Palindromic palindrome generator

Question

Create a program, that converts input string to a palindrome starting with the input string. The program itself must be a palindrome.

For example input: neverod, print neveroddoreven. You should handle multi-word, multi-line inputs as well.

Answer 1

Dyalog APL, 6 4

⌽,,⌽

Try it here.

Other solutions:

⌽,⊢⊢,⌽
⌽⊢⊢,⊢⊢⌽

Explanation

They are just:

{⌽((,⍵),(⌽⍵))}
{⌽((,⍵)⊢((⊢⍵),(⌽⍵)))}
{(⌽⍵)⊢((⊢⍵),((⊢⍵)⊢(⌽⍵)))}

Monadic , and ⊢ does nothing on strings. Dyadic , is concatenation. Dyadic ⊢ returns its right operand. And ⌽ is obviously reversion.

Answer 2

piet 19x2 = 38

Accepts input until it encounters 0x00. Doesn't terminate, but output will be correct.

Answer 3

APL, 9

⍞←Z,⌽,Z←⍞

Explanation:

       Z←⍞  ⍝ read a line from the keyboard, and store it in Z
      ,     ⍝ flatten into one-dimensional array (this has no effect here)
     ⌽      ⍝ reverse
   Z,       ⍝ concatenate Z to its reverse
⍞←         ⍝ explicit output (not necessary, but it makes it a palindrome)

Answer 4

CJam, 13 bytes

qL;_-1%1-_;Lq

qL;                 "Read the input, Put an empty array on stack and pop that array";
   _-1%             "Now the string is on top, make a copy and reverse the copy";
       1-           "Remove occurrences of integer 1 from the reverse string. [no-op]";
         _;         "Copy the reversed string and pop it";
           Lq       "Put an empty array on stack and read the remaining input. Remaining";
                    "input will be empty as we already read whole of the input";

Try it online here

---

or..

GolfScript, 9 bytes

.-1%}%1-.

.                 "Input is already on stack. Make a copy";
 -1%              "Reverse the copy";
    }             "This marks the beginning of a super comment. Anything after this in the";
                  "code is a comment";
     %1-.         "no-op comment";

Try it here

Answer 5

C++, 162 bytes

#include<cstdio>//
main(){int c=getchar();if(c>0)putchar(c),main(),putchar(c);}//};)c(rahctup,)(niam,)c(rahctup)0>c(fi;)(rahcteg=c tni{)(niam
//>oidtsc<edulcni#

C, 117 bytes

main(c){c=getchar();if(c>0)putchar(c),main(),putchar(c);}//};)c(rahctup,)(niam,)c(rahctup)0>c(fi;)(rahcteg=c{)c(niam

Answer 6

Haskell, 102+22=124 bytes

a b fa=fa<|>b
fa=reverse>>=a
main=interact fa
niam=main
af tcaretni=niam
a=>>esrever=af
b>|<af=af b a

This must be run with the Control.Applicative module in scope, which can be set via the ghci init file .ghci: :m Control.Applicative (-> +22 bytes).

No comment trick, just 7 functions where 4 of them are never called.

If functions (instead of programs) are allowed:

Haskell, 55+22=77 bytes

a b fa=fa<|>b
f=reverse>>=a
a=>>esrever=f
b>|<af=af b a

Usage f "qwer"-> "qwerrewq"

Edit: the previous version was just wrong.

Answer 7

Pyth, 11 bytes

+z_z " z_z+

In Pyth, anything preceding with a space is not printed. So we simply add the negative of the string to itself, put a space, start a string and mirror the left side of the quote"

Try it online here

Answer 8

Ruby, 44

s=gets p
s+=s.reverse||esrever.s=+s
p steg=s

Takes a multiline string as input from stdin, outputs a Ruby representation of that string concatenated to its reverse. Could trim a character by replacing || with # to comment out the dead code on the second line.

Answer 9

Jolf, 9 bytes

Newer language, non-competing

Try Here

aη+i_i+ηa

Explanation: I only just started Jolf and I don't think I'm explaining this properly.

aη         alert function, arity of the function can't be reduced by 1 so it stays at 1
  +i_i     concatenate the input with the reversed input
      +η   arity of the add reduced by 1, it just takes the following character (a)
        a  returns the input

Answer 10

PowerShell, 67

$args|%{$_+-join$_[$_.Length..0]}#}]0..htgneL._$[_$nioj-+_${%|sgra$

Try it online

As suggested by @mazzy the code can be shortened by 12 bytes when using a static range. This, however, limits the input length to 9KBytes. Theoratically 9MBytes would be possible but it would slow down the code significantly.

