ssTTsSTtRrriinInnnnNNNIiinngg












18












$begingroup$


Challenge



For each character of the string except for the last one, do the following:




  • Output the current character.



  • Followed by randomly outputting from the following list a random number of times between 1 - 5 (inclusive):




    • The current character

    • The next character of the string

    • The switchcase version of the character that you are currently on

    • The switchcase version of the next character of the string.




Test Cases



String --> SSSTSStrTrIiinIIngn



, . , . , . Hello world! --> ,,, .. , ,, .... , , .. .. . HHH HHEeelLlLllooO wwOworOOrrrRllDd!!D



Programming Puzzles and Code Golf --> PrPPrRrOooooogggRgGraAraaaMMMmmmimMIiininGGgG PPPPuZzZZzZzzZzllLLEEeEsEsssS a aANnNddD C COCoooOOdeDe E GGGoOllFFf



Notes




  • You only need to apply the switchcase version of a character if the character is part of the alphabet (A-Z and a-z).

  • Your random function does not need to be uniform but it still needs to have a chance of returning any element in the list given.

  • You are allowed to use any standard I/O format.

  • You may assume that the length of the input is greater than or equal to two.

  • You may assume that the input only consists of ASCII characters.

  • The title is not a test case (it is unintentional if it is a valid test case).

  • Switchcase means to turn the char to lowercase if it is uppercase and to turn it to uppercase if it is lowercase.










share|improve this question











$endgroup$












  • $begingroup$
    In addition to '... does not need to be uniform', I think you probably want to specify that given some input, all finite legal outputs should in principle be possible to generate (otherwise, my non-uniform random integer in [1,2,3,4,5] is always going to be 2, and I'll just output the original string).
    $endgroup$
    – Chas Brown
    Apr 3 at 1:21










  • $begingroup$
    @ChasBrown Yeah, I'll edit the question
    $endgroup$
    – MilkyWay90
    Apr 3 at 1:39






  • 2




    $begingroup$
    I find the specification confusing. Can you be more explicit? For example, work out how String produces SSSTSStrTrIiinIIngn
    $endgroup$
    – Luis Mendo
    Apr 3 at 9:02








  • 7




    $begingroup$
    @LuisMendo I'm not OP, but I think: [S]SSTSS [t]rT, [r]I, [i]inII, [n]gn, where the characters between the blocks are the first bullet points ("Output the current character"), and the other characters are 1-5 times randomly one of the four choices for that character. But I agree, some more explicit explanations would be appropriate. Apart from the test case it wasn't particularly clear we have to pick a random choice 1-5 times. Instead of picking a random choice repeated 1-5 times (as the Gaia answer currently does).
    $endgroup$
    – Kevin Cruijssen
    Apr 3 at 11:09








  • 3




    $begingroup$
    @KevinCruijssen Thanks, Your explanation fits the example, and is clear. The OP should confirm and edit that into the text
    $endgroup$
    – Luis Mendo
    Apr 3 at 11:19


















18












$begingroup$


Challenge



For each character of the string except for the last one, do the following:




  • Output the current character.



  • Followed by randomly outputting from the following list a random number of times between 1 - 5 (inclusive):




    • The current character

    • The next character of the string

    • The switchcase version of the character that you are currently on

    • The switchcase version of the next character of the string.




Test Cases



String --> SSSTSStrTrIiinIIngn



, . , . , . Hello world! --> ,,, .. , ,, .... , , .. .. . HHH HHEeelLlLllooO wwOworOOrrrRllDd!!D



Programming Puzzles and Code Golf --> PrPPrRrOooooogggRgGraAraaaMMMmmmimMIiininGGgG PPPPuZzZZzZzzZzllLLEEeEsEsssS a aANnNddD C COCoooOOdeDe E GGGoOllFFf



Notes




  • You only need to apply the switchcase version of a character if the character is part of the alphabet (A-Z and a-z).

  • Your random function does not need to be uniform but it still needs to have a chance of returning any element in the list given.

  • You are allowed to use any standard I/O format.

  • You may assume that the length of the input is greater than or equal to two.

  • You may assume that the input only consists of ASCII characters.

  • The title is not a test case (it is unintentional if it is a valid test case).

  • Switchcase means to turn the char to lowercase if it is uppercase and to turn it to uppercase if it is lowercase.










share|improve this question











$endgroup$












  • $begingroup$
    In addition to '... does not need to be uniform', I think you probably want to specify that given some input, all finite legal outputs should in principle be possible to generate (otherwise, my non-uniform random integer in [1,2,3,4,5] is always going to be 2, and I'll just output the original string).
    $endgroup$
    – Chas Brown
    Apr 3 at 1:21










  • $begingroup$
    @ChasBrown Yeah, I'll edit the question
    $endgroup$
    – MilkyWay90
    Apr 3 at 1:39






  • 2




    $begingroup$
    I find the specification confusing. Can you be more explicit? For example, work out how String produces SSSTSStrTrIiinIIngn
    $endgroup$
    – Luis Mendo
    Apr 3 at 9:02








  • 7




    $begingroup$
    @LuisMendo I'm not OP, but I think: [S]SSTSS [t]rT, [r]I, [i]inII, [n]gn, where the characters between the blocks are the first bullet points ("Output the current character"), and the other characters are 1-5 times randomly one of the four choices for that character. But I agree, some more explicit explanations would be appropriate. Apart from the test case it wasn't particularly clear we have to pick a random choice 1-5 times. Instead of picking a random choice repeated 1-5 times (as the Gaia answer currently does).
    $endgroup$
    – Kevin Cruijssen
    Apr 3 at 11:09








  • 3




    $begingroup$
    @KevinCruijssen Thanks, Your explanation fits the example, and is clear. The OP should confirm and edit that into the text
    $endgroup$
    – Luis Mendo
    Apr 3 at 11:19
















18












18








18


2



$begingroup$


Challenge



For each character of the string except for the last one, do the following:




  • Output the current character.



  • Followed by randomly outputting from the following list a random number of times between 1 - 5 (inclusive):




    • The current character

    • The next character of the string

    • The switchcase version of the character that you are currently on

    • The switchcase version of the next character of the string.




Test Cases



String --> SSSTSStrTrIiinIIngn



, . , . , . Hello world! --> ,,, .. , ,, .... , , .. .. . HHH HHEeelLlLllooO wwOworOOrrrRllDd!!D



Programming Puzzles and Code Golf --> PrPPrRrOooooogggRgGraAraaaMMMmmmimMIiininGGgG PPPPuZzZZzZzzZzllLLEEeEsEsssS a aANnNddD C COCoooOOdeDe E GGGoOllFFf



Notes




  • You only need to apply the switchcase version of a character if the character is part of the alphabet (A-Z and a-z).

  • Your random function does not need to be uniform but it still needs to have a chance of returning any element in the list given.

  • You are allowed to use any standard I/O format.

  • You may assume that the length of the input is greater than or equal to two.

  • You may assume that the input only consists of ASCII characters.

  • The title is not a test case (it is unintentional if it is a valid test case).

  • Switchcase means to turn the char to lowercase if it is uppercase and to turn it to uppercase if it is lowercase.










share|improve this question











$endgroup$




Challenge



For each character of the string except for the last one, do the following:




  • Output the current character.



  • Followed by randomly outputting from the following list a random number of times between 1 - 5 (inclusive):




    • The current character

    • The next character of the string

    • The switchcase version of the character that you are currently on

    • The switchcase version of the next character of the string.




Test Cases



String --> SSSTSStrTrIiinIIngn



, . , . , . Hello world! --> ,,, .. , ,, .... , , .. .. . HHH HHEeelLlLllooO wwOworOOrrrRllDd!!D



Programming Puzzles and Code Golf --> PrPPrRrOooooogggRgGraAraaaMMMmmmimMIiininGGgG PPPPuZzZZzZzzZzllLLEEeEsEsssS a aANnNddD C COCoooOOdeDe E GGGoOllFFf



Notes




  • You only need to apply the switchcase version of a character if the character is part of the alphabet (A-Z and a-z).

  • Your random function does not need to be uniform but it still needs to have a chance of returning any element in the list given.

  • You are allowed to use any standard I/O format.

  • You may assume that the length of the input is greater than or equal to two.

  • You may assume that the input only consists of ASCII characters.

  • The title is not a test case (it is unintentional if it is a valid test case).

  • Switchcase means to turn the char to lowercase if it is uppercase and to turn it to uppercase if it is lowercase.







code-golf string random






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 3 at 23:00







MilkyWay90

















asked Apr 2 at 23:23









MilkyWay90MilkyWay90

745316




745316












  • $begingroup$
    In addition to '... does not need to be uniform', I think you probably want to specify that given some input, all finite legal outputs should in principle be possible to generate (otherwise, my non-uniform random integer in [1,2,3,4,5] is always going to be 2, and I'll just output the original string).
    $endgroup$
    – Chas Brown
    Apr 3 at 1:21










  • $begingroup$
    @ChasBrown Yeah, I'll edit the question
    $endgroup$
    – MilkyWay90
    Apr 3 at 1:39






  • 2




    $begingroup$
    I find the specification confusing. Can you be more explicit? For example, work out how String produces SSSTSStrTrIiinIIngn
    $endgroup$
    – Luis Mendo
    Apr 3 at 9:02








  • 7




    $begingroup$
    @LuisMendo I'm not OP, but I think: [S]SSTSS [t]rT, [r]I, [i]inII, [n]gn, where the characters between the blocks are the first bullet points ("Output the current character"), and the other characters are 1-5 times randomly one of the four choices for that character. But I agree, some more explicit explanations would be appropriate. Apart from the test case it wasn't particularly clear we have to pick a random choice 1-5 times. Instead of picking a random choice repeated 1-5 times (as the Gaia answer currently does).
    $endgroup$
    – Kevin Cruijssen
    Apr 3 at 11:09








  • 3




    $begingroup$
    @KevinCruijssen Thanks, Your explanation fits the example, and is clear. The OP should confirm and edit that into the text
    $endgroup$
    – Luis Mendo
    Apr 3 at 11:19




















  • $begingroup$
    In addition to '... does not need to be uniform', I think you probably want to specify that given some input, all finite legal outputs should in principle be possible to generate (otherwise, my non-uniform random integer in [1,2,3,4,5] is always going to be 2, and I'll just output the original string).
    $endgroup$
    – Chas Brown
    Apr 3 at 1:21










  • $begingroup$
    @ChasBrown Yeah, I'll edit the question
    $endgroup$
    – MilkyWay90
    Apr 3 at 1:39






  • 2




    $begingroup$
    I find the specification confusing. Can you be more explicit? For example, work out how String produces SSSTSStrTrIiinIIngn
    $endgroup$
    – Luis Mendo
    Apr 3 at 9:02








  • 7




    $begingroup$
    @LuisMendo I'm not OP, but I think: [S]SSTSS [t]rT, [r]I, [i]inII, [n]gn, where the characters between the blocks are the first bullet points ("Output the current character"), and the other characters are 1-5 times randomly one of the four choices for that character. But I agree, some more explicit explanations would be appropriate. Apart from the test case it wasn't particularly clear we have to pick a random choice 1-5 times. Instead of picking a random choice repeated 1-5 times (as the Gaia answer currently does).
    $endgroup$
    – Kevin Cruijssen
    Apr 3 at 11:09








  • 3




    $begingroup$
    @KevinCruijssen Thanks, Your explanation fits the example, and is clear. The OP should confirm and edit that into the text
    $endgroup$
    – Luis Mendo
    Apr 3 at 11:19


















$begingroup$
In addition to '... does not need to be uniform', I think you probably want to specify that given some input, all finite legal outputs should in principle be possible to generate (otherwise, my non-uniform random integer in [1,2,3,4,5] is always going to be 2, and I'll just output the original string).
$endgroup$
– Chas Brown
Apr 3 at 1:21




$begingroup$
In addition to '... does not need to be uniform', I think you probably want to specify that given some input, all finite legal outputs should in principle be possible to generate (otherwise, my non-uniform random integer in [1,2,3,4,5] is always going to be 2, and I'll just output the original string).
$endgroup$
– Chas Brown
Apr 3 at 1:21












$begingroup$
@ChasBrown Yeah, I'll edit the question
$endgroup$
– MilkyWay90
Apr 3 at 1:39




$begingroup$
@ChasBrown Yeah, I'll edit the question
$endgroup$
– MilkyWay90
Apr 3 at 1:39




2




2




$begingroup$
I find the specification confusing. Can you be more explicit? For example, work out how String produces SSSTSStrTrIiinIIngn
$endgroup$
– Luis Mendo
Apr 3 at 9:02






$begingroup$
I find the specification confusing. Can you be more explicit? For example, work out how String produces SSSTSStrTrIiinIIngn
$endgroup$
– Luis Mendo
Apr 3 at 9:02






7




7




$begingroup$
@LuisMendo I'm not OP, but I think: [S]SSTSS [t]rT, [r]I, [i]inII, [n]gn, where the characters between the blocks are the first bullet points ("Output the current character"), and the other characters are 1-5 times randomly one of the four choices for that character. But I agree, some more explicit explanations would be appropriate. Apart from the test case it wasn't particularly clear we have to pick a random choice 1-5 times. Instead of picking a random choice repeated 1-5 times (as the Gaia answer currently does).
$endgroup$
– Kevin Cruijssen
Apr 3 at 11:09






$begingroup$
@LuisMendo I'm not OP, but I think: [S]SSTSS [t]rT, [r]I, [i]inII, [n]gn, where the characters between the blocks are the first bullet points ("Output the current character"), and the other characters are 1-5 times randomly one of the four choices for that character. But I agree, some more explicit explanations would be appropriate. Apart from the test case it wasn't particularly clear we have to pick a random choice 1-5 times. Instead of picking a random choice repeated 1-5 times (as the Gaia answer currently does).
$endgroup$
– Kevin Cruijssen
Apr 3 at 11:09






3




3




$begingroup$
@KevinCruijssen Thanks, Your explanation fits the example, and is clear. The OP should confirm and edit that into the text
$endgroup$
– Luis Mendo
Apr 3 at 11:19






$begingroup$
@KevinCruijssen Thanks, Your explanation fits the example, and is clear. The OP should confirm and edit that into the text
$endgroup$
– Luis Mendo
Apr 3 at 11:19












20 Answers
20






active

oldest

votes


















6












$begingroup$


Gaia, 25 bytes



ṇ+†ṅ⟨)₌¤:~+4ṛ⟨ṛ₌¤⟩ₓ⟩¦$


Try it online!



Thanks to Kevin Cruijssen for pointing out 2 bugs!



ṇ				| delete the last character from the input
+† | push the input again and concatenate together, so for instance
| 'abc' 'bc' becomes ['ab' 'bc' 'c']
ṅ | delete the last element
⟨ ⟩¦ | for each of the elements, do:
)₌ | take the first character and push again
¤ | swap
: | dup
~ | swap case
+ | concatenate strings
4ṛ | select a random integer from [1..5]
⟨ ⟩ₓ | and repeat that many times
ṛ₌¤ | select a random character from the string
| clean up stack
$ | convert to string


Note that 4ṛ is because is implemented for an integer z as python's random.randint(1,z+1), which returns an integer N such that 1<=N<=z+1.






share|improve this answer











$endgroup$













  • $begingroup$
    Are you sure the run-length encode is correct here? If I understand the challenge correctly: the four options should be chosen 1-5 times randomly, instead of choosing one of the four randomly, repeated 1-5 times. The first example output SSSTSStrTrIiinIIngn ([SSSTSS, trT, rI, iinII, ngn]) seems to reflect this, and currently isn't a possible output in your program (I think).
    $endgroup$
    – Kevin Cruijssen
    Apr 3 at 8:56












  • $begingroup$
    @KevinCruijssen I interpreted "output from the list a random number of times" to mean run-length decode, but you're right, the test cases do seem to indicate the other interpretation; I think it should be pretty easy to fix
    $endgroup$
    – Giuseppe
    Apr 3 at 10:50








  • 1




    $begingroup$
    5ṛ can result in 6 for some reason Try it online? PS: Isn't there an integer to ranged list, or ranged for-loop in Gaia?
    $endgroup$
    – Kevin Cruijssen
    Apr 3 at 11:32








  • 1




    $begingroup$
    @KevinCruijssen dang, Business Cat really needs to fix off-by-one errors...I really thought there was a for type construct, but I'm pretty sure it's which isn't even documented on the wiki page.
    $endgroup$
    – Giuseppe
    Apr 3 at 11:41



















3












$begingroup$


APL (dzaima/APL), 23 bytes





Anonymous tacit prefix function.



∊2(⊣,{?4⍴⍨?5}⊇,,-⍤,)/


Try it online!



