# Minimal steps down to 1

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

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

## 11l

Translation of: Python: Tabulated
```T Mintab
Set[Int] divs, subs
[Int] table
[[String]] hows

F (divs, subs)
.divs = copy(divs)
.subs = copy(subs)

F _mintab(n)
‘Tabulation, memoised minimised steps to 1’
V table = [n + 2] * (n + 1)
table[1] = 0
V how = (0 .< n + 2).map(_ -> [‘’])
how[1] = [String(‘=’)]
L(t) 1 .< n
V thisplus1 = table[t] + 1
L(d) .divs
V dt = d * t
I dt <= n & thisplus1 < table[dt]
table[dt] = thisplus1
how[dt] = how[t] [+] [‘/#.=>#2’.format(d, t)]
L(s) .subs
V st = s + t
I st <= n & thisplus1 < table[st]
table[st] = thisplus1
how[st] = how[t] [+] [‘-#.=>#2’.format(s, t)]
.table = table
.hows = how.map(h -> reversed(h)[0 .< (len)-1])
R (.table, .hows)

F ()(n)
‘Tabulation’
V (table, hows) = ._mintab(n)
R (table[n], hows[n])

L(DIVS, SUBS) [([2, 3], [1]), ([2, 3], [2])]
print("\nMINIMUM STEPS TO 1: Tabulation algorithm")
print(‘  Possible divisors:   ’DIVS)
print(‘  Possible decrements: ’SUBS)
V mintab = Mintab(Set(DIVS), Set(SUBS))
mintab(10)
L(n) 1..10
V (steps, how) = (mintab.table[n], mintab.hows[n])
print(‘    mintab(#2) in #2 by: ’.format(n, steps)‘ ’how.join(‘, ’))

L(upto) [2000, 50'000]
mintab(upto)
print("\n    Those numbers up to "upto‘ that take the maximum, "minimal steps down to 1":’)
V mx = max(mintab.table[1..])
V ans = enumerate(mintab.table).filter((n, steps) -> steps == @mx).map((n, steps) -> n)
print(‘      Taking ’mx‘ steps is/are the ’ans.len‘ numbers: ’ans.map(n -> String(n)).join(‘, ’))```
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
```

## C#

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```

## C++

```#include <cstdint>
#include <iomanip>
#include <iostream>
#include <string>
#include <vector>

const int32_t limit = 50'000;

std::vector<int32_t> divisors;
std::vector<int32_t> subtractors;
std::vector<std::vector<std::string>> minimums;

template <typename T>
void print_vector(const std::vector<T>& list) {
for ( uint64_t i = 0; i < list.size(); ++i ) {
std::cout << list[i];
if ( i < list.size() - 1 ) {
std::cout << ", ";
}
}
}

// Assumes that numbers are presented in ascending order up to 'limit'.
void minimum_steps(int32_t n) {
if ( n == 1 ) {
return;
}

int32_t minimum = limit;
int32_t p = 0;
int32_t q = 0;
std::string operator_symbol = "";
for ( int32_t divisor : divisors ) {
if ( n % divisor == 0 ) {
int32_t d = n / divisor;
int32_t steps = minimums[d].size() + 1;
if ( steps < minimum ) {
minimum = steps;
p = d;
q = divisor;
operator_symbol = "/";
}
}
}

for ( int32_t subtractor : subtractors ) {
int32_t d = n - subtractor;
if ( d >= 1 ) {
int32_t steps = minimums[d].size() + 1;
if ( steps < minimum ) {
minimum = steps;
p = d;
q = subtractor;
operator_symbol = "-";
}
}
}

minimums[n].emplace_back(operator_symbol + std::to_string(q) + " -> " + std::to_string(p));
minimums[n].insert(minimums[n].end(), minimums[p].begin(), minimums[p].end());
}

int main() {
for ( int32_t item : { 0, 1 } ) {
divisors = { 2, 3 };
subtractors = { item + 1 };
minimums = std::vector(limit + 1, std::vector<std::string>(0));
std::cout << "With: Divisors: { "; print_vector<int32_t>(divisors);
std::cout << " }, Subtractors: { "; print_vector<int32_t>(subtractors); std::cout << " } =>" << std::endl;
std::cout << "  Minimum number of steps to diminish the following numbers down to 1 is:" << std::endl;
for ( int32_t i = 1; i < limit; ++i ) {
minimum_steps(i);
if ( i <= 10 ) {
int32_t steps = minimums[i].size();
const std::string plural = ( steps == 1 ) ? " : " : "s: ";
std::cout << "    " << std::setw(2) << i << ": " << steps << " step" + plural;
print_vector<std::string>(minimums[i]); std::cout << std::endl;
}
}

for ( int32_t lim : { 2'000, 20'000, 50'000 } ) {
uint64_t max = 0;
for ( int32_t j = 1; j <= lim; ++j ) {
uint64_t m = minimums[j].size();
if ( m > max ) {
max = m;
}
}
std::vector<int32_t> maxs;
for ( int32_t j = 1; j <= lim; ++j ) {
if ( minimums[j].size() == max ) {
maxs.emplace_back(j);
}
}

int32_t size = maxs.size();
std::string verb1  = ( size == 1 ) ? "is" : "are";
std::string verb2  = ( size == 1 ) ? "has" : "have";
std::string plural = ( size == 1 ) ? "" : "s";
std::cout << "  There " << verb1 << " " << size << " number" << plural << " in the range 1 - " << lim
<< " that " << verb2 << " maximum 'minimal steps' of " << max << ":" << std::endl;
std::cout << "  [ "; print_vector<int32_t>(maxs); std::cout << " ]" << std::endl;
}
std::cout << std::endl;
}
}
```
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 ]
```

## FreeBASIC

Translation of: XPL0
```Dim Shared As Integer minPasos       'minimal number of steps to get to 1
Dim Shared As Integer Subtractor     '1 or 2
Dim Shared As Integer Ns(20000), Ops(20000), MinNs(20000), MinOps(20000)

