Same length, different string

Question

Challenge

Given a non-empty string S of length L consisting entirely of printable ASCII chars, output another string of length L that consists entirely of printable ASCII chars, but is not equal to S.

For the purposes of this challenge, a printable ASCII char is one between U+0020 and U+007E, inclusive; that is, from (space) to ~ (tilde). Newlines and tabs are not included.

For example, given "abcde", some valid outputs could be:

• "11111"
• "abcdf"
• "edcba"

But these would be invalid:

• "abcde"
• "bcde"
• "abcde0"

Test cases

"asdf"
"1111"
"       "
"~~~~~"
"abcba"
"1"
" "
"~"
" ~"
"~ "
"  0"
"!@#$%^&*()ABCDEFGhijklmnop1234567890"
" !\"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~"

Rules

• You may assume the input consists entirely of printable ASCII chars.
• You may not assume that the input does not contain all 95 printable chars.
• You may assume the input contains at least one character and is less than 256 chars long.
• The output must also consist entirely of printable ASCII chars. You could not, for example, output the byte \x7F for input "~".
• The output must be different than the input with probability 1; that is, you may generate random strings until one is different than the input, but you can't just output L random characters and hope it's different.
• Newlines are disallowed in the output, but you may output one trailing newline which is not counted toward the string.

Scoring

This is code-golf, so the shortest code in bytes in each language wins.

Answer 1

Golfscript, 3 bytes

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)5%

Take the last character's ASCII value modulo 5, and replace the last character with the result. This obviously works for non-digit characters, but if the last character is a digit, it also changes ("0" mod 5 = 3, "1" mod 5 = 4, etc).

This would also work with 7 or 9 while still keeping the same length.

Also, yay! I have a Golfscript solution as good as the best solutions here!

Answer 2

Funky, 26 22 bytes

s=>s::gsub("."a=>1&~a)

Calculates ~a which for non-digits will return NaN. Then 1& limits it to either 0 or 1, for the digit 0, this will be 1, and for 1 this will be 0. So this string is always unique.

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Answer 3

C/C++, 29 bytes

void m(char*s){(*s%=26)+=65;}

Code to test :

#include <iostream>
#include <string>
#include <cassert>

#ifdef _MSC_VER
#pragma warning(disable:4996)
#endif

void m(char*s){(*s%=26)+=65;}

int main() {
    std::initializer_list<std::string> test{
        "asdf",
        "1111",
        "       ",
        "~~~~~",
        "abcba",
        "1",
        " ",
        "~",
        " ~",
        "~ ",
        "  0",
        "!@#$%^&*()ABCDEFGhijklmnop1234567890",
        " !\"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~"
    };

    for (auto&a : test) {
        char* t = new char[a.size() + 1];
        std::strcpy(t, a.c_str());
        m(t);
        assert(a != t);
        std::cout << a << " => " << t << '\n';
        delete[] t;
    }
}

Answer 4

brainfuck, 35 27 bytes

Edit: switched to a better way of generating an odd number thanks to the brainfuck constants page

,[[[>]+<[-<]>]-[>+<---]>.,]

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Prints out a string of semi-random Vs and Us. If the first character of the input is a V it will print out a U instead and vice versa.

How it Works

,[ Gets first inputted char 
  [[>]+<[-<]>] Converts the ASCII value to binary
  -[>+<---] Adds 85 to the last digit of that binary, which is the original value mod 2. 
            This converts a V (value 86) into a U (value 85) and vice versa
  >. Print the value
,] Get a new value and repeat. 
   Note that the binary of the previous value is still in memory, 
     so it no longer prints specific chars for odd or even values

Another version with the same bytecount:

,[[>]+<[-<]>]-[>+<---]>[.,]

Only changes the first character to a V or a U, but doesn’t handle empty input (which is still valid)

Answer 5

Brain-Flak, 42 bytes

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

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Replaces the first character of the string with a space, or a ! if the first character is already a space.

Answer 6

Perl 6, 14 bytes

{.ord%7~.chop}

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Bare block, implicit lambda with input $_. The ordinal of the first character .ord modulo 7 is a different character: Non-numeric first character becomes numeric, numeric first characters map to a different character. If we append the input sans the last character (~ $_.chop), we match the length. It can never match the input because the first character never matches.

