# What base is this number in?

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

• Can I output in base 36? Aug 23 '16 at 4:42
• @LeakyNun Geez, I hope not. Aug 23 '16 at 5:22
• @LeakyNun You must output this lowest possible base in decimal. Aug 23 '16 at 6:09
• @RohanJhunjhunwala If that's your languages closest equivalent to a string, I don't see why not. Aug 23 '16 at 6:16
• Usually unary is all 1s, and leading zeros are not standard for any positional-based numeric system. Aug 23 '16 at 10:35

# 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

• If you make it Python 3, and take the input as a byte string, you can drop the ord() and win by 3 bytes. :) Aug 23 '16 at 2:22
• Nice idea! Let me ask the OP. Aug 23 '16 at 2:25
• @AleksiTorhamo NOOOOOOOOOOOO y u do dis Aug 23 '16 at 2:42

# 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.

# 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.

• It's actually 7 bytes, not 4. The first 2 chars are multi-byte. Aug 23 '16 at 9:08
• @Nicomak This answer is encoded in the Jelly code page, where all of these characters are encoded as 1 byte each. Aug 23 '16 at 10:31

f s=length['\t'..maximum s]mod39


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.

# Cheddar, 3429 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

• Or you could just put quotes around the input Aug 23 '16 at 0:26
• @DJMcMayhem .___. i didn't even know my own language could do that Aug 23 '16 at 0:37
• How about (-)&8 instead of n->n-8? Aug 23 '16 at 1:23
• @ConorO'Brien >_>_>_> I haven't gotten to that yet. I was just planning to do it and then this challenge was posted. Bassically f&n bonds n to first arg of the function. Aug 23 '16 at 1:49
• @Downgoat Oh. >_> Aug 23 '16 at 1:49

# 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

• Forgive my naivete with 05AB1E, but do you mean convert FROM base 36 (to base 10)? Aug 23 '16 at 18:30
• @Keeta You are of course correct. My bad. Aug 23 '16 at 18:33

# Actually, 6 bytes

M6²@¿u


Try it online!

# Julia, 22 bytes

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


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

• what does the -'' do? Aug 23 '16 at 0:53
• It subtracts 8 from the code point and returns an integer. Aug 23 '16 at 0:56

## JavaScript (ES6), 41 37 bytes

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


Edit: Saved 4 bytes thanks to @edc65.

• use pop() to save 4 Aug 23 '16 at 8:12
• @edc65 I can't believe that's not under JavaScript tips.
– Neil
Aug 23 '16 at 8:31

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


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

• I think you can remove f= for 38 bytes since f doesn't take explicit arguments. Aug 23 '16 at 15:00

# 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.

## Retina, 28 bytes

O.
.\B

{2
$ }T01dl_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$
}T01dl_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.

# R, 9989 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 min 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)

• I love your version; however, treating letters and numbers separately creates those pesky extra bytes. Please have a look at my solution that uses a different method. Aug 23 '16 at 13:05
• Your answer is interesting indeed. I will use the your scan method to golf some bytes ;) Aug 23 '16 at 13:10

# Scala, 25 bytes

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

Run it like:

