The best base is 10... Let's reach it!

Question

Input:

A positive integer n consisting of digits in the range 0-9.

Challenge:

If d is the highest digit in the integer, assume the base of the number is d+1. E.g. if the integer is 1256 then you shall assume it's in base-7, if it's 10110 then you shall assume it's base-2 (binary), and if it's 159 then it's decimal.

Now, do the following until you either, 1: reach a base-10 integer, or 2: reach a single digit integer.

1. Convert the integer from base-(d+1) to base-10
2. Find the base of this new integer (again, base-(d+1) where d is the highest digit in the new number)
3. Go to step 1.

---

Examples:

Assume the input is n = 413574. The highest digit d=7, so this is base-8 (octal). Convert this to decimal and get 137084. The highest digit d=8, so this is base-9. Convert this to decimal and get 83911. The highest digit is 9, so this is a decimal number and we stop. The output shall be 83911.

Assume the input is n = 13552. The highest digit is d=5, so this is base-6. Convert this to decimal and get 2156. The highest digit d=6, so this is base-7. Convert this to decimal and get 776. The highest digit is d=7, so this is base-8. Convert this to decimal and get 510. The highest digit is d=5 so this is base-6. Convert this to decimal and get 186. The highest digit is 8, so this is base-9. Convert this to decimal and get 159. The highest digit is 9, so this is a decimal number and we stop. The output shall be 159.

Assume the input is n=17. This will give us 15, then 11, then 3, which we will output since it's a single digit.

---

Test cases:

5
5

17
3

999
999

87654321  (base-9 -> 42374116 in decimal -> base-7 -> 90419978 in decimal) 
9041998

41253  (5505 -> 1265 -> 488 -> 404 -> 104 -> 29)
29

---

Notes:

• Standard rules regarding I/O, loopholes etc. You may take the input as a string
• Explanations are encouraged
• You may use builtin base-conversion commands
  • Solutions that don't use the language's builtin base-conversion functions (if they exist) are welcome, even if they end up being much longer than the obvious approach using builtin functions.

---

Apparently, this is OEIS A091047.

Answer 1

Mathematica, 56 bytes

#//.x_/;(b=Max[d=IntegerDigits@x]+1)<11:>d~FromDigits~b&

Try it online! (Using Mathics.)

I thought I'd check out what the sequence looks like:

And here is a plot of the number of steps it takes to find the result:

^{(Click for larger versions. See the revision history for plots only up to n = 1000.)}

Looks like a very interesting mixture of large scale structure and fine scale chaos. I wonder what's up with the wider gaps around 30,000 and 60,000.

Answer 2

Java 8, 172 166 163 152 151 140 138 116 114 99 bytes

s->{for(Integer b=0;b<10&s.length()>1;s=""+b.valueOf(s,b=s.chars().max().getAsInt()-47));return s;}

Takes the input as a String.

-64 bytes thanks to @OlivierGrégoire. And here I thought my initial 172 wasn't too bad.. ;)

Try it here.

Explanation:

s->{                    // Method with String as parameter and return-type
  for(Integer b=0;b<10  //  Loop as long as the largest digit + 1 is not 10
      &s.length()>1;    //  and as long as `s` contains more than 1 digit
    s=""+               //   Replace `s` with the String representation of:
         b.valueOf(s,   //    `s` as a Base-`b` integer
          b=s.chars().max().ge‌tAsInt()-47)
                        //     where `b` is the largest digit in the String + 1
  );                    //  End of loop
  return s;             //  Return result-String
}                       // End of method

Answer 3

Pyth, 9 bytes

ui`GhseS`

Test suite

Explanation:

ui`GhseS`
ui`GhseS`GQ    Implicit variable introduction
u         Q    Repeatedly apply the following function until the value repeats,
               starting with Q, the input.
        `G     Convert the working value to a string.
      eS       Take its maximum.
     s         Convert to an integer.
    h          Add one.
 i`G           Convert the working value to that base

Answer 4

JavaScript (ES6), 63 57 54 53 bytes

f=a=>a>9&(b=Math.max(...a+""))<9?f(parseInt(a,b+1)):a

Saved 8 bytes thanks to Shaggy and Dom Hastings

f=a=>a>9&(b=Math.max(...a+""))<9?f(parseInt(a,b+1)):a;

console.log(f("413574"))

Answer 5

Python 3, 91 78 76 75 73 bytes

@Emigna shaved off 5 bytes. @FelipeNardiBatista saved 1 byte. @RomanGräf saved 2 bytes

i=input()
while'9'>max(i)and~-len(i):i=str(int(i,int(max(i))+1))
print(i)

Try it online!

