Enumerate all palindromic numbers (in decimal) between 0 and n

Question

Given a non-negative integer n, enumerate all palindromic numbers (in decimal) between 0 and n (inclusive range). A palindromic number remains the same when its digits are reversed.

The first palindromic numbers (in base 10) are given here:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393, 404, 414, 424, 434, 444, 454, 464, 474, 484, 494, 505, 515, ...

This is a code golf with the prize going to fewest characters. The palindromic numbers should be output one per line to stdout. The program should read n from the commandline or stdin.

Answer 1

Perl 5.10, 29 (or 39) characters

say for grep$_==reverse,0..<>

Needs the say feature enabled. 29 chars if you consider that to be free, otherwise 39 to add use 5.010;. Argument on STDIN.

Perl, 35 characters

#!perl -l
print for grep $_==reverse,0..<>

using the old perlgolf convention that #!perl is not counted but any flags following it are.

Perl, 36 characters

print$_,$/for grep $_==reverse,0..<>

If none of the others qualify.

Answer 2

Befunge 320 313 303 characters

(including significant newlines and whitespace)

 &:#v_v#  #                  :-1<
v91:<         v          <
0     >0.@    >\25**\1-:#^_v
pv   p09+1g09<^_           >$+     v
:>:25*%\25*/:| ^:p18:+1g18\<
 :          > >90g 1-:90p  | >  ^
>|           $^     <      >-|  ^  #<
 @           > 0 81p^        >:.25*,^
            ^                      <

I wonder if I could make this smaller by rerouting the paths...

Edit: redid the top part to avoid an extra line.

Answer 3

Perl 5.10 - 27 characters

map{say if$_==reverse}0..<>

Reads the argument from stdin.

Answer 4

Golfscript, 15 chars

~),{.`-1%~=},n*

Answer 5

Ruby 1.9, 39 characters

puts (?0..gets).select{|i|i==i.reverse}

Input (must not be terminated with a newline) via stdin. Example invocation:

echo -n 500 | ruby1.9 palinenum.rb

40 characters for a version which uses commandline args:

puts (?0..$*[0]).select{|i|i==i.reverse}

Answer 6

J, 20 characters

(#~(-:|.)@":"0)>:i.n

Answer 7

Python, 57 51 chars

for i in range(input()):
 if`i`==`i`[::-1]:print i

Usage:

echo 500 | python palindromic.py

Answer 8

Perl > 5.10 : 25 characters

map$_==reverse&&say,0..<>

Answer 9

APL (25 17)

↑t/⍨t≡∘⌽¨t←⍕¨0,⍳⎕

Answer 10

Javascript 122 108 107 chars...

I'm sure this can be golfed more - I'm new to this!

n=prompt(o=[]);for(i=0;i<=n;i++)if(i+''==(i+'').split("").reverse().join(""))o.push(i);alert(o.join("\n"));

or

n=prompt(o=[]);i=-1;while(i++<n)if(i+''==(i+'').split("").reverse().join(""))o.push(i);alert(o.join("\n"));

Answer 11

Perl - 43 chars

for$i(0..<>){if($i==reverse$i){print$i,$/}}

This is my first attempt at code golf, so I'm pretty sure a Perl pro could golf it down.

Answer 12

Haskell 66 characters

main=do n<-readLn;mapM_ putStrLn[s|s<-map show[0..n],s==reverse s]

Answer 13

Mathematica 61

Column@Select[0~Range~Input[],#==Reverse@#&@IntegerDigits@#&]

Answer 14

Brachylog

With the I/O format stated in the question, 8 bytes

≥ℕA↔A≜ẉ⊥

Try it online!

With modern PPCG I/O rules, 4 bytes

≥ℕ.↔

Try it online!

This is a function that generates all the outputs, not a full program like the previous example, and so doesn't comply with the spec as written, but I thought I'd show how the program would look like if the question had been written to modern I/O standards (which permit the use of functions, and output via generators).

Explanation

≥ℕ.↔ 
 ℕ    Generate natural numbers
≥     less than or equal to the input
  .   but output only the ones
   ↔  that would produce the same output if reversed

For the full program version, we create a temporary variable A to hold the output, explicitly labelize it (this is done implicitly for the main predicate of a program), and use the well-known ẉ⊥ technique for outputting a generator's elements to standard output.

