Make a longer lenguage program












8












$begingroup$


Lenguage is a brainfuck dialect that is notorious for breaking source restriction challenges. That is because Lenguage cares only about the length of it's source and not the contents.



First, the length of the program is calculated. Then, said length is converted to binary and left-padded by zeroes to a multiple of 3. The resulting binary string is split into chunks of 3 bits each of which is translated into a brainfuck command as such:



000 -> +

001 -> -

010 -> >

011 -> <

100 -> .

101 -> ,

110 -> [

111 -> ]


Finally the program is run as brainfuck1.



From here the challenge is pretty simple, write a lenguage program that takes no input and produces an output consisting of one byte repeated $n$ times, where $n$ is strictly greater than the length of your program.



Answers will be scored in bytes with fewer bytes being better.



Here's a hacky program to calculate lenguage from brainfuck




1: For this challenge we will use wrapping cells and a non-wrapping tape.






code-golf quine brainfuck







share|improve this question









$endgroup$








  • 1




    $begingroup$
    +[.] Do I win? :P
    $endgroup$
    – Quintec
    4 hours ago










  • $begingroup$
    Perhaps it might be more interesting to score on the length of the output?
    $endgroup$
    – Jo King
    4 hours ago










  • $begingroup$
    @JoKing That's a good idea. Unfortunately it appears to be a bit late for that.
    $endgroup$
    – TRITICIMAGVS
    3 hours ago










  • $begingroup$
    @JoKing Actually since you are the only answerer do you mind if I change it?
    $endgroup$
    – TRITICIMAGVS
    2 hours ago










  • $begingroup$
    sure, i'll change my answer up
    $endgroup$
    – Jo King
    2 hours ago
















8












$begingroup$


Lenguage is a brainfuck dialect that is notorious for breaking source restriction challenges. That is because Lenguage cares only about the length of it's source and not the contents.



First, the length of the program is calculated. Then, said length is converted to binary and left-padded by zeroes to a multiple of 3. The resulting binary string is split into chunks of 3 bits each of which is translated into a brainfuck command as such:



000 -> +

001 -> -

010 -> >

011 -> <

100 -> .

101 -> ,

110 -> [

111 -> ]


Finally the program is run as brainfuck1.



From here the challenge is pretty simple, write a lenguage program that takes no input and produces an output consisting of one byte repeated $n$ times, where $n$ is strictly greater than the length of your program.



Answers will be scored in bytes with fewer bytes being better.



Here's a hacky program to calculate lenguage from brainfuck




1: For this challenge we will use wrapping cells and a non-wrapping tape.






code-golf quine brainfuck







share|improve this question









$endgroup$








  • 1




    $begingroup$
    +[.] Do I win? :P
    $endgroup$
    – Quintec
    4 hours ago










  • $begingroup$
    Perhaps it might be more interesting to score on the length of the output?
    $endgroup$
    – Jo King
    4 hours ago










  • $begingroup$
    @JoKing That's a good idea. Unfortunately it appears to be a bit late for that.
    $endgroup$
    – TRITICIMAGVS
    3 hours ago










  • $begingroup$
    @JoKing Actually since you are the only answerer do you mind if I change it?
    $endgroup$
    – TRITICIMAGVS
    2 hours ago










  • $begingroup$
    sure, i'll change my answer up
    $endgroup$
    – Jo King
    2 hours ago














8












8








8


1



$begingroup$


Lenguage is a brainfuck dialect that is notorious for breaking source restriction challenges. That is because Lenguage cares only about the length of it's source and not the contents.



First, the length of the program is calculated. Then, said length is converted to binary and left-padded by zeroes to a multiple of 3. The resulting binary string is split into chunks of 3 bits each of which is translated into a brainfuck command as such:



000 -> +

001 -> -

010 -> >

011 -> <

100 -> .

101 -> ,

110 -> [

111 -> ]


Finally the program is run as brainfuck1.



From here the challenge is pretty simple, write a lenguage program that takes no input and produces an output consisting of one byte repeated $n$ times, where $n$ is strictly greater than the length of your program.



