Just Twice Is Nice
$begingroup$
Let's have string multiplication to take a string and a positive number and concatenate that string that many times. In Haskell:
0 * s = ""
n * s = s ++ ((n - 1) * s)
Your task is to write a program that outputs a number $n > 0$. When you double your program (with the described procedure) the new program should output $2n$. When your program is tripled or quadrupled it should once again output the original $n$.
This is code-golf so your answers will be scored in bytes with fewer bytes being better.
Related, Related.
code-golf source-layout
asked Dec 20 '18 at 21:03
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
$endgroup$
add a comment |
$begingroup$
Let's have string multiplication to take a string and a positive number and concatenate that string that many times. In Haskell:
0 * s = ""
n * s = s ++ ((n - 1) * s)
Your task is to write a program that outputs a number $n > 0$. When you double your program (with the described procedure) the new program should output $2n$. When your program is tripled or quadrupled it should once again output the original $n$.
This is code-golf so your answers will be scored in bytes with fewer bytes being better.
Related, Related.
code-golf source-layout
asked Dec 20 '18 at 21:03
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
$endgroup$
6
$begingroup$
All the solutions so far are trivial modifications of solutions to existing challenges so I wouldn't be surprised to see this dupe-hammered by someone.
$endgroup$
– Shaggy
Dec 20 '18 at 22:14
$begingroup$
@Shaggy I disagree. For some the best approach is from the 3rd times the charm from some its the doubling, for some still (klein, lost etc.) a port wouldn't even work.
$endgroup$
– Sriotchilism O'Zaic
Dec 24 '18 at 22:03
add a comment |
$begingroup$
Let's have string multiplication to take a string and a positive number and concatenate that string that many times. In Haskell:
0 * s = ""
n * s = s ++ ((n - 1) * s)
Your task is to write a program that outputs a number $n > 0$. When you double your program (with the described procedure) the new program should output $2n$. When your program is tripled or quadrupled it should once again output the original $n$.
This is code-golf so your answers will be scored in bytes with fewer bytes being better.
Related, Related.
code-golf source-layout
asked Dec 20 '18 at 21:03
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
$endgroup$
Let's have string multiplication to take a string and a positive number and concatenate that string that many times. In Haskell:
0 * s = ""
n * s = s ++ ((n - 1) * s)
Your task is to write a program that outputs a number $n > 0$. When you double your program (with the described procedure) the new program should output $2n$. When your program is tripled or quadrupled it should once again output the original $n$.
This is code-golf so your answers will be scored in bytes with fewer bytes being better.
Related, Related.
code-golf source-layout
code-golf source-layout
asked Dec 20 '18 at 21:03
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
asked Dec 20 '18 at 21:03
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
asked Dec 20 '18 at 21:03
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
asked Dec 20 '18 at 21:03
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
asked Dec 20 '18 at 21:03
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
35.5k10160369
6
$begingroup$
All the solutions so far are trivial modifications of solutions to existing challenges so I wouldn't be surprised to see this dupe-hammered by someone.
$endgroup$
– Shaggy
Dec 20 '18 at 22:14
$begingroup$
@Shaggy I disagree. For some the best approach is from the 3rd times the charm from some its the doubling, for some still (klein, lost etc.) a port wouldn't even work.
$endgroup$
– Sriotchilism O'Zaic
Dec 24 '18 at 22:03
add a comment |
6
$begingroup$
All the solutions so far are trivial modifications of solutions to existing challenges so I wouldn't be surprised to see this dupe-hammered by someone.
$endgroup$
– Shaggy
Dec 20 '18 at 22:14
$begingroup$
@Shaggy I disagree. For some the best approach is from the 3rd times the charm from some its the doubling, for some still (klein, lost etc.) a port wouldn't even work.
$endgroup$
– Sriotchilism O'Zaic
Dec 24 '18 at 22:03
6
6
$begingroup$
All the solutions so far are trivial modifications of solutions to existing challenges so I wouldn't be surprised to see this dupe-hammered by someone.
$endgroup$
– Shaggy
Dec 20 '18 at 22:14
$begingroup$
All the solutions so far are trivial modifications of solutions to existing challenges so I wouldn't be surprised to see this dupe-hammered by someone.
