Doubled-letter steganography











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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.










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  • 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















up vote
18
down vote

favorite
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.










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  • 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













up vote
18
down vote

favorite
1









up vote
18
down vote

favorite
1






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.










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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






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edited Dec 8 at 20:09









Jonathan Allan

50.6k534165




50.6k534165






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asked Dec 8 at 19:23









jkpate

1914




1914




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  • 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










7 Answers
7






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up vote
4
down vote














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






























    up vote
    3
    down vote













    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


















    up vote
    2
    down vote













    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




















    up vote
    1
    down vote














    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


















    up vote
    0
    down vote














    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




























      up vote
      0
      down vote














      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




























        up vote
        0
        down vote













        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





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          7 Answers
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          7 Answers
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          up vote
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          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



























            up vote
            4
            down vote














            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

























              up vote
              4
              down vote










              up vote
              4
              down vote










              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






















                  up vote
                  3
                  down vote













                  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















                  up vote
                  3
                  down vote













                  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













                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  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

                  71.6k688299




                  71.6k688299












                  • 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










                  up vote
                  2
                  down vote













                  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

















                  up vote
                  2
                  down vote













                  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















                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  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

                  31.1k31985




                  31.1k31985












                  • 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












                  up vote
                  1
                  down vote














                  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















                  up vote
                  1
                  down vote














                  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













                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote










                  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

                  78.9k744175




                  78.9k744175












                  • 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










                  up vote
                  0
                  down vote














                  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

























                    up vote
                    0
                    down vote














                    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























                      up vote
                      0
                      down vote










                      up vote
                      0
                      down vote










                      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

                      6,28811032




                      6,28811032






















                          up vote
                          0
                          down vote














                          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

























                            up vote
                            0
                            down vote














                            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























                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote










                              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.

                              3,5051218




                              3,5051218






















                                  up vote
                                  0
                                  down vote













                                  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



























                                    up vote
                                    0
                                    down vote













                                    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

























                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      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

                                      2,0151314




                                      2,0151314






















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