Perl 6, 15 bytes

{S/./{.ord%7}/}

Performs S/// substitution (which returns the result of substituting on $_), matching . (any character) once, and replacing it with the ordinal of the first character in $_ - the same character - modulo 7. Non-numeric first character becomes numeric, numeric first characters map to a different character.

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Answer 7

str, 2 bytes

XV

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Explanation

XV
       over each character:
X         increment it  ("~" becomes "\x7f" and "@" becomes "A")
 V        wrap to printable ascii ("\x7f" becomes " ")

Alternatively, YV to decrement instead of increment.

Answer 8

PowerShell, 37 26 bytes

^{Thanks Mazzy for -11 bytes}

-join($args|% t*y|%{$_%9})

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Does alright. Uses the ToCharArray trick to save some bytes. Through a conversion chain, mods the ASCII value of each character by 9 then joins up the value.

Answer 9

JavaScript (ES6), 24 bytes

f=
s=>s.replace(/./,s[0]^1)

<input oninput=o.textContent=f(this.value)><pre id=o>

Changes the first character of the string according to this table:

1   0
2   3
3   2
4   5
5   4
6   7
7   6
8   9
9   8

Everything else becomes 1.

Answer 10

><>, 11 bytes

i:0(?;'0'=n

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Explanation

Like my MATL answer, this transforms character '0' into '1' and all other characters into '0'.

i       Input one char, or push -1 if there aren't any more chars
:       Duplicate
0(      Is it less than 0?
?;      If so, end program. Else:
'0'     Push character '0'
=       Is it equal? Gives 1 or 0
n       Output as a number. Go back to the beginning

Answer 11

Chip, 5 bytes

g*A~a

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Maps all characters to either @ or A, so that the lowest bit always differs from the input:

0, @, ~, etc. map to A
1, a, A, etc. map to @

Answer 12

x86 Assembly, 19 bytes

Alternates the first character of the string between '@' and 'A'

_diffstr:
  00000000: 8B 54 24 04        mov         edx,dword ptr [esp+4]
  00000004: 8A 02              mov         al,byte ptr [edx]
  00000006: 3C 41              cmp         al,41h
  00000008: 75 04              jne         0000000E
  0000000A: B0 40              mov         al,40h
  0000000C: EB 02              jmp         00000010
  0000000E: B0 41              mov         al,41h
  00000010: 88 02              mov         byte ptr [edx],al
  00000012: C3                 ret

Answer 13

Perl 5, 49 bytes

sub{$_="a"x length $_[0];$_++while $_ eq$_[0];$_}

Explanation:

• starts with string of letter "a" as long as input
• keeps incrementing that until it's different from input

Starting from "a"s was to avoid incrementing to point where Perl lengthens the string; with only one strings to avoid being same as, it couldn't overflow.

This is shamelessly cribbed (and simplified) from my answer to the "related" problem "The Third String".

Execute with:

perl -e '$s = ' -E 'sub{$_="a"x length $_[0];$_++while $_ eq$_[0];$_}' -E ';say $s->("z")'

Answer 14

x86 opcode, 6 bytes

    and     byte [ecx], 0xF1
    inc     byte [ecx]
    ret     

Answer 15

Julia 0.6, 26 bytes

s->map(x->x<'b'?'b':'a',s)

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Answer 16

Java 8, 39 38 bytes

s->(s.charAt(0)<49?9:0)+s.substring(1)

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Explanation:

If the first character has a unicode value below 49 (space, or one of !"#$%&'()*+'-./0), change it to a '9', else change it to a '0'.

s->                 // Method with String as both parameter and return-type
  (s.charAt(0)<49?  //  If the first character has a unicode value below 49:
    9               //   Change the first character to a '9'
  :                 //  Else
   0)               //   Change the first character to a '0'
  +s.substring(1)   //  And return the rest of the input-String (if any)

Answer 17

Pyth, 9 bytes

*h-+QGhQl

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Explanation

*h-+QGhQl
   +QG      Stick the alphabet to the end of the input.
  -   hQ    Delete all instances of the first character.
 h          Take the first remaining character.
*       lQ  Multiply by the length of the (implicit) input.

