Meta-bowling Golf












19












$begingroup$


Your task is to create the shortest program (A) that outputs another program (B) with the most unique bytes, which in turn outputs the original program (A). Essentially, you are golfing one half of a pair of ouroboros programs (a.k.a. periodic iterating quine) and bowling the other half. A and B may be in different languages.



Rules and Scoring



Your final score is <number of unique bytes in B> / <number of bytes in A>. Highest score wins. It should be noted that the theoretical maximum score is 256.




  • Program A must be at least one byte long

  • Program B cannot consist entirely of no-ops, i.e. at least one character must affect the output in some way.


  • Standard rules of quines apply to both programs. Notably, error quines are not allowed in either case.


For the sake of answer format consistency, begin your answer with something like this:



# <Language for A> & <Language for B>, Score: <B score> / <A score> = <combined score>





code-golf code-challenge quine code-bowling







share|improve this question









$endgroup$












  • $begingroup$
    In order for this to be clear I think you are going to have to clarify the second bullet point a little bit. What does it mean for a character to be a no-op exactly?
    $endgroup$
    – Sriotchilism O'Zaic
    Feb 14 at 17:48










  • $begingroup$
    By unique bytes, do you mean shared bytes, or similarity between the two?
    $endgroup$
    – KrystosTheOverlord
    Feb 14 at 18:50










  • $begingroup$
    @KrystosTheOverlord, it would be simply within program B, with no relation to program A. For example, aaaaa has one unique byte and abcde has 5 unique bytes.
    $endgroup$
    – Beefster
    Feb 14 at 18:52






  • 1




    $begingroup$
    I'm just waiting for the programs where program A is in unary or lenguage
    $endgroup$
    – Embodiment of Ignorance
    Feb 14 at 22:01






  • 1




    $begingroup$
    @EmbodimentofIgnorance Program A is scored in terms of number of bytes, not unique bytes. Lenguage/Unary would score horribly
    $endgroup$
    – Jo King
    Feb 15 at 0:48


















19












$begingroup$


Your task is to create the shortest program (A) that outputs another program (B) with the most unique bytes, which in turn outputs the original program (A). Essentially, you are golfing one half of a pair of ouroboros programs (a.k.a. periodic iterating quine) and bowling the other half. A and B may be in different languages.



Rules and Scoring



Your final score is <number of unique bytes in B> / <number of bytes in A>. Highest score wins. It should be noted that the theoretical maximum score is 256.




  • Program A must be at least one byte long

  • Program B cannot consist entirely of no-ops, i.e. at least one character must affect the output in some way.


  • Standard rules of quines apply to both programs. Notably, error quines are not allowed in either case.


For the sake of answer format consistency, begin your answer with something like this:



# <Language for A> & <Language for B>, Score: <B score> / <A score> = <combined score>





code-golf code-challenge quine code-bowling







share|improve this question









$endgroup$












  • $begingroup$
    In order for this to be clear I think you are going to have to clarify the second bullet point a little bit. What does it mean for a character to be a no-op exactly?
    $endgroup$
    – Sriotchilism O'Zaic
    Feb 14 at 17:48










  • $begingroup$
    By unique bytes, do you mean shared bytes, or similarity between the two?
    $endgroup$
    – KrystosTheOverlord
    Feb 14 at 18:50










  • $begingroup$
    @KrystosTheOverlord, it would be simply within program B, with no relation to program A. For example, aaaaa has one unique byte and abcde has 5 unique bytes.
    $endgroup$
    – Beefster
    Feb 14 at 18:52






  • 1




    $begingroup$
    I'm just waiting for the programs where program A is in unary or lenguage
    $endgroup$
    – Embodiment of Ignorance
    Feb 14 at 22:01






  • 1




    $begingroup$
    @EmbodimentofIgnorance Program A is scored in terms of number of bytes, not unique bytes. Lenguage/Unary would score horribly
    $endgroup$
    – Jo King
    Feb 15 at 0:48
















19












19








19


1



$begingroup$


Your task is to create the shortest program (A) that outputs another program (B) with the most unique bytes, which in turn outputs the original program (A). Essentially, you are golfing one half of a pair of ouroboros programs (a.k.a. periodic iterating quine) and bowling the other half. A and B may be in different languages.



Rules and Scoring



Your final score is <number of unique bytes in B> / <number of bytes in A>. Highest score wins. It should be noted that the theoretical maximum score is 256.




  • Program A must be at least one byte long

  • Program B cannot consist entirely of no-ops, i.e. at least one character must affect the output in some way.


  • Standard rules of quines apply to both programs. Notably, error quines are not allowed in either case.


For the sake of answer format consistency, begin your answer with something like this:



# <Language for A> & <Language for B>, Score: <B score> / <A score> = <combined score>





code-golf code-challenge quine code-bowling







share|improve this question









$endgroup$




Your task is to create the shortest program (A) that outputs another program (B) with the most unique bytes, which in turn outputs the original program (A). Essentially, you are golfing one half of a pair of ouroboros programs (a.k.a. periodic iterating quine) and bowling the other half. A and B may be in different languages.



Rules and Scoring



Your final score is <number of unique bytes in B> / <number of bytes in A>. Highest score wins. It should be noted that the theoretical maximum score is 256.




  • Program A must be at least one byte long

  • Program B cannot consist entirely of no-ops, i.e. at least one character must affect the output in some way.


  • Standard rules of quines apply to both programs. Notably, error quines are not allowed in either case.


