AlTeRnAtE tHe CaSe

Question

For today's challenge, you must write a program or function that alternates the case of a string. However, you must ignore non-alphabetic characters. This means that every alphabetic character must have a different case than the preceding and following alphabetic character. This is slightly more complex than uppercasing every other letter for example. If you take a string such as

hello world

and convert every other character to uppercase, you'll get:

hElLo wOrLd

As you can see, the lowercase o is followed by a lowercase w. This is invalid. Instead, you must ignore the space, giving us this result:

hElLo WoRlD

All non-alphabetic characters must be left the same. The output can start with upper or lowercase, as long as it consistently alternates. This means the following would also be an acceptable output:

HeLlO wOrLd

Your program should work regardless of the case of the input.

The input string will only ever contain printable ASCII, so you don't have to worry about unprintable characters, newlines or unicode. Your submission can be either a full program or a function, and you may take the input and output in any reasonable format. For example, function arguments/return value, STDIN/STDOUT, reading/writing a file, etc.

Examples:

ASCII                                   ->  AsCiI
42                                      ->  42
#include <iostream>                     ->  #InClUdE <iOsTrEaM>
LEAVE_my_symbols#!#&^%_ALONE!!!         ->  lEaVe_My_SyMbOlS#!#&^%_aLoNe!!!
PPCG Rocks!!! For realz.                ->  PpCg RoCkS!!! fOr ReAlZ.
This example will start with lowercase  ->  tHiS eXaMpLe WiLl StArT wItH lOwErCaSe
This example will start with uppercase  ->  ThIs ExAmPlE wIlL sTaRt WiTh UpPeRcAsE
A1B2                                    ->  A1b2

Since this is code-golf, standard loopholes apply and the shortest answer in bytes wins!

Answer 1

JavaScript (ES6), 66 63 bytes

Starts with uppercase.

s=>s.replace(/[a-z]/gi,c=>c[`to${(s=!s)?'Low':'Upp'}erCase`]())

Test cases

let f =

s=>s.replace(/[a-z]/gi,c=>c[`to${(s=!s)?'Low':'Upp'}erCase`]())

console.log(f("ASCII"))
console.log(f("42"))
console.log(f("#include <iostream>"))
console.log(f("LEAVE_my_symbols#!#&^%_ALONE!!!"))
console.log(f("PPCG Rocks!!! For realz."))
console.log(f("This example will start with lowercase"))
console.log(f("This example will start with uppercase"))
console.log(f("A1B2"))

Answer 2

05AB1E, 11 8 bytes

Code:

lvyJ¤aiš

Uses the 05AB1E encoding. Try it online!

Explanation:

l           # Lowercase the input
 vy         # For each element..
   J        #   Join the entire stack into a single string
    ¤a      #   Check if the last character is alphabetic
      iš    #   If true, swapcase the entire string

Answer 3

GNU Sed, 33

• 5 bytes saved thanks to @TobySpeight

Score includes +1 for -r flag to sed.

s/([a-z])([^a-z]*.?)/\U\1\L\2/gi

Try it online.

Answer 4

Jelly, 13 bytes

nŒsTm2
ŒlŒuǦ

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How it works

ŒlŒsǦ  Main link. Argument: s (string)

Œl      Cast to lowercase.
    Ǧ  At indices returned by the helper link...
  Œu        apply uppercase.


nŒsTm2      Helper link. Argument: s (string)

 Œs         Apply swapcase to s.
n           Perform vectorizing not-equal comparison.
   T        Compute the truthy indices.
    m2      Select every other one, starting with the first.

Answer 5

Japt, 16 14 bytes

r"%l"_m"uv"gT°

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Explanation

r              // RegEx replace input
 "%l"          // [A-Za-z] as first arg to replace
     _         // created function Z=>Z as second arg to replace
       "uv"gT° // alternates "u" & "v"
      m        // map Z to either "u" upper or "v" lower

Answer 6

Python 3, 86 76 68 66 63 bytes

^{-2 bytes thanks to DJMcMayhem
-3 bytes thanks to Cyoce}

x=0
for i in input():print(end=(2*i).title()[x]);x^=i.isalpha()

Try it online! or Try all test cases

Answer 7

C (tcc), 60 57 56 bytes

Thanks to DigitalTrauma for noticing bit 5 is the only difference for ASCII upper/lower case.

Special thanks to zch for golfing off three more bytes.

