Lowest-Base Palindrome

Question

Given a number n, write a function that finds the smallest base b ≥ 2 such that n is a palindrome in base b. For example, an input of 28 should return the base 3 since the ternary representation of 28 is 1001. Although 93 is a palindrome in both base 2 and base 5, the output should be 2 since 2<5.

Input

A positive integer n < 2^31.

Output

Return the smallest base b ≥ 2 such that the base b representation of n is a palindrome. Do not assume any leading zeros.

Samples (input => output):

11 => 10

32 => 7

59 => 4

111 => 6

Rules

The shortest code wins.

Answer 1

CJam, 19 bytes / GolfScript, 23 bytes

q~:N;1{)_N\b_W%=!}g

or

~:N;1{).N\base.-1%=!}do

Try it online:

• CJam
• GolfScript

Examples

$ cjam base.cjam <<< 11; echo
10
$ cjam base.cjam <<< 111; echo
6
$ golfscript base.gs <<< 11
10
$ golfscript base.gs <<< 111
6

How it works

q~:N;   # Read the entire input, interpret it and save the result in “N”.
1       # Push 1 (“b”).
{       #
  )     # Increment “b”.
  _N\   # Duplicate “b”, push “N” and swap.
  b     # Push the array of digits of “N” in base “b”.
  _W%   # Duplicate the array and reverse it.
  =!    # Compare the arrays.
}g      # If they're not equal, repeat the loop.

For GolfScript, q~ is ~, _ is ., b is base, W is -1 and g is do.

Answer 2

GolfScript, 20 characters

~:x,2>{x\base.-1%=}?

A different approach with GolfScript other than Dennis'. It avoids the costly explicit loop in favour of a find operator. Try online.

~:x        # interpret and save input to variable x
,2>        # make a candidate list 2 ... x-1 (note x-1 is the maximum possible base)
{          # {}? find the item on which the code block yields true
  x\       # push x under the item under investigation
  base     # perform a base conversion
  .-1%     # make a copy and reverse it
  =        # compare reversed copy and original array
}?         

Answer 3

Japt, 12 9 bytes

Unless I've missed a trick (it's late!), this should work for all numbers up to and including at least 2**53-1.

In my (admittedly limited and entirely random) testing, I've gotten results up to base 11601 310,515(!) so far. Not too shabby when you consider JavaScript only natively supports bases 2 to 36.

@ìX êê}a2

Try it

• Thanks to ETH for pointing out something new to me that saved 3 bytes and increased the efficiency considerably.

---

Explanation

Implicit input of integer U.

@     }a2

Starting with 2, return the first number that returns true when passed through the following function, with X being the current number

ìX

Convert U to an array of base X digits.

êê

Test if that array is a palindrome.

Answer 4

Mathematica, 67 66 bytes

g[n_]:=For[i=1,1>0,If[(d=n~IntegerDigits~++i)==Reverse@d,Break@i]]

Can't really compete with GolfScript here in terms of code size, but the result for 2³² is basically returned instantly.

Answer 5

Python 2 (83)

def f(n,b=2):
 l=[];m=n
 while m:l+=[m%b];m//=b
 return l==l[::-1]and b or f(n,b+1)

I'm not sure what input/output format the question wanted. I wrote a function. The code uses an optional input b to track the current base it's testing. The while loops converts the number to a list of digits in base b.

The last line returns b if l is a palindrome, and recursively tries the next b otherwise. The index-by-Boolean trick doesn't work here because it would cause both options to be evaluated regardless of the Boolean, and the recursion would never bottom out.

Answer 6

JavaScript, 88 bytes

f=function(n){for(a=b='+1';a^a.split('').reverse().join('');a=n.toString(++b));return+b}

Ungolfed:

f = function(n) {
    for(a = b = '+1'; // This is not palindrome, but equals 1 so we have at least one iteration
        a ^ a.split('').reverse().join(''); // test a is palindrome
        a = n.toString(++b));
    return+b
}

Answer 7

Husk, 11 9 bytes

ḟoS=↔`B⁰2

Thanks @Zgarb for -2!

Try it online!

Explanation

ḟ(      )2  -- find least number ≥ 2 that satisfies:
     `B⁰    --   convert input to base (` flips arguments)
  S=↔       --   is palindrome (x == ↔x)

Answer 8

Javascript, 105 bytes

function f(n){for(var b=2,c,d;d=[];++b){for(c=n;c;c=c/b^0)d.push(c%b);if(d.join()==d.reverse())return b}}

JSFiddle: http://jsfiddle.net/wR4Wf/1/

Note that this implementation also works correctly for large bases. For example, f(10014) returns 1668 (10014 is 66 in base 1668).