Sub Reduce(N As Integer, Paso As Integer) 'Reduce N to 1, recording minimum steps
If N = 1 Then
If Paso < minPasos Then
For i As Integer = 0 To Paso-1
MinOps(i) = Ops(i)
MinNs(i) = Ns(i)
Next i
minPasos = Paso
End If
End If
If Paso >= minPasos Then Exit Sub     'don't search further
If N Mod 3 = 0 Then Ops(Paso) = 3 : Ns(Paso) = N/3 : Reduce(N/3, Paso+1)
If N Mod 2 = 0 Then Ops(Paso) = 2 : Ns(Paso) = N/2 : Reduce(N/2, Paso+1)
Ops(Paso) = -Subtractor
Ns(Paso) = N-Subtractor
Reduce(N-Subtractor, Paso+1)
End Sub

Sub ShowSteps(N As Integer)     'Show minimal steps and how N steps to 1
minPasos = 50000
Reduce(N, 0)
Print "N = " & N & " takes " & minPasos & " steps: N";
For i As Integer = 0 To minPasos-1
Print Iif(Sgn(MinOps(i)) < 0, " -", " /");
Print Abs(MinOps(i)) & "=>" & MinNs(i); '" "
Next i
Print
End Sub

Sub ShowCount(Range As Integer) 'Show count of maximum minimal steps and their Ns
Dim As Integer N, MaxSteps
MaxSteps = 0          'find maximum number of minimum steps
For N = 1 To Range
minPasos = 50000
Reduce(N, 0)
If minPasos > MaxSteps Then MaxSteps = minPasos
Next N
Print "Maximum steps:"; MaxSteps; " for N =";
For N = 1 To Range    'show numbers (Ns) for Maximum steps
minPasos = 50000
Reduce(N, 0)
If minPasos = MaxSteps Then Print N; '" ";
Next N
Print
End Sub

Dim As Integer N
Subtractor = 1        '1.
For N = 1 To 10
ShowSteps(N)
Next N
ShowCount(2000)       '2.
ShowCount(20000)      '2a.

