Giới thiệu

Đối với những người không biết, một palindrom là khi một chuỗi bằng với chuỗi ngược (ngoại trừ giao thoa, khoảng trắng, v.v.). Một ví dụ về palindrom là:

abcdcba

Nếu bạn đảo ngược điều này, bạn sẽ kết thúc với:

abcdcba

Cái nào cũng vậy. Do đó, chúng tôi gọi đây là một palindrom. Để giảm nhẹ mọi thứ, chúng ta hãy xem một ví dụ về chuỗi:

adbcb

Đây không phải là một palindrom. Để giảm bớt điều này, chúng ta cần hợp nhất chuỗi đảo ngược vào chuỗi ban đầu ở bên phải chuỗi ban đầu , giữ nguyên cả hai phiên bản. Càng ngắn, càng tốt.

Điều đầu tiên chúng ta có thể thử là như sau:

adbcb
bcbda
^^ ^^

Không phải tất cả các ký tự khớp, vì vậy đây không phải là vị trí phù hợp cho chuỗi đảo ngược. Chúng tôi đi một bước về bên phải:

adbcb
 bcbda
 ^^^^

Điều này cũng không phù hợp với tất cả các nhân vật. Chúng tôi đi một bước nữa bên phải:

adbcb
  bcbda

Lần này, tất cả các nhân vật phù hợp . Chúng ta có thể hợp nhất cả hai chuỗi còn nguyên . Kết quả cuối cùng là:

adbcbda

Đây là chuỗi palindromized .

Nhiệm vụ

Cho một chuỗi (có ít nhất một ký tự) chỉ chứa các chữ cái viết thường (hoặc chữ hoa, nếu phù hợp hơn), hãy xuất chuỗi ký tự palindromized .

Các trường hợp thử nghiệm

Input     Output

abcb      abcba
hello     hellolleh
bonobo    bonobonob
radar     radar
hex       hexeh

Đây là môn đánh gôn , vì vậy bài nộp có số byte ít nhất sẽ thắng!

Jelly, 11 10 bytes

ṫỤfU$Ḣœ^;U

Try it online!

How it works

ṫỤfU$Ḣœ^;U  Main link. Argument: s (string)

 Ụ          Yield all indices of s, sorted by their corr. character values.
ṫ           Tail; for each index n, remove all characters before thr nth.
            This yields the list of suffixes of s, sorted by their first character,
            then (in descending order) by length.
    $       Combine the two links to the left into a chain:
   U        Upend; reverse all suffixes.
  f         Filter; only keep suffixes that are also reversed suffixes.
            This gives the list of all palindromic suffixes. Since all of them
            start with the same letter, they are sorted by length.
     Ḣ      Head; select the first, longest palindromic suffix.
      œ^    Multiset symmetric difference; chop the selected suffix from s.
         U  Upend; yield s, reversed.
        ;   Concatenate the results to the left and to the right.

Pyth (commit b93a874), 11 bytes

.VkI_IJ+zbB

Test suite

This code exploits a bug in the current version of Pyth, commit b93a874. The bug is that _IJ+zb is parsed as if it was q_J+zbJ+zb, which is equivalent to _I+zb+zb, when it should (by the design intention of Pyth) be parsed as q_J+zbJ, which is equivalent to _I+zb. This allows me to save a byte - after the bug is fixed, the correct code will be .VkI_IJ+zbJB. I'll explain that code instead.

Basically, the code brute forces over all possible strings until it finds the shortest string that can be appended to the input to form a palindrome, and outputs the combined string.

.VkI_IJ+zbJB
                z = input()
.Vk             For b in possible strings ordered by length,
       +zb      Add z and b,
      J         Store it in J,
    _I          Check if the result is a palindrome,
   I            If so,
          J     Print J (This line doesn't actually exist, gets added by the bug.
          B     Break.

Python, 46 bytes

f=lambda s:s*(s==s[::-1])or s[0]+f(s[1:])+s[0]

If the string is a palindrome, return it. Otherwise, sandwich the first letter around the recursive result for the remainder of the string.

Example breakdown:

f(bonobo)
b  f(onobo) b
b o f(nobo) o b 
b o n f(obo) n o b
b o n obo n o b

Haskell, 36 bytes

f s|s==reverse s=s|h:t<-s=h:f t++[h]

More readably:

f s
 |s==reverse s = s
 |(h:t)<-s     = h:(f t)++[h]

If the string is a palindrome, return it. Otherwise, sandwich the first letter around the recursive result for the tail of the string.

The string s is split into h:t in the second guard, obviating a filler 1>0 for this case. This is shorter than doing s@(h:t) for the input.

Pyth - 16 12 bytes

4 bytes saved thanks to @FryAmTheEggman.

FGITW, much golfing possible.

h_I#+R_Q+k._

Test Suite.

Brachylog, 16 6 5 bytes (Non-competing)

:Ac.r

Try it online!

