Mixing
Drawing the Peano curve
up vote
12
down vote
favorite
Introduction
In geometry, the Peano curve is the first example of a space-filling curve to be discovered, by Giuseppe Peano in 1890. Peano's curve is a surjective, continuous function from the unit interval onto the unit square, however it is not injective. Peano was motivated by an earlier result of Georg Cantor that these two sets have the same cardinality. Because of this example, some authors use the phrase "Peano curve" to refer more generally to any space-filling curve.
Challenge
The program takes an input which is an integer n, and outputs a drawing representing the nth iteration of the Peano curve, starting from the sideways 2 shown in the leftmost part of this image:
Input
An integer n giving the iteration number of the Peano curve.
Optional, additional input is described in the bonuses section.
Output
A drawing of the nth iteration of the Peano curve. The drawing can be both ASCII art or a "real" drawing, whichever is easiest or shortest.
Rules
- The input and output can be given in any convenient format (choose the most appropriate format for your language/solution).
- No need to handle negative values or invalid input
- Either a full program or a function are acceptable.
- If possible, please include a link to an online testing environment so other people can try out your code!
Standard loopholes are forbidden.- This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.
Bonuses
Since this shouldn't be a walk in the park (at least in most languages I can think of), bonus points are awarded for the following:
- -100 bytes if your code generates a gif of the construction of the Peano curves up to
n. - -100 bytes if your code draws a space-filling curve for any rectangular shape (the Peano curve only works for squares, obviously). You can assume that the input then takes on the form
n l wwherenhas the same meaning as before (the number of the iteration), but wherelandwbecome the length and width of the rectangle in which to draw the curve. Ifl == w, this becomes the regular Peano curve.
Negative scores are allowed (but are they possible...).
Edit
Please include the output of your program in the solution for n == 3 (l == w == 1).
code-golf ascii-art graphical-output geometry fractal
asked yesterday
Peiffap
726
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Peiffap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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|
show 12 more comments
up vote
12
down vote
favorite
Introduction
In geometry, the Peano curve is the first example of a space-filling curve to be discovered, by Giuseppe Peano in 1890. Peano's curve is a surjective, continuous function from the unit interval onto the unit square, however it is not injective. Peano was motivated by an earlier result of Georg Cantor that these two sets have the same cardinality. Because of this example, some authors use the phrase "Peano curve" to refer more generally to any space-filling curve.
Challenge
The program takes an input which is an integer n, and outputs a drawing representing the nth iteration of the Peano curve, starting from the sideways 2 shown in the leftmost part of this image:
Input
An integer n giving the iteration number of the Peano curve.
Optional, additional input is described in the bonuses section.
Output
A drawing of the nth iteration of the Peano curve. The drawing can be both ASCII art or a "real" drawing, whichever is easiest or shortest.
Rules
- The input and output can be given in any convenient format (choose the most appropriate format for your language/solution).
- No need to handle negative values or invalid input
- Either a full program or a function are acceptable.
- If possible, please include a link to an online testing environment so other people can try out your code!
Standard loopholes are forbidden.- This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.
Bonuses
Since this shouldn't be a walk in the park (at least in most languages I can think of), bonus points are awarded for the following:
- -100 bytes if your code generates a gif of the construction of the Peano curves up to
n. - -100 bytes if your code draws a space-filling curve for any rectangular shape (the Peano curve only works for squares, obviously). You can assume that the input then takes on the form
n l wwherenhas the same meaning as before (the number of the iteration), but wherelandwbecome the length and width of the rectangle in which to draw the curve. Ifl == w, this becomes the regular Peano curve.
Negative scores are allowed (but are they possible...).
Edit
Please include the output of your program in the solution for n == 3 (l == w == 1).
code-golf ascii-art graphical-output geometry fractal
asked yesterday
Peiffap
726
New contributor
Peiffap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Welcome to PPCG :) This, at a glance, looks like a nice first challenge. Although it feels familiar, I think the challenge I might be thinking of was an ASCII art one. Note, though, that we strongly discourage bonuses and that there will be golfing languages that can achieve this in less than 100 bytes. Also, and most importantly, you need a winning criterion. Given that your bonuses subtract bytes from a solution's score, I suspect you intend this to be code-golf.
– Shaggy
yesterday
4
Yeah, I don't think the bonuses are a good idea, especially since there at least two animation-capable ASCII-art focused golflangs
– ASCII-only
yesterday
2
Oh also what wouldnbe used for iflandware also inputs??????????? And would the Peano curve be a special case - it's not the only spacefilling curve, so some algorithms may have to specialcase it
– ASCII-only
yesterday
2
Also, what stops anyone from making a trivial spacefilling curve (just zigzagging back and forth) for non-square dimensions
– ASCII-only
yesterday
6
Bonuses in code golf is one of the most agreed-upon things to avoid when writing challenges. I suggest you remove them and decide which is to be the canonical version of the challenge.
– lirtosiast
yesterday
|
show 12 more comments
up vote
12
down vote
favorite
up vote
12
down vote
favorite
Introduction
In geometry, the Peano curve is the first example of a space-filling curve to be discovered, by Giuseppe Peano in 1890. Peano's curve is a surjective, continuous function from the unit interval onto the unit square, however it is not injective. Peano was motivated by an earlier result of Georg Cantor that these two sets have the same cardinality. Because of this example, some authors use the phrase "Peano curve" to refer more generally to any space-filling curve.
Challenge
The program takes an input which is an integer n, and outputs a drawing representing the nth iteration of the Peano curve, starting from the sideways 2 shown in the leftmost part of this image:
Input
An integer n giving the iteration number of the Peano curve.
Optional, additional input is described in the bonuses section.
Output
A drawing of the nth iteration of the Peano curve. The drawing can be both ASCII art or a "real" drawing, whichever is easiest or shortest.
Rules
- The input and output can be given in any convenient format (choose the most appropriate format for your language/solution).
- No need to handle negative values or invalid input
- Either a full program or a function are acceptable.
- If possible, please include a link to an online testing environment so other people can try out your code!
Standard loopholes are forbidden.- This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.
Bonuses
Since this shouldn't be a walk in the park (at least in most languages I can think of), bonus points are awarded for the following:
- -100 bytes if your code generates a gif of the construction of the Peano curves up to
n. - -100 bytes if your code draws a space-filling curve for any rectangular shape (the Peano curve only works for squares, obviously). You can assume that the input then takes on the form
n l wwherenhas the same meaning as before (the number of the iteration), but wherelandwbecome the length and width of the rectangle in which to draw the curve. Ifl == w, this becomes the regular Peano curve.
Negative scores are allowed (but are they possible...).
