Palindromize this string! [duplicate]

Question

Your task is to palindromize a string as follows:

Take the string.

abcde

Reverse it.

edcba

Remove the first letter.

dcba

Glue it onto the original string.

abcdedcba

But here's the catch: since this challenge is about palindromes, your code itself also has to be a palindrome.

Remember, this is code-golf, so the code with the smallest number of bytes wins.

Answer 1

Python 3, 41 bytes

lambda t:t+t[-2::-1]#]1-::2-[t+t:t adbmal

Try it here.

Answer 2

05AB1E, 1 byte

û

Try it online!

Answer 3

Dip, 1 byte

B

^{Body must be at least 30 characters; you entered 13.}

Answer 4

Pyth - 13 bytes

It kinda looks like its crying.

+Qt_Q " Q_tQ+

Try it online here.

Answer 5

아희(Aheui), 149 137 bytes (47 chars + 2 newlines)

붛뱐쎤붇쎡뻐처순의이멓희
빠본땨벌다따토먀의썩속썩의먀토따다벌땨본빠
희멓이의순처뻐쎡붇쎤뱐붛

Due to the nature of the Aheui language, the input must end in a double quotation mark (") (gets ignored).

Try it here! (copy and paste the code)

Bonus (each line is palindromic), 131 bytes (42 chars + 5 newlines)

밯빪반분반빪밯
쏜발뚝볃뚝발쏜
쏙툼닿뗘닿툼쏙
뽓첡순괆순첡뽓
숙쌱멈긕멈쌱숙
몋희익욥익희몋

Answer 6

APL -- 13 bytes.

⊣,¯1↓⌽⍝⌽↓1¯,⊣

Explanation:

⊣,¯1↓⌽
     ⌽ Reverse the argument.
  ¯1↓  Drop the last character.
⊣,     Concatenate the original to the result.

Answer 7

JavaScript (ES6), 86 bytes

s=>s+[...s].reverse().slice(1).join("")//)""(nioj.)1(ecils.)(esrever.]s...[+s>=s

I tried a smarter solution but it's still longer :/ even abusing [,...a]=s didn't seem to save bytes

Answer 8

J, 17 bytes

}:,|.NB. .BN.|,:}

Takes the easy way out by using a comment to mirror the code. In J, NB. starts a line comment. I do hope to find a method that doesn't involve a comment but the digrams in J probably do make it harder.

Also forms two smileys, }: and :}.

Usage

   f =: }:,|.NB. .BN.|,:}
   f 'abcde'
abcdedcba

Explanation

Comment part removed since it is just filler.

}:,|.  Input: string S
   |.  Reverse S
}:     Curtail, get S with its last char removed
  ,    Join them and return

Answer 9

C#, (51 50 + 1)*2 = 104 102 bytes

saved 2 bytes to the use of .Aggregate()

s=>s+s.Aggregate("",(a,b)=>b+a,s=>s.Substring(1));//;))1(gnirtsbuS.s>=s,a+b>=)b,a(,""(etagerggA.s+s>=s

You can capture this lambda with

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

and call it like this

f("abcde")

Answer 10

Ruby, 47 bytes

->s{s+s.reverse[1..-1]}#}[1-..1]esrever.s+s{s>-

Answer 11

Mathematica, 67 bytes

f=#~Join~Rest@Reverse@#//Function;noitcnuF//#@esreveR@tseR~nioJ~#=f

Defines a named function f that takes a list of characters as input and returns the appropriate palindromized list as output. The bit after the semicolon (is fortunately syntactically valid and) gives a second, way more complicated name to this same function.

It was fun trying to make Mathematica make sense without any brackets!

Answer 12

Brachylog, 11 bytes

:Lc.r
r.cL:

Try it online!

Explanation

:Lc.         Input concatenated with a string L results in the Output
   .r(.)     The Output reversed is still the Output

The second line is a new predicate declaration that never gets called in that code.