$endgroup$
– Shaggy
Dec 20 '18 at 22:14
$begingroup$
@Shaggy I disagree. For some the best approach is from the 3rd times the charm from some its the doubling, for some still (klein, lost etc.) a port wouldn't even work.
$endgroup$
– Sriotchilism O'Zaic
Dec 24 '18 at 22:03
$begingroup$
@Shaggy I disagree. For some the best approach is from the 3rd times the charm from some its the doubling, for some still (klein, lost etc.) a port wouldn't even work.
$endgroup$
– Sriotchilism O'Zaic
Dec 24 '18 at 22:03
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
R, 27 25 bytes
1+!1-length(readLines())
Try it online!
Inspired by rturnbull's answer to Third time the charm, but the simplicity of the check makes it shorter.
Doubled | Tripled | Quadrupled
Why this works:
readLines() will actually reads the source file itself rather than stdin. Hence, adding lines just increments the length() of the vector returned by readLines(). Therefore, we compute !(1-length()) to obtain 1 whenever length()==1 and 0 when length()!=1, adding one to have the desired effect.
answered Dec 20 '18 at 21:28
GiuseppeGiuseppe
17.1k31152
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add a comment |
$begingroup$
Python 2, 9 bytes
Outputs via exit code. Full credit goes to Anders Kaseorg for this answer to I double the source, you double the output!.
';exit(2)
Try it online! | Doubled | Tripled | Quadrupled
Regularly, this raises a Syntax Error because the string is not closed properly and exists with exit code 1, then, when doubled, it becomes ';exit(2)';exit(2) which simply exits with code 2 because the string literal is now quoted properly and has no effect at all, and when repeated any other arbitrary number of times, it raises Syntax Errors.
answered Dec 20 '18 at 21:29
Mr. XcoderMr. Xcoder
32.2k759200
$endgroup$
add a comment |
$begingroup$
Runic Enchantments, 29 bytes
^w '
/1@ 3
/ ~!4
2'51w
/yyy
Try it online!
Slight alteration from the Third Times A Charm entry, using Jo King's compressed version and swapping the two reflection locations.
Twice
Thrice
And frice for good measure.
answered Dec 20 '18 at 21:32
Draco18sDraco18s
1,432619
$endgroup$
$begingroup$
What did you name the language after? God of War?
$endgroup$
– Riker
Dec 25 '18 at 0:39
$begingroup$
@Riker I named it because I want to (eventually) have a RPG-esque game where enchantments on gear can be created by players using a runic script (that, to the best of my ability) mapped to the 95 keyboard printable characters. I've added some extension symbols utilizing unicode combining characters (as well as a handful of other unicode characters that more readily represent the given action, such asπand several arrows). Once I have that game I'll end up making "runic 2.0" that has those game-specific commands (and probably remaps a lot of mathy stuff).
$endgroup$
– Draco18s
Dec 25 '18 at 2:10
$begingroup$
Draco: huh, interesting, I'd love to see the final result.
$endgroup$
– Riker
Dec 25 '18 at 4:26
add a comment |
$begingroup$
Klein 000, 6 bytes
1+@
Try it online!
Explanation
The idea here is pretty simple, the first 2 copies require the ip to move off the edge of the program but the third does not, that way adding anything extra after the third doesn't change anything.
The first program reflects off of both edges the
At the begining is just a fancy noop. So we just run 1+@.
For the two copies this gets interrupted
1+@
\1+@
We hit both 1s and a + to get 2.
On the three we interrupt the path even further
1+@
\1+@
\1+@
At this point it might not be the clearest, but the ip bounces around in the mirrors a bit before it hits runs the 3rd line.
This is what it runs
\
1+@
Since this doesn't go off an edge further additions don't make any change to the program.
answered Dec 25 '18 at 2:58
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
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add a comment |
$begingroup$
Charcoal, 10 bytes
ＰＩ⊕⁼²Ｌ⊞Ｏυω
Try it online! Based on my answer to I double the source, you double the output! but compares the length to 2. Try it doubled. Try it tripled. Try it quadrupled. In verbose syntax this is Multiprint(Cast(Incremented(Equals(2, Length(PushOperator(u, w)))))));.
answered Dec 20 '18 at 21:19
NeilNeil
82k745178
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add a comment |
$begingroup$
MATL, 9 bytes
vxHQXH4=Q
Uses n = 1.