Answer 18

Gol><>, 6 bytes

iE;5%n

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Direct translation of Jo King's ><> answer.

Answer 19

Python 3, 29 26 bytes

lambda s:'a~'[s<'~']+s[1:]

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• Replacing first character with ~, if the first character is ~ already, it is replaced by a.

Answer 20

C (gcc), 22 bytes

f(char*s){*s=~1&*s^2;}

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Answer 21

SmileBASIC, 27 25 bytes

DEF D S
S[0]=@1[S>"2"]END

Creates a function named D that is called like:

STRING$="abcde"
D STRING$
'STRING$ is now "1bcde"

The first character is replaced with either @ or 1

Answer 22

MAWP, 11 bytes

0|[!5P5WA:]

Idea from Jo King's answer.

Prints the ASCII value of each letter modulo 5.

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Answer 23

Husk, 6 bytes

ṁ(s%5c

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Port of Jo King's answer - upvote it!
Outputs ASCII values MOD 5.

Answer 24

Vyxal, 3 bytes

C∷ṅ

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05AB1E port, go read that.

Answer 25

APL (Dyalog Extended), 11 8 bytes

∊⍕¨'0'=⎕

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-3 thanks to @Adám

Explanation

∊⍕¨'0'=⎕
   '0'=⎕        For each character: 1 if it's '0', 0 otherwise
 ⍕¨             Convert each to string
∊               Flatten into a single string

Answer 26

(,) N, 116 108 Chars or \$108\log_{256}(3)\approx21.4\$ Bytes

((),()()()()()()())(,,,,(()(),((()()())))(()(),(),(),,(()()),()((),,,,(),,(())))(()(),(),()(),,(()()),()()))

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Or, without flags, for 138 chars or 27.34 Bytes

((),()()()()()()()()()())(,,,,(()(),((()()())))(()(),(),,(())(())(())()(),(()()),(())(())(())(()))(()(),(),,(())(())(())(()),(()()),()()))

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Basically, it outputs 1 (with flags) or \32 (without flags) if the char code is greater than or equal to that of \50 (with flags) or \40 (without flags), and outputs 2 (with flags) or \40 (without flags) otherwise.
The reason 1 and 2 are written as 1 and 2 rather than as \49 and \50 is because they are outputted numerically, hence the need for the N flag to remove seperators. All others are encoded in the code by char code.

Answer 27

Perl 6,  48  28 bytes

{(->{[~] (' '..'~').roll(.chars)}...*ne$_).tail}

Generates a sequence of random strings, stopping when one of them doesn't match the input. Returns the last string produced

Test it

{first *ne$_,(0,1 Xx.chars)}

Generates two strings one consisting of only 0s, the other of 1s. Returns the first that is not stringwise equal to (ne) the input ($_).

Test it

Expanded

{             # bare block lambda with implicit parameter ｢$_｣

  first       # return the first value where the following is true

    * ne $_,  # WhateverCode lambda ( ｢*｣ is the parameter )

    (
      0, 1    # list of 0 and 1
      X[x]    # cross ｢X｣ that using the string repetition operator ｢x｣
      .chars  # with the number of characters in ｢$_｣
    )
}

Answer 28

Convex, 4 bytes

'0f=

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Answer 29

Aceto, 16 12 9 bytes

=01`
,'pO

Run with -F to see immediate output (but works without).

Explanation:

We read a character, then push the character 0 on the stack and test for equality:

=0
,'

If equal, we push a 1.

  1`

We now print the top element (either a one or an implicit zero), and go back to the start:

  pO

This prints zeroes for any characters that aren't zeroes, and ones for zeroes, e.g.:

Input  |  Output
----------------
abcde  |  00000
11111  |  00000
10010  |  01101

Answer 30

Braingolf, 12 bytes

.#~e1-:1+|&@

Subtracts 1 from the last char's ordinal if it is a tilde, otherwise adds one.

Explanation:

.#~e1-:1+|&@  Implicit input of string, convert each char to ordinal and push to stack
.             Duplicate last item
 #~           Push tilde ordinal
   e          If last 2 items are equal (consumes last 2 items)
    1-        ..then subtract one
      :       else
       1+     ..add one
         |    endif
          &@  Print entire stack as chars