Answer 13

Actually, 14 13 bytes

RiXßRkΣkRßXiR

Try it online!

Explanation

R          # reverse input
 i         # flatten
  X        # discard top of stack
   ßR      # push reversed input
     kΣ    # concatenate each
k          # wrap in list
 R         # reverse
  ß        # push input
   X       # discard it
    i      # flatten list to string
     R     # reverse string

Answer 14

Pyke, 1 bytes

s

Try it here!

Yes, this is the same builtin as the digital root one. This time it takes a sting and turns it into a palindrome. Added with this commit (actually on the same day as the digital root question)

Answer 15

Java 7, 146 bytes

(Enter is added for readibility and not part of the code.)

String c(String s){return s+new StringBuffer(s).reverse().substring(1);}/
/};)1(gnirtsbus.)(esrever.)s(reffuBgnirtS wen+s nruter{)s gnirtS(c gnirtS

Answer 16

Java 7, 152 bytes

String c(char[]s,int i){return s.length==i+1?s[i]+"":s[i]+c(s,++i)+s[i-1];}//};]1-i[s+)i++,s(c+]i[s:""+]i[s?1+i==htgnel.s nruter{)i tni,s][rahc(c gnirtS

Answer 17

PHP, 38+39=77 bytes

<?=($s=$argv[1]).substr(strrev($s),1);#;)1,)s$(verrts(rtsbus.)]1[vgra$=s$(=?<

stupid restriction ... for non-eso languages :)

Answer 18

Haskell, 46 44 bytes

p s=init s++reverse s--s esrever++s tini=s p

Try it on Ideone. Saved 2 bytes thanks to Damien.

Straight forward solution. init takes everything but the last character of a string (or last element of a list). -- enables a single line comment.

Answer 19

R, 107 bytes

This is the boring solution which just uses comments to make it a palindrome.

cat(x<-scan(,""),rev(strsplit(x,"")[[1]])[-1],sep="")#)""=pes,]1-[)]]1[[)"",x(tilpsrts(ver,)"",(nacs-<x(tac

R with few comments, 271 bytes

While this code is longer, only 89 of the bytes (33%) are comments, rather than the 50% in the above code. R relies extensively on parentheses for function application, so this was rather difficult.

`%q%`=function
(x,y)cat(x,y,sep=e)#
`%p%`=function
(x,y)`if`(length(y),rev(y)[-1],x)#
x=strsplit(scan(,e<-""),e)[[1]]#
n=NULL
e%q%x%p%n
n%p%x%q%e
LLUN=n
#]]1[[)e,)""-<e,(nacs(tilpsrts=x
#)x,]1-[)y(ver,)y(htgnel(`fi`)y,x(
noitcnuf=`%p%`
#)e=pes,y,x(tac)y,x(
noitcnuf=`%q%`

Answer 20

WinDbg, 142 287 bytes

db$t0 L1;.for(r$t1=@$t0;@$p;r$t1=@$t1+1){db$t1 L1};r$t3=@$t1-1;.for(r$t1=@$t1-3;@$t1>=@$t0;r$t1=@$t1-1;r$t3=@$t3+1){m$t1 L1 $t3};eb$t3 0;da$t0;*;0t$ad;0 3t$be;}3t$ 1L 1t$m{)1+3t$@=3t$r;1-1t$@=1t$r;0t$@=>1t$@;3-1t$@=1t$r(rof.;1-1t$@=3t$r;}1L 1t$bd{)1+1t$@=1t$r;p$@;0t$@=1t$r(rof.;1L 0t$bd

+145 bytes to make it a palindrome, nearly missed that requirement...

Input is passed in via an address in psuedo-register $t0. For example:

eza 2000000 "abcde"           * Write string "abcde" into memory at 0x02000000
r $t0 = 33554432              * Set $t0 = 0x02000000
* Edit: Something got messed up in my WinDB session, of course r $t0 = 2000000 should work
* not that crazy 33554432.

