Doubled-letter steganography












19














Steganography hides a given message inside a given carrier, producing a package that does not look suspicious. For this challenge, you will write a program that takes an ASCII message and an ASCII carrier as input, and return or print a package that is identical to the carrier except characters corresponding to the message are doubled, in the same order that they appear in the message.



Rules:




  1. If the carrier already contains sequences of the same character more than once, and they are not used to encode a character of the message, the program will reduce them to a single character.

  2. If the carrier does not contain the message characters in the right order, the program may return nothing, the carrier itself, or an error.

  3. You may assume that the message and carrier are non-empty ASCII strings.

  4. Capitalization matters: A is not equivalent to a.

  5. When more than one package is valid, your program may output any or all of them.

  6. Space is a character like any other character.


Test cases:




Message Carrier Package
"hi" "has it arrived?" "hhas iit arived?" OR "hhas it ariived?"
"sir" "has it arrived?" "hass iit arrived?"
"foo" "has it arrived?" "" OR "has it arrived?" OR an error.
"Car" "Cats are cool." "CCaats arre col."
"car" "Cats are cool." "" OR "Cats are cool." OR an error.
"Couch" "Couch" "CCoouucchh"
"oo" "oooooooooo" "oooo"
"o o" "oooo oooa" "oo ooa"


This is code golf, so fewest bytes wins.










share|improve this question




















  • 5




    Not suspicious at all... :P
    – Quintec
    Dec 8 at 20:59










  • Is "oooo oa" (with 2 spaces) a valid output for the last test case?
    – Arnauld
    Dec 8 at 21:36








  • 3




    It is not a valid output because the order of doubled characters in the package must match the order of the characters in the message. In the message, we have an 'o', then an ' ', then an 'o', but your package has the space after the o's
    – jkpate
    Dec 8 at 21:58












  • Ah yes, that makes sense.
    – Arnauld
    Dec 8 at 21:59






  • 1




    No. My reasoning behind this rule is that the program's output in the case of no solution should be unambiguous that no solution is possible. The three allowed outputs are unambiguous, but more extensive checking would be required for the deduplicated case.
    – jkpate
    Dec 8 at 23:46
















19














Steganography hides a given message inside a given carrier, producing a package that does not look suspicious. For this challenge, you will write a program that takes an ASCII message and an ASCII carrier as input, and return or print a package that is identical to the carrier except characters corresponding to the message are doubled, in the same order that they appear in the message.



Rules:




  1. If the carrier already contains sequences of the same character more than once, and they are not used to encode a character of the message, the program will reduce them to a single character.

  2. If the carrier does not contain the message characters in the right order, the program may return nothing, the carrier itself, or an error.

  3. You may assume that the message and carrier are non-empty ASCII strings.

  4. Capitalization matters: A is not equivalent to a.

  5. When more than one package is valid, your program may output any or all of them.

  6. Space is a character like any other character.


Test cases:




Message Carrier Package
"hi" "has it arrived?" "hhas iit arived?" OR "hhas it ariived?"
"sir" "has it arrived?" "hass iit arrived?"
"foo" "has it arrived?" "" OR "has it arrived?" OR an error.
"Car" "Cats are cool." "CCaats arre col."
"car" "Cats are cool." "" OR "Cats are cool." OR an error.
"Couch" "Couch" "CCoouucchh"
"oo" "oooooooooo" "oooo"
"o o" "oooo oooa" "oo ooa"


This is code golf, so fewest bytes wins.










share|improve this question




















  • 5




    Not suspicious at all... :P
    – Quintec
    Dec 8 at 20:59










  • Is "oooo oa" (with 2 spaces) a valid output for the last test case?
    – Arnauld
    Dec 8 at 21:36








  • 3




    It is not a valid output because the order of doubled characters in the package must match the order of the characters in the message. In the message, we have an 'o', then an ' ', then an 'o', but your package has the space after the o's
    – jkpate
    Dec 8 at 21:58












  • Ah yes, that makes sense.
    – Arnauld
    Dec 8 at 21:59






  • 1




    No. My reasoning behind this rule is that the program's output in the case of no solution should be unambiguous that no solution is possible. The three allowed outputs are unambiguous, but more extensive checking would be required for the deduplicated case.
    – jkpate
    Dec 8 at 23:46














19












19








19


1





Steganography hides a given message inside a given carrier, producing a package that does not look suspicious. For this challenge, you will write a program that takes an ASCII message and an ASCII carrier as input, and return or print a package that is identical to the carrier except characters corresponding to the message are doubled, in the same order that they appear in the message.



Rules:




  1. If the carrier already contains sequences of the same character more than once, and they are not used to encode a character of the message, the program will reduce them to a single character.

  2. If the carrier does not contain the message characters in the right order, the program may return nothing, the carrier itself, or an error.

  3. You may assume that the message and carrier are non-empty ASCII strings.

  4. Capitalization matters: A is not equivalent to a.

  5. When more than one package is valid, your program may output any or all of them.

  6. Space is a character like any other character.


Test cases:




Message Carrier Package
"hi" "has it arrived?" "hhas iit arived?" OR "hhas it ariived?"
"sir" "has it arrived?" "hass iit arrived?"
"foo" "has it arrived?" "" OR "has it arrived?" OR an error.
"Car" "Cats are cool." "CCaats arre col."
"car" "Cats are cool." "" OR "Cats are cool." OR an error.
"Couch" "Couch" "CCoouucchh"
"oo" "oooooooooo" "oooo"
"o o" "oooo oooa" "oo ooa"


This is code golf, so fewest bytes wins.










share|improve this question















Steganography hides a given message inside a given carrier, producing a package that does not look suspicious. For this challenge, you will write a program that takes an ASCII message and an ASCII carrier as input, and return or print a package that is identical to the carrier except characters corresponding to the message are doubled, in the same order that they appear in the message.



Rules:




  1. If the carrier already contains sequences of the same character more than once, and they are not used to encode a character of the message, the program will reduce them to a single character.

  2. If the carrier does not contain the message characters in the right order, the program may return nothing, the carrier itself, or an error.

  3. You may assume that the message and carrier are non-empty ASCII strings.

  4. Capitalization matters: A is not equivalent to a.

  5. When more than one package is valid, your program may output any or all of them.

  6. Space is a character like any other character.


Test cases:




Message Carrier Package
"hi" "has it arrived?" "hhas iit arived?" OR "hhas it ariived?"
"sir" "has it arrived?" "hass iit arrived?"
"foo" "has it arrived?" "" OR "has it arrived?" OR an error.
"Car" "Cats are cool." "CCaats arre col."
"car" "Cats are cool." "" OR "Cats are cool." OR an error.
"Couch" "Couch" "CCoouucchh"
"oo" "oooooooooo" "oooo"
"o o" "oooo oooa" "oo ooa"


This is code golf, so fewest bytes wins.







code-golf string steganography






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 8 at 20:09









Jonathan Allan

50.6k534165




50.6k534165










asked Dec 8 at 19:23









jkpate

1964




1964








  • 5




    Not suspicious at all... :P
    – Quintec
    Dec 8 at 20:59










  • Is "oooo oa" (with 2 spaces) a valid output for the last test case?
    – Arnauld
    Dec 8 at 21:36