Edit
Please include the output of your program in the solution for n == 3 (l == w == 1).
code-golf ascii-art graphical-output geometry fractal
asked yesterday
Peiffap
726
New contributor
Peiffap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Introduction
In geometry, the Peano curve is the first example of a space-filling curve to be discovered, by Giuseppe Peano in 1890. Peano's curve is a surjective, continuous function from the unit interval onto the unit square, however it is not injective. Peano was motivated by an earlier result of Georg Cantor that these two sets have the same cardinality. Because of this example, some authors use the phrase "Peano curve" to refer more generally to any space-filling curve.
Challenge
The program takes an input which is an integer n, and outputs a drawing representing the nth iteration of the Peano curve, starting from the sideways 2 shown in the leftmost part of this image:
Input
An integer n giving the iteration number of the Peano curve.
Optional, additional input is described in the bonuses section.
Output
A drawing of the nth iteration of the Peano curve. The drawing can be both ASCII art or a "real" drawing, whichever is easiest or shortest.
Rules
- The input and output can be given in any convenient format (choose the most appropriate format for your language/solution).
- No need to handle negative values or invalid input
- Either a full program or a function are acceptable.
- If possible, please include a link to an online testing environment so other people can try out your code!
Standard loopholes are forbidden.- This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.
Bonuses
Since this shouldn't be a walk in the park (at least in most languages I can think of), bonus points are awarded for the following:
- -100 bytes if your code generates a gif of the construction of the Peano curves up to
n. - -100 bytes if your code draws a space-filling curve for any rectangular shape (the Peano curve only works for squares, obviously). You can assume that the input then takes on the form
n l wwherenhas the same meaning as before (the number of the iteration), but wherelandwbecome the length and width of the rectangle in which to draw the curve. Ifl == w, this becomes the regular Peano curve.
Negative scores are allowed (but are they possible...).
Edit
Please include the output of your program in the solution for n == 3 (l == w == 1).
code-golf ascii-art graphical-output geometry fractal
code-golf ascii-art graphical-output geometry fractal
asked yesterday
Peiffap
726
New contributor
Peiffap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Peiffap
726
New contributor
Peiffap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
asked yesterday
Peiffap
726
New contributor
Peiffap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Peiffap
726
asked yesterday
Peiffap
726
726
New contributor
Peiffap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Peiffap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Peiffap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Welcome to PPCG :) This, at a glance, looks like a nice first challenge. Although it feels familiar, I think the challenge I might be thinking of was an ASCII art one. Note, though, that we strongly discourage bonuses and that there will be golfing languages that can achieve this in less than 100 bytes. Also, and most importantly, you need a winning criterion. Given that your bonuses subtract bytes from a solution's score, I suspect you intend this to be code-golf.
– Shaggy
yesterday
4
Yeah, I don't think the bonuses are a good idea, especially since there at least two animation-capable ASCII-art focused golflangs
– ASCII-only
yesterday
2
Oh also what wouldnbe used for iflandware also inputs??????????? And would the Peano curve be a special case - it's not the only spacefilling curve, so some algorithms may have to specialcase it
– ASCII-only
yesterday
2
Also, what stops anyone from making a trivial spacefilling curve (just zigzagging back and forth) for non-square dimensions
– ASCII-only
yesterday
6
Bonuses in code golf is one of the most agreed-upon things to avoid when writing challenges. I suggest you remove them and decide which is to be the canonical version of the challenge.
– lirtosiast
yesterday
|
show 12 more comments
1
Welcome to PPCG :) This, at a glance, looks like a nice first challenge. Although it feels familiar, I think the challenge I might be thinking of was an ASCII art one. Note, though, that we strongly discourage bonuses and that there will be golfing languages that can achieve this in less than 100 bytes. Also, and most importantly, you need a winning criterion. Given that your bonuses subtract bytes from a solution's score, I suspect you intend this to be code-golf.
– Shaggy
yesterday
4
Yeah, I don't think the bonuses are a good idea, especially since there at least two animation-capable ASCII-art focused golflangs
– ASCII-only
yesterday
2
Oh also what wouldnbe used for iflandware also inputs??????????? And would the Peano curve be a special case - it's not the only spacefilling curve, so some algorithms may have to specialcase it
– ASCII-only
yesterday
2
Also, what stops anyone from making a trivial spacefilling curve (just zigzagging back and forth) for non-square dimensions
– ASCII-only
yesterday
6
Bonuses in code golf is one of the most agreed-upon things to avoid when writing challenges. I suggest you remove them and decide which is to be the canonical version of the challenge.
– lirtosiast
yesterday
1
1
Welcome to PPCG :) This, at a glance, looks like a nice first challenge. Although it feels familiar, I think the challenge I might be thinking of was an ASCII art one. Note, though, that we strongly discourage bonuses and that there will be golfing languages that can achieve this in less than 100 bytes. Also, and most importantly, you need a winning criterion. Given that your bonuses subtract bytes from a solution's score, I suspect you intend this to be code-golf.
– Shaggy
yesterday
Welcome to PPCG :) This, at a glance, looks like a nice first challenge. Although it feels familiar, I think the challenge I might be thinking of was an ASCII art one. Note, though, that we strongly discourage bonuses and that there will be golfing languages that can achieve this in less than 100 bytes. Also, and most importantly, you need a winning criterion. Given that your bonuses subtract bytes from a solution's score, I suspect you intend this to be code-golf.
– Shaggy
yesterday
4
4
Yeah, I don't think the bonuses are a good idea, especially since there at least two animation-capable ASCII-art focused golflangs
– ASCII-only
yesterday
Yeah, I don't think the bonuses are a good idea, especially since there at least two animation-capable ASCII-art focused golflangs
– ASCII-only
yesterday
2
2
Oh also what would
n be used for if l and w are also inputs??????????? And would the Peano curve be a special case - it's not the only spacefilling curve, so some algorithms may have to specialcase it– ASCII-only
yesterday
Oh also what would
n be used for if l and w are also inputs??????????? And would the Peano curve be a special case - it's not the only spacefilling curve, so some algorithms may have to specialcase it– ASCII-only
yesterday
2
2
Also, what stops anyone from making a trivial spacefilling curve (just zigzagging back and forth) for non-square dimensions
– ASCII-only
yesterday
Also, what stops anyone from making a trivial spacefilling curve (just zigzagging back and forth) for non-square dimensions
– ASCII-only
yesterday
6
6
Bonuses in code golf is one of the most agreed-upon things to avoid when writing challenges. I suggest you remove them and decide which is to be the canonical version of the challenge.
– lirtosiast
yesterday
Bonuses in code golf is one of the most agreed-upon things to avoid when writing challenges. I suggest you remove them and decide which is to be the canonical version of the challenge.