Print
Subtractor = 2        '3.
For N = 1 To 10
ShowSteps(N)
Next N
ShowCount(2000)       '4.
ShowCount(20000)      '4a.
Sleep```
Output:
`Same as XPL0 entry.`

## Go

```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]
```

```{-# LANGUAGE DeriveFunctor #-}
import Data.List
import Data.Ord
import Data.Function (on)

------------------------------------------------------------
-- memoization utilities

data Memo a = Node a (Memo a) (Memo a)
deriving Functor

memo :: Integral a => Memo p -> a -> p
memo (Node a l r) n
| n == 0 = a
| odd n = memo l (n `div` 2)
| otherwise = memo r (n `div` 2 - 1)

nats :: Integral a => Memo a
nats = Node 0 ((+1).(*2) <\$> nats) ((*2).(+1) <\$> nats)

memoize :: Integral a => (a -> b) -> (a -> b)
memoize f = memo (f <\$> nats)

------------------------------------------------------------

data Step = Div Int | Sub Int
deriving Show

run :: Int -> Step -> [(Step, Int)]
run n s = case s of
Sub i | n > i -> [(s, n - i)]
Div d | n `mod` d == 0 -> [(s, n `div` d)]
_ -> []

minSteps :: [Step] -> Int -> (Int, [Step])
minSteps steps = go
where
go = memoize goM

goM 1 = (0, [])
goM n = minimumBy (comparing fst) \$ do
(s, k) <- steps >>= run n
let (m, ss) = go k
return (m+1, s:ss)
```
```λ> minSteps [Div 2, Div 3, Sub 1] 123
(7,[Div 3,Sub 1,Div 2,Div 2,Sub 1,Div 3,Div 3])

λ> minSteps [Div 5, Div 2, Sub 1, Sub 2] 123
(7,[Sub 1,Div 2,Sub 1,Div 5,Div 2,Div 2,Sub 2])```

```showSteps :: Int -> [Step] -> String
showSteps = foldl go . show
where
go r (Div d) = r ++ "/" ++ show d
go r (Sub s) = "(" ++ r ++ "-" ++ show s ++ ")"

task steps =  mapM_ (put . go) [1..10]
where
go n = showSteps n <\$> minSteps steps n
put (n,s) = putStrLn \$ show n ++ ":\t" ++ s

task2 steps range = mapM_ put longest
where
put (n,(l,s)) = putStrLn \$ show l ++ ": " ++
showSteps n s
longest =
head \$ groupBy ((==) `on` (fst.snd)) \$
sortOn (negate . fst . snd) \$
zip [1..] (minSteps steps <\$> range)
```
```λ> task1 [Div 2, Div 3, Sub 1]
0:	1
1:	2/2
1:	3/3
2:	4/2/2
3:	(5-1)/2/2
2:	6/2/3
3:	(7-1)/2/3
3:	8/2/2/2
2:	9/3/3
3:	(10-1)/3/3

λ> task2 [Div 2, Div 3, Sub 1] [1..2000]
14: ((((((((863-1)-1)/3-1)-1)/3-1)-1)/3-1)/3-1)/3/3
14: ((((((1079-1)/2-1)/2-1)/2/2-1)/2/3-1)-1)/3/3
14: (((((1295-1)/2-1)/2-1)/2-1)/2/2/2/2-1)/3/3
14: ((((((1439-1)-1)/3-1)-1)/3/3-1)/2/2-1)/2/2/3
14: ((((((1511-1)/2-1)/2-1)/2/2-1)/3-1)/3-1)/3/3
14: (((((((1583-1)/2-1)-1)/3-1)-1)/3/3-1)-1)/3/3/3
14: ((((1607-1)/2-1)/2-1)/2/2/2/2-1)/2/2/2/3
14: ((((1619-1)/2-1)/2/2/2-1)/2/2-1)/2/2/2/3
14: (((((((1691-1)/2-1)-1)/3-1)-1)/3/3-1)/3-1)/3/3
14: (((((((1727-1)-1)/3-1)-1)/3-1)-1)/3/3/3-1)/2/3
14: (((((1823-1)/2-1)-1)/3/3-1)/2/2-1)/2/2/2/3
14: (((((((1871-1)-1)/3-1)-1)/3/3/3-1)/2-1)-1)/3/3
14: (((((1895-1)/2-1)/2-1)/2/2-1)/3/3-1)/2/2/3
14: (((((1907-1)/2-1)/2/2/2-1)-1)/3/3-1)/2/2/3
14: (((((1919-1)-1)/3/3/3-1)/2-1)/2-1)/2/2/2/2
14: (((((1943-1)/2-1)/2-1)/2/2-1)/2/2/3-1)/3/3

λ> task1 [Div 2, Div 3, Sub 2]
0:	1
1:	2/2
1:	3/3
2:	4/2/2
2:	(5-2)/3
2:	6/2/3
3:	((7-2)-2)/3
3:	8/2/2/2
2:	9/3/3
3:	(10/2-2)/3

λ> task2 [Div 2, Div 3, Sub 2] [1..2000]
17: (((((((((((1699-2)-2)/3-2)-2)/3-2)-2)/3-2)-2)/3-2)-2)/3-2)/3```

## J

Implementation:

```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
}}
```

Counting the steps is rather trivial -- we build a step function which subtracts the possible subtractors (discarding numbers which are too small) and divides by the possible divisors (discarding numbers which are not integers), then we iterate on that until we find a 1.