Answers will be scored in bytes with fewer bytes being better.



Here's a hacky program to calculate lenguage from brainfuck




1: For this challenge we will use wrapping cells and a non-wrapping tape.






code-golf quine brainfuck







share|improve this question









$endgroup$




Lenguage is a brainfuck dialect that is notorious for breaking source restriction challenges. That is because Lenguage cares only about the length of it's source and not the contents.



First, the length of the program is calculated. Then, said length is converted to binary and left-padded by zeroes to a multiple of 3. The resulting binary string is split into chunks of 3 bits each of which is translated into a brainfuck command as such:



000 -> +

001 -> -

010 -> >

011 -> <

100 -> .

101 -> ,

110 -> [

111 -> ]


Finally the program is run as brainfuck1.



From here the challenge is pretty simple, write a lenguage program that takes no input and produces an output consisting of one byte repeated $n$ times, where $n$ is strictly greater than the length of your program.



Answers will be scored in bytes with fewer bytes being better.



Here's a hacky program to calculate lenguage from brainfuck




1: For this challenge we will use wrapping cells and a non-wrapping tape.






code-golf quine brainfuck




code-golf quine brainfuck






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 6 hours ago









TRITICIMAGVSTRITICIMAGVS

34.2k10156366




34.2k10156366








  • 1




    $begingroup$
    +[.] Do I win? :P
    $endgroup$
    – Quintec
    4 hours ago










  • $begingroup$
    Perhaps it might be more interesting to score on the length of the output?
    $endgroup$
    – Jo King
    4 hours ago










  • $begingroup$
    @JoKing That's a good idea. Unfortunately it appears to be a bit late for that.
    $endgroup$
    – TRITICIMAGVS
    3 hours ago










  • $begingroup$
    @JoKing Actually since you are the only answerer do you mind if I change it?
    $endgroup$
    – TRITICIMAGVS
    2 hours ago










  • $begingroup$
    sure, i'll change my answer up
    $endgroup$
    – Jo King
    2 hours ago














  • 1




    $begingroup$
    +[.] Do I win? :P
    $endgroup$
    – Quintec
    4 hours ago










  • $begingroup$
    Perhaps it might be more interesting to score on the length of the output?
    $endgroup$
    – Jo King
    4 hours ago










  • $begingroup$
    @JoKing That's a good idea. Unfortunately it appears to be a bit late for that.
    $endgroup$
    – TRITICIMAGVS
    3 hours ago










  • $begingroup$
    @JoKing Actually since you are the only answerer do you mind if I change it?
    $endgroup$
    – TRITICIMAGVS
    2 hours ago










  • $begingroup$
    sure, i'll change my answer up
    $endgroup$
    – Jo King
    2 hours ago








1




1




$begingroup$
+[.] Do I win? :P
$endgroup$
– Quintec
4 hours ago




$begingroup$
+[.] Do I win? :P
$endgroup$
– Quintec
4 hours ago












$begingroup$
Perhaps it might be more interesting to score on the length of the output?
$endgroup$
– Jo King
4 hours ago




$begingroup$
Perhaps it might be more interesting to score on the length of the output?
$endgroup$
– Jo King
4 hours ago












$begingroup$
@JoKing That's a good idea. Unfortunately it appears to be a bit late for that.
$endgroup$
– TRITICIMAGVS
3 hours ago




$begingroup$
@JoKing That's a good idea. Unfortunately it appears to be a bit late for that.
$endgroup$
– TRITICIMAGVS
3 hours ago












$begingroup$
@JoKing Actually since you are the only answerer do you mind if I change it?
$endgroup$
– TRITICIMAGVS
2 hours ago




$begingroup$
@JoKing Actually since you are the only answerer do you mind if I change it?
$endgroup$
– TRITICIMAGVS
2 hours ago












$begingroup$
sure, i'll change my answer up
$endgroup$
– Jo King
2 hours ago




$begingroup$
sure, i'll change my answer up
$endgroup$
– Jo King
2 hours ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

8437495638205698686671 bytes



This translates to the brainfuck program:



-[>>[>]+[->[>]+.[<]+<]<-]


Which prints exactly $231584178474632390847141970017375815706539969331281128078915168015826259279614$ SOH bytes.