For the sake of answer format consistency, begin your answer with something like this:



# <Language for A> & <Language for B>, Score: <B score> / <A score> = <combined score>





code-golf code-challenge quine code-bowling




code-golf code-challenge quine code-bowling






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Feb 14 at 17:37









BeefsterBeefster

1,647729




1,647729












  • $begingroup$
    In order for this to be clear I think you are going to have to clarify the second bullet point a little bit. What does it mean for a character to be a no-op exactly?
    $endgroup$
    – Sriotchilism O'Zaic
    Feb 14 at 17:48










  • $begingroup$
    By unique bytes, do you mean shared bytes, or similarity between the two?
    $endgroup$
    – KrystosTheOverlord
    Feb 14 at 18:50










  • $begingroup$
    @KrystosTheOverlord, it would be simply within program B, with no relation to program A. For example, aaaaa has one unique byte and abcde has 5 unique bytes.
    $endgroup$
    – Beefster
    Feb 14 at 18:52






  • 1




    $begingroup$
    I'm just waiting for the programs where program A is in unary or lenguage
    $endgroup$
    – Embodiment of Ignorance
    Feb 14 at 22:01






  • 1




    $begingroup$
    @EmbodimentofIgnorance Program A is scored in terms of number of bytes, not unique bytes. Lenguage/Unary would score horribly
    $endgroup$
    – Jo King
    Feb 15 at 0:48




















  • $begingroup$
    In order for this to be clear I think you are going to have to clarify the second bullet point a little bit. What does it mean for a character to be a no-op exactly?
    $endgroup$
    – Sriotchilism O'Zaic
    Feb 14 at 17:48










  • $begingroup$
    By unique bytes, do you mean shared bytes, or similarity between the two?
    $endgroup$
    – KrystosTheOverlord
    Feb 14 at 18:50










  • $begingroup$
    @KrystosTheOverlord, it would be simply within program B, with no relation to program A. For example, aaaaa has one unique byte and abcde has 5 unique bytes.
    $endgroup$
    – Beefster
    Feb 14 at 18:52






  • 1




    $begingroup$
    I'm just waiting for the programs where program A is in unary or lenguage
    $endgroup$
    – Embodiment of Ignorance
    Feb 14 at 22:01






  • 1




    $begingroup$
    @EmbodimentofIgnorance Program A is scored in terms of number of bytes, not unique bytes. Lenguage/Unary would score horribly
    $endgroup$
    – Jo King
    Feb 15 at 0:48


















$begingroup$
In order for this to be clear I think you are going to have to clarify the second bullet point a little bit. What does it mean for a character to be a no-op exactly?
$endgroup$
– Sriotchilism O'Zaic
Feb 14 at 17:48




$begingroup$
In order for this to be clear I think you are going to have to clarify the second bullet point a little bit. What does it mean for a character to be a no-op exactly?
$endgroup$
– Sriotchilism O'Zaic
Feb 14 at 17:48












$begingroup$
By unique bytes, do you mean shared bytes, or similarity between the two?
$endgroup$
– KrystosTheOverlord
Feb 14 at 18:50




$begingroup$
By unique bytes, do you mean shared bytes, or similarity between the two?
$endgroup$
– KrystosTheOverlord
Feb 14 at 18:50












$begingroup$
@KrystosTheOverlord, it would be simply within program B, with no relation to program A. For example, aaaaa has one unique byte and abcde has 5 unique bytes.
$endgroup$
– Beefster
Feb 14 at 18:52




$begingroup$
@KrystosTheOverlord, it would be simply within program B, with no relation to program A. For example, aaaaa has one unique byte and abcde has 5 unique bytes.
$endgroup$
– Beefster
Feb 14 at 18:52




1




1




$begingroup$
I'm just waiting for the programs where program A is in unary or lenguage
$endgroup$
– Embodiment of Ignorance
Feb 14 at 22:01




$begingroup$
I'm just waiting for the programs where program A is in unary or lenguage
$endgroup$
– Embodiment of Ignorance
Feb 14 at 22:01




1




1




$begingroup$
@EmbodimentofIgnorance Program A is scored in terms of number of bytes, not unique bytes. Lenguage/Unary would score horribly
$endgroup$
– Jo King
Feb 15 at 0:48






$begingroup$
@EmbodimentofIgnorance Program A is scored in terms of number of bytes, not unique bytes. Lenguage/Unary would score horribly
$endgroup$
– Jo King
Feb 15 at 0:48












3 Answers
3






active

oldest

votes


















11












$begingroup$


Japt & Japt, Score: 255 / 38 = 6.71



Program A:



"iQ ²¯24

G²õ@i'#+Xd"iQ ²¯24

G²õ@i'#+Xd


Program B is over 8kB long, so long that the link breaks, so I won't paste the whole thing. Here's a sample:



#þ"iQ ²¯24

G²õ@i'#+Xd"iQ ²,#ÿ"iQ ²¯24

G²õ@i'#+Xd"iQ ²,#Ā"iQ ²¯24

G²õ@i'#+Xd"iQ ²


I couldn't find a way to get a NUL byte to work, which is why program B only has 255 unique characters. Program B essentially consists of 255 copies of a single program, where a single irrelevant byte is changed each time, and the first 254 executions are ignored.



For the explanation, I'll start with this simplified version of A so that the resulting B is easier to discuss.



"iQ ²¯23

3õ@i'#+Xd"iQ ²¯23

3õ@i'#+Xd


This program is based on the basic payload-capable Japt quine. The string at the start contains a duplicate of the rest of the program, iQ ² inserts a quote and duplicates to create a string representation of the entire program, and then ¯23 trims off itself and everything after it. The resulting string is a program that outputs Program A:



"iQ ²¯23

3õ@i'#+Xd"iQ ²


I will refer to this string as U.