– lirtosiast
yesterday
|
show 12 more comments
8 Answers
8
active
oldest
votes
up vote
4
down vote
accepted
Mathematica, score 60 - 100 - 100 = -140
Graphics[PeanoCurve@a~Reverse~3~Scale~#2]~Animate~{a,1,#,1}&
Pure function. Takes n and {l, w} (width and height) as input, and gives an animated graphic as output. It first creates a nth order Peano curve with PeanoCurve. Since the l = w case still needs to create a Peano curve, we flip the expression at level 3, similar to DavidC's answer; for l ≠ w, we just Scale the curve to the rectangle. This curve will still be space-filling, satisfying the second bonus. For the first bonus, we just Animate it over all sizes. Note that OP suggested that this was sufficiently different from DavidC's to warrant its own answer. The result for n = 3, l = w = 1 is as follows:
answered 4 hours ago
LegionMammal978
15k41752
very nice! (with proper orientation too)
– DavidC
2 hours ago
add a comment |
up vote
10
down vote
GFA Basic 3.51 (Atari ST), 156 134 124 bytes
PRO f(n)
DR "MA0,199"
p(n,90)
RET
PRO p(n,a)
I n
n=n-.5
DR "RT",a
p(n,-a)
DR "FD4"
p(n,a)
DR "FD4"
p(n,-a)
DR "LT",a
EN
RET
Each line ends with CR. Once loaded, this program expands to:
PROCEDURE f(n)
DRAW "MA0,199"
p(n,90)
RETURN
PROCEDURE p(n,a)
IF n
n=n-0.5
DRAW "RT",a
p(n,-a)
DRAW "FD4"
p(n,a)
DRAW "FD4"
p(n,-a)
DRAW "LT",a
ENDIF
RETURN
Example output
answered yesterday
Arnauld
69k584291
add a comment |
up vote
10
down vote
Perl 6, 117 bytes
{map ->y{|map {(((++$+y)%2+$++)%3**(y+$^v,*/3...*%3)??$^s[$++%2]!!'│')xx$_*3},<┌ ┐>,$_,<└ ┘>,1},^$_}o*R**3
Try it online!
0-indexed. Returns a 2D array of Unicode characters. The basic idea is that for lower rows, the expression
(x + (x+y)%2) % (3 ** trailing_zeros_in_base3(3*(y+1)))
yields the pattern
|....||....||....||....||.. % 3
..||....||....||....||....| % 3
|................||........ % 9
..||....||....||....||....| % 3
|....||....||....||....||.. % 3
........||................| % 9
|....||....||....||....||.. % 3
..||....||....||....||....| % 3
|.......................... % 27
For upper rows, the expression is
(x + (x+y+1)%2) % (3 ** trailing_zeros_in_base3(3*(y+3**n)))
Explanation
{ ... }o*R**3 # Feed $_ = 3^n into block
map ->y{ ... },^$_ # Map y = 0..3^n-1
|map { ... },<┌ ┐>,$_,<└ ┘>,1 # Map pairs (('┌','┐'),3^n) for upper rows
# and (('└','┘'),1) for lower rows.
# Block takes items as s and v
( ... )xx$_*3 # Evaluate 3^(n+1) times, returning a list
(++$+y)%2 # (x+y+1)%2 for upper rows, (x+y)%2 for lower rows
( +$++) # Add x
(y+$^v,*/3...*%3) # Count trailing zeros of 3*(y+v) in base 3
3** # nth power of 3
% # Modulo
??$^s[$++%2] # If there's a remainder yield chars in s alternately
!!'│' # otherwise yield '│'
answered yesterday
nwellnhof
5,9881122
add a comment |
up vote
6
down vote
K (ngn/k), 37 27 26 bytes
{+y,(|'y:x,,>*x),x}/1,&2*
Try it online!
returns a boolean matrix
|'y is syntax specific to ngn/k. other dialects require a : to make an each-ed verb monadic: |:'y
answered yesterday
ngn
6,24312459
1
To make the output more beautiful, highlight all occurences of1(if supported by your browser)
– user202729
yesterday
3
@user202729 done - in the footer so it don't affect the byte count
– ngn
yesterday
add a comment |
up vote
5
down vote
Wolfram Language 83 36 bytes, (possibly -48 bytes with bonus)
As of version 11.1, PeanoCurve is a built-in.
My original, clumsy submission wasted many bytes on GeometricTransformation and ReflectionTransform.
This much reduced version was suggested by alephalpha.
Reverse was required to orient the output properly.
Graphics[Reverse/@#&/@PeanoCurve@#]&
Example 36 bytes
Graphics[Reverse/@#&/@PeanoCurve@#]&[3]
Bonus
If this qualifies for the 100 pt bonus, it weighs in at 52 - 100 = -48
The code [5] was not counted, only the pure function.
Table[Graphics[Reverse/@#&/@PeanoCurve@#]&@k{k,#}&[5]
answered yesterday
DavidC
23.6k243101
Graphics[Reverse/@#&/@PeanoCurve@#]&
– alephalpha
yesterday
It feels a bit like cheating to have a function which calculates the Peano curve by itself, but I'll take it as accepted answer since it's pretty impressive nonetheless ;). @LegionMammal978 I think you deserve to post your own answer, I'd argue that it's different enough to warrant accepting it as winning answer.
– Peiffap
7 hours ago
add a comment |
up vote
4
down vote
BBC BASIC, 142 ASCII characters (130 bytes tokenised)
Download interpreter at http://www.bbcbasic.co.uk/bbcwin/download.html
I.m:q=FNh(m,8,8,0)END
DEFFNh(n,x,y,i)
IFn F.i=0TO8q=FNh(n-1,x-i MOD3MOD2*2*x,y-i DIV3MOD2*2*y,i)DRAWBY(1AND36/2^i)*x,y*(12653/3^i MOD3-1)N.
=0
answered yesterday
Level River St
20.1k32579
add a comment |
up vote
4
down vote
HTML+SVG+JS, 224 213 bytes
The output is mirrored horizontally.