  • 3




    It is not a valid output because the order of doubled characters in the package must match the order of the characters in the message. In the message, we have an 'o', then an ' ', then an 'o', but your package has the space after the o's
    – jkpate
    Dec 8 at 21:58












  • Ah yes, that makes sense.
    – Arnauld
    Dec 8 at 21:59






  • 1




    No. My reasoning behind this rule is that the program's output in the case of no solution should be unambiguous that no solution is possible. The three allowed outputs are unambiguous, but more extensive checking would be required for the deduplicated case.
    – jkpate
    Dec 8 at 23:46














  • 5




    Not suspicious at all... :P
    – Quintec
    Dec 8 at 20:59










  • Is "oooo oa" (with 2 spaces) a valid output for the last test case?
    – Arnauld
    Dec 8 at 21:36








  • 3




    It is not a valid output because the order of doubled characters in the package must match the order of the characters in the message. In the message, we have an 'o', then an ' ', then an 'o', but your package has the space after the o's
    – jkpate
    Dec 8 at 21:58












  • Ah yes, that makes sense.
    – Arnauld
    Dec 8 at 21:59






  • 1




    No. My reasoning behind this rule is that the program's output in the case of no solution should be unambiguous that no solution is possible. The three allowed outputs are unambiguous, but more extensive checking would be required for the deduplicated case.
    – jkpate
    Dec 8 at 23:46








5




5




Not suspicious at all... :P
– Quintec
Dec 8 at 20:59




Not suspicious at all... :P
– Quintec
Dec 8 at 20:59












Is "oooo oa" (with 2 spaces) a valid output for the last test case?
– Arnauld
Dec 8 at 21:36






Is "oooo oa" (with 2 spaces) a valid output for the last test case?
– Arnauld
Dec 8 at 21:36






3




3




It is not a valid output because the order of doubled characters in the package must match the order of the characters in the message. In the message, we have an 'o', then an ' ', then an 'o', but your package has the space after the o's
– jkpate
Dec 8 at 21:58






It is not a valid output because the order of doubled characters in the package must match the order of the characters in the message. In the message, we have an 'o', then an ' ', then an 'o', but your package has the space after the o's
– jkpate
Dec 8 at 21:58














Ah yes, that makes sense.
– Arnauld
Dec 8 at 21:59




Ah yes, that makes sense.
– Arnauld
Dec 8 at 21:59




1




1




No. My reasoning behind this rule is that the program's output in the case of no solution should be unambiguous that no solution is possible. The three allowed outputs are unambiguous, but more extensive checking would be required for the deduplicated case.
– jkpate
Dec 8 at 23:46




No. My reasoning behind this rule is that the program's output in the case of no solution should be unambiguous that no solution is possible. The three allowed outputs are unambiguous, but more extensive checking would be required for the deduplicated case.
– jkpate
Dec 8 at 23:46










8 Answers
8






active

oldest

votes


















4















Jelly, 28 bytes



ẹⱮŒp<ƝẠ$ƇṪ
nƝ+çṬ¥a⁸ḟ0Ḥç¦ð¹ç?


A full program taking carrier and message as command line arguments which prints the result

(For a non-packable message prints the unchanged carrier).



Try it online! Or see the test-suite.



How?



ẹⱮŒp<ƝẠ$ƇṪ - Link 1, helper function to find the indices to double: carrier, message
- e.g. "programming", "rom"
Ɱ - map across message with:
ẹ - indices of [[2,5], [3], [7,8]]
Œp - Cartesian product [[2,3,7],[2,3,8],[5,3,7],[5,3,8]]
Ƈ - filter keep if:
$ - last two links as a monad:
Ɲ - for neighbours:
< - less than? [1,1] [1,1] [0,1] [0,1]
Ạ - all truthy? 1 1 0 0
- [[2,3,7],[2,3,8]]
Ṫ - tail (if empty yields 0) [2,3,8]

nƝ+çṬ¥a⁸ḟ0Ḥç¦ð¹ç? - Main Link: carrier, message
? - if...
ç - ...condition: last Link (the helper function) as a dyad
ð - ...then: perform the dyadic chain to the left (described below)
¹ - ...else: do nothing (yields carrier)
- (the then clause:)
Ɲ - for neighbours in the carrier
n - not equal?
¥ - last two links as a dyad:
ç - call last Link (the helper function) as a dyad
Ṭ - untruth (e.g. [2,5] -> [0,1,0,0,1])
+ - add (vectorises)
a⁸ - logical AND with carrier
ḟ0 - filter out zeros
¦ - sparse application...
ç - ...to indices: call last Link (the helper function) as a dyad
Ḥ - ...do: double (e.g. 'x' -> 'xx')





share|improve this answer































    3














    JavaScript (ES6), 71 bytes



    Takes input as (message)(carrier).





    s=>g=([c,...C],p)=>c?(c==s[0]?(s=s.slice(1),c)+c:p==c?'':c)+g(C,c):s&&X


    Try it online!





    Alternate version, 66 bytes



    If we can take the message as an array of characters:



    s=>g=([c,...C],p)=>c?(c==s[0]?s.shift()+c:p==c?'':c)+g(C,c):s+s&&X


    Try it online!





    Edit: Thanks to @tsh for noticing that I forgot to remove some code when switching from non-recursive to recursive versions.






    share|improve this answer























    • You could remove p= since p is passed by a parameter.
      – tsh
      Dec 11 at 3:10












    • @tsh Oops. It's some residual code from the previous, non-recursive versions that I forgot to remove. Thank you!
      – Arnauld
      Dec 11 at 11:59



















    2














    Haskell, 124 121 107 101 97 95 90 bytes



    (#).(++"ü")
    "ü"#=
    p@(m:n)#e@(c:d)|m/=c=c:p#snd(span(==c)d)|m==n!!0=m:m:n#d|1<2=m:n#e


    Raises the "Non-exhaustive patterns" exception if the carrier does not contain the message.



    Try it online!



    Edit: -5 bytes thanks to @Laikoni.






    share|improve this answer























    • I think switching the cases allows you to drop m==c: Try it online!
      – Laikoni
      Dec 9 at 9:43





















    1















    Retina 0.8.2, 67 bytes



    +`(.)(1*)1*(.*¶)(?(1)(1(2)))(.*)$(?!¶)
    $1$4$5¶$3$6
    M!s`.*¶$




    Try it online! Takes the carrier on the first line and the message on the second line. Explanation:



    +`(.)(1*)1*(.*¶)(?(1)(1(2)))(.*)$(?!¶)
    $1$4$5¶$3$6


    Process runs of 1 or more identical characters of the carrier. If there is also a run of 1 or more of the same characters in the message, then append the shorter of the two runs to the output in duplicate, otherwise append a single character of the carrier to the output. Each run of output characters is terminated with a newline to distinguish it from the input. The (?!¶) at the end prevents the regex from thinking the carrier is the message once the message is exhausted, as normally $ is allowed to match where ¶$ would match.



    M!s`.*¶$


    Delete everything if the message wasn't completely encoded.