2()/ apply the following infix tacit function between each character pair:



- the switchcase

 of

, the concatenation of the pair



,, prepend the concatenation of the pair to that



{}⊇ pick the following elements from that:



  ?5 random number in range 1…5



  4⍴⍨ that many fours



  ? random indices for those



ϵnlist (flatten)






share|improve this answer











$endgroup$





















    3












    $begingroup$


    Perl 6, 60 bytes





    {S:g{.)>(.)}=$/~[~] roll ^5 .roll+1,$/.lc,$/.uc,$0.lc,$0.uc}


    Try it online!



    The lowercase/uppercase part is kinda annoying.






    share|improve this answer









    $endgroup$













    • $begingroup$
      I don't know Perl, so I'm probably saying something stupid here. But is it somehow possible to concat the $/ and $0 together and use .lc on that string, and then create a copy of that string and use .uc, and concat those two together? Not sure if that's even possible, or shorter than your current $/.lc,$/.uc,$0.lc,$0.uc, but it would mean you'd use $/, $0, .lc, and .uc once each.
      $endgroup$
      – Kevin Cruijssen
      Apr 3 at 13:43








    • 1




      $begingroup$
      Alas, (.lc~.uc for $0~$/).comb is longer. Perl 6 really wants to distinguish strings and lists, so "abc"[0] eq "abc" (it pretends to be a single-item list).
      $endgroup$
      – Ven
      Apr 3 at 16:02












    • $begingroup$
      You can do it by slipping and an anonymous function applied to a list: {.lc,|.uc}($/,|$0) for -5 bytes, and just use the list of matches {.lc,|.uc}(@$/) for -8 bytes. tio.run/…
      $endgroup$
      – Phil H
      Apr 4 at 16:12










    • $begingroup$
      @PhilH No that doesn't work. Those solutions only capitalise one of the letters each
      $endgroup$
      – Jo King
      Apr 4 at 21:18





















    3












    $begingroup$


    Jelly, 12 bytes



    ;Œsṗ5X¤XṭṖµƝ


    Try it online!






    share|improve this answer









    $endgroup$





















      3












      $begingroup$


      Bash, 121 bytes



      -20 bytes thanks to Nahuel



      -9 bytes thanks to roblogic





      for((i=0;i<${#1};i++)){
      s=${1:i:1}
      m=${1:i:2}
      m=${m,,}${m^^}
      for((t=0;t++<RANDOM%6;)){
      s+=${m:RANDOM%4:1}
      }
      printf "$s"
      }


      Try it online!



      Original answer




      Bash, 150 bytes



      Have done very little golf bashing and trying to improve my bash, so any comments welcome.





      for((i=0;i<${#1}-1;i++));do
      c=${1:$i:1}
      n=${1:$((i+1)):1}
      a=($n ${c,} ${c^} ${n,} ${n^})
      shuf -e ${a[@]} -n "$(shuf -i 1-5 -n 1)"|xargs printf %s
      done


      Try it online!



      Code is straightforward loop through chars setting current c and next n character, then creating an array of the 4 possibilities, repeating one of them so there's exactly 5. Next we shuffle that array, and then choose n elements from it, where n itself is random between 1 and 5.






      share|improve this answer











      $endgroup$













      • $begingroup$
        seems it's missing printf %s "$c"
        $endgroup$
        – Nahuel Fouilleul
        Apr 3 at 8:06






      • 1




        $begingroup$
        do and done can be replaced with undocumented { and }
        $endgroup$
        – Nahuel Fouilleul
        Apr 3 at 8:08










      • $begingroup$
        with some changes
        $endgroup$
        – Nahuel Fouilleul
        Apr 3 at 8:25












      • $begingroup$
        @NahuelFouilleul tyvm. nice improvements and nice catch -- i'd somehow missed the requirement always to output the current char.
        $endgroup$
        – Jonah
        Apr 3 at 13:07






      • 1




        $begingroup$
        @roblogic that's clever. tyvm.
        $endgroup$
        – Jonah
        Apr 3 at 21:36



















      2












      $begingroup$


      Python 2, 107 bytes





      f=lambda s:s and s[0]+''.join(sample((s[:2]+s[:2].swapcase())*5,randint(1,5)))+f(s[1:])
      from random import*


      Try it online!






      share|improve this answer









      $endgroup$





















        2












        $begingroup$


        05AB1E, 18 17 bytes



        ü)vyн5LΩFyD.š«Ω]J


        Inspired by @Giuseppe's Gaia answer.

        -1 byte thanks to @Shaggy.



        Try it online 10 times or verify all test cases 10 times.



        Explanation:





        ü)             # Create all pairs of the (implicit) input
        # i.e. "Hello" → [["H","e"],["e","l"],["l","l"],["l","o"]]
        v # Loop over each these pairs `y`:
        yн # Push the first character of pair `y`
        5LΩ # Get a random integer in the range [1,5]
        F # Inner loop that many times:
        y # Push pair `y`
        D.š« # Duplicate it, swap the cases of the letters, and merge it with `y`
        Ω # Then pop and push a random character from this list of four
        ]J # After both loops: join the entire stack together to a single string
        # (which is output implicitly as result)





        share|improve this answer











        $endgroup$













        • $begingroup$
          I don't know 05AB1E but, instead of INè, could you save anything by pushing the first character of y?
          $endgroup$
          – Shaggy
          Apr 3 at 9:06










        • $begingroup$
          @Shaggy Yes, I indeed can.. Thanks! Maybe I should stop golfing for today, I'm a mess, lol..
          $endgroup$
          – Kevin Cruijssen
          Apr 3 at 9:08












        • $begingroup$
          You're a mess? ¨vNUy5LΩFy¹X>è«D.š«Ω?
          $endgroup$
          – Magic Octopus Urn
          Apr 3 at 12:52






        • 1




          $begingroup$
          @MagicOctopusUrn Although a pretty original approach, I'm afraid it doesn't do the first bullet point of the challenge ("Output the current character."), since the result can start with t, T, or s for input "String" in your program, while it's supposed to always start with the S.
          $endgroup$
          – Kevin Cruijssen
          Apr 3 at 13:03





















        1












        $begingroup$


        Charcoal, 27 bytes



        FLθ«F∧ι⊕‽⁵‽⭆✂θ⊖ι⊕ι¹⁺↥λ↧λ§θι


        Try it online! Link is to verbose version of code. Explanation:



        FLθ«


        Loop over all of the indices of the input string.



        F∧ι⊕‽⁵


        Except for the first index, loop over a random number from 1 to 5 inclusive...



        ‽⭆✂θ⊖ι⊕ι¹⁺↥λ↧λ


        ... extract the previous and next characters from the string, take the upper and lower case versions, and pick a random character of the four.



        §θι


        Print the character at the current index.






        share|improve this answer









        $endgroup$





















          1












          $begingroup$

          perl 5 (-p), 77 bytes



          s/(.)(?=(.))/$x=$1;'$x.=substr"U$1$2L$1$2",4*rand,1;'x(1+5*rand)/gee;s/.$//


          TIO






          share|improve this answer









          $endgroup$













          • $begingroup$
            You can save 4 bytes by using $& instead of $1, and chop + -l instead of s/.$//
            $endgroup$
            – Dada
            Apr 4 at 14:24



















          1












          $begingroup$


          Japt -P, 14 bytes



          äÈ+Zu pv ö5ö Ä


          Try it



          äÈ+Zu pv ö5ö Ä     :Implicit input of string
          ä :Take each consectutive pair of characters
          È :Pass them through the following function as Z
          + : Append to the first character of the pair
          Zu : Uppercase Z
          p : Append
          v : Lowercase
          ö : Get X random characters, where X is
          5ö : Random number in the range [0,5)
          Ä : Plus 1
          :Implicitly join and output





          share|improve this answer











          $endgroup$





















            1












            $begingroup$


            Python 3, 167 bytes





            from random import*;c=choice
            def f(s):
            i=0;r=""
            for i in range(len(s)-1):
            r+=s[i]
            for j in range(randrange(5)):r+=c([str.upper,str.lower])(c(s[i:i+2]))
            return r


            Try it online!






            share|improve this answer









            $endgroup$





















              1












              $begingroup$


              Jelly, 14 bytes



              ;;;Œs$Xɗ¥5X¤¡Ɲ


              Try it online!



              Explanation



                           Ɲ | For each overlapping pair of letters
              ; | Join the first letter to...
              5X¤¡ | Between 1 and 5 repetitions of...
              Xɗ¥ | A randomly selected character from...
              ;;Œs$ | A list of the two letters and the swapped case versions of both





              share|improve this answer











              $endgroup$





















                1












                $begingroup$

                C(GCC) 175 162 bytes



                -12 bytes from LambdaBeta



                f(s,S,i,r,a)char*s,*S,*i;{srand(time(0));for(i=S;*(s+1);++s){*i++=*s;for(r=rand()%5+1;r--;*i++=rand()&1?a>96&a<123|a>64&a<91?a^32:a:a)a=rand()&1?*s:*(s+1);}*i=0;}


                Try it online






                share|improve this answer











                $endgroup$













                • $begingroup$
                  I don't think you need the 0 in the first line.
                  $endgroup$
                  – LambdaBeta
                  Apr 3 at 21:28










                • $begingroup$
                  Also can save a lot of characters by taking the buffer S as a parameter and adding your variables to the argument list: Try it online!
                  $endgroup$
                  – LambdaBeta
                  Apr 3 at 21:31










                • $begingroup$
                  @LambdaBeta turns out you are right about the 0, which made it not worth it to have the #define anymore
                  $endgroup$
                  – rtpax
                  Apr 4 at 14:57










                • $begingroup$
                  150 bytes
                  $endgroup$
                  – ceilingcat
                  2 days ago



















                0












                $begingroup$


                Scala 2.12.8, 214 bytes



                Golfed version:



                val r=scala.util.Random;println(readLine.toList.sliding(2).flatMap{case a :: b :: Nil=>(a +: (0 to r.nextInt(5)).map{_=>((c: Char)=>if(r.nextBoolean)c.toUpper else c.toLower)(if(r.nextBoolean)a else b)})}.mkString)


                Golfed with newlines and indents:



                val r=scala.util.Random
                println(readLine.toList.sliding(2).flatMap{
                case a :: b :: Nil=>
                (a +: (0 to r.nextInt(5)).map{_=>
                ((c: Char)=>if(r.nextBoolean)c.toUpper else c.toLower)(if(r.nextBoolean)a else b)
                })
                }.mkString)


                Ungolfed:



                import scala.io.StdIn
                import scala.util.Random

                def gobble(input: String): String = {
                input.toList.sliding(2).flatMap {
                case thisChar :: nextChar :: Nil =>
                val numberOfAdditions = Random.nextInt(5)
                (thisChar +: (0 to numberOfAdditions).map { _ =>
                val char = if(Random.nextBoolean) thisChar else nextChar
                val cc = if(Random.nextBoolean) char.toUpper else char.ToLower
                cc
                })
                }.mkString
                }

                println(gobble(StdIn.readLine()))





                share|improve this answer









                $endgroup$









                • 1




                  $begingroup$
                  No way to turn a :: b :: Nil into a::b::Nil? Same for a :+, a:+() or a.:+() might work
                  $endgroup$
                  – Ven
                  Apr 4 at 14:07










                • $begingroup$
                  @Ven a::b::Nil causes a compile error. +: is a method defined on the list, so it might save space by getting rid of the outer parens?
                  $endgroup$
                  – Soren
                  Apr 4 at 22:58












                • $begingroup$
                  You only have only one elem here so it’s not autotupling anyway
                  $endgroup$
                  – Ven
                  Apr 4 at 23:41



















                0












                $begingroup$


                Perl 5 -n, 61 bytes





                s/.(?=(.))/print$&,map{(map{lc,uc}$&,$1)[rand 4]}0..rand 5/ge


                Try it online!






                share|improve this answer









                $endgroup$





















                  0












                  $begingroup$


                  C# (Visual C# Interactive Compiler), 236 213 209 bytes





                  a=>{int i=0,j;var m=new Random();var s="";var c = a.Select(x=>Char.IsLetter(x)?(char)(x^32):x).ToArray();for(;i<a.Length-1;i++)for(j=m.Next(1,5);j-->0;)s+=new{a[i],c[i],a[i+1],c[i+1]}[m.Next(0,3)];return s;}


                  Try it online!






                  share|improve this answer











                  $endgroup$













                  • $begingroup$
                    Does not work with non-alphanumeric characters. char b=a[0] -> var b=a[0], extra space in declaration of d in for-loop
                    $endgroup$
                    – Embodiment of Ignorance
                    Apr 4 at 5:11



















                  0












                  $begingroup$

                  T-SQL query, 286 bytes



                  DECLARE @ char(999)='String'

                  SELECT @=stuff(@,n+2,0,s)FROM(SELECT
                  top 999*,substring(lower(c)+upper(c),abs(v%4)+1,1)s
                  FROM(SELECT*,number n,substring(@,number+1,2)c,cast(newid()as varbinary)v
                  FROM(values(1),(2),(3),(4),(5))F(h),spt_values)D
                  WHERE'P'=type and n<len(@)-1and h>v%3+2ORDER
                  BY-n)E
                  PRINT LEFT(@,len(@)-1)


                  Try it online unfortunately the online version always show the same result for the same varchar, unlike MS SQL Server Management Studio






                  share|improve this answer











                  $endgroup$





















                    0












                    $begingroup$


                    C# (Visual C# Interactive Compiler), 156 bytes





                    n=>{var k=new Random();int i=0;foreach(var j in n.Skip(1).Zip(n,(a,b)=>a+b))for(Write(j[1]),i=0;i++<k.Next(5);)Write(k.Next()%2<1?j.ToUpper():j.ToLower());}


                    Try it online!






                    share|improve this answer











                    $endgroup$





















                      0












                      $begingroup$


                      Japt -P, 43 16 bytes



                      äÈ+(Zv +Zu)ö5ö Ä


                      Shortened by a lot now!



                      Try it






                      share|improve this answer











                      $endgroup$













                      • $begingroup$
                        This seems to return the same result every time.
                        $endgroup$
                        – Shaggy
                        Apr 3 at 10:01










                      • $begingroup$
                        @Shaggy will fix. Also, ä's description says it gives three arguments, with the last being x+y. But as you can see here, it just returns 1. Is this a bug?
                        $endgroup$
                        – Embodiment of Ignorance
                        Apr 4 at 5:21



















                      0












                      $begingroup$


                      C (gcc), 110 109 bytes





                      i,p;g(char*_){for(i=rand(putchar(*_))%1024;p=_[i%2],putchar(i&2&&p>64&~-p%32<26?p^32:p),i/=4;);_[2]&&g(_+1);}


                      Try it online!



                      -1 thanks to ceilingcat



                      i,p;g(char*_){
                      for(i=rand(putchar(*_)) //print current char
                      %1024; // and get 10 random bits
                      p=_[i%2], //1st bit => current/next char
                      putchar(i&2&& //2nd bit => toggle case
                      p>64&~-p%32<26 // if char-to-print is alphabetic
                      ?p^32:p),
                      i/=4;); //discard two bits
                      _[2]&&g(_+1); //if next isn't last char, repeat with next char
                      }




                      The number of characters printed (per input character) is not uniformly random:



                      1  if      i<   4 (  4/1024 = 1/256)
                      2 if 4<=i< 16 ( 12/1024 = 3/256)
                      3 if 16<=i< 64 ( 48/1024 = 3/ 64)
                      4 if 64<=i< 256 (192/1024 = 3/ 16)
                      5 if 256<=i<1024 (768/1024 = 3/ 4)





                      share|improve this answer











                      $endgroup$














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                        20 Answers
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                        active

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                        6












                        $begingroup$


                        Gaia, 25 bytes



                        ṇ+†ṅ⟨)₌¤:~+4ṛ⟨ṛ₌¤⟩ₓ⟩¦$


                        Try it online!



                        Thanks to Kevin Cruijssen for pointing out 2 bugs!



                        ṇ				| delete the last character from the input
                        +† | push the input again and concatenate together, so for instance
                        | 'abc' 'bc' becomes ['ab' 'bc' 'c']
                        ṅ | delete the last element
                        ⟨ ⟩¦ | for each of the elements, do:
                        )₌ | take the first character and push again
                        ¤ | swap
                        : | dup
                        ~ | swap case
                        + | concatenate strings
                        4ṛ | select a random integer from [1..5]
                        ⟨ ⟩ₓ | and repeat that many times
                        ṛ₌¤ | select a random character from the string
                        | clean up stack
                        $ | convert to string


                        Note that 4ṛ is because is implemented for an integer z as python's random.randint(1,z+1), which returns an integer N such that 1<=N<=z+1.