$args|%{$_+-join$_[9kb..0]}#}]0..bk9[_$nioj-+_${%|sgra$

Answer 11

Fuzzy Octo Guacamole, 17 bytes

_{FOG is newer than this challenge, so this is non-competing.}

^dz''sjX@Xjs''zd^

Alt solution in 19 bytes:

^Czs''.jX@Xj.''szC^

They both take input, duplicate and reverse, and join the stack.

Explanation:

^dz''sj@js''zd^
^                # Get input
 d               # Duplicate ToS (input)
  z              # Reverse ToS
   ''            # Push empty string (for joining separator)
     s           # Move the empty string to the inactive stack
      j          # Join the active stack with the top of the inactive stack as the delimiter and push the result.
       X         # Print the ToS
        @        # End the program
        Xjs''zd^  # Backwards version of the beginning.

Answer 12

tinyBF, 40

|=||==>|=|=|=+|=>==||==>=|+=|=|=|>==||=|

My first thought was Brainfuck, but it's impossible to match the braces... fortunately tinyBF has simpler flow control.

No comments, it takes a null terminated string as input and returns the result in a null terminated string. You can test it here, just be forewarned that it doesn't halt (although Firefox at least prompts to stop the unresponsive script).

Commented:

|=|                        Retrieve a byte of input.
|                          Positive (opening) bracket.
   ==                      Output the byte.
   >                       Move the pointer in positive direction.
   |=|                     Retrieve a byte of input.
   =                       Switch direction to negative.
|                          Negative (closing) bracket.
=                          Switch direction.
+                          Increment byte to execute return loop.
|                          Opening bracket.
   =>                      Move the pointer in negative direction.
   ==                      Output the byte.
|                          Closing bracket.
|=|                        Output the null terminator.
|==>|=|=|=+|=>==|          ...and keep null terminating it just to be sure.

Note that if you encode it into 2 bit instructions, it cuts the size to 10 bytes (wouldn't be a palindrome).

Answer 13

Python 3, 59 bytes

a=input()#
print(a+a[::-1])#([1-::]a+a)tnirp
#()tupni=a

I tried my best to find a solution that only used one line but I had no luck.

Python 3, 79 bytes

a=input()#()tupni=a#
print(a+a[::-1])#([1-::]a+a)tnirp
#a=input()#()tupni=a

My original attempt in which every line is a palindrome. I don't think that it is necessary for this challenge, but I included it just in case.

Answer 14

Vitsy, 9 bytes

z:Zr?rZ:z
z          Grab all string input from the command line arguments.
 :         Duplicate this stack.
  Z        Print all elements in this stack as a string.
   r       Reverse (reverses an empty stack).
    ?      Go right a stack.
     r     Reverse (reverses the input).
      Z    Print all elements in this stack as a string.
       :   Duplicate the stack (duplicates an empty stack).
        z  Grab all input from the command line (the command line arguments stack is already empty).

Try it online!

Answer 15

Befunge, 37 bytes

~:0`!#v_:,
  >:#,_@_,#:>  
,:_v#!`0:~

Try it online!

The top line pushes and prints every character of input. The second line (before the @) prints the stack in reverse, but we enter at the contional _ to consume the -1 generated when finish reading input. The other half of the code (including those ugly trailing newlines) makes the source a palindrome, but nevers runs.

Answer 16

C# (33 32 + 1) * 2 = 68 66 bytes

saved 2 bytes to the use of .Aggregate()

s=>s+s.Aggregate("",(a,b)=>b+a);//;)a+b>=)b,a(,""(etagerggA.s+s>=s

Oh the good old lambda, you can catch it with

Func<string, string> f=<lambda here>

and then call it with

f("neverod")

Answer 17

Perl, 45 bytes

;print$_=<>,~~reverse;m;esrever~~,><=_$tnirp;

Pretty straightforward, prints the input ($_=<>) followed by the reverse of it. reverse returns $_ because we're using it in scalar context by prefixing with ~~. Then we match (m// using ; as delimiter), in void context, against the reverse of the script.