    Remove the newlines from the output.






    share|improve this answer























    • I think it does not pass the second to last test case (which, to be fair, I did not have in the initial post).
      – jkpate
      Dec 8 at 20:14










    • @jkpate Thanks for pointing that out; I've had to rewrite my approach slightly.
      – Neil
      Dec 8 at 21:32



















    0















    Clean, 118 bytes



    import StdEnv,StdLib
    $=
    $[u:v]b#(_,w)=span((==)u)v
    |b%(0,0)==[u]=[u,u: $if(v%(0,0)<>b%(1,1))w v(tl b)]=[u: $w b]


    Try it online!



    Takes the carrier first, then the message.



    Errors with Run time error, rule '$;2' in module 'main' does not match if message won't fit.






    share|improve this answer





























      0















      Ruby, 73 bytes





      f=->m,c,b=p{x,*c=c;x ?(x==m[0]?x+m.shift: x==b ?'':x)+f[m,c,x]:m[0]?x:''}


      Try it online!



      Recursive function, takes inputs as array of characters.



      For once I was hoping to make use of Ruby's built-in squeeze method that contracts consecutive runs of the same character to a single instance. But unfortunately, nope - the last two test cases screwed everything so badly, that I had to resort to a completely different approach, and this turned out to be basically a port of Arnauld's answer.






      share|improve this answer





























        0














        Powershell, 134 bytes





        param($m,$c)$c-csplit"([$m])"|%{$i+=$o=$_-ceq$m[+$i]
        if($o-or$_-cne"`0$h"[-1]){$h+=($_-replace'(.)(?=1)')*($o+1)}}
        $h*!($i-$m.Length)


        The script returns the empty string if the carrier does not contain the message characters in the right order.



        Less golfed test script:



        $f = {

        param($message,$carrier)
        $carrier-csplit"([$message])"|%{ # split by chars of the message, chars itself included ()
        $offset=$_-ceq$message[+$i] # 0 or 1 if current substring is a current message char (case-sensitive equality)
        $i+=$offset # move to next message char if need it
        if($offset-or$_-cne"`0$h"[-1]){ # condition to remove redundant doubles after message char: arrrived -> arrived, ooo -> oo, etc
        # `0 to avoid exception error if $h is empty
        $h+=($_-replace'(.)(?=1)')*($offset+1) # accumulate a double message char or a single substring without inner doubles: arried -> arived, anna -> ana, etc
        }
        }
        $h*!($i-$message.Length) # repeat 0 or 1 times to return '' if the carrier does not contain the message characters in the right order

        }

        @(
        ,('hi' ,'has it arrived?' ,'hhas iit arived?', 'hhas it ariived?')
        ,('hi?' ,'has it arrived?' ,'hhas iit arived??', 'hhas it ariived??')
        ,('sir' ,'has it arrived?' ,'hass iit arrived?')
        ,('foo' ,'has it arrived?' ,'')
        ,('Car' ,'Cats are cool.' ,'CCaats arre col.')
        ,('car' ,'Cats are cool.' ,'')
        ,('Couch' ,'Couch' ,'CCoouucchh')
        ,('oo' ,'oooooooooo' ,'oooo')
        ,('o o' ,'oooo oooa' ,'oo ooa')
        ,('er' ,'error' ,'eerorr', 'eerror')
        ,('a+b' ,'anna+bob' ,'aana++bbob')
        ) | % {
        $message,$carrier,$expected = $_
        $result = &$f $message $carrier
        "$($result-in$expected): $result"
        }


        Output:





        True: hhas iit arived?
        True: hhas iit arived??
        True: hass iit arrived?
        True:
        True: CCaats arre col.
        True:
        True: CCoouucchh
        True: oooo
        True: oo ooa
        True: eerror
        True: aana++bbob





        share|improve this answer































          0















          C (gcc), 69+12 = 81 bytes





          g(char*m,char*_){for(;*_;++_)*m-*_?_[-1]-*_&&p*_):p p*m++));*m&&0/0;}


          Compile with (12 bytes)



          -Dp=putchar(


          Try it online!



          g(char*m,char*_){
          for(;*_;++_) //step through _
          *m-*_? //check if character should be encoded
          _[-1]-*_&& //no? skip duplicates
          p*_) // print non-duplicates
          :p p*m++)); //print encoded character twice
          *m&&0/0; //if m is not fully encoded, exit via Floating point exception
          }





          share|improve this answer








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            8 Answers
            8






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            8 Answers
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            4















            Jelly, 28 bytes



            ẹⱮŒp<ƝẠ$ƇṪ
            nƝ+çṬ¥a⁸ḟ0Ḥç¦ð¹ç?


            A full program taking carrier and message as command line arguments which prints the result

            (For a non-packable message prints the unchanged carrier).



            Try it online! Or see the test-suite.



            How?



            ẹⱮŒp<ƝẠ$ƇṪ - Link 1, helper function to find the indices to double: carrier, message
            - e.g. "programming", "rom"
            Ɱ - map across message with:
            ẹ - indices of [[2,5], [3], [7,8]]
            Œp - Cartesian product [[2,3,7],[2,3,8],[5,3,7],[5,3,8]]
            Ƈ - filter keep if:
            $ - last two links as a monad:
            Ɲ - for neighbours:
            < - less than? [1,1] [1,1] [0,1] [0,1]
            Ạ - all truthy? 1 1 0 0
            - [[2,3,7],[2,3,8]]
            Ṫ - tail (if empty yields 0) [2,3,8]

            nƝ+çṬ¥a⁸ḟ0Ḥç¦ð¹ç? - Main Link: carrier, message
            ? - if...
            ç - ...condition: last Link (the helper function) as a dyad
            ð - ...then: perform the dyadic chain to the left (described below)
            ¹ - ...else: do nothing (yields carrier)
            - (the then clause:)
            Ɲ - for neighbours in the carrier
            n - not equal?
            ¥ - last two links as a dyad:
            ç - call last Link (the helper function) as a dyad
            Ṭ - untruth (e.g. [2,5] -> [0,1,0,0,1])
            + - add (vectorises)
            a⁸ - logical AND with carrier
            ḟ0 - filter out zeros
            ¦ - sparse application...
            ç - ...to indices: call last Link (the helper function) as a dyad
            Ḥ - ...do: double (e.g. 'x' -> 'xx')





            share|improve this answer




























              4















              Jelly, 28 bytes



              ẹⱮŒp<ƝẠ$ƇṪ
              nƝ+çṬ¥a⁸ḟ0Ḥç¦ð¹ç?


              A full program taking carrier and message as command line arguments which prints the result

              (For a non-packable message prints the unchanged carrier).



              Try it online! Or see the test-suite.



              How?



