Code Golf: Collatz Conjecture

旧时模样 提交于 2019-11-28 16:27:29

x86 assembly, 1337 characters

;
; To assemble and link this program, just run:
;
; >> $ nasm -f elf collatz.asm && gcc -o collatz collatz.o
;
; You can then enjoy its output by passing a number to it on the command line:
;
; >> $ ./collatz 123
; >> 123 --> 370 --> 185 --> 556 --> 278 --> 139 --> 418 --> 209 --> 628 --> 314
; >> --> 157 --> 472 --> 236 --> 118 --> 59 --> 178 --> 89 --> 268 --> 134 --> 67
; >> --> 202 --> 101 --> 304 --> 152 --> 76 --> 38 --> 19 --> 58 --> 29 --> 88
; >> --> 44 --> 22 --> 11 --> 34 --> 17 --> 52 --> 26 --> 13 --> 40 --> 20 --> 10
; >> --> 5 --> 16 --> 8 --> 4 --> 2 --> 1
; 
; There's even some error checking involved:
; >> $ ./collatz
; >> Usage: ./collatz NUMBER
;
section .text
global main
extern printf
extern atoi

main:

  cmp dword [esp+0x04], 2
  jne .usage

  mov ebx, [esp+0x08]
  push dword [ebx+0x04]
  call atoi
  add esp, 4

  cmp eax, 0
  je .usage

  mov ebx, eax
  push eax
  push msg

.loop:
  mov [esp+0x04], ebx
  call printf

  test ebx, 0x01
  jz .even

.odd:
  lea ebx, [1+ebx*2+ebx]
  jmp .loop

.even:

  shr ebx, 1
  cmp ebx, 1
  jne .loop

  push ebx
  push end
  call printf

  add esp, 16
  xor eax, eax
  ret

.usage:
  mov ebx, [esp+0x08]
  push dword [ebx+0x00]
  push usage
  call printf
  add esp, 8
  mov eax, 1
  ret

msg db "%d --> ", 0
end db "%d", 10, 0
usage db "Usage: %s NUMBER", 10, 0
Josh Lee

Befunge

&>:.:1-|
  >3*^ @
  |%2: <
 v>2/>+

LOLCODE: 406 CHARAKTERZ

HAI
BTW COLLATZ SOUNDZ JUS LULZ

CAN HAS STDIO?

I HAS A NUMBAR
BTW, I WANTS UR NUMBAR
GIMMEH NUMBAR

VISIBLE NUMBAR

IM IN YR SEQUENZ
  MOD OF NUMBAR AN 2
  BOTH SAEM IT AN 0, O RLY?
    YA RLY, NUMBAR R QUOSHUNT OF NUMBAR AN 2
    NO WAI, NUMBAR R SUM OF PRODUKT OF NUMBAR AN 3 AN 1
  OIC
  VISIBLE NUMBAR
  DIFFRINT 2 AN SMALLR OF 2 AN NUMBAR, O RLY?
    YA RLY, GTFO
  OIC
IM OUTTA YR SEQUENZ

KTHXBYE

TESTD UNDR JUSTIN J. MEZA'S INTERPRETR. KTHXBYE!

makapuf

Python - 95 64 51 46 char

Obviously does not produce a stack overflow.

n=input()
while n>1:n=(n/2,n*3+1)[n%2];print n
Brad Gilbert

Perl

I decided to be a little anticompetitive, and show how you would normally code such problem in Perl.
There is also a 46 (total) char code-golf entry at the end.

These first three examples all start out with this header.