The last line of A duplicates U a bunch of times with a small change each time. Specifically, for each number X in the range [1...3] it outputs "#c" + U where c is the character with charcode X. The default behavior of Japt is to output those strings with no quotation marks and separated by commas, so this is the output of our Simplified A:



#"iQ ²¯23

3õ@i'#+Xd"iQ ²,#"iQ ²¯23

3õ@i'#+Xd"iQ ²,#"iQ ²¯23

3õ@i'#+Xd"iQ ²


We'll call this Simplified B.



Simplified B has a simple structure, alternating between #c and U. Fortunately for this answer, each #c and U is treated as separated by a comma, and in this situation the behavior of that is everything except the very last U has no effect on the output. The only portion of Simplified B which affects the output is this:



"iQ ²¯23

3õ@i'#+Xd"iQ ²


Which is identical to U that we already know outputs Simplified A.



The only difference between Simplified A and Program A is that instead of generating copies for the range [1...3] the real program generates copies for the range [1...256]. That results in 256 versions of #c each of which has a different character, though the last version "Ā" is a multi-byte character so it doesn't add any unique bytes, but everything except the last U is still ignored.






share|improve this answer











$endgroup$













  • $begingroup$
    Very nice :) Would this work for 38 bytes?
    $endgroup$
    – Shaggy
    Feb 14 at 19:12










  • $begingroup$
    @Shaggy The 27 needs to be tweaked whenever bytes are saved, but other than that it seems like it works.
    $endgroup$
    – Kamil Drakari
    Feb 14 at 19:15










  • $begingroup$
    Wait, aren't there now 256 unique characters / 257 bytes in B? If multi-byte characters aren't allowed (I've asked for clarification) then you'll need to replace G² with #ÿ. Of course, if they are then you can score much higher here.
    $endgroup$
    – Shaggy
    Feb 14 at 19:45






  • 1




    $begingroup$
    @Shaggy The scoring is based on number of unique bytes in B, and there are only 256 possible bytes. As far as I'm aware, even the new version of B does not contain the Null byte.
    $endgroup$
    – Kamil Drakari
    Feb 14 at 19:47










  • $begingroup$
    Ah, right; I was counting the bytes of unique characters, rather than the bytes themselves, which would have been silly as it would allow scores to converge on infinity.
    $endgroup$
    – Shaggy
    Feb 14 at 22:18



















4












$begingroup$

Program A, Gol><>, 256/20 bytes = 12.8



"44XFL|r2ssl3%Q4s]|H


Try it online!



Program B, Gol><>



"44XFL|r2ssl3%Q4s]|H	



 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ


Try it online!



The program alternates between outputting itself followed by every byte and just outputting itself.



Explanation:



"            Wrapping string literal pushes the first line to the stack backwards

 44X         Push 256

    FL|      Push the range from 0 to 255

       r                  Reverse the stack

        2ss               Push a quote

           l3%            If the stack length is divisible by 3

              Q4s]|       Only take the top 20 characters

                   H      And output the stack





share|improve this answer









$endgroup$





















    1












    $begingroup$

    Program A: 05AB1E, score: 256/41 256/31 bytes = 8.258...



    0"D34çý₅Ýç'q†22ǝ"D34çý₅Ýç'q†22ǝ


    Try it online.



    Program B: 05AB1E



    0"D34çý₅Ýç'q†22ǝ"D34çýq 
    

    
 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnoprstuvwxyz{|}~ ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿÝç'q†22ǝ


    Try it online.



    Explanation:



    The shortest quine for 05AB1E is this one: 0"D34çý"D34çý (14 bytes) provided by @OliverNi. My answer uses a modified version of that quine by adding ₅Ýç'q†vy27ǝD}J.





    0            # Push a 0 to the stack
    
             #  STACK: [0]
    
 "D34çý₅Ýç'q†vDy27ǝ}J"
    
             # Push the string "D34çý₅Ýç'q†vDy27ǝ}J" to the stack
    
             #  STACK: [0,"D34çý₅Ýç'q†vDy27ǝ}J"]
    
  D          # Duplicate the string
    
             #  STACK: [0,"D34çý₅Ýç'q†vDy27ǝ}J","D34çý₅Ýç'q†vDy27ǝ}J"]
    
   34çý      # Join the stack by '"'
    
             #  STACK: ['0"D34çý₅Ýç'q†vDy27ǝ}J"D34çý₅Ýç'q†vy27ǝD}J']
    
₅Ý           # Push a list in the range [0,255]
    
  ç          # Convert each integer to an ASCII character
    
   'q†      '# Filter the "q" to the front
    
      22ǝ    # Insert it at index 22 in the string (replacing the second '₅')
    
             # (and output the result implicitly)




    Program B will terminate as soon as it reaches the q, so the actual program B is:



    0"D34çý₅Ýç'q†22ǝ"D34çýq


    Everything after that is ignored, and the top of the stack (0"D34çý₅Ýç'q†22ǝ"D34çý₅Ýç'q†22ǝ) is output implicitly.






    share|improve this answer











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      11












      $begingroup$


      Japt & Japt, Score: 255 / 38 = 6.71



      Program A:



      "iQ ²¯24
      
G²õ@i'#+Xd"iQ ²¯24
      
G²õ@i'#+Xd


      Program B is over 8kB long, so long that the link breaks, so I won't paste the whole thing. Here's a sample:



      #þ"iQ ²¯24
      
G²õ@i'#+Xd"iQ ²,#ÿ"iQ ²¯24
      
G²õ@i'#+Xd"iQ ²,#Ā"iQ ²¯24
      
G²õ@i'#+Xd"iQ ²


      I couldn't find a way to get a NUL byte to work, which is why program B only has 255 unique characters. Program B essentially consists of 255 copies of a single program, where a single irrelevant byte is changed each time, and the first 254 executions are ignored.