              ẹⱮŒp<ƝẠ$ƇṪ - Link 1, helper function to find the indices to double: carrier, message
              - e.g. "programming", "rom"
              Ɱ - map across message with:
              ẹ - indices of [[2,5], [3], [7,8]]
              Œp - Cartesian product [[2,3,7],[2,3,8],[5,3,7],[5,3,8]]
              Ƈ - filter keep if:
              $ - last two links as a monad:
              Ɲ - for neighbours:
              < - less than? [1,1] [1,1] [0,1] [0,1]
              Ạ - all truthy? 1 1 0 0
              - [[2,3,7],[2,3,8]]
              Ṫ - tail (if empty yields 0) [2,3,8]

              nƝ+çṬ¥a⁸ḟ0Ḥç¦ð¹ç? - Main Link: carrier, message
              ? - if...
              ç - ...condition: last Link (the helper function) as a dyad
              ð - ...then: perform the dyadic chain to the left (described below)
              ¹ - ...else: do nothing (yields carrier)
              - (the then clause:)
              Ɲ - for neighbours in the carrier
              n - not equal?
              ¥ - last two links as a dyad:
              ç - call last Link (the helper function) as a dyad
              Ṭ - untruth (e.g. [2,5] -> [0,1,0,0,1])
              + - add (vectorises)
              a⁸ - logical AND with carrier
              ḟ0 - filter out zeros
              ¦ - sparse application...
              ç - ...to indices: call last Link (the helper function) as a dyad
              Ḥ - ...do: double (e.g. 'x' -> 'xx')





              share|improve this answer


























                4












                4








                4







                Jelly, 28 bytes



                ẹⱮŒp<ƝẠ$ƇṪ
                nƝ+çṬ¥a⁸ḟ0Ḥç¦ð¹ç?


                A full program taking carrier and message as command line arguments which prints the result

                (For a non-packable message prints the unchanged carrier).



                Try it online! Or see the test-suite.



                How?



                ẹⱮŒp<ƝẠ$ƇṪ - Link 1, helper function to find the indices to double: carrier, message
                - e.g. "programming", "rom"
                Ɱ - map across message with:
                ẹ - indices of [[2,5], [3], [7,8]]
                Œp - Cartesian product [[2,3,7],[2,3,8],[5,3,7],[5,3,8]]
                Ƈ - filter keep if:
                $ - last two links as a monad:
                Ɲ - for neighbours:
                < - less than? [1,1] [1,1] [0,1] [0,1]
                Ạ - all truthy? 1 1 0 0
                - [[2,3,7],[2,3,8]]
                Ṫ - tail (if empty yields 0) [2,3,8]

                nƝ+çṬ¥a⁸ḟ0Ḥç¦ð¹ç? - Main Link: carrier, message
                ? - if...
                ç - ...condition: last Link (the helper function) as a dyad
                ð - ...then: perform the dyadic chain to the left (described below)
                ¹ - ...else: do nothing (yields carrier)
                - (the then clause:)
                Ɲ - for neighbours in the carrier
                n - not equal?
                ¥ - last two links as a dyad:
                ç - call last Link (the helper function) as a dyad
                Ṭ - untruth (e.g. [2,5] -> [0,1,0,0,1])
                + - add (vectorises)
                a⁸ - logical AND with carrier
                ḟ0 - filter out zeros
                ¦ - sparse application...
                ç - ...to indices: call last Link (the helper function) as a dyad
                Ḥ - ...do: double (e.g. 'x' -> 'xx')





                share|improve this answer















                Jelly, 28 bytes



                ẹⱮŒp<ƝẠ$ƇṪ
                nƝ+çṬ¥a⁸ḟ0Ḥç¦ð¹ç?


                A full program taking carrier and message as command line arguments which prints the result

                (For a non-packable message prints the unchanged carrier).



                Try it online! Or see the test-suite.



                How?



                ẹⱮŒp<ƝẠ$ƇṪ - Link 1, helper function to find the indices to double: carrier, message
                - e.g. "programming", "rom"
                Ɱ - map across message with:
                ẹ - indices of [[2,5], [3], [7,8]]
                Œp - Cartesian product [[2,3,7],[2,3,8],[5,3,7],[5,3,8]]
                Ƈ - filter keep if:
                $ - last two links as a monad:
                Ɲ - for neighbours:
                < - less than? [1,1] [1,1] [0,1] [0,1]
                Ạ - all truthy? 1 1 0 0
                - [[2,3,7],[2,3,8]]
                Ṫ - tail (if empty yields 0) [2,3,8]

                nƝ+çṬ¥a⁸ḟ0Ḥç¦ð¹ç? - Main Link: carrier, message
                ? - if...
                ç - ...condition: last Link (the helper function) as a dyad
                ð - ...then: perform the dyadic chain to the left (described below)
                ¹ - ...else: do nothing (yields carrier)
                - (the then clause:)
                Ɲ - for neighbours in the carrier
                n - not equal?
                ¥ - last two links as a dyad:
                ç - call last Link (the helper function) as a dyad
                Ṭ - untruth (e.g. [2,5] -> [0,1,0,0,1])
                + - add (vectorises)
                a⁸ - logical AND with carrier
                ḟ0 - filter out zeros
                ¦ - sparse application...
                ç - ...to indices: call last Link (the helper function) as a dyad
                Ḥ - ...do: double (e.g. 'x' -> 'xx')






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Dec 8 at 21:37

























                answered Dec 8 at 21:07









                Jonathan Allan

                50.6k534165




                50.6k534165























                    3














                    JavaScript (ES6), 71 bytes



                    Takes input as (message)(carrier).





                    s=>g=([c,...C],p)=>c?(c==s[0]?(s=s.slice(1),c)+c:p==c?'':c)+g(C,c):s&&X


                    Try it online!





                    Alternate version, 66 bytes



                    If we can take the message as an array of characters:



                    s=>g=([c,...C],p)=>c?(c==s[0]?s.shift()+c:p==c?'':c)+g(C,c):s+s&&X


                    Try it online!





                    Edit: Thanks to @tsh for noticing that I forgot to remove some code when switching from non-recursive to recursive versions.






                    share|improve this answer























                    • You could remove p= since p is passed by a parameter.
                      – tsh
                      Dec 11 at 3:10












                    • @tsh Oops. It's some residual code from the previous, non-recursive versions that I forgot to remove. Thank you!
                      – Arnauld
                      Dec 11 at 11:59
















                    3














                    JavaScript (ES6), 71 bytes



                    Takes input as (message)(carrier).





                    s=>g=([c,...C],p)=>c?(c==s[0]?(s=s.slice(1),c)+c:p==c?'':c)+g(C,c):s&&X


                    Try it online!





                    Alternate version, 66 bytes



                    If we can take the message as an array of characters:



                    s=>g=([c,...C],p)=>c?(c==s[0]?s.shift()+c:p==c?'':c)+g(C,c):s+s&&X


                    Try it online!





                    Edit: Thanks to @tsh for noticing that I forgot to remove some code when switching from non-recursive to recursive versions.






                    share|improve this answer























                    • You could remove p= since p is passed by a parameter.
                      – tsh
                      Dec 11 at 3:10












                    • @tsh Oops. It's some residual code from the previous, non-recursive versions that I forgot to remove. Thank you!
                      – Arnauld
                      Dec 11 at 11:59














                    3












                    3








                    3






                    JavaScript (ES6), 71 bytes



                    Takes input as (message)(carrier).





                    s=>g=([c,...C],p)=>c?(c==s[0]?(s=s.slice(1),c)+c:p==c?'':c)+g(C,c):s&&X


                    Try it online!





                    Alternate version, 66 bytes



                    If we can take the message as an array of characters:



                    s=>g=([c,...C],p)=>c?(c==s[0]?s.shift()+c:p==c?'':c)+g(C,c):s+s&&X


                    Try it online!