Save one more byte from RJHunter's idea

l;f(char*s){for(;*s=isalpha(*s)?*s&95|++l%2<<5:*s;s++);}

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Answer 8

Alice, 18 bytes

/olZlYuN
@iy.u..//

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Explanation

This program follows a lesser-known template for odd-length programs that run entirely in ordinal mode. The linearized version of this code is:

il.l.uN.YuZyo@

Explanation of code:

i - push input onto stack            ["Hello world!"]
l - convert to lowercase             ["hello world!"]
. - duplicate                        ["hello world!", "hello world!"]
l - convert to lowercase (should be no-op, but avoids what seems to be a bug in the TIO implementation)
. - duplicate again                  ["hello world!", "hello world!", "hello world!"]
u - convert to uppercase             ["hello world!", "hello world!", "HELLO WORLD!"]
N - difference between sets          ["hello world!", "helloworld"]
. - duplicate reduced string         ["hello world!", "helloworld", "helloworld"]
Y - unzip (extract even positions)   ["hello world!", "helloworld", "hlool", "elwrd"]
u - convert to uppercase             ["hello world!", "helloworld", "hlool", "ELWRD"]
Z - zip evens back into string       ["hello world!", "helloworld", "hElLoWoRlD"]
y - perform substitution             ["hElLo WoRlD!"]
o - output                           []
@ - terminate

Without using l on the duplicate, the stack after N would be ["helloworld", "helloworld"]. I strongly suspect this is a bug.

Answer 9

MATL, 16 15 bytes

Xktkyy-f2L))5M(

Try it online! Or verify all test cases.

Explanation

Consider input 'hello world'

Xk    % To upper case
      % STACK: 'HELLO WORLD'
t     % Duplicate top element
      % STACK: 'HELLO WORLD', 'HELLO WORLD'
k     % To lower case
      % STACK: 'HELLO WORLD', 'hello word'
yy    % Duplicate top two elements
      % STACK: 'HELLO WORLD', 'hello word', 'HELLO WORLD', 'hello word'
-     % Difference (of code points; element-wise)
      % STACK: 'HELLO WORLD', 'hello word', [-32 -32 -32 -32 -32 0 -32 -32 -32 -32 -32]
f     % Indices of nonzeros
      % STACK: 'HELLO WORLD', 'hello word', [1 2 3 4 5 7 8 9 10 11]
2L)   % Keep only even-indexed values (*)
      % STACK: 'HELLO WORLD', 'hello word', [2 4 7 9 11]
)     % Reference indexing (get values at indices)
      % STACK: 'HELLO WORLD', 'elwrd'
5M    % Push (*) again
      % STACK: 'HELLO WORLD', 'elwrd', [2 4 7 9 11]
(     % Assignment indexing (write values at indices). Implicit display
      % STACK: 'HeLlO wOrLd

'

Answer 10

Ruby, 57 55 47 41 bytes

Byte count includes two bytes for command line options.
Run it for example like this: $ ruby -p0 alternate_case.rb <<< "some input"

gsub(/\p{L}/){($&.ord&95|32*$.^=1).chr}

With the p0 option, the entire input is consumed in one go, and the magical global $. is incremented to 1. This is later toggled between 0 and 1 and used for keeping the state.

Works with multiline input; Try it online!

Thanks to Ventero for amazing input -- check the comments for details.

Answer 11

Java 8, 99 bytes

a->{String r="";int i=0;for(int c:a)r+=(char)(c>64&c<91|c>96&c<123?i++%2<1?c|32:c&~32:c);return r;}

Explanation:

Try it here.

a->{                          // Lambda with char-array parameter and String return-type
  String r="";                //  Result-String
  int i=0;                    //  Flag for alteration
  for(int c:a)                //  Loop over the characters of the input
    r+=(char)                 //   And append the result-String with the following (converted to char):
      (c>64&c<91|c>96&c<123?  //    If it's a letter:
       i++%2<1?               //     And the flag states it should be lowercase:
        (c|32)                //      Convert it to lowercase
       :                      //     Else (should be uppercase):
        (c&~32)               //      Convert it to uppercase
      :                       //    Else:
       c);                    //     Simply append the non-letter character as is
                              //  End of loop (implicit / single-line body)
  return r;                   //  Return result-String
}                             // End of method

Answer 12

Brachylog, 25 bytes

{ḷ|ụ}ᵐ.{ḷ∈Ạ&}ˢ¬{s₂{∈Ạ}ᵐ}∧

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This is both long and slow.