n=>document.write(`<svg width=${w=3**n*9} height=${w}><path d="M1 ${(p=(n,z)=>n--&&(p(n,-z,a(a(p(n,-z,d+=z)),p(n,z))),d-=z))(n*2,r=d=x=y=1,a=_=>r+=`L${x+=~-(d&=3)%2*9} ${y+=(2-d)%2*9}`)&&r}"fill=#fff stroke=red>`)
Try it online! (prints the HTML)
(
n=>document.write(`<svg width=${w=3**n*9} height=${w}><path d="M1 ${(p=(n,z)=>n--&&(p(n,-z,a(a(p(n,-z,d+=z)),p(n,z))),d-=z))(n*2,r=d=x=y=1,a=_=>r+=`L${x+=~-(d&=3)%2*9} ${y+=(2-d)%2*9}`)&&r}"fill=#fff stroke=red>`)
)(3)answered yesterday
Arnauld
69k584291
add a comment |
up vote
3
down vote
Logo, 89 bytes
to p:n:a
if:n>0[rt:a
p:n-1 0-:a
fw 5
p:n-1:a
fw 5
p:n-1 0-:a
lt:a]end
to f:n
p:n*2 90
end
Port of @Arnauld's Atari BASIC answer. To use, do something like this:
reset
f 3
answered 20 hours ago
Neil
78k744175
add a comment |
8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Mathematica, score 60 - 100 - 100 = -140
Graphics[PeanoCurve@a~Reverse~3~Scale~#2]~Animate~{a,1,#,1}&
Pure function. Takes n and {l, w} (width and height) as input, and gives an animated graphic as output. It first creates a nth order Peano curve with PeanoCurve. Since the l = w case still needs to create a Peano curve, we flip the expression at level 3, similar to DavidC's answer; for l ≠ w, we just Scale the curve to the rectangle. This curve will still be space-filling, satisfying the second bonus. For the first bonus, we just Animate it over all sizes. Note that OP suggested that this was sufficiently different from DavidC's to warrant its own answer. The result for n = 3, l = w = 1 is as follows:
answered 4 hours ago
LegionMammal978
15k41752
very nice! (with proper orientation too)
– DavidC
2 hours ago
add a comment |
up vote
4
down vote
accepted
Mathematica, score 60 - 100 - 100 = -140
Graphics[PeanoCurve@a~Reverse~3~Scale~#2]~Animate~{a,1,#,1}&
Pure function. Takes n and {l, w} (width and height) as input, and gives an animated graphic as output. It first creates a nth order Peano curve with PeanoCurve. Since the l = w case still needs to create a Peano curve, we flip the expression at level 3, similar to DavidC's answer; for l ≠ w, we just Scale the curve to the rectangle. This curve will still be space-filling, satisfying the second bonus. For the first bonus, we just Animate it over all sizes. Note that OP suggested that this was sufficiently different from DavidC's to warrant its own answer. The result for n = 3, l = w = 1 is as follows:
answered 4 hours ago
LegionMammal978
15k41752
very nice! (with proper orientation too)
– DavidC
2 hours ago
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Mathematica, score 60 - 100 - 100 = -140
Graphics[PeanoCurve@a~Reverse~3~Scale~#2]~Animate~{a,1,#,1}&
Pure function. Takes n and {l, w} (width and height) as input, and gives an animated graphic as output. It first creates a nth order Peano curve with PeanoCurve. Since the l = w case still needs to create a Peano curve, we flip the expression at level 3, similar to DavidC's answer; for l ≠ w, we just Scale the curve to the rectangle. This curve will still be space-filling, satisfying the second bonus. For the first bonus, we just Animate it over all sizes. Note that OP suggested that this was sufficiently different from DavidC's to warrant its own answer. The result for n = 3, l = w = 1 is as follows:
answered 4 hours ago
LegionMammal978
15k41752
Mathematica, score 60 - 100 - 100 = -140
Graphics[PeanoCurve@a~Reverse~3~Scale~#2]~Animate~{a,1,#,1}&
Pure function. Takes n and {l, w} (width and height) as input, and gives an animated graphic as output. It first creates a nth order Peano curve with PeanoCurve. Since the l = w case still needs to create a Peano curve, we flip the expression at level 3, similar to DavidC's answer; for l ≠ w, we just Scale the curve to the rectangle. This curve will still be space-filling, satisfying the second bonus. For the first bonus, we just Animate it over all sizes. Note that OP suggested that this was sufficiently different from DavidC's to warrant its own answer. The result for n = 3, l = w = 1 is as follows:
answered 4 hours ago
LegionMammal978
15k41752
answered 4 hours ago
LegionMammal978
15k41752
answered 4 hours ago
LegionMammal978
15k41752
answered 4 hours ago
LegionMammal978
15k41752
15k41752
very nice! (with proper orientation too)
– DavidC
2 hours ago
add a comment |
very nice! (with proper orientation too)
– DavidC
2 hours ago
very nice! (with proper orientation too)
– DavidC
2 hours ago
very nice! (with proper orientation too)
– DavidC
2 hours ago
add a comment |
up vote
10
down vote
GFA Basic 3.51 (Atari ST), 156 134 124 bytes
PRO f(n)
DR "MA0,199"
p(n,90)
RET
PRO p(n,a)
I n
n=n-.5
DR "RT",a
p(n,-a)
DR "FD4"
p(n,a)
DR "FD4"
p(n,-a)
DR "LT",a
EN
RET
Each line ends with CR. Once loaded, this program expands to:
PROCEDURE f(n)
DRAW "MA0,199"
p(n,90)
RETURN
PROCEDURE p(n,a)
IF n
n=n-0.5
DRAW "RT",a
p(n,-a)
DRAW "FD4"
p(n,a)
DRAW "FD4"
p(n,-a)
DRAW "LT",a
ENDIF
RETURN
Example output
answered yesterday
Arnauld
69k584291
add a comment |
up vote
10
down vote
GFA Basic 3.51 (Atari ST), 156 134 124 bytes
PRO f(n)
DR "MA0,199"
p(n,90)
RET
PRO p(n,a)
I n
n=n-.5
DR "RT",a
p(n,-a)
DR "FD4"
p(n,a)
DR "FD4"
p(n,-a)
DR "LT",a
EN
RET
Each line ends with CR. Once loaded, this program expands to:
PROCEDURE f(n)
DRAW "MA0,199"
p(n,90)
RETURN
PROCEDURE p(n,a)
IF n
n=n-0.5
DRAW "RT",a
p(n,-a)
DRAW "FD4"
p(n,a)
DRAW "FD4"
p(n,-a)
DRAW "LT",a
ENDIF
RETURN
Example output
answered yesterday
Arnauld
69k584291
add a comment |
up vote
10
down vote
up vote
10
down vote
GFA Basic 3.51 (Atari ST), 156 134 124 bytes
PRO f(n)
DR "MA0,199"
p(n,90)
RET
PRO p(n,a)
I n
n=n-.5
DR "RT",a
p(n,-a)
DR "FD4"
p(n,a)
DR "FD4"
p(n,-a)
DR "LT",a
EN
RET
Each line ends with CR. Once loaded, this program expands to:
PROCEDURE f(n)
DRAW "MA0,199"
p(n,90)
RETURN
PROCEDURE p(n,a)
IF n
n=n-0.5
DRAW "RT",a
p(n,-a)
DRAW "FD4"
p(n,a)
DRAW "FD4"
p(n,-a)
DRAW "LT",a
ENDIF
RETURN
Example output
answered yesterday
Arnauld
69k584291
GFA Basic 3.51 (Atari ST), 156 134 124 bytes
PRO f(n)
DR "MA0,199"
p(n,90)
RET
PRO p(n,a)
I n
n=n-.5
DR "RT",a
p(n,-a)
DR "FD4"
p(n,a)
DR "FD4"
p(n,-a)
DR "LT",a
EN
RET
Each line ends with CR. Once loaded, this program expands to:
PROCEDURE f(n)
DRAW "MA0,199"
p(n,90)
RETURN
PROCEDURE p(n,a)
IF n
n=n-0.5
DRAW "RT",a
p(n,-a)
DRAW "FD4"
p(n,a)
DRAW "FD4"
p(n,-a)
DRAW "LT",a
ENDIF
RETURN
Example output
answered yesterday
Arnauld
69k584291
edited yesterday
answered yesterday
Arnauld
69k584291
answered yesterday
Arnauld
69k584291
answered yesterday
Arnauld
69k584291
69k584291
add a comment |
add a comment |
up vote
10
down vote
Perl 6, 117 bytes
{map ->y{|map {(((++$+y)%2+$++)%3**(y+$^v,*/3...*%3)??$^s[$++%2]!!'│')xx$_*3},<┌ ┐>,$_,<└ ┘>,1},^$_}o*R**3
Try it online!