                    Edit: Thanks to @tsh for noticing that I forgot to remove some code when switching from non-recursive to recursive versions.






                    share|improve this answer














                    JavaScript (ES6), 71 bytes



                    Takes input as (message)(carrier).





                    s=>g=([c,...C],p)=>c?(c==s[0]?(s=s.slice(1),c)+c:p==c?'':c)+g(C,c):s&&X


                    Try it online!





                    Alternate version, 66 bytes



                    If we can take the message as an array of characters:



                    s=>g=([c,...C],p)=>c?(c==s[0]?s.shift()+c:p==c?'':c)+g(C,c):s+s&&X


                    Try it online!





                    Edit: Thanks to @tsh for noticing that I forgot to remove some code when switching from non-recursive to recursive versions.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Dec 11 at 11:56

























                    answered Dec 8 at 21:47









                    Arnauld

                    72.3k689303




                    72.3k689303












                    • You could remove p= since p is passed by a parameter.
                      – tsh
                      Dec 11 at 3:10












                    • @tsh Oops. It's some residual code from the previous, non-recursive versions that I forgot to remove. Thank you!
                      – Arnauld
                      Dec 11 at 11:59


















                    • You could remove p= since p is passed by a parameter.
                      – tsh
                      Dec 11 at 3:10












                    • @tsh Oops. It's some residual code from the previous, non-recursive versions that I forgot to remove. Thank you!
                      – Arnauld
                      Dec 11 at 11:59
















                    You could remove p= since p is passed by a parameter.
                    – tsh
                    Dec 11 at 3:10






                    You could remove p= since p is passed by a parameter.
                    – tsh
                    Dec 11 at 3:10














                    @tsh Oops. It's some residual code from the previous, non-recursive versions that I forgot to remove. Thank you!
                    – Arnauld
                    Dec 11 at 11:59




                    @tsh Oops. It's some residual code from the previous, non-recursive versions that I forgot to remove. Thank you!
                    – Arnauld
                    Dec 11 at 11:59











                    2














                    Haskell, 124 121 107 101 97 95 90 bytes



                    (#).(++"ü")
                    "ü"#=
                    p@(m:n)#e@(c:d)|m/=c=c:p#snd(span(==c)d)|m==n!!0=m:m:n#d|1<2=m:n#e


                    Raises the "Non-exhaustive patterns" exception if the carrier does not contain the message.



                    Try it online!



                    Edit: -5 bytes thanks to @Laikoni.






                    share|improve this answer























                    • I think switching the cases allows you to drop m==c: Try it online!
                      – Laikoni
                      Dec 9 at 9:43


















                    2














                    Haskell, 124 121 107 101 97 95 90 bytes



                    (#).(++"ü")
                    "ü"#=
                    p@(m:n)#e@(c:d)|m/=c=c:p#snd(span(==c)d)|m==n!!0=m:m:n#d|1<2=m:n#e


                    Raises the "Non-exhaustive patterns" exception if the carrier does not contain the message.



                    Try it online!



                    Edit: -5 bytes thanks to @Laikoni.






                    share|improve this answer























                    • I think switching the cases allows you to drop m==c: Try it online!
                      – Laikoni
                      Dec 9 at 9:43
















                    2












                    2








                    2






                    Haskell, 124 121 107 101 97 95 90 bytes



                    (#).(++"ü")
                    "ü"#=
                    p@(m:n)#e@(c:d)|m/=c=c:p#snd(span(==c)d)|m==n!!0=m:m:n#d|1<2=m:n#e


                    Raises the "Non-exhaustive patterns" exception if the carrier does not contain the message.



                    Try it online!



                    Edit: -5 bytes thanks to @Laikoni.






                    share|improve this answer














                    Haskell, 124 121 107 101 97 95 90 bytes



                    (#).(++"ü")
                    "ü"#=
                    p@(m:n)#e@(c:d)|m/=c=c:p#snd(span(==c)d)|m==n!!0=m:m:n#d|1<2=m:n#e


                    Raises the "Non-exhaustive patterns" exception if the carrier does not contain the message.



                    Try it online!



                    Edit: -5 bytes thanks to @Laikoni.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Dec 9 at 16:03

























                    answered Dec 8 at 22:08









                    nimi

                    31.2k32085




                    31.2k32085












                    • I think switching the cases allows you to drop m==c: Try it online!
                      – Laikoni
                      Dec 9 at 9:43




















                    • I think switching the cases allows you to drop m==c: Try it online!
                      – Laikoni
                      Dec 9 at 9:43


















                    I think switching the cases allows you to drop m==c: Try it online!
                    – Laikoni
                    Dec 9 at 9:43






                    I think switching the cases allows you to drop m==c: Try it online!
                    – Laikoni
                    Dec 9 at 9:43













                    1















                    Retina 0.8.2, 67 bytes



                    +`(.)(1*)1*(.*¶)(?(1)(1(2)))(.*)$(?!¶)
                    $1$4$5¶$3$6
                    M!s`.*¶$




                    Try it online! Takes the carrier on the first line and the message on the second line. Explanation:



                    +`(.)(1*)1*(.*¶)(?(1)(1(2)))(.*)$(?!¶)
                    $1$4$5¶$3$6


                    Process runs of 1 or more identical characters of the carrier. If there is also a run of 1 or more of the same characters in the message, then append the shorter of the two runs to the output in duplicate, otherwise append a single character of the carrier to the output. Each run of output characters is terminated with a newline to distinguish it from the input. The (?!¶) at the end prevents the regex from thinking the carrier is the message once the message is exhausted, as normally $ is allowed to match where ¶$ would match.



                    M!s`.*¶$


                    Delete everything if the message wasn't completely encoded.







                    Remove the newlines from the output.






                    share|improve this answer























                    • I think it does not pass the second to last test case (which, to be fair, I did not have in the initial post).
                      – jkpate
                      Dec 8 at 20:14










                    • @jkpate Thanks for pointing that out; I've had to rewrite my approach slightly.
                      – Neil
                      Dec 8 at 21:32
















                    1















                    Retina 0.8.2, 67 bytes



                    +`(.)(1*)1*(.*¶)(?(1)(1(2)))(.*)$(?!¶)
                    $1$4$5¶$3$6
                    M!s`.*¶$




                    Try it online! Takes the carrier on the first line and the message on the second line. Explanation:



                    +`(.)(1*)1*(.*¶)(?(1)(1(2)))(.*)$(?!¶)
                    $1$4$5¶$3$6


                    Process runs of 1 or more identical characters of the carrier. If there is also a run of 1 or more of the same characters in the message, then append the shorter of the two runs to the output in duplicate, otherwise append a single character of the carrier to the output. Each run of output characters is terminated with a newline to distinguish it from the input. The (?!¶) at the end prevents the regex from thinking the carrier is the message once the message is exhausted, as normally $ is allowed to match where ¶$ would match.



                    M!s`.*¶$


                    Delete everything if the message wasn't completely encoded.