#! /usr/bin/env perl
use Modern::Perl;
# which is the same as these three lines:
# use 5.10.0;
# use strict;
# use warnings;

while( <> ){
  chomp;
  last unless $_;
  Collatz( $_ );
}
  • Simple recursive version

    use Sub::Call::Recur;
    sub Collatz{
      my( $n ) = @_;
      $n += 0; # ensure that it is numeric
      die 'invalid value' unless $n > 0;
      die 'Integer values only' unless $n == int $n;
      say $n;
      given( $n ){
        when( 1 ){}
        when( $_ % 2 != 0 ){ # odd
          recur( 3 * $n + 1 );
        }
        default{ # even
          recur( $n / 2 );
        }
      }
    }
    
  • Simple iterative version

    sub Collatz{
      my( $n ) = @_;
      $n += 0; # ensure that it is numeric
      die 'invalid value' unless $n > 0;
      die 'Integer values only' unless $n == int $n;
      say $n;
      while( $n > 1 ){
        if( $n % 2 ){ # odd
          $n = 3 * $n + 1;
        } else { #even
          $n = $n / 2;
        }
        say $n;
      }
    }
    
  • Optimized iterative version

    sub Collatz{
      my( $n ) = @_;
      $n += 0; # ensure that it is numeric
      die 'invalid value' unless $n > 0;
      die 'Integer values only' unless $n == int $n;
      #
      state @next;
      $next[1] //= 0; # sets $next[1] to 0 if it is undefined
      #
      # fill out @next until we get to a value we've already worked on
      until( defined $next[$n] ){
        say $n;
        #
        if( $n % 2 ){ # odd
          $next[$n] = 3 * $n + 1;
        } else { # even
          $next[$n] = $n / 2;
        }
        #
        $n = $next[$n];
      }
      say $n;
      # finish running until we get to 1
      say $n while $n = $next[$n];
    }
    

Now I'm going to show how you would do that last example with a version of Perl prior to v5.10.0

#! /usr/bin/env perl
use strict;
use warnings;

while( <> ){
  chomp;
  last unless $_;
  Collatz( $_ );
}
{
  my @next = (0,0); # essentially the same as a state variable
  sub Collatz{
    my( $n ) = @_;
    $n += 0; # ensure that it is numeric
    die 'invalid value' unless $n > 0;

    # fill out @next until we get to a value we've already worked on
    until( $n == 1 or defined $next[$n] ){
      print $n, "\n";

      if( $n % 2 ){ # odd
        $next[$n] = 3 * $n + 1;
      } else { # even
        $next[$n] = $n / 2;
      }
      $n = $next[$n];
    }
    print $n, "\n";

    # finish running until we get to 1
    print $n, "\n" while $n = $next[$n];
  }
}

Benchmark

First off the IO is always going to be the slow part. So if you actually benchmarked them as-is you should get about the same speed out of each one.

To test these then, I opened a file handle to /dev/null ($null), and edited every say $n to instead read say {$null} $n. This is to reduce the dependence on IO.

#! /usr/bin/env perl
use Modern::Perl;
use autodie;

open our $null, '>', '/dev/null';

use Benchmark qw':all';

cmpthese( -10,
{
  Recursive => sub{ Collatz_r( 31 ) },
  Iterative => sub{ Collatz_i( 31 ) },
  Optimized => sub{ Collatz_o( 31 ) },
});

sub Collatz_r{
  ...
  say {$null} $n;
  ...
}
sub Collatz_i{
  ...
  say {$null} $n;
  ...
}
sub Collatz_o{
  ...
  say {$null} $n;
  ...
}

After having run it 10 times, here is a representative sample output:

            Rate Recursive Iterative Optimized
Recursive 1715/s        --      -27%      -46%
Iterative 2336/s       36%        --      -27%
Optimized 3187/s       86%       36%        --

Finally, a real code-golf entry:

perl -nlE'say;say$_=$_%2?3*$_+1:$_/2while$_>1'

46 chars total

If you don't need to print the starting value, you could remove 5 more characters.

perl -nE'say$_=$_%2?3*$_+1:$_/2while$_>1'

41 chars total
31 chars for the actual code portion, but the code won't work without the -n switch. So I include the entire example in my count.