0-indexed. Returns a 2D array of Unicode characters. The basic idea is that for lower rows, the expression
(x + (x+y)%2) % (3 ** trailing_zeros_in_base3(3*(y+1)))
yields the pattern
|....||....||....||....||.. % 3
..||....||....||....||....| % 3
|................||........ % 9
..||....||....||....||....| % 3
|....||....||....||....||.. % 3
........||................| % 9
|....||....||....||....||.. % 3
..||....||....||....||....| % 3
|.......................... % 27
For upper rows, the expression is
(x + (x+y+1)%2) % (3 ** trailing_zeros_in_base3(3*(y+3**n)))
Explanation
{ ... }o*R**3 # Feed $_ = 3^n into block
map ->y{ ... },^$_ # Map y = 0..3^n-1
|map { ... },<┌ ┐>,$_,<└ ┘>,1 # Map pairs (('┌','┐'),3^n) for upper rows
# and (('└','┘'),1) for lower rows.
# Block takes items as s and v
( ... )xx$_*3 # Evaluate 3^(n+1) times, returning a list
(++$+y)%2 # (x+y+1)%2 for upper rows, (x+y)%2 for lower rows
( +$++) # Add x
(y+$^v,*/3...*%3) # Count trailing zeros of 3*(y+v) in base 3
3** # nth power of 3
% # Modulo
??$^s[$++%2] # If there's a remainder yield chars in s alternately
!!'│' # otherwise yield '│'
answered yesterday
nwellnhof
5,9881122
add a comment |
up vote
10
down vote
Perl 6, 117 bytes
{map ->y{|map {(((++$+y)%2+$++)%3**(y+$^v,*/3...*%3)??$^s[$++%2]!!'│')xx$_*3},<┌ ┐>,$_,<└ ┘>,1},^$_}o*R**3
Try it online!
0-indexed. Returns a 2D array of Unicode characters. The basic idea is that for lower rows, the expression
(x + (x+y)%2) % (3 ** trailing_zeros_in_base3(3*(y+1)))
yields the pattern
|....||....||....||....||.. % 3
..||....||....||....||....| % 3
|................||........ % 9
..||....||....||....||....| % 3
|....||....||....||....||.. % 3
........||................| % 9
|....||....||....||....||.. % 3
..||....||....||....||....| % 3
|.......................... % 27
For upper rows, the expression is
(x + (x+y+1)%2) % (3 ** trailing_zeros_in_base3(3*(y+3**n)))
Explanation
{ ... }o*R**3 # Feed $_ = 3^n into block
map ->y{ ... },^$_ # Map y = 0..3^n-1
|map { ... },<┌ ┐>,$_,<└ ┘>,1 # Map pairs (('┌','┐'),3^n) for upper rows
# and (('└','┘'),1) for lower rows.
# Block takes items as s and v
( ... )xx$_*3 # Evaluate 3^(n+1) times, returning a list
(++$+y)%2 # (x+y+1)%2 for upper rows, (x+y)%2 for lower rows
( +$++) # Add x
(y+$^v,*/3...*%3) # Count trailing zeros of 3*(y+v) in base 3
3** # nth power of 3
% # Modulo
??$^s[$++%2] # If there's a remainder yield chars in s alternately
!!'│' # otherwise yield '│'
answered yesterday
nwellnhof
5,9881122
add a comment |
up vote
10
down vote
up vote
10
down vote
Perl 6, 117 bytes
{map ->y{|map {(((++$+y)%2+$++)%3**(y+$^v,*/3...*%3)??$^s[$++%2]!!'│')xx$_*3},<┌ ┐>,$_,<└ ┘>,1},^$_}o*R**3
Try it online!
0-indexed. Returns a 2D array of Unicode characters. The basic idea is that for lower rows, the expression
(x + (x+y)%2) % (3 ** trailing_zeros_in_base3(3*(y+1)))
yields the pattern
|....||....||....||....||.. % 3
..||....||....||....||....| % 3
|................||........ % 9
..||....||....||....||....| % 3
|....||....||....||....||.. % 3
........||................| % 9
|....||....||....||....||.. % 3
..||....||....||....||....| % 3
|.......................... % 27
For upper rows, the expression is
(x + (x+y+1)%2) % (3 ** trailing_zeros_in_base3(3*(y+3**n)))
Explanation
{ ... }o*R**3 # Feed $_ = 3^n into block
map ->y{ ... },^$_ # Map y = 0..3^n-1
|map { ... },<┌ ┐>,$_,<└ ┘>,1 # Map pairs (('┌','┐'),3^n) for upper rows
# and (('└','┘'),1) for lower rows.
# Block takes items as s and v
( ... )xx$_*3 # Evaluate 3^(n+1) times, returning a list
(++$+y)%2 # (x+y+1)%2 for upper rows, (x+y)%2 for lower rows
( +$++) # Add x
(y+$^v,*/3...*%3) # Count trailing zeros of 3*(y+v) in base 3
3** # nth power of 3
% # Modulo
??$^s[$++%2] # If there's a remainder yield chars in s alternately
!!'│' # otherwise yield '│'
answered yesterday
nwellnhof
5,9881122
Perl 6, 117 bytes
{map ->y{|map {(((++$+y)%2+$++)%3**(y+$^v,*/3...*%3)??$^s[$++%2]!!'│')xx$_*3},<┌ ┐>,$_,<└ ┘>,1},^$_}o*R**3
Try it online!