Answer 15

Scala 59

(0 to readInt)filter(x=>""+x==(""+x).reverse)mkString("\n")

Answer 16

PHP, 59 55 53 chars

for($i=0;$i++<$argv[1];)if($i==strrev($i))echo"$i\n";

Usage

php palindromic.php 500

Edit : thanks Thomas

Answer 17

C, 98 characters

n,i,j,t;main(){for(scanf("%d",&n);i<=n;i-j?1:printf("%d ",i),i++)for(t=i,j=0;t;t/=10)j=j*10+t%10;}

Answer 18

k (23 chars)

{i@&{&/i=|i:$x}'i:!1+x}

Answer 19

Befunge, 97 (grid size 37x4=148)

#v&#:< ,*25-$#1._.@
:>:::01-\0v >-!#^_$1-
*\25*/:!#v_::1>\#* #*25*#\/#$:_$\25*%
   `-10:\<+_v#

Have a better Befunge answer to this question. This is Befunge-93 specifically; I could probably make it even more compact with Befunge-98. I'll include that in a future edit.

Since you can't operate on strings in Befunge, the best I could do was compute the digit-reverse of each number (which I'm surprised I was able to manage without p and g) and compare it to the original number. The digit-reverse takes up most of the code (basically the entire third and fourth lines).

Note that the program, as it stands now, prints the numbers backwards from the input down to 0. If this is a big deal, let me know. (The challenge only says to enumerate them, not specifically in increasing order.)

Answer 20

PHP, 64 58 bytes

for($i=0;$i<=$argv[1];print$i==strrev($i)?$i.'\n':'',$i++)

Changed $_GET['n'] to $argv[1] for command line input.

Answer 21

Perl 5, 24 bytes

map$_-reverse||say,0..<>

Try it online!

Answer 22

Jelly, 5 bytes

ŻŒḂƇY

Try it online!

Answer 23

05AB1E, 5 bytes

ƒNÂQ–

Explanation:

ƒ      # For N in range(0, input() + 1)
 N     #   Push N
  Â    #   Bifurcate (pushes N and N[::-1])
   Q   #   Check for equality
    –  #   If true, pop and print N

Uses CP-1252 encoding. Try it online!.

Answer 24

Python

n=raw_input('')
for a in range(0,int(n)+1):
    r=str(a)
    if str(a)==r[::-1]:
        print r

Answer 25

Groovy, 83

System.in.eachLine{(0..it.toInteger()).each{if("$it"=="$it".reverse())println(it)}}

Answer 26

Q (34 chars)

Pass n rather than n+1 as an argument for this Q solution.

{i(&)({all i=(|)i:($)x}')i:(!)1+x}

Answer 27

Q,32

{a(&)a~'((|:)')a:((-3!)')(!)1+x}

Answer 28

Q (33)

{if[x="I"$(|:) -3!x;:x]} each til

Probably a neater way to do this but anyway, sample usage (you enter n+1 to get to n):

q){if[x="I"$(|:) -3!x;:x]} each til  10
0 1 2 3 4 5 6 7 8 9

Suggestion by tmartin, gets it down to 29:

({$[x="I"$(|:) -3!x;x;]}')(!)

Same usage.

Answer 29

Python, 106 characters

import sys as a
print(type(a.argv[1]))
for x in range(int(a.argv[1])+1):
 x=str(x)
 if x==x[::-1]:print(x)

usage:

python a.py 500

Answer 30

C# (217 214 191 chars)

Golfed version:

using System;using System.Linq;class P{static void Main(){int n=int.Parse(Console.ReadLine());do{var t=(n+"").ToArray();Array.Reverse(t);Console.Write(n+""==new string(t)?n+"\n":"");}while(n-->0);Console.ReadLine();}}

Readable:

using System;
using System.Linq;
class P
{
    static void Main()
    {
        int n = int.Parse(Console.ReadLine());
        do
        {
            var t = (n + "").ToArray();
            Array.Reverse(t);
            Console.Write(n + "" == new string(t) ? n + "\n" : "");
        } while (n-->0);

        Console.ReadLine();
    }
}

This prints out palindromes in descending order making use of the n-->0 operator. (as n goes to 0).

*Edited version replaces do...while with while, saving 3 chars, but now you must input with n+1.

using System;using System.Linq;class P{static void Main(){int n=int.Parse(Console.ReadLine());while(n-->0){var t=(n+"").ToArray();Array.Reverse(t);Console.Write(n+""==new string(t)?n+"\n":"");}Console.ReadLine();}}

*edited: found a better way to reverse string without converting to array:

using System;using System.Linq;class P{static void Main(){int n=int.Parse(Console.ReadLine());while(n-->0)Console.Write(n+""==string.Join("",(""+n).Reverse())?n+"\n":"");Console.ReadLine();}}

Readable:

using System;
using System.Linq;
class P
{
    static void Main()
    {
        int n = int.Parse(Console.ReadLine());
        while (n-->0)
            Console.Write(n + "" == string.Join("", ("" + n).Reverse()) ? n + "\n" : ""); 
        Console.ReadLine();
    }
}