                        share|improve this answer











                        $endgroup$













                        • $begingroup$
                          Are you sure the run-length encode is correct here? If I understand the challenge correctly: the four options should be chosen 1-5 times randomly, instead of choosing one of the four randomly, repeated 1-5 times. The first example output SSSTSStrTrIiinIIngn ([SSSTSS, trT, rI, iinII, ngn]) seems to reflect this, and currently isn't a possible output in your program (I think).
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 3 at 8:56












                        • $begingroup$
                          @KevinCruijssen I interpreted "output from the list a random number of times" to mean run-length decode, but you're right, the test cases do seem to indicate the other interpretation; I think it should be pretty easy to fix
                          $endgroup$
                          – Giuseppe
                          Apr 3 at 10:50








                        • 1




                          $begingroup$
                          5ṛ can result in 6 for some reason Try it online? PS: Isn't there an integer to ranged list, or ranged for-loop in Gaia?
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 3 at 11:32








                        • 1




                          $begingroup$
                          @KevinCruijssen dang, Business Cat really needs to fix off-by-one errors...I really thought there was a for type construct, but I'm pretty sure it's which isn't even documented on the wiki page.
                          $endgroup$
                          – Giuseppe
                          Apr 3 at 11:41
















                        6












                        $begingroup$


                        Gaia, 25 bytes



                        ṇ+†ṅ⟨)₌¤:~+4ṛ⟨ṛ₌¤⟩ₓ⟩¦$


                        Try it online!



                        Thanks to Kevin Cruijssen for pointing out 2 bugs!



                        ṇ				| delete the last character from the input
                        +† | push the input again and concatenate together, so for instance
                        | 'abc' 'bc' becomes ['ab' 'bc' 'c']
                        ṅ | delete the last element
                        ⟨ ⟩¦ | for each of the elements, do:
                        )₌ | take the first character and push again
                        ¤ | swap
                        : | dup
                        ~ | swap case
                        + | concatenate strings
                        4ṛ | select a random integer from [1..5]
                        ⟨ ⟩ₓ | and repeat that many times
                        ṛ₌¤ | select a random character from the string
                        | clean up stack
                        $ | convert to string


                        Note that 4ṛ is because is implemented for an integer z as python's random.randint(1,z+1), which returns an integer N such that 1<=N<=z+1.






                        share|improve this answer











                        $endgroup$













                        • $begingroup$
                          Are you sure the run-length encode is correct here? If I understand the challenge correctly: the four options should be chosen 1-5 times randomly, instead of choosing one of the four randomly, repeated 1-5 times. The first example output SSSTSStrTrIiinIIngn ([SSSTSS, trT, rI, iinII, ngn]) seems to reflect this, and currently isn't a possible output in your program (I think).
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 3 at 8:56












                        • $begingroup$
                          @KevinCruijssen I interpreted "output from the list a random number of times" to mean run-length decode, but you're right, the test cases do seem to indicate the other interpretation; I think it should be pretty easy to fix
                          $endgroup$
                          – Giuseppe
                          Apr 3 at 10:50








                        • 1




                          $begingroup$
                          5ṛ can result in 6 for some reason Try it online? PS: Isn't there an integer to ranged list, or ranged for-loop in Gaia?
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 3 at 11:32








                        • 1




                          $begingroup$
                          @KevinCruijssen dang, Business Cat really needs to fix off-by-one errors...I really thought there was a for type construct, but I'm pretty sure it's which isn't even documented on the wiki page.
                          $endgroup$
                          – Giuseppe
                          Apr 3 at 11:41














                        6












                        6








                        6





                        $begingroup$


                        Gaia, 25 bytes



                        ṇ+†ṅ⟨)₌¤:~+4ṛ⟨ṛ₌¤⟩ₓ⟩¦$


                        Try it online!



                        Thanks to Kevin Cruijssen for pointing out 2 bugs!



                        ṇ				| delete the last character from the input
                        +† | push the input again and concatenate together, so for instance
                        | 'abc' 'bc' becomes ['ab' 'bc' 'c']
                        ṅ | delete the last element
                        ⟨ ⟩¦ | for each of the elements, do:
                        )₌ | take the first character and push again
                        ¤ | swap
                        : | dup
                        ~ | swap case
                        + | concatenate strings
                        4ṛ | select a random integer from [1..5]
                        ⟨ ⟩ₓ | and repeat that many times
                        ṛ₌¤ | select a random character from the string
                        | clean up stack
                        $ | convert to string


                        Note that 4ṛ is because is implemented for an integer z as python's random.randint(1,z+1), which returns an integer N such that 1<=N<=z+1.






                        share|improve this answer











                        $endgroup$




                        Gaia, 25 bytes



                        ṇ+†ṅ⟨)₌¤:~+4ṛ⟨ṛ₌¤⟩ₓ⟩¦$


                        Try it online!



                        Thanks to Kevin Cruijssen for pointing out 2 bugs!



                        ṇ				| delete the last character from the input
                        +† | push the input again and concatenate together, so for instance
                        | 'abc' 'bc' becomes ['ab' 'bc' 'c']
                        ṅ | delete the last element
                        ⟨ ⟩¦ | for each of the elements, do:
                        )₌ | take the first character and push again
                        ¤ | swap
                        : | dup
                        ~ | swap case
                        + | concatenate strings
                        4ṛ | select a random integer from [1..5]
                        ⟨ ⟩ₓ | and repeat that many times
                        ṛ₌¤ | select a random character from the string
                        | clean up stack
                        $ | convert to string


                        Note that 4ṛ is because is implemented for an integer z as python's random.randint(1,z+1), which returns an integer N such that 1<=N<=z+1.







                        share|improve this answer














                        share|improve this answer



                        share|improve this answer








                        edited Apr 3 at 18:27

























                        answered Apr 3 at 1:34









                        GiuseppeGiuseppe

                        17.6k31153




                        17.6k31153












                        • $begingroup$
                          Are you sure the run-length encode is correct here? If I understand the challenge correctly: the four options should be chosen 1-5 times randomly, instead of choosing one of the four randomly, repeated 1-5 times. The first example output SSSTSStrTrIiinIIngn ([SSSTSS, trT, rI, iinII, ngn]) seems to reflect this, and currently isn't a possible output in your program (I think).
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 3 at 8:56












                        • $begingroup$
                          @KevinCruijssen I interpreted "output from the list a random number of times" to mean run-length decode, but you're right, the test cases do seem to indicate the other interpretation; I think it should be pretty easy to fix
                          $endgroup$
                          – Giuseppe
                          Apr 3 at 10:50








                        • 1




                          $begingroup$
                          5ṛ can result in 6 for some reason Try it online? PS: Isn't there an integer to ranged list, or ranged for-loop in Gaia?
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 3 at 11:32








                        • 1




                          $begingroup$
                          @KevinCruijssen dang, Business Cat really needs to fix off-by-one errors...I really thought there was a for type construct, but I'm pretty sure it's which isn't even documented on the wiki page.
                          $endgroup$
                          – Giuseppe
                          Apr 3 at 11:41


















                        • $begingroup$
                          Are you sure the run-length encode is correct here? If I understand the challenge correctly: the four options should be chosen 1-5 times randomly, instead of choosing one of the four randomly, repeated 1-5 times. The first example output SSSTSStrTrIiinIIngn ([SSSTSS, trT, rI, iinII, ngn]) seems to reflect this, and currently isn't a possible output in your program (I think).
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 3 at 8:56












                        • $begingroup$
                          @KevinCruijssen I interpreted "output from the list a random number of times" to mean run-length decode, but you're right, the test cases do seem to indicate the other interpretation; I think it should be pretty easy to fix
                          $endgroup$
                          – Giuseppe
                          Apr 3 at 10:50








                        • 1




                          $begingroup$
                          5ṛ can result in 6 for some reason Try it online? PS: Isn't there an integer to ranged list, or ranged for-loop in Gaia?
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 3 at 11:32








                        • 1




                          $begingroup$
                          @KevinCruijssen dang, Business Cat really needs to fix off-by-one errors...I really thought there was a for type construct, but I'm pretty sure it's which isn't even documented on the wiki page.
                          $endgroup$
                          – Giuseppe
                          Apr 3 at 11:41
















                        $begingroup$
                        Are you sure the run-length encode is correct here? If I understand the challenge correctly: the four options should be chosen 1-5 times randomly, instead of choosing one of the four randomly, repeated 1-5 times. The first example output SSSTSStrTrIiinIIngn ([SSSTSS, trT, rI, iinII, ngn]) seems to reflect this, and currently isn't a possible output in your program (I think).
                        $endgroup$
                        – Kevin Cruijssen
                        Apr 3 at 8:56






                        $begingroup$
                        Are you sure the run-length encode is correct here? If I understand the challenge correctly: the four options should be chosen 1-5 times randomly, instead of choosing one of the four randomly, repeated 1-5 times. The first example output SSSTSStrTrIiinIIngn ([SSSTSS, trT, rI, iinII, ngn]) seems to reflect this, and currently isn't a possible output in your program (I think).
                        $endgroup$
                        – Kevin Cruijssen
                        Apr 3 at 8:56














                        $begingroup$
                        @KevinCruijssen I interpreted "output from the list a random number of times" to mean run-length decode, but you're right, the test cases do seem to indicate the other interpretation; I think it should be pretty easy to fix
                        $endgroup$
                        – Giuseppe
                        Apr 3 at 10:50






                        $begingroup$
                        @KevinCruijssen I interpreted "output from the list a random number of times" to mean run-length decode, but you're right, the test cases do seem to indicate the other interpretation; I think it should be pretty easy to fix
                        $endgroup$
                        – Giuseppe
                        Apr 3 at 10:50






                        1




                        1




                        $begingroup$
                        5ṛ can result in 6 for some reason Try it online? PS: Isn't there an integer to ranged list, or ranged for-loop in Gaia?
                        $endgroup$
                        – Kevin Cruijssen
                        Apr 3 at 11:32






                        $begingroup$
                        5ṛ can result in 6 for some reason Try it online? PS: Isn't there an integer to ranged list, or ranged for-loop in Gaia?
                        $endgroup$
                        – Kevin Cruijssen
                        Apr 3 at 11:32






                        1




                        1




                        $begingroup$
                        @KevinCruijssen dang, Business Cat really needs to fix off-by-one errors...I really thought there was a for type construct, but I'm pretty sure it's which isn't even documented on the wiki page.
                        $endgroup$
                        – Giuseppe
                        Apr 3 at 11:41




                        $begingroup$
                        @KevinCruijssen dang, Business Cat really needs to fix off-by-one errors...I really thought there was a for type construct, but I'm pretty sure it's which isn't even documented on the wiki page.
                        $endgroup$
                        – Giuseppe
                        Apr 3 at 11:41











                        3












                        $begingroup$


                        APL (dzaima/APL), 23 bytes





                        Anonymous tacit prefix function.



                        ∊2(⊣,{?4⍴⍨?5}⊇,,-⍤,)/


                        Try it online!



                        2()/ apply the following infix tacit function between each character pair:



                        - the switchcase

                         of

                        , the concatenation of the pair



                        ,, prepend the concatenation of the pair to that



                        {}⊇ pick the following elements from that:



                          ?5 random number in range 1…5



                          4⍴⍨ that many fours



                          ? random indices for those



                        ϵnlist (flatten)






                        share|improve this answer











                        $endgroup$


















                          3












                          $begingroup$


                          APL (dzaima/APL), 23 bytes





                          Anonymous tacit prefix function.



                          ∊2(⊣,{?4⍴⍨?5}⊇,,-⍤,)/


                          Try it online!



                          2()/ apply the following infix tacit function between each character pair:



                          - the switchcase

                           of

                          , the concatenation of the pair



                          ,, prepend the concatenation of the pair to that



                          {}⊇ pick the following elements from that:



                            ?5 random number in range 1…5



                            4⍴⍨ that many fours



                            ? random indices for those



                          ϵnlist (flatten)






                          share|improve this answer











                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$


                            APL (dzaima/APL), 23 bytes





                            Anonymous tacit prefix function.



                            ∊2(⊣,{?4⍴⍨?5}⊇,,-⍤,)/


                            Try it online!



                            2()/ apply the following infix tacit function between each character pair:



                            - the switchcase

                             of

                            , the concatenation of the pair



                            ,, prepend the concatenation of the pair to that



                            {}⊇ pick the following elements from that:



                              ?5 random number in range 1…5



                              4⍴⍨ that many fours



                              ? random indices for those



                            ϵnlist (flatten)






                            share|improve this answer











                            $endgroup$




                            APL (dzaima/APL), 23 bytes





                            Anonymous tacit prefix function.



                            ∊2(⊣,{?4⍴⍨?5}⊇,,-⍤,)/


                            Try it online!



                            2()/ apply the following infix tacit function between each character pair:



                            - the switchcase

                             of

                            , the concatenation of the pair



                            ,, prepend the concatenation of the pair to that



                            {}⊇ pick the following elements from that:



                              ?5 random number in range 1…5



                              4⍴⍨ that many fours



                              ? random indices for those



                            ϵnlist (flatten)







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Apr 3 at 8:08

























                            answered Apr 2 at 23:24









                            AdámAdám

                            29k276207




                            29k276207























                                3












                                $begingroup$


                                Perl 6, 60 bytes





                                {S:g{.)>(.)}=$/~[~] roll ^5 .roll+1,$/.lc,$/.uc,$0.lc,$0.uc}


                                Try it online!



                                The lowercase/uppercase part is kinda annoying.






                                share|improve this answer









                                $endgroup$













                                • $begingroup$
                                  I don't know Perl, so I'm probably saying something stupid here. But is it somehow possible to concat the $/ and $0 together and use .lc on that string, and then create a copy of that string and use .uc, and concat those two together? Not sure if that's even possible, or shorter than your current $/.lc,$/.uc,$0.lc,$0.uc, but it would mean you'd use $/, $0, .lc, and .uc once each.
                                  $endgroup$
                                  – Kevin Cruijssen
                                  Apr 3 at 13:43








                                • 1




                                  $begingroup$
                                  Alas, (.lc~.uc for $0~$/).comb is longer. Perl 6 really wants to distinguish strings and lists, so "abc"[0] eq "abc" (it pretends to be a single-item list).
                                  $endgroup$
                                  – Ven
                                  Apr 3 at 16:02












                                • $begingroup$
                                  You can do it by slipping and an anonymous function applied to a list: {.lc,|.uc}($/,|$0) for -5 bytes, and just use the list of matches {.lc,|.uc}(@$/) for -8 bytes. tio.run/…
                                  $endgroup$
                                  – Phil H
                                  Apr 4 at 16:12










                                • $begingroup$
                                  @PhilH No that doesn't work. Those solutions only capitalise one of the letters each
                                  $endgroup$
                                  – Jo King
                                  Apr 4 at 21:18


















                                3












                                $begingroup$


                                Perl 6, 60 bytes





                                {S:g{.)>(.)}=$/~[~] roll ^5 .roll+1,$/.lc,$/.uc,$0.lc,$0.uc}


                                Try it online!



                                The lowercase/uppercase part is kinda annoying.






                                share|improve this answer









                                $endgroup$













                                • $begingroup$
                                  I don't know Perl, so I'm probably saying something stupid here. But is it somehow possible to concat the $/ and $0 together and use .lc on that string, and then create a copy of that string and use .uc, and concat those two together? Not sure if that's even possible, or shorter than your current $/.lc,$/.uc,$0.lc,$0.uc, but it would mean you'd use $/, $0, .lc, and .uc once each.
                                  $endgroup$
                                  – Kevin Cruijssen
                                  Apr 3 at 13:43








                                • 1




                                  $begingroup$
                                  Alas, (.lc~.uc for $0~$/).comb is longer. Perl 6 really wants to distinguish strings and lists, so "abc"[0] eq "abc" (it pretends to be a single-item list).
                                  $endgroup$
                                  – Ven
                                  Apr 3 at 16:02












                                • $begingroup$
                                  You can do it by slipping and an anonymous function applied to a list: {.lc,|.uc}($/,|$0) for -5 bytes, and just use the list of matches {.lc,|.uc}(@$/) for -8 bytes. tio.run/…
                                  $endgroup$
                                  – Phil H
                                  Apr 4 at 16:12










                                • $begingroup$
                                  @PhilH No that doesn't work. Those solutions only capitalise one of the letters each
                                  $endgroup$
                                  – Jo King
                                  Apr 4 at 21:18
















                                3












                                3








                                3





                                $begingroup$


                                Perl 6, 60 bytes





                                {S:g{.)>(.)}=$/~[~] roll ^5 .roll+1,$/.lc,$/.uc,$0.lc,$0.uc}


                                Try it online!



                                The lowercase/uppercase part is kinda annoying.






                                share|improve this answer









                                $endgroup$




                                Perl 6, 60 bytes





                                {S:g{.)>(.)}=$/~[~] roll ^5 .roll+1,$/.lc,$/.uc,$0.lc,$0.uc}


                                Try it online!