Try it online! Original, doubled, tripled, quadrupled.
Explanation
This uses clipboard H to store state information. Function H pastes the clipboard contents onto the stack. Function XH copies the top of the stack into the clipboard. The clipboard initially contains the number 2.
Each time the snippet vxHQXH4=Q is run it does the following.
The stack contents, if any, are deleted (vx). The clipboard contents are pushed (H) and incremented (Q), and the result is copied back into the clipboard (XH). This gives 4 the second time, and only that time.
The number in the stack is tested for equality with 4 (4=) and incremented (Q). This gives 2 for 4 (second time), and 1 otherwise (any other time).
answered Dec 20 '18 at 22:57
Luis MendoLuis Mendo
75k888291
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add a comment |
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05AB1E, 7 6 bytes, n=3
₆.g2Qè
Try it online; try it doubled; try it tripled; try it quadrupled.
Explanation:
₆ # Push 36
.g # Push the amount of items on the stack
2Q # Check if that amount is equal to 2 (results in 1 if truthy; 0 if falsey)
è # Index that into the 36 (and output the top of the stack implicitly)
answered Dec 22 '18 at 17:50
Kevin CruijssenKevin Cruijssen
41.3k567213
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add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
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active
oldest
votes
$begingroup$
R, 27 25 bytes
1+!1-length(readLines())
Try it online!
Inspired by rturnbull's answer to Third time the charm, but the simplicity of the check makes it shorter.
Doubled | Tripled | Quadrupled
Why this works:
readLines() will actually reads the source file itself rather than stdin. Hence, adding lines just increments the length() of the vector returned by readLines(). Therefore, we compute !(1-length()) to obtain 1 whenever length()==1 and 0 when length()!=1, adding one to have the desired effect.
answered Dec 20 '18 at 21:28
GiuseppeGiuseppe
17.1k31152
$endgroup$
add a comment |
$begingroup$
R, 27 25 bytes
1+!1-length(readLines())
Try it online!
Inspired by rturnbull's answer to Third time the charm, but the simplicity of the check makes it shorter.
Doubled | Tripled | Quadrupled
Why this works:
readLines() will actually reads the source file itself rather than stdin. Hence, adding lines just increments the length() of the vector returned by readLines(). Therefore, we compute !(1-length()) to obtain 1 whenever length()==1 and 0 when length()!=1, adding one to have the desired effect.
answered Dec 20 '18 at 21:28
GiuseppeGiuseppe
17.1k31152
$endgroup$
add a comment |
$begingroup$
R, 27 25 bytes
1+!1-length(readLines())
Try it online!
Inspired by rturnbull's answer to Third time the charm, but the simplicity of the check makes it shorter.
Doubled | Tripled | Quadrupled
Why this works:
readLines() will actually reads the source file itself rather than stdin. Hence, adding lines just increments the length() of the vector returned by readLines(). Therefore, we compute !(1-length()) to obtain 1 whenever length()==1 and 0 when length()!=1, adding one to have the desired effect.
answered Dec 20 '18 at 21:28
GiuseppeGiuseppe
17.1k31152
$endgroup$
R, 27 25 bytes
1+!1-length(readLines())
Try it online!
Inspired by rturnbull's answer to Third time the charm, but the simplicity of the check makes it shorter.
Doubled | Tripled | Quadrupled
Why this works:
readLines() will actually reads the source file itself rather than stdin. Hence, adding lines just increments the length() of the vector returned by readLines(). Therefore, we compute !(1-length()) to obtain 1 whenever length()==1 and 0 when length()!=1, adding one to have the desired effect.
answered Dec 20 '18 at 21:28
GiuseppeGiuseppe
17.1k31152
edited Dec 20 '18 at 21:41
answered Dec 20 '18 at 21:28
GiuseppeGiuseppe
17.1k31152
answered Dec 20 '18 at 21:28
GiuseppeGiuseppe
17.1k31152
answered Dec 20 '18 at 21:28
GiuseppeGiuseppe
17.1k31152
17.1k31152
add a comment |
add a comment |
$begingroup$
Python 2, 9 bytes
Outputs via exit code. Full credit goes to Anders Kaseorg for this answer to I double the source, you double the output!.