Answer 9

R, 122 95 bytes

function(n)(2:n)[sapply(2:n,function(x){r={};while(n){r=c(n%%x,r);n=n%/%x};all(r==rev(r))})][1]

Three-year old solution at 122 bytes:

f=function(n)(2:n)[sapply(sapply(2:n,function(x){r=NULL;while(n){r=c(n%%x,r);n=n%/%x};r}),function(x)all(x==rev(x)))][1]

With some explanations:

f=function(n)(2:n)[sapply(
                    sapply(2:n,function(x){ #Return the decomposition of n in bases 2 to n
                                 r=NULL
                                 while(n){
                                     r=c(n%%x,r)
                                     n=n%/%x}
                                     r
                                     }
                           ),
                    function(x)all(x==rev(x))) #Check if palindrome
                   ][1] #Return the first (i. e. smallest) for which it is

Answer 10

Bash + coreutils, 100 bytes

for((b=1;b++<=$1;)){
p=`dc<<<${b}o$1p`
c=tac
((b<17))&&c=rev
[ "$p" = "`$c<<<$p`" ]&&echo $b&&exit
}

Uses dc to do base formatting. The tricky thing is dc's format is different for n > 16.

Testcases:

$ ./lowestbasepalindrome.sh 11
10
$ ./lowestbasepalindrome.sh 32
7
$ ./lowestbasepalindrome.sh 59
4
$ ./lowestbasepalindrome.sh 111
6
$ 

Answer 11

J - 28 char

#.inv~(-.@-:|.@)(1+]^:)^:_&2

Explained:

• #.inv~ - Expand left argument to the base in the right argument.

• (-.@-:|.@) - Return 0 if the expansion is palindromic, and 1 otherwise.

• (1+]^:) - Increment the right argument by one if we returned 1, else take no action.

• ^:_ - Repeat the above incrementing until it takes no action.

• &2 - Prepare the right argument as 2, making this a function of one argument.

Examples:

   #.inv~(-.@-:|.@)(1+]^:)^:_&2 (28)
3
   #.inv~(-.@-:|.@)(1+]^:)^:_&2 every 93 11 32 59 111  NB. perform on every item
2 10 7 4 6
   #.inv~(-.@-:|.@)(1+]^:)^:_&2 every 1234 2345 3456 4567 5678 6789
22 16 11 21 31 92

Answer 12

05AB1E, 8 bytes

¼[¼¾вÂQ#

Try it online!

Answer 13

Perl 5, 84 + 1 (-p) = 85 bytes

$\=1;{++$\;my@r;$n=$_;do{push@r,$n%$\}while$n=int$n/$\;@t=@r;map$_-pop@t&&redo,@r}}{

Try it online!

Answer 14

JavaScript 72 bytes

F=(n,b=2)=>eval(`for(t=n,a=c="";t;t=t/b|0)a=t%b+a,c+=t%b`)^a?F(n,b+1):b

console.log(F(11) == 10)

console.log(F(32) == 7)

console.log(F(59) == 4)

console.log(F(111) == 6)

Answer 15

Mathematica 42 bytes

A variation of Martin Ender's entry. Makes use of IntegerReverse (made available in version 10.3) which dispenses with IntegerDigits.

(i=2;While[#~IntegerReverse~i !=#,i++];i)&

Answer 16

Java 8, 103 bytes

n->{int b=1,l;for(String s;!(s=n.toString(n,++b)).equals(new StringBuffer(s).reverse()+""););return b;}

Explanation:

Try it here.

n->{                          // Method with integer as both parameter and return-type
  int b=1,                    //  Base-integer, starting at 1
      l;                      //  Length temp integer
  for(String s;               //  Temp String
      !(s=n.toString(n,++b))  //   Set the String to `n` in base `b+1`
                              //   (by first increase `b` by 1 using `++b`)
       .equals(new StringBuffer(s).reverse()+"");
                              //   And continue looping as long as it's not a palindrome
  );                          //  End of loop
  return b;                   //  Return the resulting base integer
}                             // End of method

Answer 17

Note: Pyth is newer than this question, so this answer is not eligible to win.

Pyth, 10 bytes

fq_jQTjQT2

Try it here.

Answer 18

Scala, 83 bytes

def s(n:Int)=(for(b<-2 to n;x=Integer.toString(n,b);if(x==x.reverse))yield(b)).min