Explanation

{   }ᵐ.                       The Output is the result of mapping on each char of the Input:
 ḷ                              Lowecase the char
  |                             Or
   ụ                            Uppercase the char
       {    }ˢ                In the Ouput, select the chars that:
        ḷ∈Ạ&                    when lowercased are in "abc...xyz" (ie are letters)
              ¬{       }∧     In that new string, it is impossible to find:
                s₂              a substring of 2 consecutive chars
                  {∈Ạ}ᵐ         where both of them are in the lowercase alphabet

Answer 13

Perl 6,  32  30 bytes

{S:g/<:L><-:L>*<:L>?/$/.tclc()/}

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{S:g{<:L><-:L>*<:L>?}=$/.tclc}

Try it

Expanded:

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

  S            # string replace (not in-place) implicitly against ｢$_｣

  :global

  {

    <+ :L >    # a letter
    <- :L >*   # any number of non-letters
    <+ :L >?   # an optional letter

  }

  =

  $/.tclc()    # uppercase the first letter, lowercase everything else
}

Answer 14

q/kdb+, 51 42 38 bytes

Solution:

{@[x;;upper]1#'2 cut(&)x in .Q.a}lower

Example:

q){@[x;;upper]1#'2 cut(&)x in .Q.a}lower"hello world"
"HeLlO wOrLd"

Notes:

.Q.a        // abcde...xyz lowercase alphabet
(&) x in    // where, returns indices for where x (hello world) is an alpha
2 cut       // splits list into 2-item lists
1#'         // takes first item of each 2-item list; ie the indices to uppercase
@[x;;upper] // apply (@) upper to x at these indices

Answer 15

Google Sheets, 264 bytes

=ArrayFormula(JOIN("",IF(REGEXMATCH(MID(A1,ROW(OFFSET(A1,0,0,LEN(A1))),1),"[A-Za-z]"),CHAR(CODE(UPPER(MID(A1,ROW(OFFSET(A1,0,0,LEN(A1))),1)))+MOD(LEN(REGEXREPLACE(LEFT(A1,ROW(OFFSET(A1,0,0,LEN(A1)))),"[^A-Za-z]","")),2)*32),MID(A1,ROW(OFFSET(A1,0,0,LEN(A1))),1))))

It's a big mess but it's a little easier if you expand it out:

=ArrayFormula(
  JOIN(
    "",
    IF(REGEXMATCH(MID(A1,ROW(OFFSET(A1,0,0,LEN(A1))),1),"[A-Za-z]"),
      CHAR(
        CODE(UPPER(MID(A1,ROW(OFFSET(A1,0,0,LEN(A1))),1)))
        +
        MOD(LEN(REGEXREPLACE(LEFT(A1,ROW(OFFSET(A1,0,0,LEN(A1)))),"[^A-Za-z]","")),2)*32
      ),
      MID(A1,ROW(OFFSET(A1,0,0,LEN(A1))),1)
    )
  )
) 

The pseudo-logic would run like this:

For each character {                                    // ArrayFormula()
  If (character is a letter) {                          // REGEXMATCH(MID())
    Return CHAR(                                        // CHAR()
      CODE(UPPER(letter))                               // CODE(UPPER(MID()))
      +
      If (nth letter found and n is odd) {32} else {0}  // MOD(LEN(REGEXREPLACE(LEFT())))
    )
  } else {
    Return character                                    // MID()
  }
}

Answer 16

V, 17, 13 bytes

VUÍშáü$©/ì&

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Or Verify all test cases!

HeXdUmP:

00000000: 5655 cde1 83a8 e1fc 24a9 2fec 26         VU......$./.&

Explanation:

This uses a compressed regex™️, so before explaining it, let's expand the regex out:

:%s/\v\a.{-}(\a|$)/\l&

The VU converts everything to uppercase. Then we run this:

:%                      " On every line:
  s/\v                  "   Substitute:
      \a                "     A letter
        .{-}            "     Followed by as few characters as possible
            (\a|$)      "     Followed by either another letter or an EOL
                  /     "   With:
                   \l   "     The next character is lowercased
                     &  "     The whole text we matched

---

Old/more interesting answer:

:se nows
Vuò~h2/á

Answer 17

CJam, 26 24 bytes

qeu{_'[,65>&,T^:T{el}&}%

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Explanation

q         e# Read all input.
eu        e# Uppercase it.
{         e# For each character:
 _        e#  Duplicate it.
 '[,65>&  e#  Set intersection with the uppercase alphabet.
 ,        e#  Length (either 0 or 1 in this case).
 T^:T     e#  XOR with T (T is initially 0), then store the result back in T.
 {el}&    e#  If The result of the XOR is true, lowercase the character.
}%        e# (end for)

Answer 18

PHP, 71 Bytes

for(;a&$c=$argn[$i++];)echo ctype_alpha($c)?(ul[$k++&1].cfirst)($c):$c;

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Answer 19

Retina, 32 bytes

T`l`L
01T`L`l`[A-Z][^A-Z]*[A-Z]?