Showing the paths is more complicated. The approach used here is to first find the valid step values and then, starting from 1, add the possible subtractors, and multiply by the possible divisors, discarding the values which were not valid for that step and tracking the previous values. Once we have built a representation of the valid paths (at each stage: operator, operand and net result) we reverse and format those.

```taskA=: {{
for_j. 1+i.10 do.
echo rplc&(' 1 steps';' 1 step')j,&":': ',(_1+#x steps y j),&":' steps.'
echo x show y j
echo ''
end.
}}

echo 'considering positive integers up to ',":m
tallies=. _1+#@(x steps y)every 1+i.m
echo (>./tallies) ,&": ' steps: ',&": 1+I.(=>./)tallies
echo  ''
}}

1: 0 steps.
1

2: 1 step.
2-1
2/2

3: 1 step.
3/3

4: 2 steps.
4/2-1
4/2/2
4-1/3

5: 3 steps.
5-1/2-1
5-1/2/2
5-1-1/3

6: 2 steps.
6/3-1
6/3/2
6/2/3

7: 3 steps.
7-1/3-1
7-1/3/2
7-1/2/3

8: 3 steps.
8/2/2-1
8/2/2/2
8/2-1/3

9: 2 steps.
9/3/3

10: 3 steps.
10-1/3/3

considering positive integers up to 2000
14 steps: 863 1079 1295 1439 1511 1583 1607 1619 1691 1727 1823 1871 1895 1907 1919 1943

considering positive integers up to 20000
20 steps: 12959 15551 17279 18143 19439

1: 0 steps.
1

2: 1 step.
2/2

3: 1 step.
3-2
3/3

4: 2 steps.
4-2/2
4/2/2

5: 2 steps.
5-2-2
5-2/3

6: 2 steps.
6/2-2
6/3/2
6/2/3

7: 3 steps.
7-2-2-2
7-2-2/3

8: 3 steps.
8-2/2-2
8/2-2/2
8/2/2/2
8-2/3/2
8-2/2/3

9: 2 steps.
9/3-2
9/3/3

10: 3 steps.
10/2-2-2
10/2-2/3

considering positive integers up to 2000
17 steps: 1699

considering positive integers up to 20000
24 steps: 19681
```

## Java

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) {
}

private static void runTasks(List<Function> functions) {
Map<Integer,List<String>> minPath = getInitialMap(functions, 5);

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));
}

displayMaxMin(minPath, functions, 2000);

displayMaxMin(minPath, functions, 20000);

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();
}
else if ( len == 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>();
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>();
minPath.put(i, path);
}
}
}
}
return minPath;
}

private static List<Function> getFunctions3() {
List<Function> functions = new ArrayList<>();
return functions;
}

private static List<Function> getFunctions2() {
List<Function> functions = new ArrayList<>();
return functions;
}

private static List<Function> getFunctions1() {
List<Function> functions = new ArrayList<>();
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]
```

## jq

Works with jq, the C implementation of jq

```### Generic functions

def array(\$n): . as \$in | [range(0;\$n)|\$in];

def max(stream): reduce stream as \$x (-infinite; if \$x > . then \$x else . end);

def plural: if . == 1 then " " else "s" end;

