What base is this number in?

Question

Here's a nice easy challenge:

Given a string that represents a number in an unknown base, determine the lowest possible base that number might be in. The string will only contain 0-9, a-z. If you like, you may choose to take uppercase letters instead of lowercase, but please specify this. You must output this lowest possible base in decimal.

Here is a more concrete example. If the input string was "01234", it is impossible for this number to be in binary, since 2, 3, and 4 are all undefined in binary. Similarly, this number cannot be in base 3, or base 4. Therefore, this number must be in base-5, or a higher base, so you should output '5'.

Your code must work for any base between base 1 (unary, all '0's) and base 36 ('0-9' and 'a-z').

You may take input and provide output in any reasonable format. Base-conversion builtins are allowed. As usual, standard loopholes apply, and the shortest answer in bytes is the winner!

Test IO:

#Input          #Output
00000       --> 1
123456      --> 7
ff          --> 16
4815162342  --> 9
42          --> 5
codegolf    --> 25
0123456789abcdefghijklmnopqrstuvwxyz    --> 36

Answer 1

Python, 27 22 bytes

lambda s:(max(s)-8)%39

This requires the input to be a bytestring (Python 3) or a bytearray (Python 2 and 3).

Thanks to @AleksiTorhamo for golfing off 5 bytes!

Test it on Ideone.

How it works

We begin by taking the maximum of the string. This the code points of letters are higher than the code points of digits, this maximal character is also the maximal base 36 digit.

The code point of '0' – '9' are 48 – 57, so we must subtract 48 from their code points to compute the corresponding digits, or 47 to compute the lowest possible base. Similarly, the code points of the letters 'a' – 'z' are 97 – 122. Since 'a' represents the digit with value 10, we must subtract 87 from their code points to compute the corresponding digits, or 86 to compute the lowest possible base. One way to achieve this is as follows.

The difference between 97 and 58 (':', the character after '9') is 39, so taking the code points modulo 39 can achieve the subtraction. Since 48 % 39 = 9, and the desired result for the character '0' is 1, we first subtract 8 before taking the result modulo 39. Subtracting first is necessary since otherwise 'u' % 39 = 117 % 39 = 0.

c    n    n-8    (n-8)%39
0    48    40     1
1    49    41     2
2    50    42     3
3    51    43     4
4    52    44     5
5    53    45     6
6    54    46     7
7    55    47     8
8    56    48     9
9    57    49    10
a    97    89    11
b    98    90    12
c    99    91    13
d   100    92    14
e   101    93    15
f   102    94    16
g   103    95    17
h   104    96    18
i   105    97    19
j   106    98    20
k   107    99    21
l   108   100    22
m   109   101    23
n   110   102    24
o   111   103    25
p   112   104    26
q   113   105    27
r   114   106    28
s   115   107    29
t   116   108    30
u   117   109    31
v   118   110    32
w   119   111    33
x   120   112    34
y   121   113    35
z   122   114    36

Answer 2

Python, 25 bytes

lambda x:int(max(x),36)+1

Defines a lambda that takes the string x. Finds the largest digit in the string (sorted with letters above digits, by python's default), and converts to base 36. Adds 1, because 8 is not in base 8.

Answer 3

Jelly, 4 bytes

ṀØBi

Requires uppercase. Try it online! or verify all test cases.

How it works

ṀØBi  Main link. Arguments: s (string)

Ṁ     Yield the maximum of s.
 ØB   Yield "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz".
   i  Find the 1-based index of the maximum in that string.

Answer 4

Haskell, 34 bytes

f s=length['\t'..maximum s]`mod`39

Uses the mod(ord(c)-8,39) idea from Dennis.

41 bytes

g '0'=1
g 'W'=1
g x=1+g(pred x)
g.maximum

---

45 bytes:

(`elemIndex`(['/'..'9']++['a'..'z'])).maximum

Outputs like Just 3.

Answer 5

Cheddar, 34 29 21 bytes

Saved 8 bytes thanks to Dennis!!!

s->(s.bytes.max-8)%39

Uses lowercase letters

Try it online

Explanation

s -> (      // Input is `s`
  s.bytes    // Returns array of char codes
   .max      // Get maximum item in array
) % 39      // Modulus 39

Answer 6

05AB1E, 6 bytes

{¤36ö>

Takes letters in upper case.

Explanation

{       # sort
 ¤      # take last
  36ö   # convert from base 36 to base 10
     >  # increment

Try it online

Answer 7

Actually, 6 bytes

M6²@¿u

Try it online!