                                The lowercase/uppercase part is kinda annoying.







                                share|improve this answer












                                share|improve this answer



                                share|improve this answer










                                answered Apr 3 at 9:25









                                Jo KingJo King

                                26.6k365132




                                26.6k365132












                                • $begingroup$
                                  I don't know Perl, so I'm probably saying something stupid here. But is it somehow possible to concat the $/ and $0 together and use .lc on that string, and then create a copy of that string and use .uc, and concat those two together? Not sure if that's even possible, or shorter than your current $/.lc,$/.uc,$0.lc,$0.uc, but it would mean you'd use $/, $0, .lc, and .uc once each.
                                  $endgroup$
                                  – Kevin Cruijssen
                                  Apr 3 at 13:43








                                • 1




                                  $begingroup$
                                  Alas, (.lc~.uc for $0~$/).comb is longer. Perl 6 really wants to distinguish strings and lists, so "abc"[0] eq "abc" (it pretends to be a single-item list).
                                  $endgroup$
                                  – Ven
                                  Apr 3 at 16:02












                                • $begingroup$
                                  You can do it by slipping and an anonymous function applied to a list: {.lc,|.uc}($/,|$0) for -5 bytes, and just use the list of matches {.lc,|.uc}(@$/) for -8 bytes. tio.run/…
                                  $endgroup$
                                  – Phil H
                                  Apr 4 at 16:12










                                • $begingroup$
                                  @PhilH No that doesn't work. Those solutions only capitalise one of the letters each
                                  $endgroup$
                                  – Jo King
                                  Apr 4 at 21:18




















                                • $begingroup$
                                  I don't know Perl, so I'm probably saying something stupid here. But is it somehow possible to concat the $/ and $0 together and use .lc on that string, and then create a copy of that string and use .uc, and concat those two together? Not sure if that's even possible, or shorter than your current $/.lc,$/.uc,$0.lc,$0.uc, but it would mean you'd use $/, $0, .lc, and .uc once each.
                                  $endgroup$
                                  – Kevin Cruijssen
                                  Apr 3 at 13:43








                                • 1




                                  $begingroup$
                                  Alas, (.lc~.uc for $0~$/).comb is longer. Perl 6 really wants to distinguish strings and lists, so "abc"[0] eq "abc" (it pretends to be a single-item list).
                                  $endgroup$
                                  – Ven
                                  Apr 3 at 16:02












                                • $begingroup$
                                  You can do it by slipping and an anonymous function applied to a list: {.lc,|.uc}($/,|$0) for -5 bytes, and just use the list of matches {.lc,|.uc}(@$/) for -8 bytes. tio.run/…
                                  $endgroup$
                                  – Phil H
                                  Apr 4 at 16:12










                                • $begingroup$
                                  @PhilH No that doesn't work. Those solutions only capitalise one of the letters each
                                  $endgroup$
                                  – Jo King
                                  Apr 4 at 21:18


















                                $begingroup$
                                I don't know Perl, so I'm probably saying something stupid here. But is it somehow possible to concat the $/ and $0 together and use .lc on that string, and then create a copy of that string and use .uc, and concat those two together? Not sure if that's even possible, or shorter than your current $/.lc,$/.uc,$0.lc,$0.uc, but it would mean you'd use $/, $0, .lc, and .uc once each.
                                $endgroup$
                                – Kevin Cruijssen
                                Apr 3 at 13:43






                                $begingroup$
                                I don't know Perl, so I'm probably saying something stupid here. But is it somehow possible to concat the $/ and $0 together and use .lc on that string, and then create a copy of that string and use .uc, and concat those two together? Not sure if that's even possible, or shorter than your current $/.lc,$/.uc,$0.lc,$0.uc, but it would mean you'd use $/, $0, .lc, and .uc once each.
                                $endgroup$
                                – Kevin Cruijssen
                                Apr 3 at 13:43






                                1




                                1




                                $begingroup$
                                Alas, (.lc~.uc for $0~$/).comb is longer. Perl 6 really wants to distinguish strings and lists, so "abc"[0] eq "abc" (it pretends to be a single-item list).
                                $endgroup$
                                – Ven
                                Apr 3 at 16:02






                                $begingroup$
                                Alas, (.lc~.uc for $0~$/).comb is longer. Perl 6 really wants to distinguish strings and lists, so "abc"[0] eq "abc" (it pretends to be a single-item list).
                                $endgroup$
                                – Ven
                                Apr 3 at 16:02














                                $begingroup$
                                You can do it by slipping and an anonymous function applied to a list: {.lc,|.uc}($/,|$0) for -5 bytes, and just use the list of matches {.lc,|.uc}(@$/) for -8 bytes. tio.run/…
                                $endgroup$
                                – Phil H
                                Apr 4 at 16:12




                                $begingroup$
                                You can do it by slipping and an anonymous function applied to a list: {.lc,|.uc}($/,|$0) for -5 bytes, and just use the list of matches {.lc,|.uc}(@$/) for -8 bytes. tio.run/…
                                $endgroup$
                                – Phil H
                                Apr 4 at 16:12












                                $begingroup$
                                @PhilH No that doesn't work. Those solutions only capitalise one of the letters each
                                $endgroup$
                                – Jo King
                                Apr 4 at 21:18






                                $begingroup$
                                @PhilH No that doesn't work. Those solutions only capitalise one of the letters each
                                $endgroup$
                                – Jo King
                                Apr 4 at 21:18













                                3












                                $begingroup$


                                Jelly, 12 bytes



                                ;Œsṗ5X¤XṭṖµƝ


                                Try it online!






                                share|improve this answer









                                $endgroup$


















                                  3












                                  $begingroup$


                                  Jelly, 12 bytes



                                  ;Œsṗ5X¤XṭṖµƝ


                                  Try it online!






                                  share|improve this answer









                                  $endgroup$
















                                    3












                                    3








                                    3





                                    $begingroup$


                                    Jelly, 12 bytes



                                    ;Œsṗ5X¤XṭṖµƝ


                                    Try it online!






                                    share|improve this answer









                                    $endgroup$




                                    Jelly, 12 bytes



                                    ;Œsṗ5X¤XṭṖµƝ


                                    Try it online!







                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered Apr 3 at 17:06









                                    Erik the OutgolferErik the Outgolfer

                                    33k429106




                                    33k429106























                                        3












                                        $begingroup$


                                        Bash, 121 bytes



                                        -20 bytes thanks to Nahuel



                                        -9 bytes thanks to roblogic





                                        for((i=0;i<${#1};i++)){
                                        s=${1:i:1}
                                        m=${1:i:2}
                                        m=${m,,}${m^^}
                                        for((t=0;t++<RANDOM%6;)){
                                        s+=${m:RANDOM%4:1}
                                        }
                                        printf "$s"
                                        }


                                        Try it online!



                                        Original answer




                                        Bash, 150 bytes



                                        Have done very little golf bashing and trying to improve my bash, so any comments welcome.





                                        for((i=0;i<${#1}-1;i++));do
                                        c=${1:$i:1}
                                        n=${1:$((i+1)):1}
                                        a=($n ${c,} ${c^} ${n,} ${n^})
                                        shuf -e ${a[@]} -n "$(shuf -i 1-5 -n 1)"|xargs printf %s
                                        done


                                        Try it online!



                                        Code is straightforward loop through chars setting current c and next n character, then creating an array of the 4 possibilities, repeating one of them so there's exactly 5. Next we shuffle that array, and then choose n elements from it, where n itself is random between 1 and 5.






                                        share|improve this answer











                                        $endgroup$













                                        • $begingroup$
                                          seems it's missing printf %s "$c"
                                          $endgroup$
                                          – Nahuel Fouilleul
                                          Apr 3 at 8:06






                                        • 1




                                          $begingroup$
                                          do and done can be replaced with undocumented { and }
                                          $endgroup$
                                          – Nahuel Fouilleul
                                          Apr 3 at 8:08










                                        • $begingroup$
                                          with some changes
                                          $endgroup$
                                          – Nahuel Fouilleul
                                          Apr 3 at 8:25












                                        • $begingroup$
                                          @NahuelFouilleul tyvm. nice improvements and nice catch -- i'd somehow missed the requirement always to output the current char.
                                          $endgroup$
                                          – Jonah
                                          Apr 3 at 13:07






                                        • 1




                                          $begingroup$
                                          @roblogic that's clever. tyvm.
                                          $endgroup$
                                          – Jonah
                                          Apr 3 at 21:36
















                                        3












                                        $begingroup$


                                        Bash, 121 bytes



                                        -20 bytes thanks to Nahuel



                                        -9 bytes thanks to roblogic





                                        for((i=0;i<${#1};i++)){
                                        s=${1:i:1}
                                        m=${1:i:2}
                                        m=${m,,}${m^^}
                                        for((t=0;t++<RANDOM%6;)){
                                        s+=${m:RANDOM%4:1}
                                        }
                                        printf "$s"
                                        }


                                        Try it online!



                                        Original answer




                                        Bash, 150 bytes



                                        Have done very little golf bashing and trying to improve my bash, so any comments welcome.





                                        for((i=0;i<${#1}-1;i++));do
                                        c=${1:$i:1}
                                        n=${1:$((i+1)):1}
                                        a=($n ${c,} ${c^} ${n,} ${n^})
                                        shuf -e ${a[@]} -n "$(shuf -i 1-5 -n 1)"|xargs printf %s
                                        done


                                        Try it online!



                                        Code is straightforward loop through chars setting current c and next n character, then creating an array of the 4 possibilities, repeating one of them so there's exactly 5. Next we shuffle that array, and then choose n elements from it, where n itself is random between 1 and 5.






                                        share|improve this answer











                                        $endgroup$













                                        • $begingroup$
                                          seems it's missing printf %s "$c"
                                          $endgroup$
                                          – Nahuel Fouilleul
                                          Apr 3 at 8:06






                                        • 1




                                          $begingroup$
                                          do and done can be replaced with undocumented { and }
                                          $endgroup$
                                          – Nahuel Fouilleul
                                          Apr 3 at 8:08










                                        • $begingroup$
                                          with some changes
                                          $endgroup$
                                          – Nahuel Fouilleul
                                          Apr 3 at 8:25












                                        • $begingroup$
                                          @NahuelFouilleul tyvm. nice improvements and nice catch -- i'd somehow missed the requirement always to output the current char.
                                          $endgroup$
                                          – Jonah
                                          Apr 3 at 13:07






                                        • 1




                                          $begingroup$
                                          @roblogic that's clever. tyvm.
                                          $endgroup$
                                          – Jonah
                                          Apr 3 at 21:36














                                        3












                                        3








                                        3





                                        $begingroup$


                                        Bash, 121 bytes



                                        -20 bytes thanks to Nahuel



                                        -9 bytes thanks to roblogic





                                        for((i=0;i<${#1};i++)){
                                        s=${1:i:1}
                                        m=${1:i:2}
                                        m=${m,,}${m^^}
                                        for((t=0;t++<RANDOM%6;)){
                                        s+=${m:RANDOM%4:1}
                                        }
                                        printf "$s"
                                        }


                                        Try it online!



                                        Original answer




                                        Bash, 150 bytes



                                        Have done very little golf bashing and trying to improve my bash, so any comments welcome.





                                        for((i=0;i<${#1}-1;i++));do
                                        c=${1:$i:1}
                                        n=${1:$((i+1)):1}
                                        a=($n ${c,} ${c^} ${n,} ${n^})
                                        shuf -e ${a[@]} -n "$(shuf -i 1-5 -n 1)"|xargs printf %s
                                        done


                                        Try it online!



                                        Code is straightforward loop through chars setting current c and next n character, then creating an array of the 4 possibilities, repeating one of them so there's exactly 5. Next we shuffle that array, and then choose n elements from it, where n itself is random between 1 and 5.






                                        share|improve this answer











                                        $endgroup$




                                        Bash, 121 bytes



                                        -20 bytes thanks to Nahuel



                                        -9 bytes thanks to roblogic





                                        for((i=0;i<${#1};i++)){
                                        s=${1:i:1}
                                        m=${1:i:2}
                                        m=${m,,}${m^^}
                                        for((t=0;t++<RANDOM%6;)){
                                        s+=${m:RANDOM%4:1}
                                        }
                                        printf "$s"
                                        }


                                        Try it online!



                                        Original answer




                                        Bash, 150 bytes



                                        Have done very little golf bashing and trying to improve my bash, so any comments welcome.





                                        for((i=0;i<${#1}-1;i++));do
                                        c=${1:$i:1}
                                        n=${1:$((i+1)):1}
                                        a=($n ${c,} ${c^} ${n,} ${n^})
                                        shuf -e ${a[@]} -n "$(shuf -i 1-5 -n 1)"|xargs printf %s
                                        done


                                        Try it online!



                                        Code is straightforward loop through chars setting current c and next n character, then creating an array of the 4 possibilities, repeating one of them so there's exactly 5. Next we shuffle that array, and then choose n elements from it, where n itself is random between 1 and 5.







                                        share|improve this answer














                                        share|improve this answer



                                        share|improve this answer








                                        edited Apr 3 at 21:36

























                                        answered Apr 3 at 4:26









                                        JonahJonah

                                        2,6611017




                                        2,6611017












                                        • $begingroup$
                                          seems it's missing printf %s "$c"
                                          $endgroup$
                                          – Nahuel Fouilleul
                                          Apr 3 at 8:06






                                        • 1




                                          $begingroup$
                                          do and done can be replaced with undocumented { and }
                                          $endgroup$
                                          – Nahuel Fouilleul
                                          Apr 3 at 8:08










                                        • $begingroup$
                                          with some changes
                                          $endgroup$
                                          – Nahuel Fouilleul
                                          Apr 3 at 8:25












                                        • $begingroup$
                                          @NahuelFouilleul tyvm. nice improvements and nice catch -- i'd somehow missed the requirement always to output the current char.
                                          $endgroup$
                                          – Jonah
                                          Apr 3 at 13:07






                                        • 1




                                          $begingroup$
                                          @roblogic that's clever. tyvm.
                                          $endgroup$
                                          – Jonah
                                          Apr 3 at 21:36


















                                        • $begingroup$
                                          seems it's missing printf %s "$c"
                                          $endgroup$
                                          – Nahuel Fouilleul
                                          Apr 3 at 8:06






                                        • 1




                                          $begingroup$
                                          do and done can be replaced with undocumented { and }
                                          $endgroup$
                                          – Nahuel Fouilleul
                                          Apr 3 at 8:08










                                        • $begingroup$
                                          with some changes
                                          $endgroup$
                                          – Nahuel Fouilleul
                                          Apr 3 at 8:25












                                        • $begingroup$
                                          @NahuelFouilleul tyvm. nice improvements and nice catch -- i'd somehow missed the requirement always to output the current char.
                                          $endgroup$
                                          – Jonah
                                          Apr 3 at 13:07






                                        • 1




                                          $begingroup$
                                          @roblogic that's clever. tyvm.
                                          $endgroup$
                                          – Jonah
                                          Apr 3 at 21:36
















                                        $begingroup$
                                        seems it's missing printf %s "$c"
                                        $endgroup$
                                        – Nahuel Fouilleul
                                        Apr 3 at 8:06




                                        $begingroup$
                                        seems it's missing printf %s "$c"
                                        $endgroup$
                                        – Nahuel Fouilleul
                                        Apr 3 at 8:06




                                        1




                                        1




                                        $begingroup$
                                        do and done can be replaced with undocumented { and }
                                        $endgroup$
                                        – Nahuel Fouilleul
                                        Apr 3 at 8:08




                                        $begingroup$
                                        do and done can be replaced with undocumented { and }
                                        $endgroup$
                                        – Nahuel Fouilleul
                                        Apr 3 at 8:08












                                        $begingroup$
                                        with some changes
                                        $endgroup$
                                        – Nahuel Fouilleul
                                        Apr 3 at 8:25






                                        $begingroup$
                                        with some changes
                                        $endgroup$
                                        – Nahuel Fouilleul
                                        Apr 3 at 8:25














                                        $begingroup$
                                        @NahuelFouilleul tyvm. nice improvements and nice catch -- i'd somehow missed the requirement always to output the current char.
                                        $endgroup$
                                        – Jonah
                                        Apr 3 at 13:07




                                        $begingroup$
                                        @NahuelFouilleul tyvm. nice improvements and nice catch -- i'd somehow missed the requirement always to output the current char.
                                        $endgroup$
                                        – Jonah
                                        Apr 3 at 13:07




                                        1




                                        1




                                        $begingroup$
                                        @roblogic that's clever. tyvm.
                                        $endgroup$
                                        – Jonah
                                        Apr 3 at 21:36




                                        $begingroup$
                                        @roblogic that's clever. tyvm.
                                        $endgroup$
                                        – Jonah
                                        Apr 3 at 21:36











                                        2












                                        $begingroup$


                                        Python 2, 107 bytes





                                        f=lambda s:s and s[0]+''.join(sample((s[:2]+s[:2].swapcase())*5,randint(1,5)))+f(s[1:])
                                        from random import*


                                        Try it online!