If we can guarrantee we won't have to create a palindrome of esrever,><=_$tnirp we can shorten the code to 43 bytes:

g.print$_=<>,reverse.m.esrever,><=_$tnirp.g

Usage

echo -n 'neverod' | perl -e 'g.print$_=<>,reverse.m.esrever,><=_$tnirp.g'
neveroddoreven

---

Perl, 26 bytes

Includes 25 bytes code + 1 for -p.

$_.=reverse;m;esrever=._$

I don't think this is valid since it requires the -p flag which I don't think can be easily combined into the script contents to make a true palindrome. Pretty much the same calls as above, except it relies on the fact that -p also adds a ; behind the scenes (on newer Perls...) to close the m//.

Usage

echo -n 'neverod' | perl -pe ';$_.=reverse;m;esrever=._$;'
neveroddoreven

Answer 18

Pyth, 15

 k k+ z_z +k k 

Notice the space at the beginning and at the end.

Quite annoying task in Pyth. z_z prints the desired palindrome, but it prints z (the input string) and _z the inverse on two different lines. + combines the two words, but + at the end requires two new statements at the end (and at the beginning). I choose k and k, which are just empty strings. Then a lot of white space, which suppresses printing (and printing empty spaces, which generate of course line breaks).

Since the white space suppresses every output except the +z_z, you can replace the ks and literal with arity 0. E.g. 1 2+ z_z +2 1 or T Z+ z_z +Z T.

Try it online.

Answer 19

Javascript, 137 bytes

I'm not using the "comment trick" but I am using the escaped quotation mark trick lol.

"a\"};))''(nioj.)(esrever.)''(tilps.b(tacnoc.b nruter{)b(a noitcnuf";function a(b){return b.concat(b.split('').reverse().join(''));};"\a"

Answer 20

JavaScript, 58 bytes

p=>p+[...p].reverse().join``//``nioj.)(esrever.]p...[+p>=p

Answer 21

PHP, 28+1+28=57 bytes

<?=($x=$argv[1]).strrev($x);#;)x$(verrts.)]1[vgra$=x$(=?<

takes input from command line argument. quote for multi-word, escape newlines for multi-line.

Answer 22

Python 2, 51 bytes

s=input();print s+s[::-1]#]1-::[s+s tnirp;)(tupni=s

I'm surprised no-one thought of this! Needs quoted input (' or "). If functions were allowed, I could have done this for 37 bytes instead:

lambda x:x+x[::-1]#]1-::[x+x:x adbmal

Answer 23

C++14, 152 116 bytes

As unnamed lambda, assumes s to be string

[](auto s){decltype(s)r;for(auto c:s){r=c+r;}return s+r;}//};r+s nruter};r+c=r{)s:c otua(rof;r)s(epytlced{)s otua(][

Old solution:

[](auto s){auto r=s;for(auto p=s.rbegin()-1;++p!=s.rend();r+=*p);return r;}//};r nruter;)p*=+r;)(dner.s=!p++;1-)(nigebr.s=p otua(rof;s=r otua{)s otua(][

Usage:

auto f=[](auto s){decltype(s)r;for(auto c:s){r=c+r;}return s+r;};

main(){
 string a="123456789";
 cout << f(a) << endl;
}

Answer 24

05AB1E, 5 bytes

«q«Â

Try it online.

Explanation:

        # Bifurcate (short for Duplicate & Reverse) the (implicit) input
         #  i.e. "neverod" → "neverod" and "doreven"
 «       # Concat both together
         #  i.e. "neverod" and "doreven" → "neveroddoreven"
  q      # Exit the program (and implicitly output the concatted result)
   «Â    # No-ops

Or alternatively:

R«q«R

Try it online.

Where R is reverse, and the « takes the input implicitly again to concat with.

---

NOTE: If we are allowed to output neverodoreven for the input neverod, which is still a palindrome, it can be done in 1 byte instead with the palindromize builtin:

û

Try it online.

