Make a longer lenguage program
$begingroup$
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
asked 5 hours ago
TRITICIMAGVSTRITICIMAGVS
34.2k10156366
$endgroup$
add a comment |
$begingroup$
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
asked 5 hours ago
TRITICIMAGVSTRITICIMAGVS
34.2k10156366
$endgroup$
$begingroup$
+[.]Do I win? :P
$endgroup$
– Quintec
3 hours ago
$begingroup$
Perhaps it might be more interesting to score on the length of the output?
$endgroup$
– Jo King
3 hours ago
$begingroup$
@JoKing That's a good idea. Unfortunately it appears to be a bit late for that.
$endgroup$
– 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
add a comment |
$begingroup$
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
asked 5 hours ago
TRITICIMAGVSTRITICIMAGVS
34.2k10156366
$endgroup$
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
asked 5 hours ago
TRITICIMAGVSTRITICIMAGVS
34.2k10156366
asked 5 hours ago
TRITICIMAGVSTRITICIMAGVS
34.2k10156366
asked 5 hours ago
TRITICIMAGVSTRITICIMAGVS
34.2k10156366
asked 5 hours ago
TRITICIMAGVSTRITICIMAGVS
34.2k10156366
asked 5 hours ago
TRITICIMAGVSTRITICIMAGVS
34.2k10156366
34.2k10156366
$begingroup$
+[.]Do I win? :P
$endgroup$
– Quintec
3 hours ago
$begingroup$
Perhaps it might be more interesting to score on the length of the output?
$endgroup$
– Jo King
3 hours ago
$begingroup$
@JoKing That's a good idea. Unfortunately it appears to be a bit late for that.
$endgroup$
– 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
add a comment |
$begingroup$
+[.]Do I win? :P
$endgroup$
– Quintec
3 hours ago
$begingroup$
Perhaps it might be more interesting to score on the length of the output?
$endgroup$
– Jo King
3 hours ago
$begingroup$
@JoKing That's a good idea. Unfortunately it appears to be a bit late for that.
$endgroup$
– 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
$begingroup$
+[.] Do I win? :P$endgroup$
– Quintec
3 hours ago
$begingroup$
+[.] Do I win? :P$endgroup$
– Quintec
3 hours ago
$begingroup$
Perhaps it might be more interesting to score on the length of the output?
$endgroup$
– Jo King
3 hours ago
$begingroup$
Perhaps it might be more interesting to score on the length of the output?
$endgroup$
– Jo King
3 hours ago
$begingroup$
@JoKing That's a good idea. Unfortunately it appears to be a bit late for that.
$endgroup$
– TRITICIMAGVS
2 hours ago
$begingroup$
@JoKing That's a good idea. Unfortunately it appears to be a bit late for that.
$endgroup$
– 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
$begingroup$
@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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 2 hours ago
Jo KingJo King
21.9k249113
$endgroup$
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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oldest
votes
$begingroup$
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.
answered 2 hours ago
Jo KingJo King
21.9k249113
$endgroup$
add a comment |
$begingroup$
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.
answered 2 hours ago
Jo KingJo King
21.9k249113
$endgroup$
add a comment |
$begingroup$
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.
answered 2 hours ago
Jo KingJo King
21.9k249113
$endgroup$
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.
answered 2 hours ago
Jo KingJo King
21.9k249113
edited 2 hours ago
answered 2 hours ago
Jo KingJo King
21.9k249113
answered 2 hours ago
Jo KingJo King
21.9k249113
answered 2 hours ago
Jo KingJo King
21.9k249113
21.9k249113
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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$begingroup$
+[.]Do I win? :P$endgroup$
– Quintec
3 hours ago
$begingroup$
Perhaps it might be more interesting to score on the length of the output?
$endgroup$
– Jo King
3 hours ago
$begingroup$
@JoKing That's a good idea. Unfortunately it appears to be a bit late for that.
$endgroup$
– 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