';exit(2)
Try it online! | Doubled | Tripled | Quadrupled
Regularly, this raises a Syntax Error because the string is not closed properly and exists with exit code 1, then, when doubled, it becomes ';exit(2)';exit(2) which simply exits with code 2 because the string literal is now quoted properly and has no effect at all, and when repeated any other arbitrary number of times, it raises Syntax Errors.
answered Dec 20 '18 at 21:29
Mr. XcoderMr. Xcoder
32.2k759200
$endgroup$
add a comment |
$begingroup$
Python 2, 9 bytes
Outputs via exit code. Full credit goes to Anders Kaseorg for this answer to I double the source, you double the output!.
';exit(2)
Try it online! | Doubled | Tripled | Quadrupled
Regularly, this raises a Syntax Error because the string is not closed properly and exists with exit code 1, then, when doubled, it becomes ';exit(2)';exit(2) which simply exits with code 2 because the string literal is now quoted properly and has no effect at all, and when repeated any other arbitrary number of times, it raises Syntax Errors.
answered Dec 20 '18 at 21:29
Mr. XcoderMr. Xcoder
32.2k759200
$endgroup$
add a comment |
$begingroup$
Python 2, 9 bytes
Outputs via exit code. Full credit goes to Anders Kaseorg for this answer to I double the source, you double the output!.
';exit(2)
Try it online! | Doubled | Tripled | Quadrupled
Regularly, this raises a Syntax Error because the string is not closed properly and exists with exit code 1, then, when doubled, it becomes ';exit(2)';exit(2) which simply exits with code 2 because the string literal is now quoted properly and has no effect at all, and when repeated any other arbitrary number of times, it raises Syntax Errors.
answered Dec 20 '18 at 21:29
Mr. XcoderMr. Xcoder
32.2k759200
$endgroup$
Python 2, 9 bytes
Outputs via exit code. Full credit goes to Anders Kaseorg for this answer to I double the source, you double the output!.
';exit(2)
Try it online! | Doubled | Tripled | Quadrupled
Regularly, this raises a Syntax Error because the string is not closed properly and exists with exit code 1, then, when doubled, it becomes ';exit(2)';exit(2) which simply exits with code 2 because the string literal is now quoted properly and has no effect at all, and when repeated any other arbitrary number of times, it raises Syntax Errors.
answered Dec 20 '18 at 21:29
Mr. XcoderMr. Xcoder
32.2k759200
answered Dec 20 '18 at 21:29
Mr. XcoderMr. Xcoder
32.2k759200
answered Dec 20 '18 at 21:29
Mr. XcoderMr. Xcoder
32.2k759200
answered Dec 20 '18 at 21:29
Mr. XcoderMr. Xcoder
32.2k759200
32.2k759200
add a comment |
add a comment |
$begingroup$
Runic Enchantments, 29 bytes
^w '
/1@ 3
/ ~!4
2'51w
/yyy
Try it online!
Slight alteration from the Third Times A Charm entry, using Jo King's compressed version and swapping the two reflection locations.
Twice
Thrice
And frice for good measure.
answered Dec 20 '18 at 21:32
Draco18sDraco18s
1,432619
$endgroup$
$begingroup$
What did you name the language after? God of War?
$endgroup$
– Riker
Dec 25 '18 at 0:39
$begingroup$
@Riker I named it because I want to (eventually) have a RPG-esque game where enchantments on gear can be created by players using a runic script (that, to the best of my ability) mapped to the 95 keyboard printable characters. I've added some extension symbols utilizing unicode combining characters (as well as a handful of other unicode characters that more readily represent the given action, such asπand several arrows). Once I have that game I'll end up making "runic 2.0" that has those game-specific commands (and probably remaps a lot of mathy stuff).
$endgroup$
– Draco18s
Dec 25 '18 at 2:10
$begingroup$
Draco: huh, interesting, I'd love to see the final result.
$endgroup$
– Riker
Dec 25 '18 at 4:26
add a comment |
$begingroup$
Runic Enchantments, 29 bytes
^w '
/1@ 3
/ ~!4
2'51w
/yyy
Try it online!
Slight alteration from the Third Times A Charm entry, using Jo King's compressed version and swapping the two reflection locations.
Twice
Thrice
And frice for good measure.
answered Dec 20 '18 at 21:32
Draco18sDraco18s
1,432619
$endgroup$
$begingroup$
What did you name the language after? God of War?