      For the explanation, I'll start with this simplified version of A so that the resulting B is easier to discuss.



      "iQ ²¯23
      
3õ@i'#+Xd"iQ ²¯23
      
3õ@i'#+Xd


      This program is based on the basic payload-capable Japt quine. The string at the start contains a duplicate of the rest of the program, iQ ² inserts a quote and duplicates to create a string representation of the entire program, and then ¯23 trims off itself and everything after it. The resulting string is a program that outputs Program A:



      "iQ ²¯23
      
3õ@i'#+Xd"iQ ²


      I will refer to this string as U.



      The last line of A duplicates U a bunch of times with a small change each time. Specifically, for each number X in the range [1...3] it outputs "#c" + U where c is the character with charcode X. The default behavior of Japt is to output those strings with no quotation marks and separated by commas, so this is the output of our Simplified A:



      #"iQ ²¯23
      
3õ@i'#+Xd"iQ ²,#"iQ ²¯23
      
3õ@i'#+Xd"iQ ²,#"iQ ²¯23
      
3õ@i'#+Xd"iQ ²


      We'll call this Simplified B.



      Simplified B has a simple structure, alternating between #c and U. Fortunately for this answer, each #c and U is treated as separated by a comma, and in this situation the behavior of that is everything except the very last U has no effect on the output. The only portion of Simplified B which affects the output is this:



      "iQ ²¯23
      
3õ@i'#+Xd"iQ ²


      Which is identical to U that we already know outputs Simplified A.



      The only difference between Simplified A and Program A is that instead of generating copies for the range [1...3] the real program generates copies for the range [1...256]. That results in 256 versions of #c each of which has a different character, though the last version "Ā" is a multi-byte character so it doesn't add any unique bytes, but everything except the last U is still ignored.






      share|improve this answer











      $endgroup$













      • $begingroup$
        Very nice :) Would this work for 38 bytes?
        $endgroup$
        – Shaggy
        Feb 14 at 19:12










      • $begingroup$
        @Shaggy The 27 needs to be tweaked whenever bytes are saved, but other than that it seems like it works.
        $endgroup$
        – Kamil Drakari
        Feb 14 at 19:15










      • $begingroup$
        Wait, aren't there now 256 unique characters / 257 bytes in B? If multi-byte characters aren't allowed (I've asked for clarification) then you'll need to replace G² with #ÿ. Of course, if they are then you can score much higher here.
        $endgroup$
        – Shaggy
        Feb 14 at 19:45






      • 1




        $begingroup$
        @Shaggy The scoring is based on number of unique bytes in B, and there are only 256 possible bytes. As far as I'm aware, even the new version of B does not contain the Null byte.
        $endgroup$
        – Kamil Drakari
        Feb 14 at 19:47










      • $begingroup$
        Ah, right; I was counting the bytes of unique characters, rather than the bytes themselves, which would have been silly as it would allow scores to converge on infinity.
        $endgroup$
        – Shaggy
        Feb 14 at 22:18
















      11












      $begingroup$


      Japt & Japt, Score: 255 / 38 = 6.71



      Program A:



      "iQ ²¯24
      
G²õ@i'#+Xd"iQ ²¯24
      
G²õ@i'#+Xd


      Program B is over 8kB long, so long that the link breaks, so I won't paste the whole thing. Here's a sample:



      #þ"iQ ²¯24
      
G²õ@i'#+Xd"iQ ²,#ÿ"iQ ²¯24
      
G²õ@i'#+Xd"iQ ²,#Ā"iQ ²¯24
      
G²õ@i'#+Xd"iQ ²


      I couldn't find a way to get a NUL byte to work, which is why program B only has 255 unique characters. Program B essentially consists of 255 copies of a single program, where a single irrelevant byte is changed each time, and the first 254 executions are ignored.



      For the explanation, I'll start with this simplified version of A so that the resulting B is easier to discuss.



      "iQ ²¯23
      
3õ@i'#+Xd"iQ ²¯23
      
3õ@i'#+Xd


      This program is based on the basic payload-capable Japt quine. The string at the start contains a duplicate of the rest of the program, iQ ² inserts a quote and duplicates to create a string representation of the entire program, and then ¯23 trims off itself and everything after it. The resulting string is a program that outputs Program A:



      "iQ ²¯23
      
3õ@i'#+Xd"iQ ²


      I will refer to this string as U.



      The last line of A duplicates U a bunch of times with a small change each time. Specifically, for each number X in the range [1...3] it outputs "#c" + U where c is the character with charcode X. The default behavior of Japt is to output those strings with no quotation marks and separated by commas, so this is the output of our Simplified A:



      #"iQ ²¯23
      
3õ@i'#+Xd"iQ ²,#"iQ ²¯23
      
3õ@i'#+Xd"iQ ²,#"iQ ²¯23
      
3õ@i'#+Xd"iQ ²


      We'll call this Simplified B.



      Simplified B has a simple structure, alternating between #c and U. Fortunately for this answer, each #c and U is treated as separated by a comma, and in this situation the behavior of that is everything except the very last U has no effect on the output. The only portion of Simplified B which affects the output is this:



      "iQ ²¯23
      
3õ@i'#+Xd"iQ ²


      Which is identical to U that we already know outputs Simplified A.



