Depalindromize this string!

Question

Given a palindrome generated according to this challenge, depalindromize it.

Test cases

abcdedcba -> abcde
johncenanecnhoj -> johncena
ppapapp -> ppap
codegolflogedoc -> codegolf

As this is about depalindromizing, your code cannot be a palindrome.

Remember, this is code-golf, so the code with the fewest bytes wins.

Answer 1

Pyke, 4 bytes

le>_

Try it here!

l    -    len(input)
 e   -   ^//2
  >  -  input[^:]
   _ - reversed(^)

Answer 2

IBM/Lotus Notes Formula, 21 bytes

@Left(a;@Length(a)/2)

Computed field that takes input from an editable field a. Works because Notes rounds up when an odd number is divided by 2.

Test Cases

Answer 3

Retina, 24 bytes

I basically took this from another answer of mine.

^((.)*?.??)(?<-2>.)*$
$1

Try it online

Answer 4

CJam, 7 bytes

r_,-2/<

Try it online!

Answer 5

C# 51 50 bytes

s=>string.Join("",s.Where((_,i)=>i<=s.Count()/2));

It's a lambda, woo!
You can catch it with:

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

it requires the string to be a palindrome.

Answer 6

GO, 43 bytes

func(s string)string{return s[:len(s)/2+1]}

Try it online!

Answer 7

Python 2, 22 bytes

lambda _:_[len(_)/2:]

Depalindromizes a given string. Still fails on all the test cases. Weird... ;-)

Answer 8

Racket 48 bytes

(substring s 0(+ 1(floor(/(string-length s)2))))

Ungolfed:

(define (f s)
  (substring s 0 (+ 1
                    (floor
                     (/ (string-length s)
                        2)))))

Testing:

(f "abcdedcba")
(f "johncenanecnhoj")
(f "ppapapp")
(f "codegolflogedoc")

Output:

"abcde"
"johncena"
"ppap"
"codegolf"

Answer 9

Hexagony, 56 bytes

This is awful, but I just wanted to do it in Hexagony :)

Golfed:

,<_..){/"./..\&..$@....>;{.....'..>"+"+"&"='&'&=}\}=.:(<

In a hexagon:

     , < _ . .
    ) { / " . /
   . . \ & . . $
  @ . . . . > ; {
 . . . . . ' . . >
  " + " + " & " =
   ' & ' & = } \
    } = . : ( <
     . . . . .

Way to many no-ops...

Answer 10

Python 3.5, 87 Bytes

x=list(input())
for q in range(0,int(round((len(x)/2)-0.1,0))):x.pop()
print(''.join(x))

i'm new to this codegolfing, but it's to be fun. I probably won't win but oh well.

Answer 11

><>, 7 15 bytes

l1+2,[v
 ;!?lo<

Try it here!

This is a full program. To use it simply place the palindrome on the stack, and the result will be outputted to stdout!

Explanation

l       get the stack length
 1+     add one
   2,   divide by two
     [  pop the value, create a new stack of that length

That's the meat of it, ol?!; just outputs until the stack is empty.

*><>, 11 bytes (non-competing)

l5+2,]r{[$.

Try it here! (first line is a program that calls the function, outputs the stack, then exits)

This is a function. To use it simply call it (instruction C) with the palindrome on the stack, and the result will be on the stack! This part is non-competing because *><> is younger than the challenge.

Explanation

l             get the stack length
 5+           add five
   2,         divide by two
     ]        close the stack
      r       reverse the stack
       {      shift the stack left
        [     pop the value on the end, create a new stack of that length
         $    swap the two values on the end of the stack
          .   jump back, with the result on the stack

For ><> coders who might be confused by this, a new stack is created when a function is called, so you can actually call ] right from the start if you're inside a function. The data inside this stack is the x, y coordinate of the ><> when C was called.

Answer 12

Keg, 7 bytes

^(!2/|_

That is easy but really hard to golf.