$endgroup$
– Riker
Dec 25 '18 at 0:39
$begingroup$
@Riker I named it because I want to (eventually) have a RPG-esque game where enchantments on gear can be created by players using a runic script (that, to the best of my ability) mapped to the 95 keyboard printable characters. I've added some extension symbols utilizing unicode combining characters (as well as a handful of other unicode characters that more readily represent the given action, such asπand several arrows). Once I have that game I'll end up making "runic 2.0" that has those game-specific commands (and probably remaps a lot of mathy stuff).
$endgroup$
– Draco18s
Dec 25 '18 at 2:10
$begingroup$
Draco: huh, interesting, I'd love to see the final result.
$endgroup$
– Riker
Dec 25 '18 at 4:26
add a comment |
$begingroup$
Runic Enchantments, 29 bytes
^w '
/1@ 3
/ ~!4
2'51w
/yyy
Try it online!
Slight alteration from the Third Times A Charm entry, using Jo King's compressed version and swapping the two reflection locations.
Twice
Thrice
And frice for good measure.
answered Dec 20 '18 at 21:32
Draco18sDraco18s
1,432619
$endgroup$
Runic Enchantments, 29 bytes
^w '
/1@ 3
/ ~!4
2'51w
/yyy
Try it online!
Slight alteration from the Third Times A Charm entry, using Jo King's compressed version and swapping the two reflection locations.
Twice
Thrice
And frice for good measure.
answered Dec 20 '18 at 21:32
Draco18sDraco18s
1,432619
edited Dec 21 '18 at 0:31
answered Dec 20 '18 at 21:32
Draco18sDraco18s
1,432619
answered Dec 20 '18 at 21:32
Draco18sDraco18s
1,432619
answered Dec 20 '18 at 21:32
Draco18sDraco18s
1,432619
1,432619
$begingroup$
What did you name the language after? God of War?
$endgroup$
– Riker
Dec 25 '18 at 0:39
$begingroup$
@Riker I named it because I want to (eventually) have a RPG-esque game where enchantments on gear can be created by players using a runic script (that, to the best of my ability) mapped to the 95 keyboard printable characters. I've added some extension symbols utilizing unicode combining characters (as well as a handful of other unicode characters that more readily represent the given action, such asπand several arrows). Once I have that game I'll end up making "runic 2.0" that has those game-specific commands (and probably remaps a lot of mathy stuff).
$endgroup$
– Draco18s
Dec 25 '18 at 2:10
$begingroup$
Draco: huh, interesting, I'd love to see the final result.
$endgroup$
– Riker
Dec 25 '18 at 4:26
add a comment |
$begingroup$
What did you name the language after? God of War?
$endgroup$
– Riker
Dec 25 '18 at 0:39
$begingroup$
@Riker I named it because I want to (eventually) have a RPG-esque game where enchantments on gear can be created by players using a runic script (that, to the best of my ability) mapped to the 95 keyboard printable characters. I've added some extension symbols utilizing unicode combining characters (as well as a handful of other unicode characters that more readily represent the given action, such asπand several arrows). Once I have that game I'll end up making "runic 2.0" that has those game-specific commands (and probably remaps a lot of mathy stuff).
$endgroup$
– Draco18s
Dec 25 '18 at 2:10
$begingroup$
Draco: huh, interesting, I'd love to see the final result.
$endgroup$
– Riker
Dec 25 '18 at 4:26
$begingroup$
What did you name the language after? God of War?
$endgroup$
– Riker
Dec 25 '18 at 0:39
$begingroup$
What did you name the language after? God of War?
$endgroup$
– Riker
Dec 25 '18 at 0:39
$begingroup$
@Riker I named it because I want to (eventually) have a RPG-esque game where enchantments on gear can be created by players using a runic script (that, to the best of my ability) mapped to the 95 keyboard printable characters. I've added some extension symbols utilizing unicode combining characters (as well as a handful of other unicode characters that more readily represent the given action, such as
π and several arrows). Once I have that game I'll end up making "runic 2.0" that has those game-specific commands (and probably remaps a lot of mathy stuff).$endgroup$
– Draco18s
Dec 25 '18 at 2:10
$begingroup$
@Riker I named it because I want to (eventually) have a RPG-esque game where enchantments on gear can be created by players using a runic script (that, to the best of my ability) mapped to the 95 keyboard printable characters. I've added some extension symbols utilizing unicode combining characters (as well as a handful of other unicode characters that more readily represent the given action, such as
π and several arrows). Once I have that game I'll end up making "runic 2.0" that has those game-specific commands (and probably remaps a lot of mathy stuff).$endgroup$
– Draco18s
Dec 25 '18 at 2:10
$begingroup$
Draco: huh, interesting, I'd love to see the final result.