0-indexed. Returns a 2D array of Unicode characters. The basic idea is that for lower rows, the expression
(x + (x+y)%2) % (3 ** trailing_zeros_in_base3(3*(y+1)))
yields the pattern
|....||....||....||....||.. % 3
..||....||....||....||....| % 3
|................||........ % 9
..||....||....||....||....| % 3
|....||....||....||....||.. % 3
........||................| % 9
|....||....||....||....||.. % 3
..||....||....||....||....| % 3
|.......................... % 27
For upper rows, the expression is
(x + (x+y+1)%2) % (3 ** trailing_zeros_in_base3(3*(y+3**n)))
Explanation
{ ... }o*R**3 # Feed $_ = 3^n into block
map ->y{ ... },^$_ # Map y = 0..3^n-1
|map { ... },<┌ ┐>,$_,<└ ┘>,1 # Map pairs (('┌','┐'),3^n) for upper rows
# and (('└','┘'),1) for lower rows.
# Block takes items as s and v
( ... )xx$_*3 # Evaluate 3^(n+1) times, returning a list
(++$+y)%2 # (x+y+1)%2 for upper rows, (x+y)%2 for lower rows
( +$++) # Add x
(y+$^v,*/3...*%3) # Count trailing zeros of 3*(y+v) in base 3
3** # nth power of 3
% # Modulo
??$^s[$++%2] # If there's a remainder yield chars in s alternately
!!'│' # otherwise yield '│'
answered yesterday
nwellnhof
5,9881122
edited 19 hours ago
answered yesterday
nwellnhof
5,9881122
answered yesterday
nwellnhof
5,9881122
answered yesterday
nwellnhof
5,9881122
5,9881122
add a comment |
add a comment |
up vote
6
down vote
K (ngn/k), 37 27 26 bytes
{+y,(|'y:x,,>*x),x}/1,&2*
Try it online!
returns a boolean matrix
|'y is syntax specific to ngn/k. other dialects require a : to make an each-ed verb monadic: |:'y
answered yesterday
ngn
6,24312459
1
To make the output more beautiful, highlight all occurences of1(if supported by your browser)
– user202729
yesterday
3
@user202729 done - in the footer so it don't affect the byte count
– ngn
yesterday
add a comment |
up vote
6
down vote
K (ngn/k), 37 27 26 bytes
{+y,(|'y:x,,>*x),x}/1,&2*
Try it online!
returns a boolean matrix
|'y is syntax specific to ngn/k. other dialects require a : to make an each-ed verb monadic: |:'y
answered yesterday
ngn
6,24312459
1
To make the output more beautiful, highlight all occurences of1(if supported by your browser)
– user202729
yesterday
3
@user202729 done - in the footer so it don't affect the byte count
– ngn
yesterday
add a comment |
up vote
6
down vote
up vote
6
down vote
K (ngn/k), 37 27 26 bytes
{+y,(|'y:x,,>*x),x}/1,&2*
Try it online!
returns a boolean matrix
|'y is syntax specific to ngn/k. other dialects require a : to make an each-ed verb monadic: |:'y
answered yesterday
ngn
6,24312459
K (ngn/k), 37 27 26 bytes
{+y,(|'y:x,,>*x),x}/1,&2*
Try it online!
returns a boolean matrix
|'y is syntax specific to ngn/k. other dialects require a : to make an each-ed verb monadic: |:'y
answered yesterday
ngn
6,24312459
edited yesterday
answered yesterday
ngn
6,24312459
answered yesterday
ngn
6,24312459
answered yesterday
ngn
6,24312459
6,24312459
1
To make the output more beautiful, highlight all occurences of1(if supported by your browser)
– user202729
yesterday
3
@user202729 done - in the footer so it don't affect the byte count
– ngn
yesterday
add a comment |
1
To make the output more beautiful, highlight all occurences of1(if supported by your browser)
– user202729
yesterday
3
@user202729 done - in the footer so it don't affect the byte count
– ngn
yesterday
1
1
To make the output more beautiful, highlight all occurences of
1 (if supported by your browser)– user202729
yesterday
To make the output more beautiful, highlight all occurences of
1 (if supported by your browser)– user202729
yesterday
3
3
@user202729 done - in the footer so it don't affect the byte count
– ngn
yesterday
@user202729 done - in the footer so it don't affect the byte count
– ngn
yesterday
add a comment |
up vote
5
down vote
Wolfram Language 83 36 bytes, (possibly -48 bytes with bonus)
As of version 11.1, PeanoCurve is a built-in.
My original, clumsy submission wasted many bytes on GeometricTransformation and ReflectionTransform.
This much reduced version was suggested by alephalpha.
Reverse was required to orient the output properly.
Graphics[Reverse/@#&/@PeanoCurve@#]&
Example 36 bytes
Graphics[Reverse/@#&/@PeanoCurve@#]&[3]
Bonus
If this qualifies for the 100 pt bonus, it weighs in at 52 - 100 = -48
The code [5] was not counted, only the pure function.
Table[Graphics[Reverse/@#&/@PeanoCurve@#]&@k{k,#}&[5]
answered yesterday
DavidC
23.6k243101
Graphics[Reverse/@#&/@PeanoCurve@#]&
– alephalpha
yesterday
It feels a bit like cheating to have a function which calculates the Peano curve by itself, but I'll take it as accepted answer since it's pretty impressive nonetheless ;). @LegionMammal978 I think you deserve to post your own answer, I'd argue that it's different enough to warrant accepting it as winning answer.
– Peiffap
7 hours ago
add a comment |
up vote
5
down vote
Wolfram Language 83 36 bytes, (possibly -48 bytes with bonus)
As of version 11.1, PeanoCurve is a built-in.
My original, clumsy submission wasted many bytes on GeometricTransformation and ReflectionTransform.
This much reduced version was suggested by alephalpha.
Reverse was required to orient the output properly.
Graphics[Reverse/@#&/@PeanoCurve@#]&
Example 36 bytes
Graphics[Reverse/@#&/@PeanoCurve@#]&[3]
Bonus
If this qualifies for the 100 pt bonus, it weighs in at 52 - 100 = -48
The code [5] was not counted, only the pure function.
Table[Graphics[Reverse/@#&/@PeanoCurve@#]&@k{k,#}&[5]
answered yesterday
DavidC
23.6k243101
Graphics[Reverse/@#&/@PeanoCurve@#]&
– alephalpha
yesterday
It feels a bit like cheating to have a function which calculates the Peano curve by itself, but I'll take it as accepted answer since it's pretty impressive nonetheless ;). @LegionMammal978 I think you deserve to post your own answer, I'd argue that it's different enough to warrant accepting it as winning answer.
– Peiffap
7 hours ago
add a comment |
up vote
5
down vote
up vote
5
down vote
Wolfram Language 83 36 bytes, (possibly -48 bytes with bonus)
As of version 11.1, PeanoCurve is a built-in.
My original, clumsy submission wasted many bytes on GeometricTransformation and ReflectionTransform.
This much reduced version was suggested by alephalpha.
Reverse was required to orient the output properly.