                                        share|improve this answer









                                        $endgroup$


















                                          2












                                          $begingroup$


                                          Python 2, 107 bytes





                                          f=lambda s:s and s[0]+''.join(sample((s[:2]+s[:2].swapcase())*5,randint(1,5)))+f(s[1:])
                                          from random import*


                                          Try it online!






                                          share|improve this answer









                                          $endgroup$
















                                            2












                                            2








                                            2





                                            $begingroup$


                                            Python 2, 107 bytes





                                            f=lambda s:s and s[0]+''.join(sample((s[:2]+s[:2].swapcase())*5,randint(1,5)))+f(s[1:])
                                            from random import*


                                            Try it online!






                                            share|improve this answer









                                            $endgroup$




                                            Python 2, 107 bytes





                                            f=lambda s:s and s[0]+''.join(sample((s[:2]+s[:2].swapcase())*5,randint(1,5)))+f(s[1:])
                                            from random import*


                                            Try it online!







                                            share|improve this answer












                                            share|improve this answer



                                            share|improve this answer










                                            answered Apr 3 at 5:30









                                            Chas BrownChas Brown

                                            5,2191523




                                            5,2191523























                                                2












                                                $begingroup$


                                                05AB1E, 18 17 bytes



                                                ü)vyн5LΩFyD.š«Ω]J


                                                Inspired by @Giuseppe's Gaia answer.

                                                -1 byte thanks to @Shaggy.



                                                Try it online 10 times or verify all test cases 10 times.



                                                Explanation:





                                                ü)             # Create all pairs of the (implicit) input
                                                # i.e. "Hello" → [["H","e"],["e","l"],["l","l"],["l","o"]]
                                                v # Loop over each these pairs `y`:
                                                yн # Push the first character of pair `y`
                                                5LΩ # Get a random integer in the range [1,5]
                                                F # Inner loop that many times:
                                                y # Push pair `y`
                                                D.š« # Duplicate it, swap the cases of the letters, and merge it with `y`
                                                Ω # Then pop and push a random character from this list of four
                                                ]J # After both loops: join the entire stack together to a single string
                                                # (which is output implicitly as result)





                                                share|improve this answer











                                                $endgroup$













                                                • $begingroup$
                                                  I don't know 05AB1E but, instead of INè, could you save anything by pushing the first character of y?
                                                  $endgroup$
                                                  – Shaggy
                                                  Apr 3 at 9:06










                                                • $begingroup$
                                                  @Shaggy Yes, I indeed can.. Thanks! Maybe I should stop golfing for today, I'm a mess, lol..
                                                  $endgroup$
                                                  – Kevin Cruijssen
                                                  Apr 3 at 9:08












                                                • $begingroup$
                                                  You're a mess? ¨vNUy5LΩFy¹X>è«D.š«Ω?
                                                  $endgroup$
                                                  – Magic Octopus Urn
                                                  Apr 3 at 12:52






                                                • 1




                                                  $begingroup$
                                                  @MagicOctopusUrn Although a pretty original approach, I'm afraid it doesn't do the first bullet point of the challenge ("Output the current character."), since the result can start with t, T, or s for input "String" in your program, while it's supposed to always start with the S.
                                                  $endgroup$
                                                  – Kevin Cruijssen
                                                  Apr 3 at 13:03


















                                                2












                                                $begingroup$


                                                05AB1E, 18 17 bytes



                                                ü)vyн5LΩFyD.š«Ω]J


                                                Inspired by @Giuseppe's Gaia answer.

                                                -1 byte thanks to @Shaggy.



                                                Try it online 10 times or verify all test cases 10 times.



                                                Explanation:





                                                ü)             # Create all pairs of the (implicit) input
                                                # i.e. "Hello" → [["H","e"],["e","l"],["l","l"],["l","o"]]
                                                v # Loop over each these pairs `y`:
                                                yн # Push the first character of pair `y`
                                                5LΩ # Get a random integer in the range [1,5]
                                                F # Inner loop that many times:
                                                y # Push pair `y`
                                                D.š« # Duplicate it, swap the cases of the letters, and merge it with `y`
                                                Ω # Then pop and push a random character from this list of four
                                                ]J # After both loops: join the entire stack together to a single string
                                                # (which is output implicitly as result)





                                                share|improve this answer











                                                $endgroup$













                                                • $begingroup$
                                                  I don't know 05AB1E but, instead of INè, could you save anything by pushing the first character of y?
                                                  $endgroup$
                                                  – Shaggy
                                                  Apr 3 at 9:06










                                                • $begingroup$
                                                  @Shaggy Yes, I indeed can.. Thanks! Maybe I should stop golfing for today, I'm a mess, lol..
                                                  $endgroup$
                                                  – Kevin Cruijssen
                                                  Apr 3 at 9:08












                                                • $begingroup$
                                                  You're a mess? ¨vNUy5LΩFy¹X>è«D.š«Ω?
                                                  $endgroup$
                                                  – Magic Octopus Urn
                                                  Apr 3 at 12:52






                                                • 1




                                                  $begingroup$
                                                  @MagicOctopusUrn Although a pretty original approach, I'm afraid it doesn't do the first bullet point of the challenge ("Output the current character."), since the result can start with t, T, or s for input "String" in your program, while it's supposed to always start with the S.
                                                  $endgroup$
                                                  – Kevin Cruijssen
                                                  Apr 3 at 13:03
















                                                2












                                                2








                                                2





                                                $begingroup$


                                                05AB1E, 18 17 bytes



                                                ü)vyн5LΩFyD.š«Ω]J


                                                Inspired by @Giuseppe's Gaia answer.

                                                -1 byte thanks to @Shaggy.



                                                Try it online 10 times or verify all test cases 10 times.



                                                Explanation:





                                                ü)             # Create all pairs of the (implicit) input
                                                # i.e. "Hello" → [["H","e"],["e","l"],["l","l"],["l","o"]]
                                                v # Loop over each these pairs `y`:
                                                yн # Push the first character of pair `y`
                                                5LΩ # Get a random integer in the range [1,5]
                                                F # Inner loop that many times:
                                                y # Push pair `y`
                                                D.š« # Duplicate it, swap the cases of the letters, and merge it with `y`
                                                Ω # Then pop and push a random character from this list of four
                                                ]J # After both loops: join the entire stack together to a single string
                                                # (which is output implicitly as result)





                                                share|improve this answer











                                                $endgroup$




                                                05AB1E, 18 17 bytes



                                                ü)vyн5LΩFyD.š«Ω]J


                                                Inspired by @Giuseppe's Gaia answer.

                                                -1 byte thanks to @Shaggy.



                                                Try it online 10 times or verify all test cases 10 times.



                                                Explanation:





                                                ü)             # Create all pairs of the (implicit) input
                                                # i.e. "Hello" → [["H","e"],["e","l"],["l","l"],["l","o"]]
                                                v # Loop over each these pairs `y`:
                                                yн # Push the first character of pair `y`
                                                5LΩ # Get a random integer in the range [1,5]
                                                F # Inner loop that many times:
                                                y # Push pair `y`
                                                D.š« # Duplicate it, swap the cases of the letters, and merge it with `y`
                                                Ω # Then pop and push a random character from this list of four
                                                ]J # After both loops: join the entire stack together to a single string
                                                # (which is output implicitly as result)






                                                share|improve this answer














                                                share|improve this answer



                                                share|improve this answer








                                                edited Apr 3 at 9:08

























                                                answered Apr 3 at 8:14









                                                Kevin CruijssenKevin Cruijssen

                                                42.6k571217




                                                42.6k571217












                                                • $begingroup$
                                                  I don't know 05AB1E but, instead of INè, could you save anything by pushing the first character of y?
                                                  $endgroup$
                                                  – Shaggy
                                                  Apr 3 at 9:06










                                                • $begingroup$
                                                  @Shaggy Yes, I indeed can.. Thanks! Maybe I should stop golfing for today, I'm a mess, lol..
                                                  $endgroup$
                                                  – Kevin Cruijssen
                                                  Apr 3 at 9:08












                                                • $begingroup$
                                                  You're a mess? ¨vNUy5LΩFy¹X>è«D.š«Ω?
                                                  $endgroup$
                                                  – Magic Octopus Urn
                                                  Apr 3 at 12:52






                                                • 1




                                                  $begingroup$
                                                  @MagicOctopusUrn Although a pretty original approach, I'm afraid it doesn't do the first bullet point of the challenge ("Output the current character."), since the result can start with t, T, or s for input "String" in your program, while it's supposed to always start with the S.
                                                  $endgroup$
                                                  – Kevin Cruijssen
                                                  Apr 3 at 13:03




















                                                • $begingroup$
                                                  I don't know 05AB1E but, instead of INè, could you save anything by pushing the first character of y?
                                                  $endgroup$
                                                  – Shaggy
                                                  Apr 3 at 9:06










                                                • $begingroup$
                                                  @Shaggy Yes, I indeed can.. Thanks! Maybe I should stop golfing for today, I'm a mess, lol..
                                                  $endgroup$
                                                  – Kevin Cruijssen
                                                  Apr 3 at 9:08












                                                • $begingroup$
                                                  You're a mess? ¨vNUy5LΩFy¹X>è«D.š«Ω?
                                                  $endgroup$
                                                  – Magic Octopus Urn
                                                  Apr 3 at 12:52






                                                • 1




                                                  $begingroup$
                                                  @MagicOctopusUrn Although a pretty original approach, I'm afraid it doesn't do the first bullet point of the challenge ("Output the current character."), since the result can start with t, T, or s for input "String" in your program, while it's supposed to always start with the S.
                                                  $endgroup$
                                                  – Kevin Cruijssen
                                                  Apr 3 at 13:03


















                                                $begingroup$
                                                I don't know 05AB1E but, instead of INè, could you save anything by pushing the first character of y?
                                                $endgroup$
                                                – Shaggy
                                                Apr 3 at 9:06




                                                $begingroup$
                                                I don't know 05AB1E but, instead of INè, could you save anything by pushing the first character of y?
                                                $endgroup$
                                                – Shaggy
                                                Apr 3 at 9:06












                                                $begingroup$
                                                @Shaggy Yes, I indeed can.. Thanks! Maybe I should stop golfing for today, I'm a mess, lol..
                                                $endgroup$
                                                – Kevin Cruijssen
                                                Apr 3 at 9:08






                                                $begingroup$
                                                @Shaggy Yes, I indeed can.. Thanks! Maybe I should stop golfing for today, I'm a mess, lol..
                                                $endgroup$
                                                – Kevin Cruijssen
                                                Apr 3 at 9:08














                                                $begingroup$
                                                You're a mess? ¨vNUy5LΩFy¹X>è«D.š«Ω?
                                                $endgroup$
                                                – Magic Octopus Urn
                                                Apr 3 at 12:52




                                                $begingroup$
                                                You're a mess? ¨vNUy5LΩFy¹X>è«D.š«Ω?
                                                $endgroup$
                                                – Magic Octopus Urn
                                                Apr 3 at 12:52




                                                1




                                                1




                                                $begingroup$
                                                @MagicOctopusUrn Although a pretty original approach, I'm afraid it doesn't do the first bullet point of the challenge ("Output the current character."), since the result can start with t, T, or s for input "String" in your program, while it's supposed to always start with the S.
                                                $endgroup$
                                                – Kevin Cruijssen
                                                Apr 3 at 13:03






                                                $begingroup$
                                                @MagicOctopusUrn Although a pretty original approach, I'm afraid it doesn't do the first bullet point of the challenge ("Output the current character."), since the result can start with t, T, or s for input "String" in your program, while it's supposed to always start with the S.
                                                $endgroup$
                                                – Kevin Cruijssen
                                                Apr 3 at 13:03













                                                1












                                                $begingroup$


                                                Charcoal, 27 bytes



                                                FLθ«F∧ι⊕‽⁵‽⭆✂θ⊖ι⊕ι¹⁺↥λ↧λ§θι


                                                Try it online! Link is to verbose version of code. Explanation:



                                                FLθ«


                                                Loop over all of the indices of the input string.



                                                F∧ι⊕‽⁵


                                                Except for the first index, loop over a random number from 1 to 5 inclusive...



                                                ‽⭆✂θ⊖ι⊕ι¹⁺↥λ↧λ


                                                ... extract the previous and next characters from the string, take the upper and lower case versions, and pick a random character of the four.



                                                §θι


                                                Print the character at the current index.






                                                share|improve this answer









                                                $endgroup$


















                                                  1












                                                  $begingroup$


                                                  Charcoal, 27 bytes



                                                  FLθ«F∧ι⊕‽⁵‽⭆✂θ⊖ι⊕ι¹⁺↥λ↧λ§θι


                                                  Try it online! Link is to verbose version of code. Explanation:



                                                  FLθ«


                                                  Loop over all of the indices of the input string.



                                                  F∧ι⊕‽⁵


                                                  Except for the first index, loop over a random number from 1 to 5 inclusive...



                                                  ‽⭆✂θ⊖ι⊕ι¹⁺↥λ↧λ


                                                  ... extract the previous and next characters from the string, take the upper and lower case versions, and pick a random character of the four.



                                                  §θι


                                                  Print the character at the current index.






                                                  share|improve this answer









                                                  $endgroup$
















                                                    1












                                                    1








                                                    1





                                                    $begingroup$


                                                    Charcoal, 27 bytes



                                                    FLθ«F∧ι⊕‽⁵‽⭆✂θ⊖ι⊕ι¹⁺↥λ↧λ§θι


                                                    Try it online! Link is to verbose version of code. Explanation:



                                                    FLθ«


                                                    Loop over all of the indices of the input string.



                                                    F∧ι⊕‽⁵


                                                    Except for the first index, loop over a random number from 1 to 5 inclusive...



                                                    ‽⭆✂θ⊖ι⊕ι¹⁺↥λ↧λ


                                                    ... extract the previous and next characters from the string, take the upper and lower case versions, and pick a random character of the four.



                                                    §θι


                                                    Print the character at the current index.






                                                    share|improve this answer









                                                    $endgroup$




                                                    Charcoal, 27 bytes



                                                    FLθ«F∧ι⊕‽⁵‽⭆✂θ⊖ι⊕ι¹⁺↥λ↧λ§θι


                                                    Try it online! Link is to verbose version of code. Explanation:



                                                    FLθ«


                                                    Loop over all of the indices of the input string.



                                                    F∧ι⊕‽⁵


                                                    Except for the first index, loop over a random number from 1 to 5 inclusive...



                                                    ‽⭆✂θ⊖ι⊕ι¹⁺↥λ↧λ


                                                    ... extract the previous and next characters from the string, take the upper and lower case versions, and pick a random character of the four.



                                                    §θι


                                                    Print the character at the current index.