$endgroup$
– Riker
Dec 25 '18 at 4:26
$begingroup$
Draco: huh, interesting, I'd love to see the final result.
$endgroup$
– Riker
Dec 25 '18 at 4:26
add a comment |
$begingroup$
Klein 000, 6 bytes
1+@
Try it online!
Explanation
The idea here is pretty simple, the first 2 copies require the ip to move off the edge of the program but the third does not, that way adding anything extra after the third doesn't change anything.
The first program reflects off of both edges the
At the begining is just a fancy noop. So we just run 1+@.
For the two copies this gets interrupted
1+@
\1+@
We hit both 1s and a + to get 2.
On the three we interrupt the path even further
1+@
\1+@
\1+@
At this point it might not be the clearest, but the ip bounces around in the mirrors a bit before it hits runs the 3rd line.
This is what it runs
\
1+@
Since this doesn't go off an edge further additions don't make any change to the program.
answered Dec 25 '18 at 2:58
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
$endgroup$
add a comment |
$begingroup$
Klein 000, 6 bytes
1+@
Try it online!
Explanation
The idea here is pretty simple, the first 2 copies require the ip to move off the edge of the program but the third does not, that way adding anything extra after the third doesn't change anything.
The first program reflects off of both edges the
At the begining is just a fancy noop. So we just run 1+@.
For the two copies this gets interrupted
1+@
\1+@
We hit both 1s and a + to get 2.
On the three we interrupt the path even further
1+@
\1+@
\1+@
At this point it might not be the clearest, but the ip bounces around in the mirrors a bit before it hits runs the 3rd line.
This is what it runs
\
1+@
Since this doesn't go off an edge further additions don't make any change to the program.
answered Dec 25 '18 at 2:58
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
$endgroup$
add a comment |
$begingroup$
Klein 000, 6 bytes
1+@
Try it online!
Explanation
The idea here is pretty simple, the first 2 copies require the ip to move off the edge of the program but the third does not, that way adding anything extra after the third doesn't change anything.
The first program reflects off of both edges the
At the begining is just a fancy noop. So we just run 1+@.
For the two copies this gets interrupted
1+@
\1+@
We hit both 1s and a + to get 2.
On the three we interrupt the path even further
1+@
\1+@
\1+@
At this point it might not be the clearest, but the ip bounces around in the mirrors a bit before it hits runs the 3rd line.
This is what it runs
\
1+@
Since this doesn't go off an edge further additions don't make any change to the program.
answered Dec 25 '18 at 2:58
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
$endgroup$
Klein 000, 6 bytes
1+@
Try it online!
Explanation
The idea here is pretty simple, the first 2 copies require the ip to move off the edge of the program but the third does not, that way adding anything extra after the third doesn't change anything.
The first program reflects off of both edges the
At the begining is just a fancy noop. So we just run 1+@.
For the two copies this gets interrupted
1+@
\1+@
We hit both 1s and a + to get 2.
On the three we interrupt the path even further
1+@
\1+@
\1+@
At this point it might not be the clearest, but the ip bounces around in the mirrors a bit before it hits runs the 3rd line.