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First converts the input to uppercase, and then groups the input into matches containing up to two capital letters. The only time it will contain only one letter is if the last letter doesn't have a pair. Then it lowercases the first letter of each of these matches.

The 01 in the second stage translates roughly to: do not change the behaviour of this stage based on the match number, but only apply the changes to the first character of each match.

Answer 20

Pyth, 11 bytes

srR~xZ}dGrZ

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Explanation

              # Z = 0; Q = eval(input())
srR~xZ}dGrZQ  # Auto-fill variables
         rZQ  # lowercase the input
 rR           # Apply the r function to each letter of the input with
   ~xZ}dG     # ... this as the other argument
   ~          # use the old value of the variable Z, then update it with the value of ...
    xZ        # Z xor ...
      }dG     # the variable d is a lowercase letter
              # because of how mapping works in pyth, d will contain the current letter
              # This causes Z to flip between 0 and 1, alternately upper and lower casing
              # the current character if it is a letter

Answer 21

PowerShell, 86 bytes

-join($args[0]|%{if($_-match"[a-z]"-and($i=!$i)){"$_".toupper()}else{"$_".tolower()}})

Input is a [char[]] array.

Comments in code for explanation

# Join the array of string and char back together.
-join
    # Take the first argument and pass each element ([char]) down the pipe. 
    ($args[0]|%{
        # Check if this is a letter. Second condition is a boolean that changes at every pass 
        # but only if the current element is a letter. If not the condition never fires
        if($_-match"[a-z]"-and($i=!$i)){
            # Change the character to uppercase
            "$_".toupper()
        }else{
            # Output the character to lowercase. 
            # Special characters are not affected by this method
            "$_".tolower()
        }
    })

Answer 22

Haskell, 105 83 + 2 4 + 1 byte of separator = 108 86 88 Bytes

import Data.Char
f#(x:y)|isLetter x=([toUpper,toLower]!!f)x:(1-f)#y|1>0=x:f#y
_#l=l

Function is (1#), starts lowercase. Try it online!

The sad thing is that this is longer than the Java and C# answers Thanks to Ørjan Johansen for saving 22 bytes by merging three lines into one!

Answer 23

Perl 5, 24 bytes

23 bytes + 1 byte for -p.

Thanks to @Dada for -2 bytes.

s/\pl/--$|?uc$&:lc$&/eg

Try it online!

Answer 24

Retina, 46 bytes

T`L`l
T`l`L`.(?=([^a-z]*|[a-z][^a-z]*[a-z])*$)

Try it online! Includes test cases.

Answer 25

C 64 bytes

B;R(char *s){for(;*s=isalpha(*s)?(B=!B)?*s|=32:*s&=~32:*s;s++);}

Takes advantage of ascii encoding where upper and lower case letters are offset by 0x20.

Answer 26

PHP 5, 54 bytes

<?=preg_filter('/\pL/e','($0|" ")^a^aA[$i^=1]',$argn);

Answer 27

C#, 100 bytes

s=>{var r="";int m=0;foreach(var c in s)r+=char.IsLetter(c)?(char)(++m%2>0?c|32:c&~32):c;return r;};

Answer 28

Python 3, 192 bytes

x=list(input())
s=[]
for i in x[1::2]:
 s.append(i)
 x.remove(i)
s.reverse()
while len(x)<len(s):
 x.append("")
while len(x)>len(s):
 s.append("")
for i in range(len(x)):
 print(end=x[i]+s[i])

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Answer 29

Groovy, 79 bytes

{x=0;it.toUpperCase().collect{(it==~/\w/)?x++%2?it:it.toLowerCase():it}.join()}

Answer 30

Convex, 16 bytes

¯{_U&,R^:R{¬}&}%

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Convex port of @Business Cat's answer.

Answer 31

Python 3, 95 bytes

Not nearly as golfed as @Rod's, but I'm posting it nevertheless.

lambda s:"".join([s[i],[s[i].lower(),s[i].upper()][i%2]][s[i].isalpha()]for i in range(len(s)))

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Answer 32

J, 40 bytes

(2|+/\@:e.&Alpha_j_)`(tolower,:toupper)}

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This follows almost directly from the definition of Composite Item (}):

We stack the upper and lowercase version of the input for our two possibilities using (tolower,:toupper).

Next we create a boolean list to indicate whether each character is alphabetic: e.&Alpha_j_ and take the scan sum of that: +/\@: which creates a monotonically increasing list which increases only on alphabetic characters. Finally we turn that into a boolean list where evens are 0 and odds are 1: 2|.

Putting those parts together means we alternate between the lower and uppercase versions of our input whenever we encounter a new alphabetic character, and only then.