                    Remove the newlines from the output.






                    share|improve this answer























                    • I think it does not pass the second to last test case (which, to be fair, I did not have in the initial post).
                      – jkpate
                      Dec 8 at 20:14










                    • @jkpate Thanks for pointing that out; I've had to rewrite my approach slightly.
                      – Neil
                      Dec 8 at 21:32














                    1












                    1








                    1







                    Retina 0.8.2, 67 bytes



                    +`(.)(1*)1*(.*¶)(?(1)(1(2)))(.*)$(?!¶)
                    $1$4$5¶$3$6
                    M!s`.*¶$




                    Try it online! Takes the carrier on the first line and the message on the second line. Explanation:



                    +`(.)(1*)1*(.*¶)(?(1)(1(2)))(.*)$(?!¶)
                    $1$4$5¶$3$6


                    Process runs of 1 or more identical characters of the carrier. If there is also a run of 1 or more of the same characters in the message, then append the shorter of the two runs to the output in duplicate, otherwise append a single character of the carrier to the output. Each run of output characters is terminated with a newline to distinguish it from the input. The (?!¶) at the end prevents the regex from thinking the carrier is the message once the message is exhausted, as normally $ is allowed to match where ¶$ would match.



                    M!s`.*¶$


                    Delete everything if the message wasn't completely encoded.







                    Remove the newlines from the output.






                    share|improve this answer















                    Retina 0.8.2, 67 bytes



                    +`(.)(1*)1*(.*¶)(?(1)(1(2)))(.*)$(?!¶)
                    $1$4$5¶$3$6
                    M!s`.*¶$




                    Try it online! Takes the carrier on the first line and the message on the second line. Explanation:



                    +`(.)(1*)1*(.*¶)(?(1)(1(2)))(.*)$(?!¶)
                    $1$4$5¶$3$6


                    Process runs of 1 or more identical characters of the carrier. If there is also a run of 1 or more of the same characters in the message, then append the shorter of the two runs to the output in duplicate, otherwise append a single character of the carrier to the output. Each run of output characters is terminated with a newline to distinguish it from the input. The (?!¶) at the end prevents the regex from thinking the carrier is the message once the message is exhausted, as normally $ is allowed to match where ¶$ would match.



                    M!s`.*¶$


                    Delete everything if the message wasn't completely encoded.







                    Remove the newlines from the output.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Dec 8 at 21:31

























                    answered Dec 8 at 20:10









                    Neil

                    79.2k744177




                    79.2k744177












                    • I think it does not pass the second to last test case (which, to be fair, I did not have in the initial post).
                      – jkpate
                      Dec 8 at 20:14










                    • @jkpate Thanks for pointing that out; I've had to rewrite my approach slightly.
                      – Neil
                      Dec 8 at 21:32


















                    • I think it does not pass the second to last test case (which, to be fair, I did not have in the initial post).
                      – jkpate
                      Dec 8 at 20:14










                    • @jkpate Thanks for pointing that out; I've had to rewrite my approach slightly.
                      – Neil
                      Dec 8 at 21:32
















                    I think it does not pass the second to last test case (which, to be fair, I did not have in the initial post).
                    – jkpate
                    Dec 8 at 20:14




                    I think it does not pass the second to last test case (which, to be fair, I did not have in the initial post).
                    – jkpate
                    Dec 8 at 20:14












                    @jkpate Thanks for pointing that out; I've had to rewrite my approach slightly.
                    – Neil
                    Dec 8 at 21:32




                    @jkpate Thanks for pointing that out; I've had to rewrite my approach slightly.
                    – Neil
                    Dec 8 at 21:32











                    0















                    Clean, 118 bytes



                    import StdEnv,StdLib
                    $=
                    $[u:v]b#(_,w)=span((==)u)v
                    |b%(0,0)==[u]=[u,u: $if(v%(0,0)<>b%(1,1))w v(tl b)]=[u: $w b]


                    Try it online!



                    Takes the carrier first, then the message.



                    Errors with Run time error, rule '$;2' in module 'main' does not match if message won't fit.






                    share|improve this answer


























                      0















                      Clean, 118 bytes



                      import StdEnv,StdLib
                      $=
                      $[u:v]b#(_,w)=span((==)u)v
                      |b%(0,0)==[u]=[u,u: $if(v%(0,0)<>b%(1,1))w v(tl b)]=[u: $w b]


                      Try it online!



                      Takes the carrier first, then the message.



                      Errors with Run time error, rule '$;2' in module 'main' does not match if message won't fit.






                      share|improve this answer
























                        0












                        0








                        0







                        Clean, 118 bytes



                        import StdEnv,StdLib
                        $=
                        $[u:v]b#(_,w)=span((==)u)v
                        |b%(0,0)==[u]=[u,u: $if(v%(0,0)<>b%(1,1))w v(tl b)]=[u: $w b]


                        Try it online!



                        Takes the carrier first, then the message.



                        Errors with Run time error, rule '$;2' in module 'main' does not match if message won't fit.






                        share|improve this answer













                        Clean, 118 bytes



                        import StdEnv,StdLib
                        $=
                        $[u:v]b#(_,w)=span((==)u)v
                        |b%(0,0)==[u]=[u,u: $if(v%(0,0)<>b%(1,1))w v(tl b)]=[u: $w b]


                        Try it online!



                        Takes the carrier first, then the message.



                        Errors with Run time error, rule '$;2' in module 'main' does not match if message won't fit.







                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered Dec 10 at 4:38









                        Οurous

                        6,42311033




                        6,42311033























                            0















                            Ruby, 73 bytes





                            f=->m,c,b=p{x,*c=c;x ?(x==m[0]?x+m.shift: x==b ?'':x)+f[m,c,x]:m[0]?x:''}


                            Try it online!



                            Recursive function, takes inputs as array of characters.



                            For once I was hoping to make use of Ruby's built-in squeeze method that contracts consecutive runs of the same character to a single instance. But unfortunately, nope - the last two test cases screwed everything so badly, that I had to resort to a completely different approach, and this turned out to be basically a port of Arnauld's answer.






                            share|improve this answer


























                              0















                              Ruby, 73 bytes





                              f=->m,c,b=p{x,*c=c;x ?(x==m[0]?x+m.shift: x==b ?'':x)+f[m,c,x]:m[0]?x:''}


                              Try it online!



                              Recursive function, takes inputs as array of characters.



                              For once I was hoping to make use of Ruby's built-in squeeze method that contracts consecutive runs of the same character to a single instance. But unfortunately, nope - the last two test cases screwed everything so badly, that I had to resort to a completely different approach, and this turned out to be basically a port of Arnauld's answer.






                              share|improve this answer
























                                0












                                0








                                0







                                Ruby, 73 bytes





                                f=->m,c,b=p{x,*c=c;x ?(x==m[0]?x+m.shift: x==b ?'':x)+f[m,c,x]:m[0]?x:''}


                                Try it online!



                                Recursive function, takes inputs as array of characters.



                                For once I was hoping to make use of Ruby's built-in squeeze method that contracts consecutive runs of the same character to a single instance. But unfortunately, nope - the last two test cases screwed everything so badly, that I had to resort to a completely different approach, and this turned out to be basically a port of Arnauld's answer.






                                share|improve this answer













                                Ruby, 73 bytes





                                f=->m,c,b=p{x,*c=c;x ?(x==m[0]?x+m.shift: x==b ?'':x)+f[m,c,x]:m[0]?x:''}


                                Try it online!



                                Recursive function, takes inputs as array of characters.



                                For once I was hoping to make use of Ruby's built-in squeeze method that contracts consecutive runs of the same character to a single instance. But unfortunately, nope - the last two test cases screwed everything so badly, that I had to resort to a completely different approach, and this turned out to be basically a port of Arnauld's answer.







