Minimal steps down to 1: Difference between revisions

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Given:

A starting, positive integer (greater than one), N.
A selection of possible integer perfect divisors, D.
And a selection of possible subtractors, S.

The goal is find the minimum number of steps necessary to reduce N down to one.

At any step, the number may be:

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

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

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

Examples

No divisors, D. a single subtractor of 1.

Obviousely N will take N-1 subtractions of 1 to reach 1

Single divisor of 2; single subtractor of 1:

N = 7 Takes 4 steps N -1=> 6, /2=> 3, -1=> 2, /2=> 1

N = 23 Takes 7 steps N -1=>22, /2=>11, -1=>10, /2=> 5, -1=> 4, /2=> 2, /2=> 1

Divisors 2 and 3; subtractor 1:

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

Task

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

1. Show the number of steps and possible way of diminishing the numbers 1 to 10 down to 1.

2. Show a count of, and the numbers that: have the maximum minimal_steps_to_1, in the range 1 to 2,000.

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

3. Show the number of steps and possible way of diminishing the numbers 1 to 10 down to 1.

4. Show a count of, and the numbers that: have the maximum minimal_steps_to_1, in the range 1 to 2,000.

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

Reference

Learn Dynamic Programming (Memoization & Tabulation) Video of similar task.

Go

<lang 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()
   }

}</lang>

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]

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. <lang julia>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
       print("There are ", length(solutions), " solutions 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])

</lang>

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

There are 1 solutions 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 are 1 solutions 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]

Perl 6

Works with: Rakudo version 2019.11

<lang perl6>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 $op = @ops.min( {%min{.[0]}} );
      %min{$n} = {:op($op[1]), :v($op[0]), :s(1 + %min{$op[0]})};
   });

~~my $max = %min.max( {.value~~} ).value~~; my @max = %min.grep( {.value. ~~== $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} ~~{ @op.push: "{%min{$n}<op>}=>{%min{$n}<v>}"; $n = %min{$n}<v>; } say "($m) {%min{$m}~~} steps: ", @op.join(', '); } }~~~~
~~}</lang>~~

~~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
~~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.
<lang python> 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()</lang>

~~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 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.
<lang python>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))</lang>

~~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
~~zkl~~
~~<lang 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())~~
}</lang> <lang zkl>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);~~
~~}</lang>~~

~~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

Revision as of 09:02, 20 December 2019

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Julia

Perl 6

Python

Python: Recursive, with memoization

Python: Tabulated

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