MtnViewMark

Haskell, 62 chars 63 76 83, 86, 97, 137

c 1=[1]
c n=n:c(div(n`mod`2*(5*n+2)+n)2)
main=readLn>>=print.c

User input, printed output, uses constant memory and stack, works with arbitrarily big integers.

A sample run of this code, given an 80 digit number of all '1's (!) as input, is pretty fun to look at.


Original, function only version:

Haskell 51 chars

f n=n:[[],f([n`div`2,3*n+1]!!(n`mod`2))]!!(1`mod`n)

Who the @&^# needs conditionals, anyway?

(edit: I was being "clever" and used fix. Without it, the code dropped to 54 chars. edit2: dropped to 51 by factoring out f())

KennyTM

Golfscript : 20 chars

  ~{(}{3*).1&5*)/}/1+`
# 
# Usage: echo 21 | ruby golfscript.rb collatz.gs

This is equivalent to

stack<int> s;
s.push(21);
while (s.top() - 1) {
  int x = s.top();
  int numerator = x*3+1;
  int denominator = (numerator&1) * 5 + 1;
  s.push(numerator/denominator);
}
s.push(1);
return s;
Carlos Gutiérrez

bc 41 chars

I guess this kind of problems is what bc was invented for:

for(n=read();n>1;){if(n%2)n=n*6+2;n/=2;n}

Test:

bc1 -q collatz.bc
21
64
32
16
8
4
2
1

Proper code:

for(n=read();n>1;){if(n%2)n=n*3+1else n/=2;print n,"\n"}

bc handles numbers with up to INT_MAX digits

Edit: The Wikipedia article mentions this conjecture has been checked for all values up to 20x258 (aprox. 5.76e18). This program:

c=0;for(n=2^20000+1;n>1;){if(n%2)n=n*6+2;n/=2;c+=1};n;c

tests 220,000+1 (aprox. 3.98e6,020) in 68 seconds, 144,404 cycles.

a'r

Perl : 31 chars

perl -nE 'say$_=$_%2?$_*3+1:$_/2while$_>1'
#         123456789 123456789 123456789 1234567

Edited to remove 2 unnecessary spaces.

Edited to remove 1 unnecessary space.

MS Excel, 35 chars

=IF(A1/2=ROUND(A1/2,0),A1/2,A1*3+1)

Taken straight from Wikipedia:

In cell A1, place the starting number.
In cell A2 enter this formula =IF(A1/2=ROUND(A1/2,0),A1/2,A1*3+1) 
Drag and copy the formula down until 4, 2, 1

It only took copy/pasting the formula 111 times to get the result for a starting number of 1000. ;)

KennyTM

C : 64 chars

main(x){for(scanf("%d",&x);x>=printf("%d,",x);x=x&1?3*x+1:x/2);}

With big integer support: 431 (necessary) chars

#include <stdlib.h>
#define B (w>=m?d=realloc(d,m=m+m):0)
#define S(a,b)t=a,a=b,b=t
main(m,w,i,t){char*d=malloc(m=9);for(w=0;(i=getchar()+2)/10==5;)
B,d[w++]=i%10;for(i=0;i<w/2;i++)S(d[i],d[w-i-1]);for(;;w++){
while(w&&!d[w-1])w--;for(i=w+1;i--;)putchar(i?d[i-1]+48:10);if(
w==1&&*d==1)break;if(*d&1){for(i=w;i--;)d[i]*=3;*d+=1;}else{
for(i=w;i-->1;)d[i-1]+=d[i]%2*10,d[i]/=2;*d/=2;}B,d[w]=0;for(i=0
;i<w;i++)d[i+1]+=d[i]/10,d[i]%=10;}}

Note: Do not remove #include <stdlib.h> without at least prototyping malloc/realloc, as doing so will not be safe on 64-bit platforms (64-bit void* will be converted to 32-bit int).

This one hasn't been tested vigorously yet. It could use some shortening as well.