Answer 33

C# (Visual C# Interactive Compiler), 84 bytes

int m,c;for(;(c=Read())>0;)Write((char)(c>64&c<91|c>96&c<123?m++%2>0?c|32:c&~32:c));

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Modified version of The Lethal Coder's answer that takes advantage of some of the features of the interactive compiler.

Answer 34

Japt v2.0a0, 11 bytes

r\lÈc^H*°Tv

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Answer 35

Acc!!, 110 bytes

N*8
Count i while _/256 {
_+_/520*(727/_)*2+_/776*(983/_)*4
Write _/8+32*(_%4/3-_%8%5/4)
(_+(_%8+2)/4)%2+N*8
}

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Algorithm

Set toLower flag to 0
Read a character
While not EOL:
  Calculate isUpper and isLower flags
  If isUpper && toLower, output (character + 32)
  If isLower && !toLower, output (character - 32)
  Else, output character unchanged
  If isUpper or isLower, toggle toLower
  Reset isUpper and isLower, and read a new character

Accumulator partitioning

We reserve the lowest three bits of the accumulator for storing the flags: toLower in the 1's place, isUpper in the 2's place, and isLower in the 4's place. The rest of the accumulator, at the 8's place and above, stores the input character.

For example, with input of Hello World, after reading the H and calculating flags, the accumulator has a value of 578, or in binary:

1001000 0 1 0
   |    | | \-> toLower (false)
   |    | \---> isUpper (true)
   |    \-----> isLower (false)
   \----------> ASCII code for 'H'

After reading the space, the accumulator is 257:

0100000 0 0 1
   |    | | \-> toLower (true)
   |    | \---> isUpper (false)
   |    \-----> isLower (false)
   \----------> ASCII code for ' '

Code

Line 1

N*8

Stores the first input character and sets all the flags to 0.

Line 2

Count i while _/256 {

Loops while the accumulator value is >= 256 (i.e. the character's ASCII code is at least 32).

Line 3

_+_/520*(727/_)*2+_/776*(983/_)*4

Calculates the isUpper and isLower flags. The main trick here is using integer division for the inequalities. For example, if x is a printable-ASCII code, then 65 <= x is equivalent to x/65 with integer division: values smaller than 65 become 0, values between 65 and 126 become 1, and we don't have to worry about anything larger than 126. We can use this approach for all four inequalities (keeping in mind that the ASCII codes are all multiplied by 8 in the accumulator).

isUpper:

65 <= char <= 90
65*8 <= acc <= 90*8+7
acc/520 == 1 && 727/acc == 1

Expression: _/520*(727/_)

isLower:

97 <= char <= 122
97*8 <= acc <= 122*8+7
acc/776 == 1 && 983/acc == 1

Expression: _/776*(983/_)

This line, then, takes the accumulator and adds isUpper*2 + isLower*4 to it.

Line 4

Write _/8+32*(_%4/3-_%8%5/4)

Outputs the character, swapping case if necessary.

There are two instances in which we need to modify the ASCII value before output:

1. isUpper is 1, toLower is 1
   • We need to convert the uppercase letter to lowercase by adding 32
   • The three flag bits are 011, i.e. acc%4 == 3
2. isLower is 1, toLower is 0
   • We need to convert the lowercase letter to uppercase by subtracting 32
   • The three flag bits are 100, i.e. acc%8 == 4

We can simulate the equality test using mod and int-div. For example, acc%8 == 4 is equivalent to acc%8%5/4:

acc%8      0 1 2 3 4 5 6 7
     %5    0 1 2 3 4 0 1 2
       /4  0 0 0 0 1 0 0 0

This line, therefore, takes 1 if we need to convert upper-to-lower, subtracts 1 if we need to convert lower-to-upper, multiplies that by 32, adds it to the ASCII code, and outputs it.

Line 5

(_+(_%8+2)/4)%2+N*8

Toggle the toLower flag if necessary; inputs and stores a new character.

We need to toggle toLower if the character was a letter, i.e. either isUpper or isLower was set. The possible flag bit values are 010, 011, 100, and 101--i.e. 2 through 5 (note that 6 and 7 are not possible because a letter can't be both upper- and lowercase). This expression gives 1 for those values and 0 otherwise:

 _%8       0 1 2 3 4 5
    +2     2 3 4 5 6 7
(     )/4  0 0 1 1 1 1

Call this value isLetter. Then we want the new value of toLower to be toLower XOR isLetter (if isLetter is true, swap the value of toLower; otherwise, leave it unchanged). We can do XOR by adding the two values mod 2. Finally, we also read another character with N, multiply it by 8, and add it in.