                                                    share|improve this answer












                                                    share|improve this answer



                                                    share|improve this answer










                                                    answered Apr 2 at 23:58









                                                    NeilNeil

                                                    82.7k745179




                                                    82.7k745179























                                                        1












                                                        $begingroup$

                                                        perl 5 (-p), 77 bytes



                                                        s/(.)(?=(.))/$x=$1;'$x.=substr"U$1$2L$1$2",4*rand,1;'x(1+5*rand)/gee;s/.$//


                                                        TIO






                                                        share|improve this answer









                                                        $endgroup$













                                                        • $begingroup$
                                                          You can save 4 bytes by using $& instead of $1, and chop + -l instead of s/.$//
                                                          $endgroup$
                                                          – Dada
                                                          Apr 4 at 14:24
















                                                        1












                                                        $begingroup$

                                                        perl 5 (-p), 77 bytes



                                                        s/(.)(?=(.))/$x=$1;'$x.=substr"U$1$2L$1$2",4*rand,1;'x(1+5*rand)/gee;s/.$//


                                                        TIO






                                                        share|improve this answer









                                                        $endgroup$













                                                        • $begingroup$
                                                          You can save 4 bytes by using $& instead of $1, and chop + -l instead of s/.$//
                                                          $endgroup$
                                                          – Dada
                                                          Apr 4 at 14:24














                                                        1












                                                        1








                                                        1





                                                        $begingroup$

                                                        perl 5 (-p), 77 bytes



                                                        s/(.)(?=(.))/$x=$1;'$x.=substr"U$1$2L$1$2",4*rand,1;'x(1+5*rand)/gee;s/.$//


                                                        TIO






                                                        share|improve this answer









                                                        $endgroup$



                                                        perl 5 (-p), 77 bytes



                                                        s/(.)(?=(.))/$x=$1;'$x.=substr"U$1$2L$1$2",4*rand,1;'x(1+5*rand)/gee;s/.$//


                                                        TIO







                                                        share|improve this answer












                                                        share|improve this answer



                                                        share|improve this answer










                                                        answered Apr 3 at 8:01









                                                        Nahuel FouilleulNahuel Fouilleul

                                                        3,015211




                                                        3,015211












                                                        • $begingroup$
                                                          You can save 4 bytes by using $& instead of $1, and chop + -l instead of s/.$//
                                                          $endgroup$
                                                          – Dada
                                                          Apr 4 at 14:24


















                                                        • $begingroup$
                                                          You can save 4 bytes by using $& instead of $1, and chop + -l instead of s/.$//
                                                          $endgroup$
                                                          – Dada
                                                          Apr 4 at 14:24
















                                                        $begingroup$
                                                        You can save 4 bytes by using $& instead of $1, and chop + -l instead of s/.$//
                                                        $endgroup$
                                                        – Dada
                                                        Apr 4 at 14:24




                                                        $begingroup$
                                                        You can save 4 bytes by using $& instead of $1, and chop + -l instead of s/.$//
                                                        $endgroup$
                                                        – Dada
                                                        Apr 4 at 14:24











                                                        1












                                                        $begingroup$


                                                        Japt -P, 14 bytes



                                                        äÈ+Zu pv ö5ö Ä


                                                        Try it



                                                        äÈ+Zu pv ö5ö Ä     :Implicit input of string
                                                        ä :Take each consectutive pair of characters
                                                        È :Pass them through the following function as Z
                                                        + : Append to the first character of the pair
                                                        Zu : Uppercase Z
                                                        p : Append
                                                        v : Lowercase
                                                        ö : Get X random characters, where X is
                                                        5ö : Random number in the range [0,5)
                                                        Ä : Plus 1
                                                        :Implicitly join and output





                                                        share|improve this answer











                                                        $endgroup$


















                                                          1












                                                          $begingroup$


                                                          Japt -P, 14 bytes



                                                          äÈ+Zu pv ö5ö Ä


                                                          Try it



                                                          äÈ+Zu pv ö5ö Ä     :Implicit input of string
                                                          ä :Take each consectutive pair of characters
                                                          È :Pass them through the following function as Z
                                                          + : Append to the first character of the pair
                                                          Zu : Uppercase Z
                                                          p : Append
                                                          v : Lowercase
                                                          ö : Get X random characters, where X is
                                                          5ö : Random number in the range [0,5)
                                                          Ä : Plus 1
                                                          :Implicitly join and output





                                                          share|improve this answer











                                                          $endgroup$
















                                                            1












                                                            1








                                                            1





                                                            $begingroup$


                                                            Japt -P, 14 bytes



                                                            äÈ+Zu pv ö5ö Ä


                                                            Try it



                                                            äÈ+Zu pv ö5ö Ä     :Implicit input of string
                                                            ä :Take each consectutive pair of characters
                                                            È :Pass them through the following function as Z
                                                            + : Append to the first character of the pair
                                                            Zu : Uppercase Z
                                                            p : Append
                                                            v : Lowercase
                                                            ö : Get X random characters, where X is
                                                            5ö : Random number in the range [0,5)
                                                            Ä : Plus 1
                                                            :Implicitly join and output





                                                            share|improve this answer











                                                            $endgroup$




                                                            Japt -P, 14 bytes



                                                            äÈ+Zu pv ö5ö Ä


                                                            Try it



                                                            äÈ+Zu pv ö5ö Ä     :Implicit input of string
                                                            ä :Take each consectutive pair of characters
                                                            È :Pass them through the following function as Z
                                                            + : Append to the first character of the pair
                                                            Zu : Uppercase Z
                                                            p : Append
                                                            v : Lowercase
                                                            ö : Get X random characters, where X is
                                                            5ö : Random number in the range [0,5)
                                                            Ä : Plus 1
                                                            :Implicitly join and output






                                                            share|improve this answer














                                                            share|improve this answer



                                                            share|improve this answer








                                                            edited Apr 3 at 14:07

























                                                            answered Apr 3 at 10:07









                                                            ShaggyShaggy

                                                            18.9k21768




                                                            18.9k21768























                                                                1












                                                                $begingroup$


                                                                Python 3, 167 bytes





                                                                from random import*;c=choice
                                                                def f(s):
                                                                i=0;r=""
                                                                for i in range(len(s)-1):
                                                                r+=s[i]
                                                                for j in range(randrange(5)):r+=c([str.upper,str.lower])(c(s[i:i+2]))
                                                                return r


                                                                Try it online!






                                                                share|improve this answer









                                                                $endgroup$


















                                                                  1












                                                                  $begingroup$


                                                                  Python 3, 167 bytes





                                                                  from random import*;c=choice
                                                                  def f(s):
                                                                  i=0;r=""
                                                                  for i in range(len(s)-1):
                                                                  r+=s[i]
                                                                  for j in range(randrange(5)):r+=c([str.upper,str.lower])(c(s[i:i+2]))
                                                                  return r


                                                                  Try it online!






                                                                  share|improve this answer









                                                                  $endgroup$
















                                                                    1












                                                                    1








                                                                    1





                                                                    $begingroup$


                                                                    Python 3, 167 bytes





                                                                    from random import*;c=choice
                                                                    def f(s):
                                                                    i=0;r=""
                                                                    for i in range(len(s)-1):
                                                                    r+=s[i]
                                                                    for j in range(randrange(5)):r+=c([str.upper,str.lower])(c(s[i:i+2]))
                                                                    return r


                                                                    Try it online!






                                                                    share|improve this answer









                                                                    $endgroup$




                                                                    Python 3, 167 bytes





                                                                    from random import*;c=choice
                                                                    def f(s):
                                                                    i=0;r=""
                                                                    for i in range(len(s)-1):
                                                                    r+=s[i]
                                                                    for j in range(randrange(5)):r+=c([str.upper,str.lower])(c(s[i:i+2]))
                                                                    return r


                                                                    Try it online!







                                                                    share|improve this answer












                                                                    share|improve this answer



                                                                    share|improve this answer










                                                                    answered Apr 3 at 15:27









                                                                    Sara JSara J

                                                                    565210




                                                                    565210























                                                                        1












                                                                        $begingroup$


                                                                        Jelly, 14 bytes



                                                                        ;;;Œs$Xɗ¥5X¤¡Ɲ


                                                                        Try it online!



                                                                        Explanation



                                                                                     Ɲ | For each overlapping pair of letters
                                                                        ; | Join the first letter to...
                                                                        5X¤¡ | Between 1 and 5 repetitions of...
                                                                        Xɗ¥ | A randomly selected character from...
                                                                        ;;Œs$ | A list of the two letters and the swapped case versions of both





                                                                        share|improve this answer











                                                                        $endgroup$


















                                                                          1












                                                                          $begingroup$


                                                                          Jelly, 14 bytes



                                                                          ;;;Œs$Xɗ¥5X¤¡Ɲ


                                                                          Try it online!



                                                                          Explanation



                                                                                       Ɲ | For each overlapping pair of letters
                                                                          ; | Join the first letter to...
                                                                          5X¤¡ | Between 1 and 5 repetitions of...
                                                                          Xɗ¥ | A randomly selected character from...
                                                                          ;;Œs$ | A list of the two letters and the swapped case versions of both





                                                                          share|improve this answer











                                                                          $endgroup$
















                                                                            1












                                                                            1








                                                                            1





                                                                            $begingroup$


                                                                            Jelly, 14 bytes



                                                                            ;;;Œs$Xɗ¥5X¤¡Ɲ


                                                                            Try it online!



                                                                            Explanation



                                                                                         Ɲ | For each overlapping pair of letters
                                                                            ; | Join the first letter to...
                                                                            5X¤¡ | Between 1 and 5 repetitions of...
                                                                            Xɗ¥ | A randomly selected character from...
                                                                            ;;Œs$ | A list of the two letters and the swapped case versions of both





                                                                            share|improve this answer











                                                                            $endgroup$




                                                                            Jelly, 14 bytes



                                                                            ;;;Œs$Xɗ¥5X¤¡Ɲ


                                                                            Try it online!



                                                                            Explanation



                                                                                         Ɲ | For each overlapping pair of letters
                                                                            ; | Join the first letter to...
                                                                            5X¤¡ | Between 1 and 5 repetitions of...
                                                                            Xɗ¥ | A randomly selected character from...
                                                                            ;;Œs$ | A list of the two letters and the swapped case versions of both






                                                                            share|improve this answer














                                                                            share|improve this answer



                                                                            share|improve this answer








                                                                            edited Apr 3 at 16:58

























                                                                            answered Apr 3 at 6:25









                                                                            Nick KennedyNick Kennedy

                                                                            1,45649




                                                                            1,45649























                                                                                1












                                                                                $begingroup$

                                                                                C(GCC) 175 162 bytes



                                                                                -12 bytes from LambdaBeta



                                                                                f(s,S,i,r,a)char*s,*S,*i;{srand(time(0));for(i=S;*(s+1);++s){*i++=*s;for(r=rand()%5+1;r--;*i++=rand()&1?a>96&a<123|a>64&a<91?a^32:a:a)a=rand()&1?*s:*(s+1);}*i=0;}


                                                                                Try it online






                                                                                share|improve this answer











                                                                                $endgroup$













                                                                                • $begingroup$
                                                                                  I don't think you need the 0 in the first line.
                                                                                  $endgroup$
                                                                                  – LambdaBeta
                                                                                  Apr 3 at 21:28










                                                                                • $begingroup$
                                                                                  Also can save a lot of characters by taking the buffer S as a parameter and adding your variables to the argument list: Try it online!
                                                                                  $endgroup$
                                                                                  – LambdaBeta
                                                                                  Apr 3 at 21:31










                                                                                • $begingroup$
                                                                                  @LambdaBeta turns out you are right about the 0, which made it not worth it to have the #define anymore
                                                                                  $endgroup$
                                                                                  – rtpax
                                                                                  Apr 4 at 14:57










                                                                                • $begingroup$
                                                                                  150 bytes
                                                                                  $endgroup$
                                                                                  – ceilingcat
                                                                                  2 days ago
















                                                                                1












                                                                                $begingroup$

                                                                                C(GCC) 175 162 bytes



                                                                                -12 bytes from LambdaBeta



                                                                                f(s,S,i,r,a)char*s,*S,*i;{srand(time(0));for(i=S;*(s+1);++s){*i++=*s;for(r=rand()%5+1;r--;*i++=rand()&1?a>96&a<123|a>64&a<91?a^32:a:a)a=rand()&1?*s:*(s+1);}*i=0;}


                                                                                Try it online






                                                                                share|improve this answer











                                                                                $endgroup$













                                                                                • $begingroup$
                                                                                  I don't think you need the 0 in the first line.
                                                                                  $endgroup$
                                                                                  – LambdaBeta
                                                                                  Apr 3 at 21:28










                                                                                • $begingroup$
                                                                                  Also can save a lot of characters by taking the buffer S as a parameter and adding your variables to the argument list: Try it online!
                                                                                  $endgroup$
                                                                                  – LambdaBeta
                                                                                  Apr 3 at 21:31










                                                                                • $begingroup$
                                                                                  @LambdaBeta turns out you are right about the 0, which made it not worth it to have the #define anymore
                                                                                  $endgroup$
                                                                                  – rtpax
                                                                                  Apr 4 at 14:57










                                                                                • $begingroup$
                                                                                  150 bytes
                                                                                  $endgroup$
                                                                                  – ceilingcat
                                                                                  2 days ago














                                                                                1












                                                                                1








                                                                                1





                                                                                $begingroup$

                                                                                C(GCC) 175 162 bytes



                                                                                -12 bytes from LambdaBeta



                                                                                f(s,S,i,r,a)char*s,*S,*i;{srand(time(0));for(i=S;*(s+1);++s){*i++=*s;for(r=rand()%5+1;r--;*i++=rand()&1?a>96&a<123|a>64&a<91?a^32:a:a)a=rand()&1?*s:*(s+1);}*i=0;}


                                                                                Try it online






                                                                                share|improve this answer











                                                                                $endgroup$



                                                                                C(GCC) 175 162 bytes



                                                                                -12 bytes from LambdaBeta



                                                                                f(s,S,i,r,a)char*s,*S,*i;{srand(time(0));for(i=S;*(s+1);++s){*i++=*s;for(r=rand()%5+1;r--;*i++=rand()&1?a>96&a<123|a>64&a<91?a^32:a:a)a=rand()&1?*s:*(s+1);}*i=0;}


                                                                                Try it online







                                                                                share|improve this answer














                                                                                share|improve this answer



                                                                                share|improve this answer








                                                                                edited Apr 4 at 14:57

























                                                                                answered Apr 3 at 18:02









                                                                                rtpaxrtpax

                                                                                3165




                                                                                3165












                                                                                • $begingroup$
                                                                                  I don't think you need the 0 in the first line.
                                                                                  $endgroup$
                                                                                  – LambdaBeta
                                                                                  Apr 3 at 21:28










                                                                                • $begingroup$
                                                                                  Also can save a lot of characters by taking the buffer S as a parameter and adding your variables to the argument list: Try it online!
                                                                                  $endgroup$
                                                                                  – LambdaBeta
                                                                                  Apr 3 at 21:31










                                                                                • $begingroup$
                                                                                  @LambdaBeta turns out you are right about the 0, which made it not worth it to have the #define anymore
                                                                                  $endgroup$
                                                                                  – rtpax
                                                                                  Apr 4 at 14:57










                                                                                • $begingroup$
                                                                                  150 bytes
                                                                                  $endgroup$
                                                                                  – ceilingcat
                                                                                  2 days ago


















                                                                                • $begingroup$
                                                                                  I don't think you need the 0 in the first line.
                                                                                  $endgroup$
                                                                                  – LambdaBeta
                                                                                  Apr 3 at 21:28










                                                                                • $begingroup$
                                                                                  Also can save a lot of characters by taking the buffer S as a parameter and adding your variables to the argument list: Try it online!
                                                                                  $endgroup$
                                                                                  – LambdaBeta
                                                                                  Apr 3 at 21:31










                                                                                • $begingroup$
                                                                                  @LambdaBeta turns out you are right about the 0, which made it not worth it to have the #define anymore
                                                                                  $endgroup$
                                                                                  – rtpax
                                                                                  Apr 4 at 14:57










                                                                                • $begingroup$
                                                                                  150 bytes
                                                                                  $endgroup$
                                                                                  – ceilingcat
                                                                                  2 days ago
















                                                                                $begingroup$
                                                                                I don't think you need the 0 in the first line.
                                                                                $endgroup$
                                                                                – LambdaBeta
                                                                                Apr 3 at 21:28




                                                                                $begingroup$
                                                                                I don't think you need the 0 in the first line.
                                                                                $endgroup$
                                                                                – LambdaBeta
                                                                                Apr 3 at 21:28












                                                                                $begingroup$
                                                                                Also can save a lot of characters by taking the buffer S as a parameter and adding your variables to the argument list: Try it online!
                                                                                $endgroup$
                                                                                – LambdaBeta
                                                                                Apr 3 at 21:31




                                                                                $begingroup$
                                                                                Also can save a lot of characters by taking the buffer S as a parameter and adding your variables to the argument list: Try it online!
                                                                                $endgroup$
                                                                                – LambdaBeta
                                                                                Apr 3 at 21:31












                                                                                $begingroup$
                                                                                @LambdaBeta turns out you are right about the 0, which made it not worth it to have the #define anymore
                                                                                $endgroup$
                                                                                – rtpax
                                                                                Apr 4 at 14:57




                                                                                $begingroup$
                                                                                @LambdaBeta turns out you are right about the 0, which made it not worth it to have the #define anymore
                                                                                $endgroup$
                                                                                – rtpax
                                                                                Apr 4 at 14:57












                                                                                $begingroup$
                                                                                150 bytes
                                                                                $endgroup$
                                                                                – ceilingcat
                                                                                2 days ago




                                                                                $begingroup$
                                                                                150 bytes
                                                                                $endgroup$
                                                                                – ceilingcat
                                                                                2 days ago











                                                                                0












                                                                                $begingroup$


                                                                                Scala 2.12.8, 214 bytes



                                                                                Golfed version:



                                                                                val r=scala.util.Random;println(readLine.toList.sliding(2).flatMap{case a :: b :: Nil=>(a +: (0 to r.nextInt(5)).map{_=>((c: Char)=>if(r.nextBoolean)c.toUpper else c.toLower)(if(r.nextBoolean)a else b)})}.mkString)


                                                                                Golfed with newlines and indents:



                                                                                val r=scala.util.Random
                                                                                println(readLine.toList.sliding(2).flatMap{
                                                                                case a :: b :: Nil=>
                                                                                (a +: (0 to r.nextInt(5)).map{_=>
                                                                                ((c: Char)=>if(r.nextBoolean)c.toUpper else c.toLower)(if(r.nextBoolean)a else b)
                                                                                })
                                                                                }.mkString)


                                                                                Ungolfed:



                                                                                import scala.io.StdIn
                                                                                import scala.util.Random

                                                                                def gobble(input: String): String = {
                                                                                input.toList.sliding(2).flatMap {
                                                                                case thisChar :: nextChar :: Nil =>
                                                                                val numberOfAdditions = Random.nextInt(5)
                                                                                (thisChar +: (0 to numberOfAdditions).map { _ =>
                                                                                val char = if(Random.nextBoolean) thisChar else nextChar
                                                                                val cc = if(Random.nextBoolean) char.toUpper else char.ToLower
                                                                                cc
                                                                                })
                                                                                }.mkString
                                                                                }

                                                                                println(gobble(StdIn.readLine()))





                                                                                share|improve this answer









                                                                                $endgroup$









                                                                                • 1




                                                                                  $begingroup$
                                                                                  No way to turn a :: b :: Nil into a::b::Nil? Same for a :+, a:+() or a.:+() might work
                                                                                  $endgroup$
                                                                                  – Ven
                                                                                  Apr 4 at 14:07










                                                                                • $begingroup$
                                                                                  @Ven a::b::Nil causes a compile error. +: is a method defined on the list, so it might save space by getting rid of the outer parens?
                                                                                  $endgroup$
                                                                                  – Soren
                                                                                  Apr 4 at 22:58












                                                                                • $begingroup$
                                                                                  You only have only one elem here so it’s not autotupling anyway
                                                                                  $endgroup$
                                                                                  – Ven
                                                                                  Apr 4 at 23:41
















                                                                                0












                                                                                $begingroup$


                                                                                Scala 2.12.8, 214 bytes



                                                                                Golfed version:



                                                                                val r=scala.util.Random;println(readLine.toList.sliding(2).flatMap{case a :: b :: Nil=>(a +: (0 to r.nextInt(5)).map{_=>((c: Char)=>if(r.nextBoolean)c.toUpper else c.toLower)(if(r.nextBoolean)a else b)})}.mkString)


                                                                                Golfed with newlines and indents:



                                                                                val r=scala.util.Random
                                                                                println(readLine.toList.sliding(2).flatMap{
                                                                                case a :: b :: Nil=>
                                                                                (a +: (0 to r.nextInt(5)).map{_=>
                                                                                ((c: Char)=>if(r.nextBoolean)c.toUpper else c.toLower)(if(r.nextBoolean)a else b)
                                                                                })
                                                                                }.mkString)


                                                                                Ungolfed:



                                                                                import scala.io.StdIn
                                                                                import scala.util.Random

                                                                                def gobble(input: String): String = {
                                                                                input.toList.sliding(2).flatMap {
                                                                                case thisChar :: nextChar :: Nil =>
                                                                                val numberOfAdditions = Random.nextInt(5)
                                                                                (thisChar +: (0 to numberOfAdditions).map { _ =>
                                                                                val char = if(Random.nextBoolean) thisChar else nextChar
                                                                                val cc = if(Random.nextBoolean) char.toUpper else char.ToLower
                                                                                cc
                                                                                })
                                                                                }.mkString
                                                                                }

                                                                                println(gobble(StdIn.readLine()))





                                                                                share|improve this answer









                                                                                $endgroup$









                                                                                • 1




                                                                                  $begingroup$
                                                                                  No way to turn a :: b :: Nil into a::b::Nil? Same for a :+, a:+() or a.:+() might work
                                                                                  $endgroup$
                                                                                  – Ven
                                                                                  Apr 4 at 14:07










                                                                                • $begingroup$
                                                                                  @Ven a::b::Nil causes a compile error. +: is a method defined on the list, so it might save space by getting rid of the outer parens?
                                                                                  $endgroup$
                                                                                  – Soren
                                                                                  Apr 4 at 22:58












                                                                                • $begingroup$
                                                                                  You only have only one elem here so it’s not autotupling anyway
                                                                                  $endgroup$
                                                                                  – Ven
                                                                                  Apr 4 at 23:41














                                                                                0












                                                                                0








                                                                                0





                                                                                $begingroup$


                                                                                Scala 2.12.8, 214 bytes



                                                                                Golfed version:



                                                                                val r=scala.util.Random;println(readLine.toList.sliding(2).flatMap{case a :: b :: Nil=>(a +: (0 to r.nextInt(5)).map{_=>((c: Char)=>if(r.nextBoolean)c.toUpper else c.toLower)(if(r.nextBoolean)a else b)})}.mkString)


                                                                                Golfed with newlines and indents:



                                                                                val r=scala.util.Random
                                                                                println(readLine.toList.sliding(2).flatMap{
                                                                                case a :: b :: Nil=>
                                                                                (a +: (0 to r.nextInt(5)).map{_=>
                                                                                ((c: Char)=>if(r.nextBoolean)c.toUpper else c.toLower)(if(r.nextBoolean)a else b)
                                                                                })
                                                                                }.mkString)


                                                                                Ungolfed:



                                                                                import scala.io.StdIn
                                                                                import scala.util.Random

                                                                                def gobble(input: String): String = {
                                                                                input.toList.sliding(2).flatMap {
                                                                                case thisChar :: nextChar :: Nil =>
                                                                                val numberOfAdditions = Random.nextInt(5)
                                                                                (thisChar +: (0 to numberOfAdditions).map { _ =>
                                                                                val char = if(Random.nextBoolean) thisChar else nextChar
                                                                                val cc = if(Random.nextBoolean) char.toUpper else char.ToLower
                                                                                cc
                                                                                })
                                                                                }.mkString
                                                                                }

                                                                                println(gobble(StdIn.readLine()))





                                                                                share|improve this answer









                                                                                $endgroup$




                                                                                Scala 2.12.8, 214 bytes



                                                                                Golfed version:



                                                                                val r=scala.util.Random;println(readLine.toList.sliding(2).flatMap{case a :: b :: Nil=>(a +: (0 to r.nextInt(5)).map{_=>((c: Char)=>if(r.nextBoolean)c.toUpper else c.toLower)(if(r.nextBoolean)a else b)})}.mkString)


                                                                                Golfed with newlines and indents:



                                                                                val r=scala.util.Random
                                                                                println(readLine.toList.sliding(2).flatMap{
                                                                                case a :: b :: Nil=>
                                                                                (a +: (0 to r.nextInt(5)).map{_=>
                                                                                ((c: Char)=>if(r.nextBoolean)c.toUpper else c.toLower)(if(r.nextBoolean)a else b)
                                                                                })
                                                                                }.mkString)


                                                                                Ungolfed:



                                                                                import scala.io.StdIn
                                                                                import scala.util.Random

                                                                                def gobble(input: String): String = {
                                                                                input.toList.sliding(2).flatMap {
                                                                                case thisChar :: nextChar :: Nil =>
                                                                                val numberOfAdditions = Random.nextInt(5)
                                                                                (thisChar +: (0 to numberOfAdditions).map { _ =>
                                                                                val char = if(Random.nextBoolean) thisChar else nextChar
                                                                                val cc = if(Random.nextBoolean) char.toUpper else char.ToLower
                                                                                cc
                                                                                })
                                                                                }.mkString
                                                                                }

                                                                                println(gobble(StdIn.readLine()))






                                                                                share|improve this answer












                                                                                share|improve this answer



                                                                                share|improve this answer










                                                                                answered Apr 3 at 22:32









                                                                                SorenSoren

                                                                                2008




                                                                                2008








                                                                                • 1




                                                                                  $begingroup$
                                                                                  No way to turn a :: b :: Nil into a::b::Nil? Same for a :+, a:+() or a.:+() might work
                                                                                  $endgroup$
                                                                                  – Ven
                                                                                  Apr 4 at 14:07










                                                                                • $begingroup$
                                                                                  @Ven a::b::Nil causes a compile error. +: is a method defined on the list, so it might save space by getting rid of the outer parens?
                                                                                  $endgroup$
                                                                                  – Soren
                                                                                  Apr 4 at 22:58












                                                                                • $begingroup$
                                                                                  You only have only one elem here so it’s not autotupling anyway
                                                                                  $endgroup$
                                                                                  – Ven
                                                                                  Apr 4 at 23:41














                                                                                • 1




                                                                                  $begingroup$
                                                                                  No way to turn a :: b :: Nil into a::b::Nil? Same for a :+, a:+() or a.:+() might work
                                                                                  $endgroup$
                                                                                  – Ven
                                                                                  Apr 4 at 14:07










                                                                                • $begingroup$
                                                                                  @Ven a::b::Nil causes a compile error. +: is a method defined on the list, so it might save space by getting rid of the outer parens?
                                                                                  $endgroup$
                                                                                  – Soren
                                                                                  Apr 4 at 22:58












                                                                                • $begingroup$
                                                                                  You only have only one elem here so it’s not autotupling anyway
                                                                                  $endgroup$
                                                                                  – Ven
                                                                                  Apr 4 at 23:41








                                                                                1




                                                                                1




                                                                                $begingroup$
                                                                                No way to turn a :: b :: Nil into a::b::Nil? Same for a :+, a:+() or a.:+() might work
                                                                                $endgroup$
                                                                                – Ven
                                                                                Apr 4 at 14:07




                                                                                $begingroup$
                                                                                No way to turn a :: b :: Nil into a::b::Nil? Same for a :+, a:+() or a.:+() might work
                                                                                $endgroup$
                                                                                – Ven
                                                                                Apr 4 at 14:07












                                                                                $begingroup$
                                                                                @Ven a::b::Nil causes a compile error. +: is a method defined on the list, so it might save space by getting rid of the outer parens?
                                                                                $endgroup$
                                                                                – Soren
                                                                                Apr 4 at 22:58






                                                                                $begingroup$
                                                                                @Ven a::b::Nil causes a compile error. +: is a method defined on the list, so it might save space by getting rid of the outer parens?
                                                                                $endgroup$
                                                                                – Soren
                                                                                Apr 4 at 22:58














                                                                                $begingroup$
                                                                                You only have only one elem here so it’s not autotupling anyway
                                                                                $endgroup$
                                                                                – Ven
                                                                                Apr 4 at 23:41




                                                                                $begingroup$
                                                                                You only have only one elem here so it’s not autotupling anyway
                                                                                $endgroup$
                                                                                – Ven
                                                                                Apr 4 at 23:41











                                                                                0












                                                                                $begingroup$


                                                                                Perl 5 -n, 61 bytes





                                                                                s/.(?=(.))/print$&,map{(map{lc,uc}$&,$1)[rand 4]}0..rand 5/ge


                                                                                Try it online!






                                                                                share|improve this answer









                                                                                $endgroup$


















                                                                                  0












                                                                                  $begingroup$


                                                                                  Perl 5 -n, 61 bytes





                                                                                  s/.(?=(.))/print$&,map{(map{lc,uc}$&,$1)[rand 4]}0..rand 5/ge


                                                                                  Try it online!






                                                                                  share|improve this answer









                                                                                  $endgroup$
















                                                                                    0












                                                                                    0








                                                                                    0





                                                                                    $begingroup$


                                                                                    Perl 5 -n, 61 bytes





                                                                                    s/.(?=(.))/print$&,map{(map{lc,uc}$&,$1)[rand 4]}0..rand 5/ge


                                                                                    Try it online!






                                                                                    share|improve this answer









                                                                                    $endgroup$




                                                                                    Perl 5 -n, 61 bytes





                                                                                    s/.(?=(.))/print$&,map{(map{lc,uc}$&,$1)[rand 4]}0..rand 5/ge


                                                                                    Try it online!







                                                                                    share|improve this answer












                                                                                    share|improve this answer



                                                                                    share|improve this answer










                                                                                    answered Apr 4 at 13:59









                                                                                    DadaDada

                                                                                    7,77411140




                                                                                    7,77411140























                                                                                        0












                                                                                        $begingroup$


                                                                                        C# (Visual C# Interactive Compiler), 236 213 209 bytes





                                                                                        a=>{int i=0,j;var m=new Random();var s="";var c = a.Select(x=>Char.IsLetter(x)?(char)(x^32):x).ToArray();for(;i<a.Length-1;i++)for(j=m.Next(1,5);j-->0;)s+=new{a[i],c[i],a[i+1],c[i+1]}[m.Next(0,3)];return s;}


                                                                                        Try it online!






                                                                                        share|improve this answer











                                                                                        $endgroup$













                                                                                        • $begingroup$
                                                                                          Does not work with non-alphanumeric characters. char b=a[0] -> var b=a[0], extra space in declaration of d in for-loop
                                                                                          $endgroup$
                                                                                          – Embodiment of Ignorance
                                                                                          Apr 4 at 5:11
















                                                                                        0












                                                                                        $begingroup$


                                                                                        C# (Visual C# Interactive Compiler), 236 213 209 bytes





                                                                                        a=>{int i=0,j;var m=new Random();var s="";var c = a.Select(x=>Char.IsLetter(x)?(char)(x^32):x).ToArray();for(;i<a.Length-1;i++)for(j=m.Next(1,5);j-->0;)s+=new{a[i],c[i],a[i+1],c[i+1]}[m.Next(0,3)];return s;}


                                                                                        Try it online!






                                                                                        share|improve this answer











                                                                                        $endgroup$













                                                                                        • $begingroup$
                                                                                          Does not work with non-alphanumeric characters. char b=a[0] -> var b=a[0], extra space in declaration of d in for-loop
                                                                                          $endgroup$
                                                                                          – Embodiment of Ignorance
                                                                                          Apr 4 at 5:11














                                                                                        0












                                                                                        0








                                                                                        0





                                                                                        $begingroup$


                                                                                        C# (Visual C# Interactive Compiler), 236 213 209 bytes





                                                                                        a=>{int i=0,j;var m=new Random();var s="";var c = a.Select(x=>Char.IsLetter(x)?(char)(x^32):x).ToArray();for(;i<a.Length-1;i++)for(j=m.Next(1,5);j-->0;)s+=new{a[i],c[i],a[i+1],c[i+1]}[m.Next(0,3)];return s;}


                                                                                        Try it online!






                                                                                        share|improve this answer











                                                                                        $endgroup$




                                                                                        C# (Visual C# Interactive Compiler), 236 213 209 bytes





                                                                                        a=>{int i=0,j;var m=new Random();var s="";var c = a.Select(x=>Char.IsLetter(x)?(char)(x^32):x).ToArray();for(;i<a.Length-1;i++)for(j=m.Next(1,5);j-->0;)s+=new{a[i],c[i],a[i+1],c[i+1]}[m.Next(0,3)];return s;}


                                                                                        Try it online!