      The only difference between Simplified A and Program A is that instead of generating copies for the range [1...3] the real program generates copies for the range [1...256]. That results in 256 versions of #c each of which has a different character, though the last version "Ā" is a multi-byte character so it doesn't add any unique bytes, but everything except the last U is still ignored.






      share|improve this answer











      $endgroup$













      • $begingroup$
        Very nice :) Would this work for 38 bytes?
        $endgroup$
        – Shaggy
        Feb 14 at 19:12










      • $begingroup$
        @Shaggy The 27 needs to be tweaked whenever bytes are saved, but other than that it seems like it works.
        $endgroup$
        – Kamil Drakari
        Feb 14 at 19:15










      • $begingroup$
        Wait, aren't there now 256 unique characters / 257 bytes in B? If multi-byte characters aren't allowed (I've asked for clarification) then you'll need to replace G² with #ÿ. Of course, if they are then you can score much higher here.
        $endgroup$
        – Shaggy
        Feb 14 at 19:45






      • 1




        $begingroup$
        @Shaggy The scoring is based on number of unique bytes in B, and there are only 256 possible bytes. As far as I'm aware, even the new version of B does not contain the Null byte.
        $endgroup$
        – Kamil Drakari
        Feb 14 at 19:47










      • $begingroup$
        Ah, right; I was counting the bytes of unique characters, rather than the bytes themselves, which would have been silly as it would allow scores to converge on infinity.
        $endgroup$
        – Shaggy
        Feb 14 at 22:18














      11












      11








      11





      $begingroup$


      Japt & Japt, Score: 255 / 38 = 6.71



      Program A:



      "iQ ²¯24
      
G²õ@i'#+Xd"iQ ²¯24
      
G²õ@i'#+Xd


      Program B is over 8kB long, so long that the link breaks, so I won't paste the whole thing. Here's a sample:



      #þ"iQ ²¯24
      
G²õ@i'#+Xd"iQ ²,#ÿ"iQ ²¯24
      
G²õ@i'#+Xd"iQ ²,#Ā"iQ ²¯24
      
G²õ@i'#+Xd"iQ ²


      I couldn't find a way to get a NUL byte to work, which is why program B only has 255 unique characters. Program B essentially consists of 255 copies of a single program, where a single irrelevant byte is changed each time, and the first 254 executions are ignored.



      For the explanation, I'll start with this simplified version of A so that the resulting B is easier to discuss.



      "iQ ²¯23
      
3õ@i'#+Xd"iQ ²¯23
      
3õ@i'#+Xd


      This program is based on the basic payload-capable Japt quine. The string at the start contains a duplicate of the rest of the program, iQ ² inserts a quote and duplicates to create a string representation of the entire program, and then ¯23 trims off itself and everything after it. The resulting string is a program that outputs Program A:



      "iQ ²¯23
      
3õ@i'#+Xd"iQ ²


      I will refer to this string as U.



      The last line of A duplicates U a bunch of times with a small change each time. Specifically, for each number X in the range [1...3] it outputs "#c" + U where c is the character with charcode X. The default behavior of Japt is to output those strings with no quotation marks and separated by commas, so this is the output of our Simplified A:



      #"iQ ²¯23
      
3õ@i'#+Xd"iQ ²,#"iQ ²¯23
      
3õ@i'#+Xd"iQ ²,#"iQ ²¯23
      
3õ@i'#+Xd"iQ ²


      We'll call this Simplified B.



      Simplified B has a simple structure, alternating between #c and U. Fortunately for this answer, each #c and U is treated as separated by a comma, and in this situation the behavior of that is everything except the very last U has no effect on the output. The only portion of Simplified B which affects the output is this:



      "iQ ²¯23
      
3õ@i'#+Xd"iQ ²


      Which is identical to U that we already know outputs Simplified A.



      The only difference between Simplified A and Program A is that instead of generating copies for the range [1...3] the real program generates copies for the range [1...256]. That results in 256 versions of #c each of which has a different character, though the last version "Ā" is a multi-byte character so it doesn't add any unique bytes, but everything except the last U is still ignored.






      share|improve this answer











      $endgroup$




      Japt & Japt, Score: 255 / 38 = 6.71



      Program A:



      "iQ ²¯24
      
G²õ@i'#+Xd"iQ ²¯24
      
G²õ@i'#+Xd


      Program B is over 8kB long, so long that the link breaks, so I won't paste the whole thing. Here's a sample:



      #þ"iQ ²¯24
      
G²õ@i'#+Xd"iQ ²,#ÿ"iQ ²¯24
      
G²õ@i'#+Xd"iQ ²,#Ā"iQ ²¯24
      
G²õ@i'#+Xd"iQ ²


      I couldn't find a way to get a NUL byte to work, which is why program B only has 255 unique characters. Program B essentially consists of 255 copies of a single program, where a single irrelevant byte is changed each time, and the first 254 executions are ignored.



      For the explanation, I'll start with this simplified version of A so that the resulting B is easier to discuss.



      "iQ ²¯23
      
3õ@i'#+Xd"iQ ²¯23
      
3õ@i'#+Xd


      This program is based on the basic payload-capable Japt quine. The string at the start contains a duplicate of the rest of the program, iQ ² inserts a quote and duplicates to create a string representation of the entire program, and then ¯23 trims off itself and everything after it. The resulting string is a program that outputs Program A:



      "iQ ²¯23
      
3õ@i'#+Xd"iQ ²


      I will refer to this string as U.



      The last line of A duplicates U a bunch of times with a small change each time. Specifically, for each number X in the range [1...3] it outputs "#c" + U where c is the character with charcode X. The default behavior of Japt is to output those strings with no quotation marks and separated by commas, so this is the output of our Simplified A:



      #"iQ ²¯23
      
3õ@i'#+Xd"iQ ²,#"iQ ²¯23
      
3õ@i'#+Xd"iQ ²,#"iQ ²¯23
      
3õ@i'#+Xd"iQ ²


      We'll call this Simplified B.