$scala whatbase.scala 0123456789abcdefghijklmnopqrstuvwxyz # PHP, 51 38 bytes (From Dennis) ^^ <?=(ord(max(str_split($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 $argv[1]; • Take max character (using ASCII) values • If it is a number (inferior to < 'a' ascii value) then output number+1 • Else output ascii value -86 (97 for 'a' in ascii, -11 for 'a' is 11th base-digit) • It's too bad PHP has such verbose function names: <?=base_convert(max(str_split($argv[1])),36,10)+1 is an elegant solution, but at 49 bytes!
– user15259
Aug 23 '16 at 18:15
• @YiminRong you can use intval() instead of base_convert() which shortens down to 38 bytes <?=intval(max(str_split($argn)),36)+1; tio: tio.run/##K8go@P/… Jun 26 '19 at 15:53 # Octave, 20 bytes @(a)mod(max(a)-8,39)  ## Pyke, 6 bytes Seb36h  Try it here! Se - sorted(input)[-1] b36 - base(^, 36) h - ^ + 1  # 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  • Instead of Math.max() you can use m = m>c?m:c Aug 23 '16 at 9:53 • @RobAu Ah of course, thanks. Completely forgot about it.. Sometimes I forget the easiest codegolfing things in Java that are even mentioned multiple times in the Tips for Codegolfing in Java post. Thanks for the reminder. Aug 23 '16 at 12:30 • If you switch to Java 8 you can replace this whole function with a lambda that does a single reduce Aug 23 '16 at 12:32 • @BlueRaja-DannyPflughoeft I know, which is why I specifically mentioned it as Java 7. Feel free to post a Java 8 lambda as a separate answer. Aug 23 '16 at 12:33 • @BlueRaja-DannyPflughoeft I wonder if that would end up with less bytes.. Aug 23 '16 at 13:46 ## C89, 555352 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 • You can save 2 bytes with the non-prototype declaration: f(char*s,int b) becomes f(s,b)char*s;. Aug 23 '16 at 14:13 • You can save 3 bytes by removing the unnecessary parenthesis and space: f(s,b)char*s;{return*s?f(s+1,*s>b?*s:b):(b-8)%39;} Aug 23 '16 at 14:24 • @KevinCruijssen thx – YSC Aug 23 '16 at 14:46 # 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  • Couldn't you remove the m=0? If m appears at the top level of the file, its extern which implies static which implies it is initialized to zero. Aug 24 '16 at 3:47 • @Batman - yes, but only if you won't call f() more than once. I know that almost anything's fair game in golf, but my professional instincts regard that as too fragile! Aug 24 '16 at 7:46 • On further thought, I could make it an external requirement to reset m between calls to f(). Then my test program could still work. Aug 24 '16 at 7:47 • @Batman: on Code Golf Meta, the majority opinion on the question "Do function submissions have to be reusable?" seems to be against allowing single-use functions. So I'll stick with what I have. Thanks for the suggestion, anyway. Aug 24 '16 at 8:03 # 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 • Use some prefix notation: Max@Mod[ToCharacterCode@#-8,39]& (same goes for your other answer) Aug 23 '16 at 7:43 • Also, you need to add & to the end to indicate an anonymous function. Aug 23 '16 at 11:43 • You forgot one @ in both of your answers (ToCharacterCode@# and Characters@#). Aug 23 '16 at 16:37 # 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. • Max@BaseForm[Characters@#,36]+1& Aug 23 '16 at 14:19 ## C# REPL, 17 bytes x=>(x.Max()-8)%39  Just ported @Dennis's answer to C#. ## 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  ## 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. # Vyxal, 6 bytes kr$Gḟ›


Try it Online!

# 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!

grep -o .|sort -r|head -c1|od -An -tuC|sed s/$/-86/|bc|sed s/-/39-/|bc  Input letters are lowercase. # JavaScript, 5750 48 bytes 7 bytes saved thnks to @kamaroso97 2 bytes saved thanks to @Neil n=>Math.max(...[...n].map(a=>parseInt(a,36))+1)  Original answer: n=>n.split.map(a=>parseInt(a,36)).sort((a,b)=>b-a)[0]+1  • You can knock off 7 bytes with n=>Math.max(...n.split.map(a=>parseInt(a,36)+1)). Aug 23 '16 at 5:45 • @kamoroso94 I didn't realize Math.max existed. Thanks for telling me about it! Aug 23 '16 at 5:52 • [...s] is shorter than s.split. – Neil Aug 23 '16 at 7:31 # Perl, 30 27 bytes Includes +1 for -p Run with the input on STDIN, e.g. base.pl <<< codegolf  base.pl: #!/usr/bin/perl -p \@F[unpack"W*"];$_=@F%39-9