Graphics[Reverse/@#&/@PeanoCurve@#]&
Example 36 bytes
Graphics[Reverse/@#&/@PeanoCurve@#]&[3]
Bonus
If this qualifies for the 100 pt bonus, it weighs in at 52 - 100 = -48
The code [5] was not counted, only the pure function.
Table[Graphics[Reverse/@#&/@PeanoCurve@#]&@k{k,#}&[5]
answered yesterday
DavidC
23.6k243101
Wolfram Language 83 36 bytes, (possibly -48 bytes with bonus)
As of version 11.1, PeanoCurve is a built-in.
My original, clumsy submission wasted many bytes on GeometricTransformation and ReflectionTransform.
This much reduced version was suggested by alephalpha.
Reverse was required to orient the output properly.
Graphics[Reverse/@#&/@PeanoCurve@#]&
Example 36 bytes
Graphics[Reverse/@#&/@PeanoCurve@#]&[3]
Bonus
If this qualifies for the 100 pt bonus, it weighs in at 52 - 100 = -48
The code [5] was not counted, only the pure function.
Table[Graphics[Reverse/@#&/@PeanoCurve@#]&@k{k,#}&[5]
answered yesterday
DavidC
23.6k243101
edited 3 hours ago
answered yesterday
DavidC
23.6k243101
answered yesterday
DavidC
23.6k243101
answered yesterday
DavidC
23.6k243101
23.6k243101
Graphics[Reverse/@#&/@PeanoCurve@#]&
– alephalpha
yesterday
It feels a bit like cheating to have a function which calculates the Peano curve by itself, but I'll take it as accepted answer since it's pretty impressive nonetheless ;). @LegionMammal978 I think you deserve to post your own answer, I'd argue that it's different enough to warrant accepting it as winning answer.
– Peiffap
7 hours ago
add a comment |
Graphics[Reverse/@#&/@PeanoCurve@#]&
– alephalpha
yesterday
It feels a bit like cheating to have a function which calculates the Peano curve by itself, but I'll take it as accepted answer since it's pretty impressive nonetheless ;). @LegionMammal978 I think you deserve to post your own answer, I'd argue that it's different enough to warrant accepting it as winning answer.
– Peiffap
7 hours ago
Graphics[Reverse/@#&/@PeanoCurve@#]&– alephalpha
yesterday
Graphics[Reverse/@#&/@PeanoCurve@#]&– alephalpha
yesterday
It feels a bit like cheating to have a function which calculates the Peano curve by itself, but I'll take it as accepted answer since it's pretty impressive nonetheless ;). @LegionMammal978 I think you deserve to post your own answer, I'd argue that it's different enough to warrant accepting it as winning answer.
– Peiffap
7 hours ago
It feels a bit like cheating to have a function which calculates the Peano curve by itself, but I'll take it as accepted answer since it's pretty impressive nonetheless ;). @LegionMammal978 I think you deserve to post your own answer, I'd argue that it's different enough to warrant accepting it as winning answer.
– Peiffap
7 hours ago
add a comment |
up vote
4
down vote
BBC BASIC, 142 ASCII characters (130 bytes tokenised)
Download interpreter at http://www.bbcbasic.co.uk/bbcwin/download.html
I.m:q=FNh(m,8,8,0)END
DEFFNh(n,x,y,i)
IFn F.i=0TO8q=FNh(n-1,x-i MOD3MOD2*2*x,y-i DIV3MOD2*2*y,i)DRAWBY(1AND36/2^i)*x,y*(12653/3^i MOD3-1)N.
=0
answered yesterday
Level River St
20.1k32579
add a comment |
up vote
4
down vote
BBC BASIC, 142 ASCII characters (130 bytes tokenised)
Download interpreter at http://www.bbcbasic.co.uk/bbcwin/download.html
I.m:q=FNh(m,8,8,0)END
DEFFNh(n,x,y,i)
IFn F.i=0TO8q=FNh(n-1,x-i MOD3MOD2*2*x,y-i DIV3MOD2*2*y,i)DRAWBY(1AND36/2^i)*x,y*(12653/3^i MOD3-1)N.
=0
answered yesterday
Level River St
20.1k32579
add a comment |
up vote
4
down vote
up vote
4
down vote
BBC BASIC, 142 ASCII characters (130 bytes tokenised)
Download interpreter at http://www.bbcbasic.co.uk/bbcwin/download.html
I.m:q=FNh(m,8,8,0)END
DEFFNh(n,x,y,i)
IFn F.i=0TO8q=FNh(n-1,x-i MOD3MOD2*2*x,y-i DIV3MOD2*2*y,i)DRAWBY(1AND36/2^i)*x,y*(12653/3^i MOD3-1)N.
=0
answered yesterday
Level River St
20.1k32579
BBC BASIC, 142 ASCII characters (130 bytes tokenised)
Download interpreter at http://www.bbcbasic.co.uk/bbcwin/download.html
I.m:q=FNh(m,8,8,0)END
DEFFNh(n,x,y,i)
IFn F.i=0TO8q=FNh(n-1,x-i MOD3MOD2*2*x,y-i DIV3MOD2*2*y,i)DRAWBY(1AND36/2^i)*x,y*(12653/3^i MOD3-1)N.
=0
answered yesterday
Level River St
20.1k32579
answered yesterday
Level River St
20.1k32579
answered yesterday
Level River St
20.1k32579
answered yesterday
Level River St
20.1k32579
20.1k32579
add a comment |
add a comment |
up vote
4
down vote
HTML+SVG+JS, 224 213 bytes
The output is mirrored horizontally.
n=>document.write(`<svg width=${w=3**n*9} height=${w}><path d="M1 ${(p=(n,z)=>n--&&(p(n,-z,a(a(p(n,-z,d+=z)),p(n,z))),d-=z))(n*2,r=d=x=y=1,a=_=>r+=`L${x+=~-(d&=3)%2*9} ${y+=(2-d)%2*9}`)&&r}"fill=#fff stroke=red>`)
Try it online! (prints the HTML)
(
n=>document.write(`<svg width=${w=3**n*9} height=${w}><path d="M1 ${(p=(n,z)=>n--&&(p(n,-z,a(a(p(n,-z,d+=z)),p(n,z))),d-=z))(n*2,r=d=x=y=1,a=_=>r+=`L${x+=~-(d&=3)%2*9} ${y+=(2-d)%2*9}`)&&r}"fill=#fff stroke=red>`)
)(3)answered yesterday
Arnauld
69k584291
add a comment |
up vote
4
down vote
HTML+SVG+JS, 224 213 bytes
The output is mirrored horizontally.