      Simplified B has a simple structure, alternating between #c and U. Fortunately for this answer, each #c and U is treated as separated by a comma, and in this situation the behavior of that is everything except the very last U has no effect on the output. The only portion of Simplified B which affects the output is this:



      "iQ ²¯23
      
3õ@i'#+Xd"iQ ²


      Which is identical to U that we already know outputs Simplified A.



      The only difference between Simplified A and Program A is that instead of generating copies for the range [1...3] the real program generates copies for the range [1...256]. That results in 256 versions of #c each of which has a different character, though the last version "Ā" is a multi-byte character so it doesn't add any unique bytes, but everything except the last U is still ignored.







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Feb 15 at 14:00

























      answered Feb 14 at 18:53









      Kamil DrakariKamil Drakari

      3,391417




      3,391417












      • $begingroup$
        Very nice :) Would this work for 38 bytes?
        $endgroup$
        – Shaggy
        Feb 14 at 19:12










      • $begingroup$
        @Shaggy The 27 needs to be tweaked whenever bytes are saved, but other than that it seems like it works.
        $endgroup$
        – Kamil Drakari
        Feb 14 at 19:15










      • $begingroup$
        Wait, aren't there now 256 unique characters / 257 bytes in B? If multi-byte characters aren't allowed (I've asked for clarification) then you'll need to replace G² with #ÿ. Of course, if they are then you can score much higher here.
        $endgroup$
        – Shaggy
        Feb 14 at 19:45






      • 1




        $begingroup$
        @Shaggy The scoring is based on number of unique bytes in B, and there are only 256 possible bytes. As far as I'm aware, even the new version of B does not contain the Null byte.
        $endgroup$
        – Kamil Drakari
        Feb 14 at 19:47










      • $begingroup$
        Ah, right; I was counting the bytes of unique characters, rather than the bytes themselves, which would have been silly as it would allow scores to converge on infinity.
        $endgroup$
        – Shaggy
        Feb 14 at 22:18


















      • $begingroup$
        Very nice :) Would this work for 38 bytes?
        $endgroup$
        – Shaggy
        Feb 14 at 19:12










      • $begingroup$
        @Shaggy The 27 needs to be tweaked whenever bytes are saved, but other than that it seems like it works.
        $endgroup$
        – Kamil Drakari
        Feb 14 at 19:15










      • $begingroup$
        Wait, aren't there now 256 unique characters / 257 bytes in B? If multi-byte characters aren't allowed (I've asked for clarification) then you'll need to replace G² with #ÿ. Of course, if they are then you can score much higher here.
        $endgroup$
        – Shaggy
        Feb 14 at 19:45






      • 1




        $begingroup$
        @Shaggy The scoring is based on number of unique bytes in B, and there are only 256 possible bytes. As far as I'm aware, even the new version of B does not contain the Null byte.
        $endgroup$
        – Kamil Drakari
        Feb 14 at 19:47










      • $begingroup$
        Ah, right; I was counting the bytes of unique characters, rather than the bytes themselves, which would have been silly as it would allow scores to converge on infinity.
        $endgroup$
        – Shaggy
        Feb 14 at 22:18
















      $begingroup$
      Very nice :) Would this work for 38 bytes?
      $endgroup$
      – Shaggy
      Feb 14 at 19:12




      $begingroup$
      Very nice :) Would this work for 38 bytes?
      $endgroup$
      – Shaggy
      Feb 14 at 19:12












      $begingroup$
      @Shaggy The 27 needs to be tweaked whenever bytes are saved, but other than that it seems like it works.
      $endgroup$
      – Kamil Drakari
      Feb 14 at 19:15




      $begingroup$
      @Shaggy The 27 needs to be tweaked whenever bytes are saved, but other than that it seems like it works.
      $endgroup$
      – Kamil Drakari
      Feb 14 at 19:15












      $begingroup$
      Wait, aren't there now 256 unique characters / 257 bytes in B? If multi-byte characters aren't allowed (I've asked for clarification) then you'll need to replace G² with #ÿ. Of course, if they are then you can score much higher here.
      $endgroup$
      – Shaggy
      Feb 14 at 19:45




      $begingroup$
      Wait, aren't there now 256 unique characters / 257 bytes in B? If multi-byte characters aren't allowed (I've asked for clarification) then you'll need to replace G² with #ÿ. Of course, if they are then you can score much higher here.
      $endgroup$
      – Shaggy
      Feb 14 at 19:45




      1




      1




      $begingroup$
      @Shaggy The scoring is based on number of unique bytes in B, and there are only 256 possible bytes. As far as I'm aware, even the new version of B does not contain the Null byte.
      $endgroup$
      – Kamil Drakari
      Feb 14 at 19:47




      $begingroup$
      @Shaggy The scoring is based on number of unique bytes in B, and there are only 256 possible bytes. As far as I'm aware, even the new version of B does not contain the Null byte.
      $endgroup$
      – Kamil Drakari
      Feb 14 at 19:47












      $begingroup$
      Ah, right; I was counting the bytes of unique characters, rather than the bytes themselves, which would have been silly as it would allow scores to converge on infinity.
      $endgroup$
      – Shaggy
      Feb 14 at 22:18




      $begingroup$
      Ah, right; I was counting the bytes of unique characters, rather than the bytes themselves, which would have been silly as it would allow scores to converge on infinity.
      $endgroup$
      – Shaggy
      Feb 14 at 22:18











      4












      $begingroup$

      Program A, Gol><>, 256/20 bytes = 12.8



      "44XFL|r2ssl3%Q4s]|H


      Try it online!