### Minimal Steps
# Input: { divs, subs, limit, mins }
# where .divs and .subs are the arrays of divisors and subtractors respectively,
# and .limit the value specified for task/1;
# .mins should be an array of arrays.
def minsteps(\$n):
if \$n == 1
then .mins[1] = []
else .min = .limit
| .p = 0
| .q = 0
| .op = ""
| reduce .divs[] as \$div (.;
if (\$n % \$div == 0)
then ((\$n/\$div)|floor) as \$d
| (.mins[\$d]|length + 1) as \$steps
| if \$steps < .min
then .min = \$steps
| .p = \$d
| .q = \$div
| .op = "/"
end
end )
| reduce .subs[] as \$sub (.;
(\$n - \$sub) as \$d
| if \$d >= 1
then (.mins[\$d]|length + 1) as \$steps
| if \$steps < .min
then .min = \$steps
| .p = \$d
| .q = \$sub
| .op = "-"
end
end)
| .mins[\$n] = ["\(.op)\(.q) -> \(.p)"] + .mins[.p]
end ;

range(0;2) as \$r
| { divs: [2, 3],
subs: (if \$r == 0 then [1] else [2] end),
mins: [],
\$limit
}
| "\nWith: Divisors: \(.divs), Subtractors: \(.subs) =>",
"  Minimum number of steps to diminish the following numbers down to 1 is:",
( foreach range(1; 1+\$limit) as \$i (.;
minsteps(\$i)  # sets .min[\$i]
| if (\$i <= 10)
then (.mins[\$i]|length) as \$steps
| .emit = "    \(\$i): \(\$steps) step\(\$steps|plural): \(.mins[\$i] | join(", "))"
else .emit = null
end;
select(.emit).emit,

# ... and when all is done:
(select(\$i==\$limit)
| foreach (2000, 20000, \$limit) as \$lim (.;
.max = max( .mins[0: 1 + \$lim][] | length)
| .maxs = []
| .i = 0
| reduce .mins[0: 1 + \$lim][] as \$min (.;
if (\$min|length) == .max then .maxs += [.i] end
| .i += 1 )
| .nums = (.maxs|length)
| (if .nums == 1 then "is" else "are" end) as \$are
| (if .nums == 1 then "has" else "have" end) as \$have
| .emit =  "  There \(\$are) \(.nums) number\(.nums|plural) in the range 1-\(\$lim)"
| .emit += " that \(\$have) maximum 'minimal steps' of \(.max):"
| .emit += "\n   \(.maxs)";
.emit) ) ) )
;

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]
```

## Julia

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) &&
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]
```

## Kotlin

Translation of: Java
```fun main() {
}

val minPath = getInitialMap(functions, 5)

val max = 10
populateMap(minPath, functions, max)
println("\nWith functions: \$functions")
println("  Minimum steps to 1:")
for (n in 2..max) {
val steps = minPath[n]?.size ?: 0
println("    %2d: %d step%s: %s".format(n, steps, if (steps == 1) "" else "s", minPath[n]))
}

displayMaxMin(minPath, functions, 2000)

displayMaxMin(minPath, functions, 20000)

displayMaxMin(minPath, functions, 100000)
}

fun displayMaxMin(minPath: MutableMap<Int, List<String>>, functions: List<Function>, max: Int) {
populateMap(minPath, functions, max)
val maxIntegers = getMaxMin(minPath, max)
val maxSteps = maxIntegers.removeAt(0)
val numCount = maxIntegers.size
println("  There \${if (numCount == 1) "is" else "are"} \$numCount number\${if (numCount == 1) "" else "s"} in the range 1-\$max that have maximum 'minimal steps' of \$maxSteps:\n    \$maxIntegers")
}

fun getMaxMin(minPath: Map<Int, List<String>>, max: Int): MutableList<Int> {
var maxSteps = Int.MIN_VALUE
val maxIntegers = mutableListOf<Int>()
for (n in 2..max) {
val len = minPath[n]?.size ?: 0
if (len > maxSteps) {
maxSteps = len
maxIntegers.clear()
} else if (len == maxSteps) {
}
}
return maxIntegers
}

fun populateMap(minPath: MutableMap<Int, List<String>>, functions: List<Function>, max: Int) {
for (n in 2..max) {
if (n in minPath) continue
var minFunction: Function? = null
var minSteps = Int.MAX_VALUE
for (f in functions) {
if (f.actionOk(n)) {
val result = f.action(n)
val steps = 1 + (minPath[result]?.size ?: 0)
if (steps < minSteps) {
minFunction = f
minSteps = steps
}
}
}
minFunction?.let {
val result = it.action(n)
val path = mutableListOf(it.toString(n))
minPath[n] = path
}
}
}

fun getInitialMap(functions: List<Function>, max: Int): MutableMap<Int, List<String>> {
val minPath = mutableMapOf<Int, List<String>>()
for (i in 2..max) {
for (f in functions) {
if (f.actionOk(i)) {
val result = f.action(i)
if (result == 1) {
minPath[i] = listOf(f.toString(i))
}
}
}
}
return minPath
}

fun getFunctions1(): List<Function> = listOf(
Divide3Function(),
Divide2Function(),
Subtract1Function()
)

fun getFunctions2(): List<Function> = listOf(
Divide3Function(),
Divide2Function(),
Subtract2Function()
)

fun getFunctions3(): List<Function> = listOf(
Divide2Function(),
Divide3Function(),
Subtract2Function(),
Subtract1Function()
)

abstract class Function {
abstract fun action(n: Int): Int
abstract fun actionOk(n: Int): Boolean
abstract fun toString(n: Int): String
}

class Divide2Function : Function() {
override fun action(n: Int) = n / 2
override fun actionOk(n: Int) = n % 2 == 0
override fun toString(n: Int) = "/2 -> \${n / 2}"
override fun toString() = "Divisor 2"
}

class Divide3Function : Function() {
override fun action(n: Int) = n / 3
override fun actionOk(n: Int) = n % 3 == 0
override fun toString(n: Int) = "/3 -> \${n / 3}"
override fun toString() = "Divisor 3"
}

class Subtract1Function : Function() {
override fun action(n: Int) = n - 1
override fun actionOk(n: Int) = true
override fun toString(n: Int) = "-1 -> \${n - 1}"
override fun toString() = "Subtractor 1"
}

class Subtract2Function : Function() {
override fun action(n: Int) = n - 2
override fun actionOk(n: Int) = n > 2
override fun toString(n: Int) = "-2 -> \${n - 2}"
override fun toString() = "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]