Previous versions:

main(x){for(scanf("%d",&x);printf("%d,",x),x-1;x=x&1?3*x+1:x/2);} // 66

(removed 12 chars because no one follows the output format... :| )

Another assembler version. This one is not limited to 32 bit numbers, it can handle numbers up to 1065534 although the ".com" format MS-DOS uses is limited to 80 digit numbers. Written for A86 assembler and requires a Win-XP DOS box to run. Assembles to 180 bytes:

    mov ax,cs
    mov si,82h
    add ah,10h
    mov es,ax
    mov bh,0
    mov bl,byte ptr [80h]
    cmp bl,1
    jbe ret
    dec bl
    mov cx,bx
    dec bl
    xor di,di
 p1:lodsb
    sub al,'0'
    cmp al,10
    jae ret
    stosb
    loop p1
    xor bp,bp
    push es
    pop ds
 p2:cmp byte ptr ds:[bp],0
    jne p3
    inc bp
    jmp p2
    ret
 p3:lea si,[bp-1]
    cld
 p4:inc si
    mov dl,[si]
    add dl,'0'
    mov ah,2
    int 21h
    cmp si,bx
    jne p4
    cmp bx,bp
    jne p5
    cmp byte ptr [bx],1
    je ret
 p5:mov dl,'-'
    mov ah,2
    int 21h
    mov dl,'>'
    int 21h
    test byte ptr [bx],1
    jz p10
    ;odd
    mov si,bx
    mov di,si
    mov dx,3
    dec bp
    std
 p6:lodsb
    mul dl
    add al,dh
    aam
    mov dh,ah
    stosb
    cmp si,bp
    jnz p6
    or dh,dh
    jz p7
    mov al,dh
    stosb
    dec bp
 p7:mov si,bx
    mov di,si
 p8:lodsb
    inc al
    xor ah,ah
    aaa
    stosb
    or ah,ah
    jz p9
    cmp si,bp
    jne p8
    mov al,1
    stosb
    jmp p2
 p9:inc bp
    jmp p2
    p10:mov si,bp
    mov di,bp
    xor ax,ax
p11:lodsb
    test ah,1
    jz p12
    add al,10
p12:mov ah,al
    shr al,1
    cmp di,bx
    stosb
    jne p11
    jmp p2
Carlos Gutiérrez

dc - 24 chars 25 28

dc is a good tool for this sequence:

?[d5*2+d2%*+2/pd1<L]dsLx
dc -f collatz.dc
21
64
32
16
8
4
2
1

Also 24 chars using the formula from the Golfscript entry:

?[3*1+d2%5*1+/pd1<L]dsLx

57 chars to meet the specs:

[Number: ]n?[Results: ]ndn[d5*2+d2%*+2/[ -> ]ndnd1<L]dsLx
dc -f collatz-spec.dc
Number: 3
Results: 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
Jerry Coffin

Scheme: 72

(define(c n)(if(= n 1)`(1)(cons n(if(odd? n)(c(+(* n 3)1))(c(/ n 2))))))

This uses recursion, but the calls are tail-recursive so I think they'll be optimized to iteration. In some quick testing, I haven't been able to find a number for which the stack overflows anyway. Just for example:

(c 9876543219999999999000011234567898888777766665555444433332222 7777777777777777777777777777777798797657657651234143375987342987 5398709812374982529830983743297432985230985739287023987532098579 058095873098753098370938753987)

...runs just fine. [that's all one number -- I've just broken it to fit on screen.]