Answer 25

x86-64 Assembly (Microsoft x64 calling convention), 89 bytes:

80 39 00 48 8B D1 4C 8B C1 74 0B 48 FF C2 49 FF C0 80 3A 00 75 F5 48 FF CA 8A 02 41 88 00 48 8B C2 48 FF CA 49 FF C0 48 3B C1 77 ED C3 ED 77 C1 3B 48 C0 FF 49 CA FF 48 C2 8B 48 00 88 41 02 8A CA FF 48 F5 75 00 3A 80 C0 FF 49 C2 FF 48 0B 74 C1 8B 4C D1 8B 48 00 39 80

Disassembled:

 0000000000000000: 80 39 00           cmp         byte ptr [rcx],0
 0000000000000003: 48 8B D1           mov         rdx,rcx
 0000000000000006: 4C 8B C1           mov         r8,rcx
 0000000000000009: 74 0B              je          0000000000000016
 000000000000000B: 48 FF C2           inc         rdx
 000000000000000E: 49 FF C0           inc         r8
 0000000000000011: 80 3A 00           cmp         byte ptr [rdx],0
 0000000000000014: 75 F5              jne         000000000000000B
 0000000000000016: 48 FF CA           dec         rdx
 0000000000000019: 8A 02              mov         al,byte ptr [rdx]
 000000000000001B: 41 88 00           mov         byte ptr [r8],al
 000000000000001E: 48 8B C2           mov         rax,rdx
 0000000000000021: 48 FF CA           dec         rdx
 0000000000000024: 49 FF C0           inc         r8
 0000000000000027: 48 3B C1           cmp         rax,rcx
 000000000000002A: 77 ED              ja          0000000000000019
 000000000000002C: C3                 ret
 000000000000002D: ED                 in          eax,dx
 000000000000002E: 77 C1              ja          FFFFFFFFFFFFFFF1
 0000000000000030: 3B 48 C0           cmp         ecx,dword ptr [rax-40h]
 0000000000000033: FF 49 CA           dec         dword ptr [rcx-36h]
 0000000000000036: FF 48 C2           dec         dword ptr [rax-3Eh]
 0000000000000039: 8B 48 00           mov         ecx,dword ptr [rax]
 000000000000003C: 88 41 02           mov         byte ptr [rcx+2],al
 000000000000003F: 8A CA              mov         cl,dl
 0000000000000041: FF 48 F5           dec         dword ptr [rax-0Bh]
 0000000000000044: 75 00              jne         0000000000000046
 0000000000000046: 3A 80 C0 FF 49 C2  cmp         al,byte ptr [rax+FFFFFFFFC249FFC0h]
 000000000000004C: FF 48 0B           dec         dword ptr [rax+0Bh]
 000000000000004F: 74 C1              je          0000000000000012
 0000000000000051: 8B 4C D1 8B        mov         ecx,dword ptr [rcx+rdx*8-75h]
 0000000000000055: 48 00 39           add         byte ptr [rcx],dil
 0000000000000058: 80

Note that the code after the ret instruction at 2C is unreachable so it doesn't matter that it's nonsense

Answer 26

Japt, 4 bytes

êêêê

Try it online!

How it works

U.ê("ê".ê("ê"))  Transpiled to JS

       .ê("ê")   String.ê(string): true if `this` is palindrome
    "ê".ê("ê")   true (treated same as 1)
U.ê(          )  String.ê(number): palindromify
                   "abc"->"abccba" if `number` is odd, "abcba" otherwise
                 `true` is odd number, so we achieve the desired function

Alternative 4 bytes

pwwp

Try it online!

How it works

U.p("w".w("p"))  Transpiled to JS
    "w".w(   )   Reverse of "w" ("p" is ignored)
U.p("w")         Append U.w(), which is reverse of U, to the right of U

Answer 27

Backhand, 33 27 bytes

iH~0}|{<:: oi]io ::<{|}0~Hi

Try it online!

Unlike a lot of the solutions here, this one actually does use the palindromised code!

Explanation:

i  0 |{      Get the first character and enter the loop
        :  o    Output the character while preserving it
              i  :     Get input and duplicate it
                   <{  Turn around
             ]         Increment the copy to check if EOF   
    }| <    Loop again if not EOF
  ~   If EOF, pop the extra copy of EOF
 H    Terminate, printing the contents of the stack.

Altogether, the unexecuted instructions are:

       :   i  o :   |}0~Hi