Smallest Zeroless Base

Question

Given a positive integer n, output the smallest base b >= 2 where the representation of n in base b with no leading zeroes does not contain a 0. You may assume that b <= 256 for all inputs.

Test Cases

1 -> 2 (1)
2 -> 3 (2)
3 -> 2 (11)
4 -> 3 (11)
5 -> 3 (12)
6 -> 4 (12)
7 -> 2 (111)
10 -> 4 (22)
17 -> 3 (122)
20 -> 6 (32)
50 -> 3 (1212)
100 -> 6 (244)
777 -> 6 (3333)
999 -> 4 (33213)
1000 -> 6 (4344)
1179360 -> 23 ([12, 9, 21, 4, 4])
232792560 -> 23 ([15, 12, 2, 20, 3, 13, 1])
2329089562800 -> 31 ([20, 3, 18, 2, 24, 9, 20, 22, 2])
69720375229712477164533808935312303556800 -> 101 ([37, 17, 10, 60, 39, 32, 21, 87, 80, 71, 82, 14, 68, 99, 95, 4, 53, 44, 10, 72, 5])
8337245403447921335829504375888192675135162254454825924977726845769444687965016467695833282339504042669808000 -> 256 ([128, 153, 236, 224, 97, 21, 177, 119, 159, 45, 133, 161, 113, 172, 138, 130, 229, 183, 58, 35, 99, 184, 186, 197, 207, 20, 183, 191, 181, 250, 130, 153, 230, 61, 136, 142, 35, 54, 199, 213, 170, 214, 139, 202, 140, 3])

Answer 1

Pyth, 6 bytes

f*FjQT

Verify all the test cases.

How it works

f*FjQT  ~ Full program.

f       ~ First positive integer where the condition is truthy.
   jQT  ~ The input converted to base of the current element.
 *F     ~ Product. If the list contains 0, then it's 0, else it is strictly positive.
          0 -> Falsy; > 0 -> Truthy.
        ~ Output the result implicitly.

Although Pyth's f operates on 1, 2, 3, 4, ... (starting at 1), Pyth treats numbers in base 1 (unary) as a bunch of zeros, so base 1 is ignored.

Answer 2

C,  52  50 bytes

i,k;f(n){for(i=2,k=n;k;)k=k%i++?k/--i:n;return i;}

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C (gcc),  47  45 bytes

i,k;f(n){for(i=2,k=n;k;)k=k%i++?k/--i:n;n=i;}

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Two bytes saved thanks to @Nevay's suggestion on @Kevin Cruijssen's answer!

Answer 3

Haskell, 56 52 48 bytes

b#n=n<1||mod n b>0&&b#div n b
f n=until(#n)(+1)2

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Pretty basic but can't think of any good ways to shorten it

EDIT: Thanks to Laikoni for saving me 4 bytes! Don't know why I never thought of !!0. I probably should've tried removing those parentheses but I have vague memories of some weird error when you try to use || and && together. Maybe I'm confusing it with the equality operators.

EDIT 2: Thanks @Lynn for shaving another 4 bytes! Don't know how I never knew about until before now.

Answer 4

Wolfram Language (Mathematica), 33 bytes

2//.i_/;DigitCount[#,i,0]>0:>i+1&

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Answer 5

Husk, 7 bytes

→V▼MBtN

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Explanation

→V▼MBtN
     tN    list of natural numbers starting from 2
   MB      convert the (implicit) input to each of those bases
 V▼        find the (1-based) index of the first result where the minimum digit is truthy
→          add 1 to this index

Answer 6

Jelly, 7 bytes

2b@Ạ¥1#

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Answer 7

Python 2, 57 bytes

n=x=input()
b=2
while x:z=x%b<1;b+=z;x=[x/b,n][z]
print b

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This is one byte shorter than a recursive function:

f=lambda n,b=1,x=1:b*(x<1)or f(n,b+(x%b<1),[x/b,n][x%b<1])

Answer 8

05AB1E, 6 bytes

-4 bytes thanks to Adnan

1µNвPĀ

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Answer 9

Husk, 9 bytes

←foΠ`B⁰tN

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Explanation

            -- input N
        tN  -- tail of [1..] == [2..]
←f(    )    -- filter with the following:
    `B⁰     --   convert N to that base
   Π        --   product (0 if it contains 0)
←           -- only keep first element