Pillsy

Mathematica, 45 50 chars

c=NestWhileList[If[OddQ@#,3#+1,#/2]&,#,#>1&]&
Jordan Running

Ruby, 50 chars, no stack overflow

Basically a direct rip of makapuf's Python solution:

def c(n)while n>1;n=n.odd?? n*3+1: n/2;p n end end

Ruby, 45 chars, will overflow

Basically a direct rip of the code provided in the question:

def c(n)p n;n.odd?? c(3*n+1):c(n/2)if n>1 end
import java.math.BigInteger;
public class SortaJava {

    static final BigInteger THREE = new BigInteger("3");
    static final BigInteger TWO = new BigInteger("2");

    interface BiFunc<R, A, B> {
      R call(A a, B b);
    }

    interface Cons<A, B> {
      <R> R apply(BiFunc<R, A, B> func);
    }

    static class Collatz implements Cons<BigInteger, Collatz> {
      BigInteger value;
      public Collatz(BigInteger value) { this.value = value; }
      public <R> R apply(BiFunc<R, BigInteger, Collatz> func) {
        if(BigInteger.ONE.equals(value))
          return func.call(value, null);
        if(value.testBit(0))
          return func.call(value, new Collatz((value.multiply(THREE)).add(BigInteger.ONE)));
        return func.call(value, new Collatz(value.divide(TWO)));
      }
    }

    static class PrintAReturnB<A, B> implements BiFunc<B, A, B> {
      boolean first = true;
      public B call(A a, B b) {
        if(first)
          first = false;
        else
          System.out.print(" -> ");
        System.out.print(a);
        return b;
      }
    }

    public static void main(String[] args) {
      BiFunc<Collatz, BigInteger, Collatz> printer = new PrintAReturnB<BigInteger, Collatz>();
      Collatz collatz = new Collatz(new BigInteger(args[0]));
      while(collatz != null)
        collatz = collatz.apply(printer);
    }
}

Python 45 Char

Shaved a char off of makapuf's answer.

n=input()
while~-n:n=(n/2,n*3+1)[n%2];print n

TI-BASIC

Not the shortest, but a novel approach. Certain to slow down considerably with large sequences, but it shouldn't overflow.

PROGRAM:COLLATZ
:ClrHome
:Input X
:Lbl 1
:While X≠1
:If X/2=int(X/2)
:Then
:Disp X/2→X
:Else
:Disp X*3+1→X
:End
:Goto 1
:End
Josh Lee

Haskell : 50

c 1=[1];c n=n:(c$if odd n then 3*n+1 else n`div`2)
cobbal

not the shortest, but an elegant clojure solution

(defn collatz [n]
 (print n "")
 (if (> n 1)
  (recur
   (if (odd? n)
    (inc (* 3 n))
    (/ n 2)))))
Venr

C#: 216 Characters

using C=System.Console;class P{static void Main(){var p="start:";System.Action<object> o=C.Write;o(p);ulong i;while(ulong.TryParse(C.ReadLine(),out i)){o(i);while(i > 1){i=i%2==0?i/2:i*3+1;o(" -> "+i);}o("\n"+p);}}}

in long form:

using C = System.Console;
class P
{
    static void Main()
    {
        var p = "start:"; 
        System.Action<object> o = C.Write; 
        o(p); 
        ulong i; 
        while (ulong.TryParse(C.ReadLine(), out i))
        {
            o(i); 
            while (i > 1)
            {
                i = i % 2 == 0 ? i / 2 : i * 3 + 1; 
                o(" -> " + i);
            } 
            o("\n" + p);
        }
    }
}

New Version, accepts one number as input provided through the command line, no input validation. 173 154 characters.

using System;class P{static void Main(string[]a){Action<object>o=Console.Write;var i=ulong.Parse(a[0]);o(i);while(i>1){i=i%2==0?i/2:i*3+1;o(" -> "+i);}}}

in long form:

using System;
class P
{
    static void Main(string[]a)
    {
        Action<object>o=Console.Write;
        var i=ulong.Parse(a[0]);
        o(i);
        while(i>1)
        {
            i=i%2==0?i/2:i*3+1;
            o(" -> "+i);
        }
    }
}

I am able to shave a few characters by ripping off the idea in this answer to use a for loop rather than a while. 150 characters.