                                share|improve this answer












                                share|improve this answer



                                share|improve this answer










                                answered Dec 10 at 12:47









                                Kirill L.

                                3,6451318




                                3,6451318























                                    0














                                    Powershell, 134 bytes





                                    param($m,$c)$c-csplit"([$m])"|%{$i+=$o=$_-ceq$m[+$i]
                                    if($o-or$_-cne"`0$h"[-1]){$h+=($_-replace'(.)(?=1)')*($o+1)}}
                                    $h*!($i-$m.Length)


                                    The script returns the empty string if the carrier does not contain the message characters in the right order.



                                    Less golfed test script:



                                    $f = {

                                    param($message,$carrier)
                                    $carrier-csplit"([$message])"|%{ # split by chars of the message, chars itself included ()
                                    $offset=$_-ceq$message[+$i] # 0 or 1 if current substring is a current message char (case-sensitive equality)
                                    $i+=$offset # move to next message char if need it
                                    if($offset-or$_-cne"`0$h"[-1]){ # condition to remove redundant doubles after message char: arrrived -> arrived, ooo -> oo, etc
                                    # `0 to avoid exception error if $h is empty
                                    $h+=($_-replace'(.)(?=1)')*($offset+1) # accumulate a double message char or a single substring without inner doubles: arried -> arived, anna -> ana, etc
                                    }
                                    }
                                    $h*!($i-$message.Length) # repeat 0 or 1 times to return '' if the carrier does not contain the message characters in the right order

                                    }

                                    @(
                                    ,('hi' ,'has it arrived?' ,'hhas iit arived?', 'hhas it ariived?')
                                    ,('hi?' ,'has it arrived?' ,'hhas iit arived??', 'hhas it ariived??')
                                    ,('sir' ,'has it arrived?' ,'hass iit arrived?')
                                    ,('foo' ,'has it arrived?' ,'')
                                    ,('Car' ,'Cats are cool.' ,'CCaats arre col.')
                                    ,('car' ,'Cats are cool.' ,'')
                                    ,('Couch' ,'Couch' ,'CCoouucchh')
                                    ,('oo' ,'oooooooooo' ,'oooo')
                                    ,('o o' ,'oooo oooa' ,'oo ooa')
                                    ,('er' ,'error' ,'eerorr', 'eerror')
                                    ,('a+b' ,'anna+bob' ,'aana++bbob')
                                    ) | % {
                                    $message,$carrier,$expected = $_
                                    $result = &$f $message $carrier
                                    "$($result-in$expected): $result"
                                    }


                                    Output:





                                    True: hhas iit arived?
                                    True: hhas iit arived??
                                    True: hass iit arrived?
                                    True:
                                    True: CCaats arre col.
                                    True:
                                    True: CCoouucchh
                                    True: oooo
                                    True: oo ooa
                                    True: eerror
                                    True: aana++bbob





                                    share|improve this answer




























                                      0














                                      Powershell, 134 bytes





                                      param($m,$c)$c-csplit"([$m])"|%{$i+=$o=$_-ceq$m[+$i]
                                      if($o-or$_-cne"`0$h"[-1]){$h+=($_-replace'(.)(?=1)')*($o+1)}}
                                      $h*!($i-$m.Length)


                                      The script returns the empty string if the carrier does not contain the message characters in the right order.



                                      Less golfed test script:



                                      $f = {

                                      param($message,$carrier)
                                      $carrier-csplit"([$message])"|%{ # split by chars of the message, chars itself included ()
                                      $offset=$_-ceq$message[+$i] # 0 or 1 if current substring is a current message char (case-sensitive equality)
                                      $i+=$offset # move to next message char if need it
                                      if($offset-or$_-cne"`0$h"[-1]){ # condition to remove redundant doubles after message char: arrrived -> arrived, ooo -> oo, etc
                                      # `0 to avoid exception error if $h is empty
                                      $h+=($_-replace'(.)(?=1)')*($offset+1) # accumulate a double message char or a single substring without inner doubles: arried -> arived, anna -> ana, etc
                                      }
                                      }
                                      $h*!($i-$message.Length) # repeat 0 or 1 times to return '' if the carrier does not contain the message characters in the right order

                                      }

                                      @(
                                      ,('hi' ,'has it arrived?' ,'hhas iit arived?', 'hhas it ariived?')
                                      ,('hi?' ,'has it arrived?' ,'hhas iit arived??', 'hhas it ariived??')
                                      ,('sir' ,'has it arrived?' ,'hass iit arrived?')
                                      ,('foo' ,'has it arrived?' ,'')
                                      ,('Car' ,'Cats are cool.' ,'CCaats arre col.')
                                      ,('car' ,'Cats are cool.' ,'')
                                      ,('Couch' ,'Couch' ,'CCoouucchh')
                                      ,('oo' ,'oooooooooo' ,'oooo')
                                      ,('o o' ,'oooo oooa' ,'oo ooa')
                                      ,('er' ,'error' ,'eerorr', 'eerror')
                                      ,('a+b' ,'anna+bob' ,'aana++bbob')
                                      ) | % {
                                      $message,$carrier,$expected = $_
                                      $result = &$f $message $carrier
                                      "$($result-in$expected): $result"
                                      }


                                      Output:





                                      True: hhas iit arived?
                                      True: hhas iit arived??
                                      True: hass iit arrived?
                                      True:
                                      True: CCaats arre col.
                                      True:
                                      True: CCoouucchh
                                      True: oooo
                                      True: oo ooa
                                      True: eerror
                                      True: aana++bbob





                                      share|improve this answer


























                                        0












                                        0








                                        0






                                        Powershell, 134 bytes





                                        param($m,$c)$c-csplit"([$m])"|%{$i+=$o=$_-ceq$m[+$i]
                                        if($o-or$_-cne"`0$h"[-1]){$h+=($_-replace'(.)(?=1)')*($o+1)}}
                                        $h*!($i-$m.Length)


                                        The script returns the empty string if the carrier does not contain the message characters in the right order.



                                        Less golfed test script:



                                        $f = {

                                        param($message,$carrier)
                                        $carrier-csplit"([$message])"|%{ # split by chars of the message, chars itself included ()
                                        $offset=$_-ceq$message[+$i] # 0 or 1 if current substring is a current message char (case-sensitive equality)
                                        $i+=$offset # move to next message char if need it
                                        if($offset-or$_-cne"`0$h"[-1]){ # condition to remove redundant doubles after message char: arrrived -> arrived, ooo -> oo, etc
                                        # `0 to avoid exception error if $h is empty
                                        $h+=($_-replace'(.)(?=1)')*($offset+1) # accumulate a double message char or a single substring without inner doubles: arried -> arived, anna -> ana, etc
                                        }
                                        }
                                        $h*!($i-$message.Length) # repeat 0 or 1 times to return '' if the carrier does not contain the message characters in the right order