Answer 8

Julia, 22 bytes

!s=(maximum(s)-'')%39

There's a BS character (0x08) between the quotes. Try it online!

Answer 9

JavaScript (ES6), 41 37 bytes

s=>parseInt([...s].sort().pop(),36)+1

Edit: Saved 4 bytes thanks to @edc65.

Answer 10

Haskell, 55 40 bytes

f=(\y->mod(y-8)39).Data.Char.ord.maximum

Thanks @Dennis for his approach. (take that, @xnor ;))

Answer 11

Perl 6: 18 bytes

{:36(.comb.max)+1}

Defines a lambda that takes a single string argument, and returns an integer. It splits the string into characters, finds the "highest" one, converts it to base 36, adds 1.

{(.ords.max-8)%39}

This one uses the modulo approach from Dennis. Same length.

Answer 12

Retina, 28 bytes

O`.
.\B

{2`
$`
}T01`dl`_o
.

Try it online! (The first line enables a linefeed-separated test suite.)

Explanation

O`.

This sorts the characters of the input.

.\B

This removes all characters except the last, so the first two stages find the maximum character.

{2`
$`
}T01`dl`_o

These are two stages which form a loop. The first one duplicates the first character and the second one "decrements" it (replacing e.g. x with w, a with 9 and 1 with 0). The latter stage encounters a zero as the first character, it removes it instead. This is a standard technique for generating a range of characters, given the upper end. Hence, this generates all "digits" from 0 to the maximum digit.

.

Finally, we simply count the number of digits, which gives us the base.

Answer 13

R, 99 89 85 bytes

Look ! Less than 100 bytes !
Look ! 10 bytes off !
Look ! 4 bytes off !

ifelse((m=max(strsplit(scan(,''),"")[[1]]))%in%(l=letters),match(m,l)+10,strtoi(m)+1)

Ungolfed :

l=letters                  #R's built-in vector of lowercase letters

n=scan(what=characters())  #Takes an input from STDIN and convert it to characters

m=max(strsplit(n,"")[[1]]) #Splits the input and takes to max. 
                           #`letters` are considered > to numbers (i.e. a>1)


ifelse(m%in%l,match(m,l)+10,strtoi(m)+1) #If the max is in `letters`,
                                             #outputs the matching position of `m`in `letters` + 10 (because of [0-9]). 
                                             #Else, outputs `m` (as a number) + 1.

As often, this answer makes use of the ifelse function : ifelse(Condition, WhatToDoIfTrue, WhatToDoElse)

Answer 14

Scala, 25 bytes

print((args(0).max-8)%39)

Run it like:

$ scala whatbase.scala 0123456789abcdefghijklmnopqrstuvwxyz

Answer 15

PHP, 51 38 bytes

(From Dennis) ^^

<?=(ord(max(str_split($argv[1])))-8)%39;

Other proposal without Dennis' trick

<?=($a=max(str_split($argv[1])))<a?$a+1:ord($a)-86;

• Takes input as argument $argv[1];
• Take max character (using ASCII) values
• If it is a number (inferior to < 'a' ascii value) then output number+1
• Else output ascii value -86 (97 for 'a' in ascii, -11 for 'a' is 11th base-digit)

Answer 16

Octave, 20 bytes

@(a)mod(max(a)-8,39)

Answer 17

Pyke, 6 bytes

Seb36h

Try it here!

Se     -   sorted(input)[-1]
  b36  -  base(^, 36)
     h - ^ + 1

Answer 18

Java 7, 67 61 bytes

int c(char[]i){int m=0;for(int c:i)m=m>c?m:c;return(m-8)%39;}

(m-8)%39 is thanks to @Dennis' amazing answer.

Ungolfed & test code:

Try it here.

class Main{
  static int c(char[] i){
    int m = 0;
    for(int c : i){
      m = m > c
           ? m
           : c;
    }
    return (m-8) % 39;
  }

  public static void main(String[] a){
    System.out.println(c("00000".toCharArray()));
    System.out.println(c("123456".toCharArray()));
    System.out.println(c("ff".toCharArray()));
    System.out.println(c("4815162342".toCharArray()));
    System.out.println(c("42".toCharArray()));
    System.out.println(c("codegolf".toCharArray()));
    System.out.println(c("0123456789abcdefghijklmnopqrstuvwxyz".toCharArray()));
  }
}

Output:

1
7
16
9
5
25
36

Answer 19

C89, 55 53 52 50 bytes

f(s,b)char*s;{return*s?f(s+1,*s>b?*s:b):(b-8)%39;}

-8%39 shamelessly stolen from Dennis

Test

test(const char* input)
{
    printf("%36s -> %u\n", input, f((char*)input,0));
}

main()
{
    test("00000");
    test("123456");
    test("ff");
    test("4815162342");
    test("42");
    test("codegolf");
    test("0123456789abcdefghijklmnopqrstuvwxyz");
}

Output

                               00000 -> 1
                              123456 -> 7
                                  ff -> 16
                          4815162342 -> 9
                                  42 -> 5
                            codegolf -> 25
0123456789abcdefghijklmnopqrstuvwxyz -> 36

Saved 2 bytes thanks to Toby Speight

Saved 2 bytes thanks to Kevin Cruijssen

Answer 20

C, 55 bytes

This answer assumes that the input is in ASCII (or identical in the numbers and letters, e.g. ISO-8859 or UTF-8):

m;f(char*s){for(m=0;*s;++s)m=m>*s?m:*s;return(m-8)%39;}

We simply iterate along the string, remembering the largest value seen, then use the well-known modulo-39 conversion from base-{11..36}.

Test program

int printf(char*,...);
int main(int c,char **v){while(*++v)printf("%s -> ",*v),printf("%d\n",f(*v));}

Test results

00000 -> 1
123456 -> 7
ff -> 16
4815162342 -> 9
42 -> 5
codegolf -> 25
0123456789abcdefghijklmnopqrstuvwxyz -> 36

Answer 21

Mathematica, 34 32 bytes

2 bytes saved thanks to Martin Ender

Max@Mod[ToCharacterCode@#-8,39]&

I decided the different method deserved a new answer.

_{method stolen inspired by Dennis' solution}

Answer 22

Mathematica, 34 32 bytes

saved 2 bytes thanks to Martin Ender

Max@BaseForm[Characters@#,36]+1&

Defines a pure function that takes a string as input.

Splits the input into characters, converts them to base 36 numbers, and returns the maximum +1.

Answer 23

C# REPL, 17 bytes

x=>(x.Max()-8)%39

Just ported @Dennis's answer to C#.

Answer 24

CJam, 10 bytes

Thanks to Martin Ender for saving me a few bytes!

Uses Dennis's formula

q:e>8-i39%

Try it online

CJam, 18 16 btyes

Alternative solution:

A,s'{,97>+q:e>#)

Try it online

A,s'{,97>+       e# Push the string "0123456789abcdefghijklmnopqrstuvwxyz"
          q      e# Get the input
           :e>   e# Find the highest character in the input
              #  e# Find the index of that character in the string
               ) e# Increment

Answer 25

R, 62 54 bytes

max(match(strsplit(scan(,''),"")[[1]],c(0:9,letters)))

Ungolfed:

max(
  match( # 2: Finds the respective positions of these characters
    strsplit(scan(,''),"")[[1]], # 1: Breaks the input into characters
                                c(0:9,letters)) # 3: In the vector "0123...yz"
                                                )

Update: shaved off 8 bytes due to the redundancy of na.rm=T under the assumption of input validity.

A 39% improvement in size compared to Frédéric's answer. Besides that, it runs a wee bit faster: 0.86 seconds for 100000 replications versus 1.09 seconds for the competing answer. So the one of mine is both smaller and more efficient.

Answer 26

Vyxal, 6 bytes

kr$Gḟ›

Try it Online!

Answer 27

Dyalog APL, 10 bytes

Prompts for uppercase input.

⌈/⍞⍳⍨⎕D,⎕A

⌈/ maximum

⍞ characters of input

⍳⍨ 1-indexed into

⎕D, all digits followed by

⎕A all characters

TryAPL online!

Answer 28

BASH 70

grep -o .|sort -r|head -c1|od -An -tuC|sed s/$/-86/|bc|sed s/-/39-/|bc

Input letters are lowercase.

Answer 29

JavaScript, 57 50 48 bytes

7 bytes saved thnks to @kamaroso97 2 bytes saved thanks to @Neil

n=>Math.max(...[...n].map(a=>parseInt(a,36))+1)

Original answer:

n=>n.split``.map(a=>parseInt(a,36)).sort((a,b)=>b-a)[0]+1

Answer 30

Powershell, 62 bytes

function t($s){(($s[0..$s.Length]|measure -Max).Maximum-8)%39}

%39 @Dennis, of course :)