using System;class P{static void Main(string[]a){Action<object>o=Console.Write;for(var i=ulong.Parse(a[0]);i>1;i=i%2==0?i/2:i*3+1)o(i+" -> ");o(1);}}
Sniggerfardimungus

Ruby, 43 characters

bignum supported, with stack overflow susceptibility:

def c(n)p n;n%2>0?c(3*n+1):c(n/2)if n>1 end

...and 50 characters, bignum supported, without stack overflow:

def d(n)while n>1 do p n;n=n%2>0?3*n+1:n/2 end end

Kudos to Jordan. I didn't know about 'p' as a replacement for puts.

nroff1

Run with nroff -U hail.g

.warn
.pl 1
.pso (printf "Enter a number: " 1>&2); read x; echo .nr x $x
.while \nx>1 \{\
.  ie \nx%2 .nr x \nx*3+1
.  el .nr x \nx/2
\nx
.\}

1. groff version

Ben Lings

Scala + Scalaz

import scalaz._
import Scalaz._
val collatz = 
   (_:Int).iterate[Stream](a=>Seq(a/2,3*a+1)(a%2)).takeWhile(1<) // This line: 61 chars

And in action:

scala> collatz(7).toList
res15: List[Int] = List(7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2)

Scala 2.8

val collatz = 
   Stream.iterate(_:Int)(a=>Seq(a/2,3*a+1)(a%2)).takeWhile(1<) :+ 1

This also includes the trailing 1.

scala> collatz(7)
res12: scala.collection.immutable.Stream[Int] = Stream(7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1)

With the following implicit

implicit def intToEven(i:Int) = new {
  def ~(even: Int=>Int, odd: Int=>Int) = { 
    if (i%2==0) { even(i) } else { odd(i) }
  }
}

this can be shortened to

val collatz = Stream.iterate(_:Int)(_~(_/2,3*_+1)).takeWhile(1<) :+ 1

Edit - 58 characters (including input and output, but not including initial number)

var n=readInt;while(n>1){n=Seq(n/2,n*3+1)(n%2);println(n)}

Could be reduced by 2 if you don't need newlines...

F#, 90 characters

let c=Seq.unfold(function|n when n<=1->None|n when n%2=0->Some(n,n/2)|n->Some(n,(3*n)+1))

> c 21;;
val it : seq<int> = seq [21; 64; 32; 16; ...]

Or if you're not using F# interactive to display the result, 102 characters:

let c=Seq.unfold(function|n when n<=1->None|n when n%2=0->Some(n,n/2)|n->Some(n,(3*n)+1))>>printf"%A"

Common Lisp, 141 characters:

(defun c ()
  (format t"Number: ")
  (loop for n = (read) then (if(oddp n)(+ 1 n n n)(/ n 2))
     until (= n 1)
     do (format t"~d -> "n))
  (format t"1~%"))

Test run:

Number: 171
171 -> 514 -> 257 -> 772 -> 386 -> 193 -> 580 -> 290 -> 145 -> 436 ->
218 -> 109 -> 328 -> 164 -> 82 -> 41 -> 124 -> 62 -> 31 -> 94 -> 47 ->
142 -> 71 -> 214 -> 107 -> 322 -> 161 -> 484 -> 242 -> 121 -> 364 ->
182 -> 91 -> 274 -> 137 -> 412 -> 206 -> 103 -> 310 -> 155 -> 466 ->
233 -> 700 -> 350 -> 175 -> 526 -> 263 -> 790 -> 395 -> 1186 -> 593 ->
1780 -> 890 -> 445 -> 1336 -> 668 -> 334 -> 167 -> 502 -> 251 -> 754 ->
377 -> 1132 -> 566 -> 283 -> 850 -> 425 -> 1276 -> 638 -> 319 ->
958 -> 479 -> 1438 -> 719 -> 2158 -> 1079 -> 3238 -> 1619 -> 4858 ->
2429 -> 7288 -> 3644 -> 1822 -> 911 -> 2734 -> 1367 -> 4102 -> 2051 ->
6154 -> 3077 -> 9232 -> 4616 -> 2308 -> 1154 -> 577 -> 1732 -> 866 ->
433 -> 1300 -> 650 -> 325 -> 976 -> 488 -> 244 -> 122 -> 61 -> 184 ->
92 -> 46 -> 23 -> 70 -> 35 -> 106 -> 53 -> 160 -> 80 -> 40 -> 20 ->
10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 
Soumen Sarkar