      Program B, Gol><>



      "44XFL|r2ssl3%Q4s]|H	
      

      
 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ


      Try it online!



      The program alternates between outputting itself followed by every byte and just outputting itself.



      Explanation:



      "            Wrapping string literal pushes the first line to the stack backwards
      
 44X         Push 256
      
    FL|      Push the range from 0 to 255
      
       r                  Reverse the stack
      
        2ss               Push a quote
      
           l3%            If the stack length is divisible by 3
      
              Q4s]|       Only take the top 20 characters
      
                   H      And output the stack





      share|improve this answer









      $endgroup$


















        4












        $begingroup$

        Program A, Gol><>, 256/20 bytes = 12.8



        "44XFL|r2ssl3%Q4s]|H


        Try it online!



        Program B, Gol><>



        "44XFL|r2ssl3%Q4s]|H	
        

        
 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ


        Try it online!



        The program alternates between outputting itself followed by every byte and just outputting itself.



        Explanation:



        "            Wrapping string literal pushes the first line to the stack backwards
        
 44X         Push 256
        
    FL|      Push the range from 0 to 255
        
       r                  Reverse the stack
        
        2ss               Push a quote
        
           l3%            If the stack length is divisible by 3
        
              Q4s]|       Only take the top 20 characters
        
                   H      And output the stack





        share|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Program A, Gol><>, 256/20 bytes = 12.8



          "44XFL|r2ssl3%Q4s]|H


          Try it online!



          Program B, Gol><>



          "44XFL|r2ssl3%Q4s]|H	
          

          
 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ


          Try it online!



          The program alternates between outputting itself followed by every byte and just outputting itself.



          Explanation:



          "            Wrapping string literal pushes the first line to the stack backwards
          
 44X         Push 256
          
    FL|      Push the range from 0 to 255
          
       r                  Reverse the stack
          
        2ss               Push a quote
          
           l3%            If the stack length is divisible by 3
          
              Q4s]|       Only take the top 20 characters
          
                   H      And output the stack





          share|improve this answer









          $endgroup$



          Program A, Gol><>, 256/20 bytes = 12.8



          "44XFL|r2ssl3%Q4s]|H


          Try it online!



          Program B, Gol><>



          "44XFL|r2ssl3%Q4s]|H	
          

          
 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ


          Try it online!



          The program alternates between outputting itself followed by every byte and just outputting itself.



          Explanation:



          "            Wrapping string literal pushes the first line to the stack backwards
          
 44X         Push 256
          
    FL|      Push the range from 0 to 255
          
       r                  Reverse the stack
          
        2ss               Push a quote
          
           l3%            If the stack length is divisible by 3
          
              Q4s]|       Only take the top 20 characters
          
                   H      And output the stack






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Feb 15 at 2:23









          Jo KingJo King

          23.3k254121




          23.3k254121























              1












              $begingroup$

              Program A: 05AB1E, score: 256/41 256/31 bytes = 8.258...



              0"D34çý₅Ýç'q†22ǝ"D34çý₅Ýç'q†22ǝ


              Try it online.



              Program B: 05AB1E



              0"D34çý₅Ýç'q†22ǝ"D34çýq 
              

              
 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnoprstuvwxyz{|}~ ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿÝç'q†22ǝ


              Try it online.



              Explanation:



              The shortest quine for 05AB1E is this one: 0"D34çý"D34çý (14 bytes) provided by @OliverNi. My answer uses a modified version of that quine by adding ₅Ýç'q†vy27ǝD}J.





              0            # Push a 0 to the stack
              
             #  STACK: [0]
              
 "D34çý₅Ýç'q†vDy27ǝ}J"
              
             # Push the string "D34çý₅Ýç'q†vDy27ǝ}J" to the stack
              
             #  STACK: [0,"D34çý₅Ýç'q†vDy27ǝ}J"]
              
  D          # Duplicate the string
              
             #  STACK: [0,"D34çý₅Ýç'q†vDy27ǝ}J","D34çý₅Ýç'q†vDy27ǝ}J"]
              
   34çý      # Join the stack by '"'
              
             #  STACK: ['0"D34çý₅Ýç'q†vDy27ǝ}J"D34çý₅Ýç'q†vy27ǝD}J']
              
₅Ý           # Push a list in the range [0,255]
              
  ç          # Convert each integer to an ASCII character
              
   'q†      '# Filter the "q" to the front
              
      22ǝ    # Insert it at index 22 in the string (replacing the second '₅')
              
             # (and output the result implicitly)




              Program B will terminate as soon as it reaches the q, so the actual program B is:



              0"D34çý₅Ýç'q†22ǝ"D34çýq


              Everything after that is ignored, and the top of the stack (0"D34çý₅Ýç'q†22ǝ"D34çý₅Ýç'q†22ǝ) is output implicitly.






              share|improve this answer











              $endgroup$


















                1












                $begingroup$

                Program A: 05AB1E, score: 256/41 256/31 bytes = 8.258...



                0"D34çý₅Ýç'q†22ǝ"D34çý₅Ýç'q†22ǝ


                Try it online.



                Program B: 05AB1E



                0"D34çý₅Ýç'q†22ǝ"D34çýq 
                

                
 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnoprstuvwxyz{|}~ ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿÝç'q†22ǝ


                Try it online.



                Explanation:



                The shortest quine for 05AB1E is this one: 0"D34çý"D34çý (14 bytes) provided by @OliverNi. My answer uses a modified version of that quine by adding ₅Ýç'q†vy27ǝD}J.