When I posted my initial answer, it was still on the old implementation in Java. Since I have reprogrammed everything in Prolog, it now works as it should have in the first place.

Explanation

(?):Ac.        Output is the concatenation of Input with another unknown string A
      .r(.)    The reverse of the Output is the Output

Backpropagation makes it so that the first valid value for A it will find will be the shortest that you can concatenate to Input to make it a palindrome.

Alternate solution, 5 bytes

~@[.r

This is roughly the same as the above answer, except that instead of stating "Output is the concatenation of the Input with a string A", we state that "Output is a string for which the Input is a prefix of the Output".

JavaScript (ES6), 92 bytes

(s,a=[...s],r=a.reverse().join``)=>s.slice(0,a.findIndex((_,i)=>r.startsWith(s.slice(i))))+r

Computes and slices away the overlap between the original string and its reversal.

Retina, 29 25

$
¶$_
O^#r`.\G
(.+)¶\1
$1

Try it online!

Big thanks to Martin for 11 bytes saved!

This just creates a reversed copy of the string and smooshes them together. The only really fancy part of this is the reversing method: O^#r`.\G, which is done by using sort mode. We sort the letters of the second string (the ones that aren't newlines and are consecutive from the end of the string, thanks to the \G) by their numeric value, which, since there are no numbers, is 0. Then we reverse the order of the results of this stable sort with the ^ option. All credit for the fancy use of \G belongs to Martin :)

CJam, 18

q__,,{1$>_W%=}#<W%

Try it online

Explanation:

q         read the input
__        make 2 copies
,,        convert the last one to a range [0 … length-1]
{…}#      find the first index that satisfies the condition:
  1$>     copy the input string and take the suffix from that position
  _W%=    duplicate, reverse and compare (palindrome check)
<         take the prefix before the found index
W%        reverse that prefix
          at the end, the stack contains the input string and that reversed prefix

Lua, 89 88 Bytes

I beat the Javascript! \o/ Saved 1 byte thanks to @LeakyNun ^^

It's a full program, takes its input as command-line argument.

i=1s=...r=s:reverse()while s:sub(i)~=r:sub(0,#r-i+1)do i=i+1 end print(s..r:sub(#r-i+2))

ungolfed

i=1                             -- initialise i at 1 as string are 1-indexed in lua
s=...                           -- use s as a shorthand for the first argument
r=s:reverse()                   -- reverse the string s and save it into r
while(s:sub(i)~=r:sub(0,#r-i+1))-- iterate while the last i characters of s
do                              -- aren't equals to the first i characters of r
  i=i+1                         -- increment the number of character to skip
end
print(s..r:sub(#r-i+2))         -- output the merged string

Prolog, 43 bytes

a(S):-append(S,_,U),reverse(U,U),writef(U).

This expects a codes string as input, e.g. on SWI-Prolog 7: a(`hello`).

Explanation

This is basically a port of my Brachylog answer.

a(S) :-               % S is the input string as a list of codes
    append(S,_,U),    % U is a list of codes resulting in appending an unknown list to S
    reverse(U,U),     % The reverse of U is U
    writef(U).        % Write U to STDOUT as a list of codes

Octave, 78 75 bytes

Saved 3 bytes thanks to Eʀɪᴋ ᴛʜᴇ Gᴏʟғᴇʀ!

function p=L(s)d=0;while~all(diag(s==rot90(s),d++))p=[s fliplr(s(1:d))];end

ideone still fails for named functions, but here is a test run of the code as a program.

Perl, 37 bytes

Based on xnor's answer.

Includes +2 for -lp

Run with input on STDIN, e.g.

palindromize.pl <<< bonobo

palindromize.pl:

#!/usr/bin/perl -lp
s/.//,do$0,$_=$&.$_.$&if$_!~reverse

05AB1E, 18 bytes

Code:

©gF¹ðN×®«ðñD®åi,q

Uses CP-1252 encoding. Try it online!

Pyke, 15 bytes

D_Dlhm>m+#D_q)e

Try it here!

J, 20 bytes

,[|.@{.~(-:|.)\.i.1:

This is a monadic verb. Try it here. Usage:

   f =: ,[|.@{.~(-:|.)\.i.1:
   f 'race'
'racecar'

Explanation

I'm using the fact that the palindromization of S is S + reverse(P), where P is the shortest prefix of S whose removal results in a palindrome. In J, it's a little clunky to do a search for the first element of an array that satisfies a predicate; hence the indexing.

,[|.@{.~(-:|.)\.i.1:  Input is S.
        (    )\.      Map over suffixes of S:
         -:             Does it match
           |.           its reversal? This gives 1 for palindromic suffixes and 0 for others.
                i.1:  Take the first (0-based) index of 1 in that array.
 [   {.~              Take a prefix of S of that length: this is P.
  |.@                 Reverse of P.
,                     Concatenate it to S.

Haskell, 68 bytes

import Data.List
f i=[i++r x|x<-inits i,i++r x==x++r i]!!0
r=reverse

Usage example: f "abcb" -> "abcba".