The program frm Jerry Coffin has integer over flow, try this one:

#include <iostream>

int main(unsigned long long i)
{
    int j = 0;
    for(  std::cin>>i; i>1; i = i&1? i*3+1:i/2, ++j)
        std::cout<<i<<" -> ";

    std::cout<<"\n"<<j << " iterations\n";
}

tested with

The number less than 100 million with the longest total stopping time is 63,728,127, with 949 steps.

The number less than 1 billion with the longest total stopping time is 670,617,279, with 986 steps.

ruby, 43, possibly meeting the I/O requirement


Run with ruby -n hail

n=$_.to_i
(n=n%2>0?n*3+1: n/2
p n)while n>1

C# : 659 chars with BigInteger support

using System.Linq;using C=System.Console;class Program{static void Main(){var v=C.ReadLine();C.Write(v);while(v!="1"){C.Write("->");if(v[v.Length-1]%2==0){v=v.Aggregate(new{s="",o=0},(r,c)=>new{s=r.s+(char)((c-48)/2+r.o+48),o=(c%2)*5}).s.TrimStart('0');}else{var q=v.Reverse().Aggregate(new{s="",o=0},(r, c)=>new{s=(char)((c-48)*3+r.o+(c*3+r.o>153?c*3+r.o>163?28:38:48))+r.s,o=c*3+r.o>153?c*3+r.o>163?2:1:0});var t=(q.o+q.s).TrimStart('0').Reverse();var x=t.First();q=t.Skip(1).Aggregate(new{s=x>56?(x-57).ToString():(x-47).ToString(),o=x>56?1:0},(r,c)=>new{s=(char)(c-48+r.o+(c+r.o>57?38:48))+r.s,o=c+r.o>57?1:0});v=(q.o+q.s).TrimStart('0');}C.Write(v);}}}

Ungolfed

using System.Linq;
using C = System.Console;
class Program
{
    static void Main()
    {
        var v = C.ReadLine();
        C.Write(v);
        while (v != "1")
        {
            C.Write("->");
            if (v[v.Length - 1] % 2 == 0)
            {
                v = v
                    .Aggregate(
                        new { s = "", o = 0 }, 
                        (r, c) => new { s = r.s + (char)((c - 48) / 2 + r.o + 48), o = (c % 2) * 5 })
                    .s.TrimStart('0');
            }
            else
            {
                var q = v
                    .Reverse()
                    .Aggregate(
                        new { s = "", o = 0 }, 
                        (r, c) => new { s = (char)((c - 48) * 3 + r.o + (c * 3 + r.o > 153 ? c * 3 + r.o > 163 ? 28 : 38 : 48)) + r.s, o = c * 3 + r.o > 153 ? c * 3 + r.o > 163 ? 2 : 1 : 0 });
                var t = (q.o + q.s)
                    .TrimStart('0')
                    .Reverse();
                var x = t.First();
                q = t
                    .Skip(1)
                    .Aggregate(
                        new { s = x > 56 ? (x - 57).ToString() : (x - 47).ToString(), o = x > 56 ? 1 : 0 }, 
                        (r, c) => new { s = (char)(c - 48 + r.o + (c + r.o > 57 ? 38 : 48)) + r.s, o = c + r.o > 57 ? 1 : 0 });
                v = (q.o + q.s)
                    .TrimStart('0');
            }
            C.Write(v);
        }
    }
}
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