This is what it runs
\
1+@
Since this doesn't go off an edge further additions don't make any change to the program.
answered Dec 25 '18 at 2:58
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
edited Dec 25 '18 at 3:04
answered Dec 25 '18 at 2:58
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
answered Dec 25 '18 at 2:58
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
answered Dec 25 '18 at 2:58
Sriotchilism O'ZaicSriotchilism O'Zaic
35.5k10160369
35.5k10160369
add a comment |
add a comment |
$begingroup$
Charcoal, 10 bytes
ＰＩ⊕⁼²Ｌ⊞Ｏυω
Try it online! Based on my answer to I double the source, you double the output! but compares the length to 2. Try it doubled. Try it tripled. Try it quadrupled. In verbose syntax this is Multiprint(Cast(Incremented(Equals(2, Length(PushOperator(u, w)))))));.
answered Dec 20 '18 at 21:19
NeilNeil
82k745178
$endgroup$
add a comment |
$begingroup$
Charcoal, 10 bytes
ＰＩ⊕⁼²Ｌ⊞Ｏυω
Try it online! Based on my answer to I double the source, you double the output! but compares the length to 2. Try it doubled. Try it tripled. Try it quadrupled. In verbose syntax this is Multiprint(Cast(Incremented(Equals(2, Length(PushOperator(u, w)))))));.
answered Dec 20 '18 at 21:19
NeilNeil
82k745178
$endgroup$
add a comment |
$begingroup$
Charcoal, 10 bytes
ＰＩ⊕⁼²Ｌ⊞Ｏυω
Try it online! Based on my answer to I double the source, you double the output! but compares the length to 2. Try it doubled. Try it tripled. Try it quadrupled. In verbose syntax this is Multiprint(Cast(Incremented(Equals(2, Length(PushOperator(u, w)))))));.
answered Dec 20 '18 at 21:19
NeilNeil
82k745178
$endgroup$
Charcoal, 10 bytes
ＰＩ⊕⁼²Ｌ⊞Ｏυω
Try it online! Based on my answer to I double the source, you double the output! but compares the length to 2. Try it doubled. Try it tripled. Try it quadrupled. In verbose syntax this is Multiprint(Cast(Incremented(Equals(2, Length(PushOperator(u, w)))))));.
answered Dec 20 '18 at 21:19
NeilNeil
82k745178
answered Dec 20 '18 at 21:19
NeilNeil
82k745178
answered Dec 20 '18 at 21:19
NeilNeil
82k745178
answered Dec 20 '18 at 21:19
NeilNeil
82k745178
82k745178
add a comment |
add a comment |
$begingroup$
MATL, 9 bytes
vxHQXH4=Q
Uses n = 1.
Try it online! Original, doubled, tripled, quadrupled.
Explanation
This uses clipboard H to store state information. Function H pastes the clipboard contents onto the stack. Function XH copies the top of the stack into the clipboard. The clipboard initially contains the number 2.
Each time the snippet vxHQXH4=Q is run it does the following.
The stack contents, if any, are deleted (vx). The clipboard contents are pushed (H) and incremented (Q), and the result is copied back into the clipboard (XH). This gives 4 the second time, and only that time.
The number in the stack is tested for equality with 4 (4=) and incremented (Q). This gives 2 for 4 (second time), and 1 otherwise (any other time).
answered Dec 20 '18 at 22:57
Luis MendoLuis Mendo
75k888291
$endgroup$
add a comment |
$begingroup$
MATL, 9 bytes
vxHQXH4=Q
Uses n = 1.
Try it online! Original, doubled, tripled, quadrupled.
Explanation
This uses clipboard H to store state information. Function H pastes the clipboard contents onto the stack. Function XH copies the top of the stack into the clipboard. The clipboard initially contains the number 2.
Each time the snippet vxHQXH4=Q is run it does the following.
The stack contents, if any, are deleted (vx). The clipboard contents are pushed (H) and incremented (Q), and the result is copied back into the clipboard (XH). This gives 4 the second time, and only that time.
The number in the stack is tested for equality with 4 (4=) and incremented (Q). This gives 2 for 4 (second time), and 1 otherwise (any other time).
answered Dec 20 '18 at 22:57
Luis MendoLuis Mendo
75k888291
$endgroup$
add a comment |
$begingroup$
MATL, 9 bytes
vxHQXH4=Q
Uses n = 1.
Try it online! Original, doubled, tripled, quadrupled.
Explanation
This uses clipboard H to store state information. Function H pastes the clipboard contents onto the stack. Function XH copies the top of the stack into the clipboard. The clipboard initially contains the number 2.
Each time the snippet vxHQXH4=Q is run it does the following.
The stack contents, if any, are deleted (vx). The clipboard contents are pushed (H) and incremented (Q), and the result is copied back into the clipboard (XH). This gives 4 the second time, and only that time.