n=>document.write(`<svg width=${w=3**n*9} height=${w}><path d="M1 ${(p=(n,z)=>n--&&(p(n,-z,a(a(p(n,-z,d+=z)),p(n,z))),d-=z))(n*2,r=d=x=y=1,a=_=>r+=`L${x+=~-(d&=3)%2*9} ${y+=(2-d)%2*9}`)&&r}"fill=#fff stroke=red>`)
Try it online! (prints the HTML)
(
n=>document.write(`<svg width=${w=3**n*9} height=${w}><path d="M1 ${(p=(n,z)=>n--&&(p(n,-z,a(a(p(n,-z,d+=z)),p(n,z))),d-=z))(n*2,r=d=x=y=1,a=_=>r+=`L${x+=~-(d&=3)%2*9} ${y+=(2-d)%2*9}`)&&r}"fill=#fff stroke=red>`)
)(3)answered yesterday
Arnauld
69k584291
add a comment |
up vote
4
down vote
up vote
4
down vote
HTML+SVG+JS, 224 213 bytes
The output is mirrored horizontally.
n=>document.write(`<svg width=${w=3**n*9} height=${w}><path d="M1 ${(p=(n,z)=>n--&&(p(n,-z,a(a(p(n,-z,d+=z)),p(n,z))),d-=z))(n*2,r=d=x=y=1,a=_=>r+=`L${x+=~-(d&=3)%2*9} ${y+=(2-d)%2*9}`)&&r}"fill=#fff stroke=red>`)
Try it online! (prints the HTML)
(
n=>document.write(`<svg width=${w=3**n*9} height=${w}><path d="M1 ${(p=(n,z)=>n--&&(p(n,-z,a(a(p(n,-z,d+=z)),p(n,z))),d-=z))(n*2,r=d=x=y=1,a=_=>r+=`L${x+=~-(d&=3)%2*9} ${y+=(2-d)%2*9}`)&&r}"fill=#fff stroke=red>`)
)(3)answered yesterday
Arnauld
69k584291
HTML+SVG+JS, 224 213 bytes
The output is mirrored horizontally.
n=>document.write(`<svg width=${w=3**n*9} height=${w}><path d="M1 ${(p=(n,z)=>n--&&(p(n,-z,a(a(p(n,-z,d+=z)),p(n,z))),d-=z))(n*2,r=d=x=y=1,a=_=>r+=`L${x+=~-(d&=3)%2*9} ${y+=(2-d)%2*9}`)&&r}"fill=#fff stroke=red>`)
Try it online! (prints the HTML)
(
n=>document.write(`<svg width=${w=3**n*9} height=${w}><path d="M1 ${(p=(n,z)=>n--&&(p(n,-z,a(a(p(n,-z,d+=z)),p(n,z))),d-=z))(n*2,r=d=x=y=1,a=_=>r+=`L${x+=~-(d&=3)%2*9} ${y+=(2-d)%2*9}`)&&r}"fill=#fff stroke=red>`)
)(3)(
n=>document.write(`<svg width=${w=3**n*9} height=${w}><path d="M1 ${(p=(n,z)=>n--&&(p(n,-z,a(a(p(n,-z,d+=z)),p(n,z))),d-=z))(n*2,r=d=x=y=1,a=_=>r+=`L${x+=~-(d&=3)%2*9} ${y+=(2-d)%2*9}`)&&r}"fill=#fff stroke=red>`)
)(3)(
n=>document.write(`<svg width=${w=3**n*9} height=${w}><path d="M1 ${(p=(n,z)=>n--&&(p(n,-z,a(a(p(n,-z,d+=z)),p(n,z))),d-=z))(n*2,r=d=x=y=1,a=_=>r+=`L${x+=~-(d&=3)%2*9} ${y+=(2-d)%2*9}`)&&r}"fill=#fff stroke=red>`)
)(3)answered yesterday
Arnauld
69k584291
edited yesterday
answered yesterday
Arnauld
69k584291
answered yesterday
Arnauld
69k584291
answered yesterday
Arnauld
69k584291
69k584291
add a comment |
add a comment |
up vote
3
down vote
Logo, 89 bytes
to p:n:a
if:n>0[rt:a
p:n-1 0-:a
fw 5
p:n-1:a
fw 5
p:n-1 0-:a
lt:a]end
to f:n
p:n*2 90
end
Port of @Arnauld's Atari BASIC answer. To use, do something like this:
reset
f 3
answered 20 hours ago
Neil
78k744175
add a comment |
up vote
3
down vote
Logo, 89 bytes
to p:n:a
if:n>0[rt:a
p:n-1 0-:a
fw 5
p:n-1:a
fw 5
p:n-1 0-:a
lt:a]end
to f:n
p:n*2 90
end
Port of @Arnauld's Atari BASIC answer. To use, do something like this:
reset
f 3
answered 20 hours ago
Neil
78k744175
add a comment |
up vote
3
down vote
up vote
3
down vote
Logo, 89 bytes
to p:n:a
if:n>0[rt:a
p:n-1 0-:a
fw 5
p:n-1:a
fw 5
p:n-1 0-:a
lt:a]end
to f:n
p:n*2 90
end
Port of @Arnauld's Atari BASIC answer. To use, do something like this:
reset
f 3
answered 20 hours ago
Neil
78k744175
Logo, 89 bytes
to p:n:a
if:n>0[rt:a
p:n-1 0-:a
fw 5
p:n-1:a
fw 5
p:n-1 0-:a
lt:a]end
to f:n
p:n*2 90
end
Port of @Arnauld's Atari BASIC answer. To use, do something like this:
reset
f 3
answered 20 hours ago
Neil
78k744175
answered 20 hours ago
Neil
78k744175
answered 20 hours ago
Neil
78k744175
answered 20 hours ago
Neil
78k744175
78k744175
add a comment |
add a comment |
Peiffap is a new contributor. Be nice, and check out our Code of Conduct.
Peiffap is a new contributor. Be nice, and check out our Code of Conduct.
Peiffap is a new contributor. Be nice, and check out our Code of Conduct.
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1
Welcome to PPCG :) This, at a glance, looks like a nice first challenge. Although it feels familiar, I think the challenge I might be thinking of was an ASCII art one. Note, though, that we strongly discourage bonuses and that there will be golfing languages that can achieve this in less than 100 bytes. Also, and most importantly, you need a winning criterion. Given that your bonuses subtract bytes from a solution's score, I suspect you intend this to be code-golf.
– Shaggy
yesterday
4
Yeah, I don't think the bonuses are a good idea, especially since there at least two animation-capable ASCII-art focused golflangs
– ASCII-only
yesterday
2
Oh also what would
nbe used for iflandware also inputs??????????? And would the Peano curve be a special case - it's not the only spacefilling curve, so some algorithms may have to specialcase it– ASCII-only
yesterday
2
Also, what stops anyone from making a trivial spacefilling curve (just zigzagging back and forth) for non-square dimensions
– ASCII-only
yesterday
6
Bonuses in code golf is one of the most agreed-upon things to avoid when writing challenges. I suggest you remove them and decide which is to be the canonical version of the challenge.
– lirtosiast
yesterday