This may be more golfable, for example I feel like there should be an easier way to convert a memory address in a register to the value at that address.

It works by concatenating the chars from the second-to-last to the first to the end of the string.

db $t0 L1;                                      * Set $p = memory-at($t0)
.for (r $t1 = @$t0; @$p; r $t1 = @$t1 + 1)      * Set $t1 = $t0 and increment until $p == 0
{
    db $t1 L1                                   * Set $p = memory-at($t1)
};
r $t3 = @$t1 - 1;                               * Point $t3 at end of string

* From the second-to-last char, reverse through the string with $t1 back to the start ($t0)
* and continue to increment $t3 as chars are appended to the string.
.for (r $t1 = @$t1 - 3; @$t1 >= @$t0; r $t1 = @$t1 - 1; r $t3 = @$t3 + 1)
{
    m $t1 L1 $t3                                * Copy char at $t1 to end of string ($t3)
};
eb $t3 0;                                       * Null terminate the new string
da $t0;                                         * Print the palindrome string

* Comment of the previous code in reverse, making the whole thing a palindrome
*;0t$ad;0 3t$be;}3t$ 1L 1t$m{)1+3t$@=3t$r;1-1t$@=1t$r;0t$@=>1t$@;3-1t$@=1t$r(rof.;1-1t$@=3t$r;}1L 1t$bd{)1+1t$@=1t$r;p$@;0t$@=1t$r(rof.;1L 0t$bd

Output:

0:000> eza 2000000 "abcde"
0:000> r $t0 = 33554432
0:000> db$t0 L1;.for(r$t1=@$t0;@$p;r$t1=@$t1+1){db$t1 L1};r$t3=@$t1-1;.for(r$t1=@$t1-3;@$t1>=@$t0;r$t1=@$t1-1;r$t3=@$t3+1){m$t1 L1 $t3};eb$t3 0;da$t0;*;0t$ad;0 3t$be;}3t$ 1L 1t$m{)1+3t$@=3t$r;1-1t$@=1t$r;0t$@=>1t$@;3-1t$@=1t$r(rof.;1-1t$@=3t$r;}1L 1t$bd{)1+1t$@=1t$r;p$@;0t$@=1t$r(rof.;1L 0t$bd
02000000  61                                               a
02000000  61                                               a
02000001  62                                               b
02000002  63                                               c
02000003  64                                               d
02000004  65                                               e
02000005  00                                               .
02000000  "abcdedcba"

Answer 21

C, 162 bytes

f(char*c,char*d){char*e=c;while(*d++=*c++);c-=2;d-=2;do{*d++=*c--;}while(e<=c);}//};)c=<e(elihw};--c*=++d*{od;2=-d;2=-c;)++c*=++d*(elihw;c=e*rahc{)d*rahc,c*rahc(f

Ungolfed, unpalindromized:

f(char* c,char* d){
 char*e=c;
 while(*d++=*c++);
 c-=2;d-=2;
 do{*d++=*c--;}while(e<=c);
}

c is is input string, assumes output string d has sufficient length.

Usage:

int main() {
 char a[] = "abcde";
 char* b=malloc(strlen(a)*2);
 f(a,b);
 printf("%s\n",b);
}

Answer 22

Perl 5, 44 bytes

43 bytes, plus 1 for -pe instead of -e

s#(.*).#$&.reverse$1#e#1$esrever.&$#.)*.(#s

Answer 23

𝔼𝕊𝕄𝕚𝕟, 16 chars / 20 bytes

ï+ïĦ⬮Đ1//1Đ⬮Ħï+ï

Try it here (ES6 browsers only).

Generated from the interpreter console using this:

c.value=`ï+ï${alias(String.prototype,'reverse')}⬮${alias(String.prototype,'slice')}1//1${alias(String.prototype,'slice')}⬮${alias(String.prototype,'reverse')}ï+ï`