                                        }

                                        @(
                                        ,('hi' ,'has it arrived?' ,'hhas iit arived?', 'hhas it ariived?')
                                        ,('hi?' ,'has it arrived?' ,'hhas iit arived??', 'hhas it ariived??')
                                        ,('sir' ,'has it arrived?' ,'hass iit arrived?')
                                        ,('foo' ,'has it arrived?' ,'')
                                        ,('Car' ,'Cats are cool.' ,'CCaats arre col.')
                                        ,('car' ,'Cats are cool.' ,'')
                                        ,('Couch' ,'Couch' ,'CCoouucchh')
                                        ,('oo' ,'oooooooooo' ,'oooo')
                                        ,('o o' ,'oooo oooa' ,'oo ooa')
                                        ,('er' ,'error' ,'eerorr', 'eerror')
                                        ,('a+b' ,'anna+bob' ,'aana++bbob')
                                        ) | % {
                                        $message,$carrier,$expected = $_
                                        $result = &$f $message $carrier
                                        "$($result-in$expected): $result"
                                        }


                                        Output:





                                        True: hhas iit arived?
                                        True: hhas iit arived??
                                        True: hass iit arrived?
                                        True:
                                        True: CCaats arre col.
                                        True:
                                        True: CCoouucchh
                                        True: oooo
                                        True: oo ooa
                                        True: eerror
                                        True: aana++bbob





                                        share|improve this answer














                                        Powershell, 134 bytes





                                        param($m,$c)$c-csplit"([$m])"|%{$i+=$o=$_-ceq$m[+$i]
                                        if($o-or$_-cne"`0$h"[-1]){$h+=($_-replace'(.)(?=1)')*($o+1)}}
                                        $h*!($i-$m.Length)


                                        The script returns the empty string if the carrier does not contain the message characters in the right order.



                                        Less golfed test script:



                                        $f = {

                                        param($message,$carrier)
                                        $carrier-csplit"([$message])"|%{ # split by chars of the message, chars itself included ()
                                        $offset=$_-ceq$message[+$i] # 0 or 1 if current substring is a current message char (case-sensitive equality)
                                        $i+=$offset # move to next message char if need it
                                        if($offset-or$_-cne"`0$h"[-1]){ # condition to remove redundant doubles after message char: arrrived -> arrived, ooo -> oo, etc
                                        # `0 to avoid exception error if $h is empty
                                        $h+=($_-replace'(.)(?=1)')*($offset+1) # accumulate a double message char or a single substring without inner doubles: arried -> arived, anna -> ana, etc
                                        }
                                        }
                                        $h*!($i-$message.Length) # repeat 0 or 1 times to return '' if the carrier does not contain the message characters in the right order

                                        }

                                        @(
                                        ,('hi' ,'has it arrived?' ,'hhas iit arived?', 'hhas it ariived?')
                                        ,('hi?' ,'has it arrived?' ,'hhas iit arived??', 'hhas it ariived??')
                                        ,('sir' ,'has it arrived?' ,'hass iit arrived?')
                                        ,('foo' ,'has it arrived?' ,'')
                                        ,('Car' ,'Cats are cool.' ,'CCaats arre col.')
                                        ,('car' ,'Cats are cool.' ,'')
                                        ,('Couch' ,'Couch' ,'CCoouucchh')
                                        ,('oo' ,'oooooooooo' ,'oooo')
                                        ,('o o' ,'oooo oooa' ,'oo ooa')
                                        ,('er' ,'error' ,'eerorr', 'eerror')
                                        ,('a+b' ,'anna+bob' ,'aana++bbob')
                                        ) | % {
                                        $message,$carrier,$expected = $_
                                        $result = &$f $message $carrier
                                        "$($result-in$expected): $result"
                                        }


                                        Output:





                                        True: hhas iit arived?
                                        True: hhas iit arived??
                                        True: hass iit arrived?
                                        True:
                                        True: CCaats arre col.
                                        True:
                                        True: CCoouucchh
                                        True: oooo
                                        True: oo ooa
                                        True: eerror
                                        True: aana++bbob






                                        share|improve this answer














                                        share|improve this answer



                                        share|improve this answer








                                        edited Dec 10 at 22:17

























                                        answered Dec 10 at 22:00









                                        mazzy

                                        2,0551315




                                        2,0551315























                                            0















                                            C (gcc), 69+12 = 81 bytes





                                            g(char*m,char*_){for(;*_;++_)*m-*_?_[-1]-*_&&p*_):p p*m++));*m&&0/0;}


                                            Compile with (12 bytes)



                                            -Dp=putchar(


                                            Try it online!



                                            g(char*m,char*_){
                                            for(;*_;++_) //step through _
                                            *m-*_? //check if character should be encoded
                                            _[-1]-*_&& //no? skip duplicates
                                            p*_) // print non-duplicates
                                            :p p*m++)); //print encoded character twice
                                            *m&&0/0; //if m is not fully encoded, exit via Floating point exception
                                            }





                                            share|improve this answer








                                            New contributor




                                            attinat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                            Check out our Code of Conduct.























                                              0















                                              C (gcc), 69+12 = 81 bytes





                                              g(char*m,char*_){for(;*_;++_)*m-*_?_[-1]-*_&&p*_):p p*m++));*m&&0/0;}


                                              Compile with (12 bytes)



                                              -Dp=putchar(


                                              Try it online!



                                              g(char*m,char*_){
                                              for(;*_;++_) //step through _
                                              *m-*_? //check if character should be encoded
                                              _[-1]-*_&& //no? skip duplicates
                                              p*_) // print non-duplicates
                                              :p p*m++)); //print encoded character twice
                                              *m&&0/0; //if m is not fully encoded, exit via Floating point exception
                                              }





                                              share|improve this answer








                                              New contributor




                                              attinat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.





















                                                0












                                                0








                                                0







                                                C (gcc), 69+12 = 81 bytes





                                                g(char*m,char*_){for(;*_;++_)*m-*_?_[-1]-*_&&p*_):p p*m++));*m&&0/0;}


                                                Compile with (12 bytes)



                                                -Dp=putchar(


                                                Try it online!



                                                g(char*m,char*_){
                                                for(;*_;++_) //step through _
                                                *m-*_? //check if character should be encoded
                                                _[-1]-*_&& //no? skip duplicates
                                                p*_) // print non-duplicates
                                                :p p*m++)); //print encoded character twice
                                                *m&&0/0; //if m is not fully encoded, exit via Floating point exception
                                                }





                                                share|improve this answer








                                                New contributor




                                                attinat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.










                                                C (gcc), 69+12 = 81 bytes





                                                g(char*m,char*_){for(;*_;++_)*m-*_?_[-1]-*_&&p*_):p p*m++));*m&&0/0;}


                                                Compile with (12 bytes)



                                                -Dp=putchar(


                                                Try it online!



                                                g(char*m,char*_){
                                                for(;*_;++_) //step through _
                                                *m-*_? //check if character should be encoded
                                                _[-1]-*_&& //no? skip duplicates
                                                p*_) // print non-duplicates
                                                :p p*m++)); //print encoded character twice
                                                *m&&0/0; //if m is not fully encoded, exit via Floating point exception
                                                }






                                                share|improve this answer








                                                New contributor




                                                attinat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.









                                                share|improve this answer



                                                share|improve this answer






                                                New contributor




                                                attinat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.









                                                answered Dec 18 at 9:01









                                                attinat

                                                914




                                                914




                                                New contributor




                                                attinat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.





                                                New contributor





                                                attinat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.






                                                attinat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.






























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