                0            # Push a 0 to the stack
                
             #  STACK: [0]
                
 "D34çý₅Ýç'q†vDy27ǝ}J"
                
             # Push the string "D34çý₅Ýç'q†vDy27ǝ}J" to the stack
                
             #  STACK: [0,"D34çý₅Ýç'q†vDy27ǝ}J"]
                
  D          # Duplicate the string
                
             #  STACK: [0,"D34çý₅Ýç'q†vDy27ǝ}J","D34çý₅Ýç'q†vDy27ǝ}J"]
                
   34çý      # Join the stack by '"'
                
             #  STACK: ['0"D34çý₅Ýç'q†vDy27ǝ}J"D34çý₅Ýç'q†vy27ǝD}J']
                
₅Ý           # Push a list in the range [0,255]
                
  ç          # Convert each integer to an ASCII character
                
   'q†      '# Filter the "q" to the front
                
      22ǝ    # Insert it at index 22 in the string (replacing the second '₅')
                
             # (and output the result implicitly)




                Program B will terminate as soon as it reaches the q, so the actual program B is:



                0"D34çý₅Ýç'q†22ǝ"D34çýq


                Everything after that is ignored, and the top of the stack (0"D34çý₅Ýç'q†22ǝ"D34çý₅Ýç'q†22ǝ) is output implicitly.






                share|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Program A: 05AB1E, score: 256/41 256/31 bytes = 8.258...



                  0"D34çý₅Ýç'q†22ǝ"D34çý₅Ýç'q†22ǝ


                  Try it online.



                  Program B: 05AB1E



                  0"D34çý₅Ýç'q†22ǝ"D34çýq 
                  

                  
 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnoprstuvwxyz{|}~ ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿÝç'q†22ǝ


                  Try it online.



                  Explanation:



                  The shortest quine for 05AB1E is this one: 0"D34çý"D34çý (14 bytes) provided by @OliverNi. My answer uses a modified version of that quine by adding ₅Ýç'q†vy27ǝD}J.





                  0            # Push a 0 to the stack
                  
             #  STACK: [0]
                  
 "D34çý₅Ýç'q†vDy27ǝ}J"
                  
             # Push the string "D34çý₅Ýç'q†vDy27ǝ}J" to the stack
                  
             #  STACK: [0,"D34çý₅Ýç'q†vDy27ǝ}J"]
                  
  D          # Duplicate the string
                  
             #  STACK: [0,"D34çý₅Ýç'q†vDy27ǝ}J","D34çý₅Ýç'q†vDy27ǝ}J"]
                  
   34çý      # Join the stack by '"'
                  
             #  STACK: ['0"D34çý₅Ýç'q†vDy27ǝ}J"D34çý₅Ýç'q†vy27ǝD}J']
                  
₅Ý           # Push a list in the range [0,255]
                  
  ç          # Convert each integer to an ASCII character
                  
   'q†      '# Filter the "q" to the front
                  
      22ǝ    # Insert it at index 22 in the string (replacing the second '₅')
                  
             # (and output the result implicitly)




                  Program B will terminate as soon as it reaches the q, so the actual program B is:



                  0"D34çý₅Ýç'q†22ǝ"D34çýq


                  Everything after that is ignored, and the top of the stack (0"D34çý₅Ýç'q†22ǝ"D34çý₅Ýç'q†22ǝ) is output implicitly.






                  share|improve this answer











                  $endgroup$



                  Program A: 05AB1E, score: 256/41 256/31 bytes = 8.258...



                  0"D34çý₅Ýç'q†22ǝ"D34çý₅Ýç'q†22ǝ


                  Try it online.



                  Program B: 05AB1E



                  0"D34çý₅Ýç'q†22ǝ"D34çýq 
                  

                  
 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnoprstuvwxyz{|}~ ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿÝç'q†22ǝ


                  Try it online.



                  Explanation:



                  The shortest quine for 05AB1E is this one: 0"D34çý"D34çý (14 bytes) provided by @OliverNi. My answer uses a modified version of that quine by adding ₅Ýç'q†vy27ǝD}J.





                  0            # Push a 0 to the stack
                  
             #  STACK: [0]
                  
 "D34çý₅Ýç'q†vDy27ǝ}J"
                  
             # Push the string "D34çý₅Ýç'q†vDy27ǝ}J" to the stack
                  
             #  STACK: [0,"D34çý₅Ýç'q†vDy27ǝ}J"]
                  
  D          # Duplicate the string
                  
             #  STACK: [0,"D34çý₅Ýç'q†vDy27ǝ}J","D34çý₅Ýç'q†vDy27ǝ}J"]
                  
   34çý      # Join the stack by '"'
                  
             #  STACK: ['0"D34çý₅Ýç'q†vDy27ǝ}J"D34çý₅Ýç'q†vy27ǝD}J']
                  
₅Ý           # Push a list in the range [0,255]
                  
  ç          # Convert each integer to an ASCII character
                  
   'q†      '# Filter the "q" to the front
                  
      22ǝ    # Insert it at index 22 in the string (replacing the second '₅')
                  
             # (and output the result implicitly)




                  Program B will terminate as soon as it reaches the q, so the actual program B is:



                  0"D34çý₅Ýç'q†22ǝ"D34çýq


                  Everything after that is ignored, and the top of the stack (0"D34çý₅Ýç'q†22ǝ"D34çý₅Ýç'q†22ǝ) is output implicitly.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Feb 15 at 9:50

























                  answered Feb 15 at 9:37









                  Kevin CruijssenKevin Cruijssen

                  38.5k557200




                  38.5k557200






























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