Answer 10

Java 8, 61 56 54 bytes

n->{int b=2,t=n;for(;t>0;)t=t%b++<1?n:t/--b;return b;}

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Explanation:

n->{            // Method with integer as both parameter and return-type
  int b=2,      //  Base-integer, starting at 2
      t=n;      //  Temp-integer, copy of the input
  for(;t>0;)    //  Loop as long as `t` is not 0
    t=t%b++<1?  //   If `t` is divisible by the base `b`
                //   (and increase the base `b` by 1 afterwards with `b++`)
       n        //    Set `t` to the input `n`
      :         //   Else:
       t/--b;   //    Divide `t` by the `b-1`
                //    (by decreasing the base `b` by 1 first with `--b`)
                //  End of loop (implicit / single-line body)
  return b;     //  Return the resulting base
}               // End of method

I have the feeling this can be golfed by using an arithmetic approach. It indeed can, with a port of @Steadybox' C answer, and then golfed by 2 bytes thanks to @Nevay.

Old (61 bytes) answer:

n->{int b=1;for(;n.toString(n,++b).contains("0"););return b;}

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Explanation:

n->{         // Method with Integer as both parameter and return-type
  int b=1;   //  Base-integer, starting at 1
  for(;n.toString(n,++b).contains("0"););
             //  Loop as long as the input in base-`b` does contain a 0,
             //  after we've first increased `b` by 1 before every iteration with `++b`
  return b;  //  Return the resulting base
}            // End of method

Answer 11

APL (Dyalog), 20 19 bytes

1+⍣{~0∊⍺⊥⍣¯1⊢n}≢n←⎕

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As usual, thanks to @Adám for helping out in chat and getting the code to work in TIO. Also, saving 1 byte.

This is tradfn (traditional function) body. To use it, you need to assign it a name (which is in TIO's header field), enclose it in ∇s (one before the name and one in TIO's footer field), and then call it using its name. Since it uses a quad (⎕) to take the user's input, it's called as f \n input instead of the usual f input

How?

1+⍣{~0∊⍺⊥⍣¯1⊢n}≢n←⎕  ⍝ Main function.
                  n←⎕ ⍝ Assigns the input to the variable n
1+⍣{           }≢     ⍝ Starting with 1, add 1 until the expression in braces is truthy
    ~0∊               ⍝ returns falsy if 0 "is in"
        ⊥             ⍝ convert
            ⊢n        ⍝ the input
         ⍣¯1          ⍝ to base
       ⍺              ⍝ left argument (which starts at 1 and increments by 1)

The function then returns the resulting base.

Answer 12

Japt, 8 bytes

@ìX e}a2

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Explanation

@    }a2

Return the first number (X) to pass the function, starting at 2

ìX

Convert the input number to an array of base-X digits.

e

Check if all digits are truthy.

Answer 13

JavaScript (ES6), 43 41 37 bytes

n=>(g=x=>x?g(x%b++?x/--b|0:n):b)(b=1)

Test cases

let f =

n=>(g=x=>x?g(x%b++?x/--b|0:n):b)(b=1)

console.log(f(1   )) // 2 (1)
console.log(f(2   )) // 3 (2)
console.log(f(3   )) // 2 (11)
console.log(f(4   )) // 3 (11)
console.log(f(5   )) // 3 (12)
console.log(f(6   )) // 4 (12)
console.log(f(7   )) // 2 (111)
console.log(f(10  )) // 4 (22)
console.log(f(17  )) // 3 (122)
console.log(f(20  )) // 6 (32)
console.log(f(50  )) // 3 (1212)
console.log(f(100 )) // 6 (244)
console.log(f(777 )) // 6 (3333)
console.log(f(999 )) // 4 (33213)
console.log(f(1000)) // 6 (4344)

Answer 14

Brachylog, 11 bytes

;.≜ḃ₎¬∋0∧1¬

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Explanation

;.≜ḃ₎           The Input represented in base Output…
     ¬∋0        …contains no 0
        ∧1¬     And Output ≠ 1

Answer 15

Python 2, 57 bytes

n=m=input()
b=2
while m:c=m%b<1;b+=c;m=(m/b,n)[c]
print b

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-1 thanks to Felipe Nardi Batista.
-2 thanks to Lynn (and now this is a dupe of her solution :D)

Answer 16

Proton, 40 bytes

x=>filter(y=>all(digits(y,x)),2..x+3)[0]

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Answer 17

R, 79 71 66 63 65 bytes

function(n){while(!{T=T+1;all(n%/%T^(0:floor(log(n,T)))%%T)})T
T}

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This answer is based on Giuseppe's re-arrangement in one single loop.

Saved 8 bytes thanks to JDL, and 6 bytes thanks to Giuseppe.