Search through the inits of the input i (e.g. inits "abcb" -> ["", "a", "ab", "abc", "abcb"]) until you find one where it's reverse appended to i builds a palindrome.

MATL, 17 16 bytes

Loosely inspired in @aditsu's CJam answer.

`xGt@q:)PhttP=A~

Try it online!

Explanation

`        % Do...while loop
  x      %   Delete top of stack, which contains a not useful result from the
         %   iteration. Takes input implicitly on first iteration, and deletes it
  G      %   Push input
  t      %   Duplicate
  @q:    %   Generate range [1,...,n-1], where n is iteration index. On the  first
         %   iteration this is an empty array
  )      %   Use that as index into copy of input string: get its first n elements
  Ph     %   Flip and concatenate to input string
  t      %   Duplicate. This will be the final result, or will be deleted at the
         %   beginning of next iteration
  tP     %   Duplicate and flip
  =A~    %   Compare element-wise. Is there some element different? If so, the
         %   result is true. This is the loop condition, so go there will be a 
         %   new iteration. Else the loop is exited with the stack containing
         %   the contatenated string
         % End loop implicitly
         % Display stack contents implicitly

Ruby, 44 bytes

This answer is based on xnor's Python and Haskell solutions.

f=->s{s.reverse==s ?s:s[0]+f[s[1..-1]]+s[0]}

Oracle SQL 11.2, 195 bytes

SELECT MIN(p)KEEP(DENSE_RANK FIRST ORDER BY LENGTH(p))FROM(SELECT:1||SUBSTR(REVERSE(:1),LEVEL+1)p FROM DUAL WHERE SUBSTR(:1,-LEVEL,LEVEL)=SUBSTR(REVERSE(:1),1,LEVEL)CONNECT BY LEVEL<=LENGTH(:1));

Un-golfed

SELECT MIN(p)KEEP(DENSE_RANK FIRST ORDER BY LENGTH(p))
FROM (
       SELECT :1||SUBSTR(REVERSE(:1),LEVEL+1)p 
       FROM   DUAL 
       WHERE  SUBSTR(:1,-LEVEL,LEVEL)=SUBSTR(REVERSE(:1),1,LEVEL)
       CONNECT BY LEVEL<=LENGTH(:1)
     );

Seriously, 34 bytes

╩╜lur`╜╨"Σ╜+;;R=*"£M`MΣ;░p╜;;R=I.

The last character is a non-breaking space (ASCII 127 or 0x7F).

Try it online!

Explanation:

╩╜lur`╜╨"Σ╜+;;R=*"£M`MΣ;░p╜;;R=I.<NBSP>
╩                                        push inputs to registers (call the value in register 0 "s" for this explanation)
 ╜lur                                    push range(0, len(s)+1)
     `              `M                   map (for i in a):
      ╜╨                                   push all i-length permutations of s
        "        "£M                       map (for k in perms):
         Σ╜+                                 push s+''.join(k) (call it p)
            ;;R=                             palindrome test
                *                            multiply (push p if palindrome else '')
                      Σ                  summation (flatten lists into list of strings)
                       ;░                filter truthy values
                         p               pop first element (guaranteed to be shortest, call it x)
                          ╜;;R=I         pop x, push s if s is palindromic else x
                                .<NBSP>  print and quit

C#, 202 bytes

I tried.

class P{static void Main(string[]a){string s=Console.ReadLine(),o=new string(s.Reverse().ToArray()),w=s;for(int i=0;w!=new string(w.Reverse().ToArray());){w=s.Substring(0,i++)+o;}Console.WriteLine(w);}}

Ungolfed:

class P
{
    static void Main(string[] a)
    {
        string s = Console.ReadLine(), o = new string(s.Reverse().ToArray()), w = s;
        for(int i = 0; w!=new string(w.Reverse().ToArray());)
        {
            w = s.Substring(0, i++) + o;
        }
        Console.WriteLine(w);
        Console.ReadKey();
    }

}

Can anyone provide me with any ideas to group the two calls to .Reverse().ToArray() ? A separate method is more bytes.

QBIC, 41 bytes

;_FA|C=A{a=a+1~C=_fC||_XC\C=A+right$(B,a)

Explanation:

;_FA|    Read A$ from the cmd line, then flip it to create B$
C=A      Set C$ to be A$
{        Start an infinite DO-loop
a=a+1    Increment a (not to be confused with A$...)
~C=_fC|  If C$ is equal to its own reversed version
|_XC     THEN end, printing C$
\C=A+    ELSE, C$ is reset to the base A$, with
right$(B the right part of its own reversal
,a)      for length a (remember, we increment this each iteration
         DO and IF implicitly closed at EOF

Haskell, 46 bytes

f l|l==reverse l=l|(h:t)<-l=l!!0:(f$tail l)++[l!!0]

I'm wondering if there's a way to remove the parenthesis in (f$tail l)++[l!!0]...