---

Explanation

i=input()                                  - takes input and assigns it to a variable i
while '9'>max(i)and~-len(i):               - repeatedly checks if the number is still base-9 or lower and has a length greater than 1
    i=str(...)                             - assigns i to the string representation of ...
          int(i,int(max(i))+1)             - converts the current number to the real base 10 and loops back again
print(i)                                   - prints the mutated i

Answer 6

05AB1E, 10 5 bytes

5 bytes saved thanks to Magic Octopus Urn

F§Z>ö

As this grows slow very quickly for large input, I'm leaving the old much faster version here for testing. The algorithm is the same, only the number of iterations differ.

[©Z>öD®Q#§

Try it online!

Explanation

[             # start a loop
 ©            # store a copy of the current value in the register
  Z>          # get the maximum (digit) and increment
    ö         # convert the current value to base-10 from this base
     D        # duplicate
      ®Q#     # break loop if the new value is the same as the stored value
         §    # convert to string (to prevent Z from taking the maximum int on the stack)

Answer 7

APL (Dyalog), 20 16 bytes

Takes and returns a string

((⍕⊢⊥⍨1+⌈/)⍎¨)⍣≡

(…)⍣≡ apply the following function until two consecutive terms are identical:

 ⍎¨ execute each character (turns the string into a list of numbers)

 (…) apply the following tacit function to that:

  ⌈/ find the maximum of the argument

  1+ add one

  ⊢⊥⍨ evaluate the argument in that base

  ⍕ format (stringify, in preparation for another application of the outer function)

Try it online!

Answer 8

Ruby, 60 56 bytes

->n{n=(s=n.to_s).to_i(s.bytes.max-47)while/9/!~s&&n>9;n}

Try it online!

Answer 9

Mathematica, 52 bytes

FromDigits[s=IntegerDigits@#,Max@s+1]&~FixedPoint~#&

Pure function taking a nonnegative integer as input and returning a nonnegative integer. Uses the same core mechanic FromDigits[s=IntegerDigits@#,Max@s+1] as Jenny_mathy's answer, but exploits FixedPoint to do the iteration.

Answer 10

Perl 6, 49 bytes

{($_,{.Str.parse-base(1+.comb.max)}...*==*).tail}

Test it

Expanded:

{
  (

    $_,                 # seed the sequence with the input

    {
      .Str
      .parse-base(      # parse in base:
        1 + .comb.max   # largest digit + 1
      )
    }

    ...                 # keep generating values until

    * == *              # two of them match (should only be in base 10)

  ).tail                # get the last value from the sequence
}

Answer 11

Python 2, 60 59 56 53 bytes

Saved 4 byte thanks to Felipe Nardi Batista
Saved 3 bytes thanks to ovs

f=lambda x,y=0:x*(x==y)or f(`int(x,int(max(x))+1)`,x)

Try it online!

Using a recursive lambda, comparing the result of the base conversion to the previous iteration.

Answer 12

PHP, 71 bytes

for(;$b<10&9<$n=&$argn;)$n=intval("$n",$b=max(str_split($n))+1);echo$n;

Try it online!

Answer 13

Pip, 17 bytes

Wa>9>YMXaaFB:1+ya

Takes input as a command-line argument. Try it online!

Explanation

This was fun--I got to pull out the chaining comparison operators.

We want to loop until the number is a single digit OR contains a 9. Equivalently, we want to loop while the number is multiple digits AND does not contain a 9. Equivalently, loop while the number is greater than 9 AND the maximum digit is less than 9: a>9>MXa.