```

## Mathematica/Wolfram Language

```\$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

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:
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

```#!/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

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

```with javascript_semantics
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
string step = sprintf("%s%d -> %d",{"/-"[j],dsk,m})
steps = prepend(deep_copy(cache[cdx][m]),step)
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,"%2d 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, shorten(mc,"",3)}
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,"...",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

### Python: Recursive, with memoization

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

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

(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 {+@max} number{+@max == 1 ?? '' !! 's'} " ~
"that require{+@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```

## Swift

Translation of: Python
```func minToOne(divs: [Int], subs: [Int], upTo n: Int) -> ([Int], [[String]]) {
var table = Array(repeating: n + 2, count: n + 1)
var how = Array(repeating: [""], count: n + 2)

table[1] = 0
how[1] = ["="]

for t in 1..<n {
let thisPlus1 = table[t] + 1

for div in divs {
let dt = div * t

if dt <= n && thisPlus1 < table[dt] {
table[dt] = thisPlus1
how[dt] = how[t] + ["/\(div)=>  \(t)"]
}
}

for sub in subs {
let st = sub + t

if st <= n && thisPlus1 < table[st] {
table[st] = thisPlus1
how[st] = how[t] + ["-\(sub)=> \(t)"]
}
}
}

return (table, how.map({ \$0.reversed().dropLast() }))
}

for (divs, subs) in [([2, 3], [1]), ([2, 3], [2])] {
print("\nMINIMUM STEPS TO 1:")
print("  Possible divisors:  \(divs)")
print("  Possible decrements: \(subs)")

let (table, hows) = minToOne(divs: divs, subs: subs, upTo: 10)

for n in 1...10 {
print("    mintab(  \(n)) in {  \(table[n])} by: ", hows[n].joined(separator: ", "))
}

for upTo in [2_000, 50_000] {
print("\n    Those numbers up to \(upTo) that take the maximum, \"minimal steps down to 1\":")
let (table, _) = minToOne(divs: divs, subs: subs, upTo: upTo)
let max = table.dropFirst().max()!
let maxNs = table.enumerated().filter({ \$0.element == max })

print(
"      Taking", max, "steps are the \(maxNs.count) numbers:",
maxNs.map({ String(\$0.offset) }).joined(separator: ", ")
)
}
}
```
Output:
```MINIMUM STEPS TO 1:
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 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 are the 3 numbers: 25919, 31103, 38879

MINIMUM STEPS TO 1:
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 are the 1 numbers: 1699

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

## Wren

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 = "-"
}
}
}
}

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]) {
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

```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

```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
```