Answer 18

MATL, 13 12 bytes

`G@Q_YAA~}@Q

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-1 byte thanks to Luis Mendo. This program does not handle testcases bigger than 2^53 (flintmax, the maximum consecutive integer representable by a floating point type), as the default datatype is double in MATL. However, it should be able to find any arbitrary zeroless base below that number.

`            % Do while
 G           %  Push input
  @ _        %  Outputs the iteration number, negate.
     YA      %  Convert input to base given by the iteration number, the negative number is to instruct MATL we want an arbitrary high base with a integer vector rather than the default character vector we know from hexadecimal
       A~    %  If they're not all ones, repeat
         }   % But if they are equal, we finally
          @  %  Push the last base
   Q       Q %  As base 1 makes no sense, to prevent MATL from errors we always increase the iteration number by one.

Answer 19

><>, 56 52 51 49 bytes

:2&v
1(?\:&:&:@%@,:1%-:
 v~<
l<*v?=2
&v?/:&+1
n>&

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The interpreter bugs out on the last 2 testcases.
It only reads the first 16 digits of the input correct and then replaces the rest with other digits.

Answer 20

PHP, 59+1 bytes

using builtins, max base 36:

for($b=1;strpos(_.base_convert($argn,10,++$b),48););echo$b;

no builtins, 6360+1 bytes, any base:

for($n=$b=1;$n&&++$b;)for($n=$argn;$n%$b;$n=$n/$b|0);echo$b;

Run as pipe with -nR or try them online.

Answer 21

Actually,  12  11 bytes

╗1⌠╜¡m'0<⌡╓

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Uses this consensus. Thanks to Mego for byte-saving help in chat.

Answer 22

J, 26 bytes

]>:@]^:(0 e.]#.inv[)^:_ 2:

Would love to know if this can be improved.

The main verb is a the dyadic phrase:

>:@]^:(0 e.]#.inv[)^:_

which is given the input on the left and the constant 2 on the right. That main verb phrase then uses J's Do..While construct, incrementing the right y argument as long as 0 is an element of e. the original argument in base y.

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Answer 23

Lua, 77 76 bytes

b=2n=...N=n
while~~N>0 do
if N%b<1then
b=b+1N=n
else
N=N//b
end
end
print(b)

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Answer 24

Milky Way, 38 bytes

^^'%{255£2+>:>R&{~^?{_>:<;m_+¡}}^^^}

usage: ./mw base.mwg -i 3

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Explanation

code                                 explanation                    stack layout

^^                                   clear the preinitialized stack []
  '                                  push the input                 [input]
   %{                              } for loop
     255£                             push next value from 0..254   [input, base-2]
         2+                           add 2 to the get the base     [input, base]
           >                          rotate stack right            [base, input]
            :                         duplicate ToS                 [base, input, input]
             >                        rotate stack right            [input, base, input]
              R                       push 1                        [input, base, input, 1]
               &{~             }      while ToS (=remainder) is true ...
                  ^                    pop ToS                      [input, base, number]
                   ?{         }        if ToS (=quotient) ...
                     _>:<;              modify stack                [input, base, number, base]
                           m            divmod                      [input, base, quotient, remainder]
                           _+¡         else: output ToS (0) + SoS and exit
                                ^^^   pop everything but the input.

I'm sure this can be shortened using a while-loop instead of a for loop, but I couldn't get it to work.

Answer 25

Stacked, 23 bytes

{!2[1+][n\tb all]until}

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This increments ([1+]) J starting from two (2) while the base_J representation of the input has no zeroes (all and until).

Answer 26

Perl 5, 52 + 2 (-pa) = 54 bytes

$\=1;while($_&&=$F[0]){$\++;$_=int$_/$\while$_%$\}}{

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Answer 27

Ruby, 41 bytes

->n{(2..).find{!n.digits(_1).include? 0}}

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Answer 28

Stax, 7 bytes

Ç╔ü$■╖Ö

Run and debug it

Trivial brute force approach

Answer 29

Pip, 12 bytes

T NyYaTD Uoo

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Explanation

An iterative solution:

T NyYaTD Uoo
              a is command-line input; y is ""; o is 1 (implicit)
T             Loop until
  Ny          Min(y) is truthy:
         Uo     Increment o
     aTD        Convert a to list of digits in that base
    Y           Store that list in y
           o  After the loop, output o

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Alternate solution using filter, also 12 bytes:

@:NaTD_FI2,m
              a is command-line input; m is 1000 (implicit)
         2,m  Range from 2 to 999 (could be 256, but using m is shorter)
       FI     Filter, keeping elements for which this function returns truthy:
   aTD_         Convert a to list of digits in base (function argument)
  N             Get the minimum
@:            Take the first element of the filtered list

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