Wa>9>YMXaaFB:1+ya
                   a is 1st cmdline arg (implicit)
     YMXa          Yank a's maximum digit into the y variable
Wa>9>              While a is greater than 9 and 9 is greater than a's max digit:
         aFB:1+y    Convert a from base 1+y to decimal and assign back to a
                a  At the end of the program, print a

Answer 14

C#, 257 244 243 244 233 222 bytes

using System.Linq;z=m=>{int b=int.Parse(m.OrderBy(s=>int.Parse(s+"")).Last()+""),n=0,p=0;if(b<9&m.Length>1){for(int i=m.Length-1;i>=0;i--)n+=int.Parse(m[i]+"")*(int)System.Math.Pow(b+1,p++);return z(n+"");}else return m;};

Well, C# always takes a lot of bytes but this is just ridiculous. None of the built-ins can handle an arbitrary base, so I had to calculate the conversion myself. Ungolfed:

using System.Linq;
z = m => {
int b = int.Parse(m.OrderBy(s => int.Parse(s + "")).Last()+""), n = 0, p = 0; //Get the max digit in the string
if (b < 9 & m.Length > 1) //if conditions not satisfied, process and recurse
{
    for (int i = m.Length - 1; i >= 0; i--)
        n += int.Parse(m[i] + "") * (int)System.Math.Pow(b+1, p++); //convert each base-b+1 representation to base-10
    return z(n + ""); //recurse
}
else return m; };

Answer 15

Mathematica, 92 bytes

f[x_]:=FromDigits[s=IntegerDigits@x,d=Max@s+1];(z=f@#;While[d!=10&&Length@s>1,h=f@z;z=h];z)&

Answer 16

Javascript (ES6) with 0 arrow function, 74 bytes

function f(a){a>9&&b=Math.max(...a)<9&&f(parseInt(a,b+1));alert(a)}f('11')

Answer 17

K4, 19 bytes

Solution:

{(1+|/x)/:x:10\:x}/

Examples:

q)\
  {(1+|/x)/:x:10\:x}/413574
83911
  {(1+|/x)/:x:10\:x}/13552
159
  {(1+|/x)/:x:10\:x}/17
3
  {(1+|/x)/:x:10\:x}/41253
29    

Explanation:

Use /: built-in to convert base.

{(1+|/x)/:x:10\:x}/ / the solution
{                }/ / converge lambda, repeat until same result returned
            10\:x   / convert input x to base 10 (.:'$x would do the same)
          x:        / store in variable x
 (     )/:          / convert to base given by the result of the brackets
    |/x             / max (|) over (/) input, ie return largest
  1+                / add 1 to this

Answer 18

Kotlin, 97 bytes

fun String.f():String=if(length==1||contains("9"))this else "${toInt(map{it-'0'}.max()!!+1)}".f()

Beautified

fun String.f(): String = if (length == 1 || contains("9")) this else "${toInt(map { it - '0' }.max()!! + 1)}".f()

Test

fun String.f():String=if(length==1||contains("9"))this else "${toInt(map{it-'0'}.max()!!+1)}".f()
val tests = listOf(
        5 to 5,
        17 to 3,
        999 to 999,
        87654321 to 9041998,
        41253 to 29
)

fun main(args: Array<String>) {
    tests.forEach { (input, output) ->
        val answer = input.toString().f()
        if (answer != output.toString()) {
            throw AssertionError()
        }
    }
}

TIO

TryItOnline

Answer 19

Japt, 25 bytes

>9©(A=Uì rw)<9?ßUs nAÄ):U

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Answer 20

Jelly, 9 bytes

DḅṀ‘$$µÐL

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Answer 21

C, 159 157 bytes

#include <stdlib.h>
char b[99];i=0;m=-1;c(n){do{m=-1;sprintf(b,"%d",n);i=0;while(b[i]){m=max(b[i]-48,m);i++;}n=strtol(b,0,++m);}while(b[1]&&m<10);return n;}

Answer 22

Scala, 119 bytes

if(x.max==57|x.length==1)x else{val l=x.length-1
f((0 to l).map(k=>(((x(k)-48)*math.pow(x.max-47,l-k))) toInt).sum+"")}

Try it online!

Scala, 119 bytes

if(x.max==57|x.length==1)x else f((0 to x.length-1).map(k=>(((x(k)-48)*math.pow(x.max-47,x.length-1-k))) toInt).sum+"")

Try it online!

Both methods work the same way, but in the first one I put x.length-1 in a variable, and in the second one I don't.