This is calculated by the function



f(n)=2*f(n-1)+n

f(0)=0


(submission pending change from code-golf to the shortest output)



664613997892457936451903530140172168 bytes



The program is $190096884071251415869242504183037647$ bytes long, which translates to the brainfuck program:



----[>+++++<--]>[>>[>]+[->[>]+.[<]+<]<-]


This is calculated by the function



f(n)=2*f(n-1)+n

f(0)=0


With an input of 118.






share|improve this answer











$endgroup$













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    1 Answer
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    active

    oldest

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    5












    $begingroup$

    8437495638205698686671 bytes



    This translates to the brainfuck program:



    -[>>[>]+[->[>]+.[<]+<]<-]


    Which prints exactly $231584178474632390847141970017375815706539969331281128078915168015826259279614$ SOH bytes.



    This is calculated by the function



    f(n)=2*f(n-1)+n
    
f(0)=0


    (submission pending change from code-golf to the shortest output)



    664613997892457936451903530140172168 bytes



    The program is $190096884071251415869242504183037647$ bytes long, which translates to the brainfuck program:



    ----[>+++++<--]>[>>[>]+[->[>]+.[<]+<]<-]


    This is calculated by the function



    f(n)=2*f(n-1)+n
    
f(0)=0


    With an input of 118.






    share|improve this answer











    $endgroup$


















      5












      $begingroup$

      8437495638205698686671 bytes



      This translates to the brainfuck program:



      -[>>[>]+[->[>]+.[<]+<]<-]


      Which prints exactly $231584178474632390847141970017375815706539969331281128078915168015826259279614$ SOH bytes.



      This is calculated by the function



      f(n)=2*f(n-1)+n
      
f(0)=0


      (submission pending change from code-golf to the shortest output)



      664613997892457936451903530140172168 bytes



      The program is $190096884071251415869242504183037647$ bytes long, which translates to the brainfuck program:



      ----[>+++++<--]>[>>[>]+[->[>]+.[<]+<]<-]


      This is calculated by the function



      f(n)=2*f(n-1)+n
      
f(0)=0


      With an input of 118.






      share|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        8437495638205698686671 bytes



        This translates to the brainfuck program:



        -[>>[>]+[->[>]+.[<]+<]<-]


        Which prints exactly $231584178474632390847141970017375815706539969331281128078915168015826259279614$ SOH bytes.



        This is calculated by the function



        f(n)=2*f(n-1)+n
        
f(0)=0


        (submission pending change from code-golf to the shortest output)



        664613997892457936451903530140172168 bytes



        The program is $190096884071251415869242504183037647$ bytes long, which translates to the brainfuck program:



        ----[>+++++<--]>[>>[>]+[->[>]+.[<]+<]<-]


        This is calculated by the function



        f(n)=2*f(n-1)+n
        
f(0)=0


        With an input of 118.






        share|improve this answer











        $endgroup$



        8437495638205698686671 bytes



        This translates to the brainfuck program:



        -[>>[>]+[->[>]+.[<]+<]<-]


        Which prints exactly $231584178474632390847141970017375815706539969331281128078915168015826259279614$ SOH bytes.



        This is calculated by the function



        f(n)=2*f(n-1)+n
        
f(0)=0


        (submission pending change from code-golf to the shortest output)



        664613997892457936451903530140172168 bytes



        The program is $190096884071251415869242504183037647$ bytes long, which translates to the brainfuck program:



        ----[>+++++<--]>[>>[>]+[->[>]+.[<]+<]<-]


        This is calculated by the function



        f(n)=2*f(n-1)+n
        
f(0)=0


        With an input of 118.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 2 hours ago

























        answered 3 hours ago









        Jo KingJo King

        21.9k249113




        21.9k249113






























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