn stepsToOne(N){ // D & S are determined by minCache
ops,steps := Sink(String).write(N), minCache[N][2];
do(steps){ v,o,s := minCache[N]; ops.write(" ",o,"-->",N=v); }
return(steps,ops.close())
}
MAX, D,S := 50_000, T(2,3), T(1);
buildCache(MAX,D,S);
 
do(2){
println("\nDivisors: %s, subtracters: %s".fmt(D.concat(","), S.concat(",")));
foreach n in ([1..10]){ println("%2d: %d steps: %s".fmt(n,stepsToOne(n).xplode())) }
 
maxSteps:=minCache.reduce(fcn(mkv,kv){ if(mkv[1][2]>kv[1][2]) mkv else kv })[1][2];
biggies :=minCache.filter('wrap(kv){ kv[1][2]==maxSteps }).pump(List,fcn(kv){ kv[0].toInt() }).sort();
println("\nBelow %,d, found %d numbers that require %d steps (the mostest)."
.fmt(MAX,biggies.len(),maxSteps));
foreach n in (biggies){ println("%,6d: %d steps: %s".fmt(n,stepsToOne(n).xplode())) }
 
S=T(2); buildCache(MAX,D,S);
}
Output:
Divisors: 2,3, subtracters: 1
 1: 0 steps: 1
 2: 1 steps: 2 -1-->1
 3: 1 steps: 3 /3-->1
 4: 2 steps: 4 -1-->3 /3-->1
 5: 3 steps: 5 -1-->4 -1-->3 /3-->1
 6: 2 steps: 6 /3-->2 -1-->1
 7: 3 steps: 7 -1-->6 /3-->2 -1-->1
 8: 3 steps: 8 /2-->4 -1-->3 /3-->1
 9: 2 steps: 9 /3-->3 /3-->1
10: 3 steps: 10 -1-->9 /3-->3 /3-->1

Below 50,000, found 3 numbers that require 22 steps (the mostest).
25,919: 22 steps: 25919 -1-->25918 -1-->25917 /3-->8639 -1-->8638 -1-->8637 /3-->2879 -1-->2878 -1-->2877 /3-->959 -1-->958 -1-->957 /3-->319 -1-->318 /3-->106 /2-->53 -1-->52 /2-->26 /2-->13 -1-->12 /3-->4 -1-->3 /3-->1
31,103: 22 steps: 31103 -1-->31102 -1-->31101 /3-->10367 -1-->10366 -1-->10365 /3-->3455 -1-->3454 -1-->3453 /3-->1151 -1-->1150 -1-->1149 /3-->383 -1-->382 -1-->381 /3-->127 -1-->126 /3-->42 /3-->14 /2-->7 -1-->6 /3-->2 -1-->1
38,879: 22 steps: 38879 -1-->38878 /2-->19439 -1-->19438 /2-->9719 -1-->9718 /2-->4859 -1-->4858 /2-->2429 -1-->2428 /2-->1214 /2-->607 -1-->606 /3-->202 -1-->201 /3-->67 -1-->66 /3-->22 -1-->21 /3-->7 -1-->6 /3-->2 -1-->1

Divisors: 2,3, subtracters: 2
 1: 0 steps: 1
 2: 1 steps: 2 /2-->1
 3: 1 steps: 3 -2-->1
 4: 2 steps: 4 -2-->2 /2-->1
 5: 2 steps: 5 -2-->3 -2-->1
 6: 2 steps: 6 /3-->2 /2-->1
 7: 3 steps: 7 -2-->5 -2-->3 -2-->1
 8: 3 steps: 8 -2-->6 /3-->2 /2-->1
 9: 2 steps: 9 /3-->3 -2-->1
10: 3 steps: 10 /2-->5 -2-->3 -2-->1

Below 50,000, found 1 numbers that require 26 steps (the mostest).
45,925: 26 steps: 45925 -2-->45923 -2-->45921 /3-->15307 -2-->15305 -2-->15303 /3-->5101 -2-->5099 -2-->5097 /3-->1699 -2-->1697 -2-->1695 /3-->565 -2-->563 -2-->561 /3-->187 -2-->185 -2-->183 /3-->61 -2-->59 -2-->57 /3-->19 -2-->17 -2-->15 /3-->5 -2-->3 -2-->1