The number in the stack is tested for equality with 4 (4=) and incremented (Q). This gives 2 for 4 (second time), and 1 otherwise (any other time).
answered Dec 20 '18 at 22:57
Luis MendoLuis Mendo
75k888291
$endgroup$
MATL, 9 bytes
vxHQXH4=Q
Uses n = 1.
Try it online! Original, doubled, tripled, quadrupled.
Explanation
This uses clipboard H to store state information. Function H pastes the clipboard contents onto the stack. Function XH copies the top of the stack into the clipboard. The clipboard initially contains the number 2.
Each time the snippet vxHQXH4=Q is run it does the following.
The stack contents, if any, are deleted (vx). The clipboard contents are pushed (H) and incremented (Q), and the result is copied back into the clipboard (XH). This gives 4 the second time, and only that time.
The number in the stack is tested for equality with 4 (4=) and incremented (Q). This gives 2 for 4 (second time), and 1 otherwise (any other time).
answered Dec 20 '18 at 22:57
Luis MendoLuis Mendo
75k888291
edited Dec 20 '18 at 23:47
answered Dec 20 '18 at 22:57
Luis MendoLuis Mendo
75k888291
answered Dec 20 '18 at 22:57
Luis MendoLuis Mendo
75k888291
answered Dec 20 '18 at 22:57
Luis MendoLuis Mendo
75k888291
75k888291
add a comment |
add a comment |
$begingroup$
05AB1E, 7 6 bytes, n=3
₆.g2Qè
Try it online; try it doubled; try it tripled; try it quadrupled.
Explanation:
₆ # Push 36
.g # Push the amount of items on the stack
2Q # Check if that amount is equal to 2 (results in 1 if truthy; 0 if falsey)
è # Index that into the 36 (and output the top of the stack implicitly)
answered Dec 22 '18 at 17:50
Kevin CruijssenKevin Cruijssen
41.3k567213
$endgroup$
add a comment |
$begingroup$
05AB1E, 7 6 bytes, n=3
₆.g2Qè
Try it online; try it doubled; try it tripled; try it quadrupled.
Explanation:
₆ # Push 36
.g # Push the amount of items on the stack
2Q # Check if that amount is equal to 2 (results in 1 if truthy; 0 if falsey)
è # Index that into the 36 (and output the top of the stack implicitly)
answered Dec 22 '18 at 17:50
Kevin CruijssenKevin Cruijssen
41.3k567213
$endgroup$
add a comment |
$begingroup$
05AB1E, 7 6 bytes, n=3
₆.g2Qè
Try it online; try it doubled; try it tripled; try it quadrupled.
Explanation:
₆ # Push 36
.g # Push the amount of items on the stack
2Q # Check if that amount is equal to 2 (results in 1 if truthy; 0 if falsey)
è # Index that into the 36 (and output the top of the stack implicitly)
answered Dec 22 '18 at 17:50
Kevin CruijssenKevin Cruijssen
41.3k567213
$endgroup$
05AB1E, 7 6 bytes, n=3
₆.g2Qè
Try it online; try it doubled; try it tripled; try it quadrupled.
Explanation:
₆ # Push 36
.g # Push the amount of items on the stack
2Q # Check if that amount is equal to 2 (results in 1 if truthy; 0 if falsey)
è # Index that into the 36 (and output the top of the stack implicitly)
answered Dec 22 '18 at 17:50
Kevin CruijssenKevin Cruijssen
41.3k567213
edited Dec 22 '18 at 18:09
answered Dec 22 '18 at 17:50
Kevin CruijssenKevin Cruijssen
41.3k567213
answered Dec 22 '18 at 17:50
Kevin CruijssenKevin Cruijssen
41.3k567213
answered Dec 22 '18 at 17:50
Kevin CruijssenKevin Cruijssen
41.3k567213
41.3k567213
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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$begingroup$
All the solutions so far are trivial modifications of solutions to existing challenges so I wouldn't be surprised to see this dupe-hammered by someone.
$endgroup$
– Shaggy
Dec 20 '18 at 22:14
$begingroup$
@Shaggy I disagree. For some the best approach is from the 3rd times the charm from some its the doubling, for some still (klein, lost etc.) a port wouldn't even work.
$endgroup$
– Sriotchilism O'Zaic
Dec 24 '18 at 22:03