Explanation

^       # Input in the right order
 (!2/   # Repeat half of length times:
     |_ # Pop the stack

TIO

Answer 13

R, 35 bytes

substr(x<-scan(,''),1,nchar(x)/2+1)

Try it online!

Answer 14

Vyxal, 2 bytes

½h

Try it Online!

Makes a list of each half of the string and takes the first item. Vyxal rounds up when splitting odd numbers in half. Don't think this can be golfed to 1 byte because the code can't be a palindrome :p

Answer 15

Haskell, 27 bytes

f[x]=[x]
f(a:b)=a:f(init b)

No indexing. Recursively takes from the start and end, keeping letters from the end.

Answer 16

Perl 6, 21 bytes

*.substr(0,1+* div 2)

Expanded:

*\            # Whatever lambda
.substr(
  0,
  1 + * div 2 # Whatever lambda
)

Answer 17

Brain-Flak, 45 bytes

([][()]<>){({}[()()])<>{}<>}<>{({}<>)<>}<>

Try it online!

42 bytes of code, and +3 for the -c flag which enables ASCII input and output.

Answer 18

Python 3, 26 bytes

lambda c:c[:(len(c)+1)//2]

This squeezes 2 chars out of the other Python3 answer by replacing the int() with division with //.

Answer 19

Java 7, 57 bytes

String c(String s){return s.substring(0,-~s.length()/2);}

At first I misinterpreted the question and also made code that supported even palindromes (which isn't necessary due to this being the reverse of that other challenge, and therefore there aren't any even palindrome test-cases). So I deleted my answer.
After seeing @KarlNapf's great answer however, I undeleted and edited this. It's the same byte count as the other Java 7 answer, but works for both odd and even length palindromes (although only odd were asked for this challenge).

Ungolfed & test code:

Try it here.

class M{
  static String c(String s){
    return s.substring(0, -~s.length()/2);
  }

  public static void main(String[] a){
    System.out.println(c("abcdedcba"));
    System.out.println(c("abcddcba"));
  }
}

Output:

abcde
abcd

Answer 20

R, 59 bytes

Not pretty but it gets the job done.

cat(strsplit(scan(,""),"")[[1]][0:(nchar(x)%/%2)+1],sep="")

Answer 21

GolfSharp, 20 bytes

v=>v.R(0,v.L()/2+1);

Answer 22

PowerShell, 46 36 bytes

Unfortunately there's no PowerShell on Try it Online. Alas, here's my attempt.

$args[0][0..(($args[0].Length-1)/2)]

Answer 23

Python 3, 28 26 24 bytes

Python 3 strings don't like non-int indexes...

lambda S:S[:len(S)//2+1]

Alternatively, with the same byte count,

lambda S:S[:-~len(S)//2]

Answer 24

QBIC, 18 bytes

;?left$$|(A,_lA|/2)

; gets a command line parameter, then we print the left half of it using _l. Works fine for even and odd-length strings.

Answer 25

Ruby, 21 19 bytes

->s{s[0..s.size/2]}

Answer 26

Lua, 28 bytes

x=...print(x:sub(1,#x//2+1))

Answer 27

Common Lisp, 43

(lambda(s)(subseq s 0(ceiling(length s)2)))

Truly hard to golf without accidentally writing a palindrome, will all those ( matching opposite ).

_{I know, this is not what "palindrome" means.}

Answer 28

Element, 11 bytes

_2:$'[(`)#]

Try it online!

_2:$'[(`)#]
_             take input
 2:           make a second copy
   $          length of string
    '[    ]   FOR loop
     [(`  ]   chop off then print the first character
     [  )#]   chop off then discard the last character

The FOR loop is actually executed about twice as many times as it needs to be, but those extra iterations do not affect the output since it runs out of data to print.

Answer 29

Scala, 21 bytes

a=>a.take(a.size/2+1)

requires an assignment with a type annotation: val f:(Seq[_]=>Seq[_])...

Answer 30

Clojure, 30 bytes

Clojure port of basically every answer here:

#(subs % 0(inc(/(count %)2)))

Should be self explanatory.