                                                                                        share|improve this answer














                                                                                        share|improve this answer



                                                                                        share|improve this answer








                                                                                        edited Apr 4 at 14:10

























                                                                                        answered Apr 3 at 16:10









                                                                                        Expired DataExpired Data

                                                                                        57314




                                                                                        57314












                                                                                        • $begingroup$
                                                                                          Does not work with non-alphanumeric characters. char b=a[0] -> var b=a[0], extra space in declaration of d in for-loop
                                                                                          $endgroup$
                                                                                          – Embodiment of Ignorance
                                                                                          Apr 4 at 5:11


















                                                                                        • $begingroup$
                                                                                          Does not work with non-alphanumeric characters. char b=a[0] -> var b=a[0], extra space in declaration of d in for-loop
                                                                                          $endgroup$
                                                                                          – Embodiment of Ignorance
                                                                                          Apr 4 at 5:11
















                                                                                        $begingroup$
                                                                                        Does not work with non-alphanumeric characters. char b=a[0] -> var b=a[0], extra space in declaration of d in for-loop
                                                                                        $endgroup$
                                                                                        – Embodiment of Ignorance
                                                                                        Apr 4 at 5:11




                                                                                        $begingroup$
                                                                                        Does not work with non-alphanumeric characters. char b=a[0] -> var b=a[0], extra space in declaration of d in for-loop
                                                                                        $endgroup$
                                                                                        – Embodiment of Ignorance
                                                                                        Apr 4 at 5:11











                                                                                        0












                                                                                        $begingroup$

                                                                                        T-SQL query, 286 bytes



                                                                                        DECLARE @ char(999)='String'

                                                                                        SELECT @=stuff(@,n+2,0,s)FROM(SELECT
                                                                                        top 999*,substring(lower(c)+upper(c),abs(v%4)+1,1)s
                                                                                        FROM(SELECT*,number n,substring(@,number+1,2)c,cast(newid()as varbinary)v
                                                                                        FROM(values(1),(2),(3),(4),(5))F(h),spt_values)D
                                                                                        WHERE'P'=type and n<len(@)-1and h>v%3+2ORDER
                                                                                        BY-n)E
                                                                                        PRINT LEFT(@,len(@)-1)


                                                                                        Try it online unfortunately the online version always show the same result for the same varchar, unlike MS SQL Server Management Studio






                                                                                        share|improve this answer











                                                                                        $endgroup$


















                                                                                          0












                                                                                          $begingroup$

                                                                                          T-SQL query, 286 bytes



                                                                                          DECLARE @ char(999)='String'

                                                                                          SELECT @=stuff(@,n+2,0,s)FROM(SELECT
                                                                                          top 999*,substring(lower(c)+upper(c),abs(v%4)+1,1)s
                                                                                          FROM(SELECT*,number n,substring(@,number+1,2)c,cast(newid()as varbinary)v
                                                                                          FROM(values(1),(2),(3),(4),(5))F(h),spt_values)D
                                                                                          WHERE'P'=type and n<len(@)-1and h>v%3+2ORDER
                                                                                          BY-n)E
                                                                                          PRINT LEFT(@,len(@)-1)


                                                                                          Try it online unfortunately the online version always show the same result for the same varchar, unlike MS SQL Server Management Studio






                                                                                          share|improve this answer











                                                                                          $endgroup$
















                                                                                            0












                                                                                            0








                                                                                            0





                                                                                            $begingroup$

                                                                                            T-SQL query, 286 bytes



                                                                                            DECLARE @ char(999)='String'

                                                                                            SELECT @=stuff(@,n+2,0,s)FROM(SELECT
                                                                                            top 999*,substring(lower(c)+upper(c),abs(v%4)+1,1)s
                                                                                            FROM(SELECT*,number n,substring(@,number+1,2)c,cast(newid()as varbinary)v
                                                                                            FROM(values(1),(2),(3),(4),(5))F(h),spt_values)D
                                                                                            WHERE'P'=type and n<len(@)-1and h>v%3+2ORDER
                                                                                            BY-n)E
                                                                                            PRINT LEFT(@,len(@)-1)


                                                                                            Try it online unfortunately the online version always show the same result for the same varchar, unlike MS SQL Server Management Studio






                                                                                            share|improve this answer











                                                                                            $endgroup$



                                                                                            T-SQL query, 286 bytes



                                                                                            DECLARE @ char(999)='String'

                                                                                            SELECT @=stuff(@,n+2,0,s)FROM(SELECT
                                                                                            top 999*,substring(lower(c)+upper(c),abs(v%4)+1,1)s
                                                                                            FROM(SELECT*,number n,substring(@,number+1,2)c,cast(newid()as varbinary)v
                                                                                            FROM(values(1),(2),(3),(4),(5))F(h),spt_values)D
                                                                                            WHERE'P'=type and n<len(@)-1and h>v%3+2ORDER
                                                                                            BY-n)E
                                                                                            PRINT LEFT(@,len(@)-1)


                                                                                            Try it online unfortunately the online version always show the same result for the same varchar, unlike MS SQL Server Management Studio







                                                                                            share|improve this answer














                                                                                            share|improve this answer



                                                                                            share|improve this answer








                                                                                            edited Apr 4 at 15:34

























                                                                                            answered Apr 4 at 15:21









                                                                                            t-clausen.dkt-clausen.dk

                                                                                            2,074314




                                                                                            2,074314























                                                                                                0












                                                                                                $begingroup$


                                                                                                C# (Visual C# Interactive Compiler), 156 bytes





                                                                                                n=>{var k=new Random();int i=0;foreach(var j in n.Skip(1).Zip(n,(a,b)=>a+b))for(Write(j[1]),i=0;i++<k.Next(5);)Write(k.Next()%2<1?j.ToUpper():j.ToLower());}


                                                                                                Try it online!






                                                                                                share|improve this answer











                                                                                                $endgroup$


















                                                                                                  0












                                                                                                  $begingroup$


                                                                                                  C# (Visual C# Interactive Compiler), 156 bytes





                                                                                                  n=>{var k=new Random();int i=0;foreach(var j in n.Skip(1).Zip(n,(a,b)=>a+b))for(Write(j[1]),i=0;i++<k.Next(5);)Write(k.Next()%2<1?j.ToUpper():j.ToLower());}


                                                                                                  Try it online!






                                                                                                  share|improve this answer











                                                                                                  $endgroup$
















                                                                                                    0












                                                                                                    0








                                                                                                    0





                                                                                                    $begingroup$


                                                                                                    C# (Visual C# Interactive Compiler), 156 bytes





                                                                                                    n=>{var k=new Random();int i=0;foreach(var j in n.Skip(1).Zip(n,(a,b)=>a+b))for(Write(j[1]),i=0;i++<k.Next(5);)Write(k.Next()%2<1?j.ToUpper():j.ToLower());}


                                                                                                    Try it online!






                                                                                                    share|improve this answer











                                                                                                    $endgroup$




                                                                                                    C# (Visual C# Interactive Compiler), 156 bytes





                                                                                                    n=>{var k=new Random();int i=0;foreach(var j in n.Skip(1).Zip(n,(a,b)=>a+b))for(Write(j[1]),i=0;i++<k.Next(5);)Write(k.Next()%2<1?j.ToUpper():j.ToLower());}


                                                                                                    Try it online!







                                                                                                    share|improve this answer














                                                                                                    share|improve this answer



                                                                                                    share|improve this answer








                                                                                                    edited Apr 5 at 3:52

























                                                                                                    answered Apr 5 at 3:15









                                                                                                    Embodiment of IgnoranceEmbodiment of Ignorance

                                                                                                    2,886127




                                                                                                    2,886127























                                                                                                        0












                                                                                                        $begingroup$


                                                                                                        Japt -P, 43 16 bytes



                                                                                                        äÈ+(Zv +Zu)ö5ö Ä


                                                                                                        Shortened by a lot now!



                                                                                                        Try it






                                                                                                        share|improve this answer











                                                                                                        $endgroup$













                                                                                                        • $begingroup$
                                                                                                          This seems to return the same result every time.
                                                                                                          $endgroup$
                                                                                                          – Shaggy
                                                                                                          Apr 3 at 10:01










                                                                                                        • $begingroup$
                                                                                                          @Shaggy will fix. Also, ä's description says it gives three arguments, with the last being x+y. But as you can see here, it just returns 1. Is this a bug?
                                                                                                          $endgroup$
                                                                                                          – Embodiment of Ignorance
                                                                                                          Apr 4 at 5:21
















                                                                                                        0












                                                                                                        $begingroup$


                                                                                                        Japt -P, 43 16 bytes



                                                                                                        äÈ+(Zv +Zu)ö5ö Ä


                                                                                                        Shortened by a lot now!



                                                                                                        Try it






                                                                                                        share|improve this answer











                                                                                                        $endgroup$













                                                                                                        • $begingroup$
                                                                                                          This seems to return the same result every time.
                                                                                                          $endgroup$
                                                                                                          – Shaggy
                                                                                                          Apr 3 at 10:01










                                                                                                        • $begingroup$
                                                                                                          @Shaggy will fix. Also, ä's description says it gives three arguments, with the last being x+y. But as you can see here, it just returns 1. Is this a bug?
                                                                                                          $endgroup$
                                                                                                          – Embodiment of Ignorance
                                                                                                          Apr 4 at 5:21














                                                                                                        0












                                                                                                        0








                                                                                                        0





                                                                                                        $begingroup$


                                                                                                        Japt -P, 43 16 bytes



                                                                                                        äÈ+(Zv +Zu)ö5ö Ä


                                                                                                        Shortened by a lot now!



                                                                                                        Try it






                                                                                                        share|improve this answer











                                                                                                        $endgroup$




                                                                                                        Japt -P, 43 16 bytes



                                                                                                        äÈ+(Zv +Zu)ö5ö Ä


                                                                                                        Shortened by a lot now!



                                                                                                        Try it







                                                                                                        share|improve this answer














                                                                                                        share|improve this answer



                                                                                                        share|improve this answer








                                                                                                        edited Apr 5 at 3:55

























                                                                                                        answered Apr 3 at 5:18









                                                                                                        Embodiment of IgnoranceEmbodiment of Ignorance

                                                                                                        2,886127




                                                                                                        2,886127












                                                                                                        • $begingroup$
                                                                                                          This seems to return the same result every time.
                                                                                                          $endgroup$
                                                                                                          – Shaggy
                                                                                                          Apr 3 at 10:01










                                                                                                        • $begingroup$
                                                                                                          @Shaggy will fix. Also, ä's description says it gives three arguments, with the last being x+y. But as you can see here, it just returns 1. Is this a bug?
                                                                                                          $endgroup$
                                                                                                          – Embodiment of Ignorance
                                                                                                          Apr 4 at 5:21


















                                                                                                        • $begingroup$
                                                                                                          This seems to return the same result every time.
                                                                                                          $endgroup$
                                                                                                          – Shaggy
                                                                                                          Apr 3 at 10:01










                                                                                                        • $begingroup$
                                                                                                          @Shaggy will fix. Also, ä's description says it gives three arguments, with the last being x+y. But as you can see here, it just returns 1. Is this a bug?
                                                                                                          $endgroup$
                                                                                                          – Embodiment of Ignorance
                                                                                                          Apr 4 at 5:21
















                                                                                                        $begingroup$
                                                                                                        This seems to return the same result every time.
                                                                                                        $endgroup$
                                                                                                        – Shaggy
                                                                                                        Apr 3 at 10:01




                                                                                                        $begingroup$
                                                                                                        This seems to return the same result every time.
                                                                                                        $endgroup$
                                                                                                        – Shaggy
                                                                                                        Apr 3 at 10:01












                                                                                                        $begingroup$
                                                                                                        @Shaggy will fix. Also, ä's description says it gives three arguments, with the last being x+y. But as you can see here, it just returns 1. Is this a bug?
                                                                                                        $endgroup$
                                                                                                        – Embodiment of Ignorance
                                                                                                        Apr 4 at 5:21




                                                                                                        $begingroup$
                                                                                                        @Shaggy will fix. Also, ä's description says it gives three arguments, with the last being x+y. But as you can see here, it just returns 1. Is this a bug?
                                                                                                        $endgroup$
                                                                                                        – Embodiment of Ignorance
                                                                                                        Apr 4 at 5:21











                                                                                                        0












                                                                                                        $begingroup$


                                                                                                        C (gcc), 110 109 bytes





                                                                                                        i,p;g(char*_){for(i=rand(putchar(*_))%1024;p=_[i%2],putchar(i&2&&p>64&~-p%32<26?p^32:p),i/=4;);_[2]&&g(_+1);}


                                                                                                        Try it online!



                                                                                                        -1 thanks to ceilingcat



                                                                                                        i,p;g(char*_){
                                                                                                        for(i=rand(putchar(*_)) //print current char
                                                                                                        %1024; // and get 10 random bits
                                                                                                        p=_[i%2], //1st bit => current/next char
                                                                                                        putchar(i&2&& //2nd bit => toggle case
                                                                                                        p>64&~-p%32<26 // if char-to-print is alphabetic
                                                                                                        ?p^32:p),
                                                                                                        i/=4;); //discard two bits
                                                                                                        _[2]&&g(_+1); //if next isn't last char, repeat with next char
                                                                                                        }




                                                                                                        The number of characters printed (per input character) is not uniformly random:



                                                                                                        1  if      i<   4 (  4/1024 = 1/256)
                                                                                                        2 if 4<=i< 16 ( 12/1024 = 3/256)
                                                                                                        3 if 16<=i< 64 ( 48/1024 = 3/ 64)
                                                                                                        4 if 64<=i< 256 (192/1024 = 3/ 16)
                                                                                                        5 if 256<=i<1024 (768/1024 = 3/ 4)





                                                                                                        share|improve this answer











                                                                                                        $endgroup$


















                                                                                                          0












                                                                                                          $begingroup$


                                                                                                          C (gcc), 110 109 bytes





                                                                                                          i,p;g(char*_){for(i=rand(putchar(*_))%1024;p=_[i%2],putchar(i&2&&p>64&~-p%32<26?p^32:p),i/=4;);_[2]&&g(_+1);}


                                                                                                          Try it online!



                                                                                                          -1 thanks to ceilingcat



                                                                                                          i,p;g(char*_){
                                                                                                          for(i=rand(putchar(*_)) //print current char
                                                                                                          %1024; // and get 10 random bits
                                                                                                          p=_[i%2], //1st bit => current/next char
                                                                                                          putchar(i&2&& //2nd bit => toggle case
                                                                                                          p>64&~-p%32<26 // if char-to-print is alphabetic
                                                                                                          ?p^32:p),
                                                                                                          i/=4;); //discard two bits
                                                                                                          _[2]&&g(_+1); //if next isn't last char, repeat with next char
                                                                                                          }




                                                                                                          The number of characters printed (per input character) is not uniformly random:



                                                                                                          1  if      i<   4 (  4/1024 = 1/256)
                                                                                                          2 if 4<=i< 16 ( 12/1024 = 3/256)
                                                                                                          3 if 16<=i< 64 ( 48/1024 = 3/ 64)
                                                                                                          4 if 64<=i< 256 (192/1024 = 3/ 16)
                                                                                                          5 if 256<=i<1024 (768/1024 = 3/ 4)





                                                                                                          share|improve this answer











                                                                                                          $endgroup$
















                                                                                                            0












                                                                                                            0








                                                                                                            0





                                                                                                            $begingroup$


                                                                                                            C (gcc), 110 109 bytes





                                                                                                            i,p;g(char*_){for(i=rand(putchar(*_))%1024;p=_[i%2],putchar(i&2&&p>64&~-p%32<26?p^32:p),i/=4;);_[2]&&g(_+1);}


                                                                                                            Try it online!



                                                                                                            -1 thanks to ceilingcat



                                                                                                            i,p;g(char*_){
                                                                                                            for(i=rand(putchar(*_)) //print current char
                                                                                                            %1024; // and get 10 random bits
                                                                                                            p=_[i%2], //1st bit => current/next char
                                                                                                            putchar(i&2&& //2nd bit => toggle case
                                                                                                            p>64&~-p%32<26 // if char-to-print is alphabetic
                                                                                                            ?p^32:p),
                                                                                                            i/=4;); //discard two bits
                                                                                                            _[2]&&g(_+1); //if next isn't last char, repeat with next char
                                                                                                            }




                                                                                                            The number of characters printed (per input character) is not uniformly random:



                                                                                                            1  if      i<   4 (  4/1024 = 1/256)
                                                                                                            2 if 4<=i< 16 ( 12/1024 = 3/256)
                                                                                                            3 if 16<=i< 64 ( 48/1024 = 3/ 64)
                                                                                                            4 if 64<=i< 256 (192/1024 = 3/ 16)
                                                                                                            5 if 256<=i<1024 (768/1024 = 3/ 4)





                                                                                                            share|improve this answer











                                                                                                            $endgroup$




                                                                                                            C (gcc), 110 109 bytes





                                                                                                            i,p;g(char*_){for(i=rand(putchar(*_))%1024;p=_[i%2],putchar(i&2&&p>64&~-p%32<26?p^32:p),i/=4;);_[2]&&g(_+1);}


                                                                                                            Try it online!



                                                                                                            -1 thanks to ceilingcat



                                                                                                            i,p;g(char*_){
                                                                                                            for(i=rand(putchar(*_)) //print current char
                                                                                                            %1024; // and get 10 random bits
                                                                                                            p=_[i%2], //1st bit => current/next char
                                                                                                            putchar(i&2&& //2nd bit => toggle case
                                                                                                            p>64&~-p%32<26 // if char-to-print is alphabetic
                                                                                                            ?p^32:p),
                                                                                                            i/=4;); //discard two bits
                                                                                                            _[2]&&g(_+1); //if next isn't last char, repeat with next char
                                                                                                            }




                                                                                                            The number of characters printed (per input character) is not uniformly random:



                                                                                                            1  if      i<   4 (  4/1024 = 1/256)
                                                                                                            2 if 4<=i< 16 ( 12/1024 = 3/256)
                                                                                                            3 if 16<=i< 64 ( 48/1024 = 3/ 64)
                                                                                                            4 if 64<=i< 256 (192/1024 = 3/ 16)
                                                                                                            5 if 256<=i<1024 (768/1024 = 3/ 4)






                                                                                                            share|improve this answer














                                                                                                            share|improve this answer



                                                                                                            share|improve this answer








                                                                                                            edited yesterday

























                                                                                                            answered Apr 6 at 9:16









                                                                                                            attinatattinat

                                                                                                            4897




                                                                                                            4897






























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