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:
Three iterations of the Peano curve

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 w where n has the same meaning as before (the number of the iteration), but where l and w become the length and width of the rectangle in which to draw the curve. If l == 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).

edited yesterday

asked yesterday

Peiffap

726

New contributor

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 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

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

Introduction

Challenge

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 w where n has the same meaning as before (the number of the iteration), but where l and w become the length and width of the rectangle in which to draw the curve. If l == 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).

edited yesterday

asked yesterday

Peiffap

726

New contributor

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 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

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

Introduction

Challenge

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 w where n has the same meaning as before (the number of the iteration), but where l and w become the length and width of the rectangle in which to draw the curve. If l == 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).

edited yesterday

asked yesterday

Peiffap

726

New contributor

Introduction

Challenge

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 w where n has the same meaning as before (the number of the iteration), but where l and w become the length and width of the rectangle in which to draw the curve. If l == 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

edited yesterday

asked yesterday

Peiffap

726

New contributor

edited yesterday

asked yesterday

Peiffap

726

New contributor

edited yesterday

asked yesterday

Peiffap

726

New contributor

asked yesterday

Peiffap

726

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 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

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 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

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

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

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

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

Also, what stops anyone from making a trivial spacefilling curve (just zigzagging back and forth) for non-square dimensions
– ASCII-only
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

peano-gfa

edited yesterday

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 '│'

edited 19 hours ago

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

edited yesterday

answered yesterday

ngn

6,24312459

1

To make the output more beautiful, highlight all occurences of 1 (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]

Peano curve

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]

sequence

edited 3 hours ago

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

enter image description here

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)

edited yesterday

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 |

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8 Answers
8

active

oldest

votes

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}&

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}&

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}&

answered 4 hours ago

LegionMammal978

15k41752

Mathematica, score 60 - 100 - 100 = -140

Graphics[PeanoCurve@a~Reverse~3~Scale~#2]~Animate~{a,1,#,1}&

answered 4 hours ago

LegionMammal978

15k41752

answered 4 hours ago

LegionMammal978

15k41752

answered 4 hours ago

LegionMammal978

15k41752

answered 4 hours ago

LegionMammal978

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

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

peano-gfa

edited yesterday

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

peano-gfa

edited yesterday

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

peano-gfa

edited yesterday

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

peano-gfa

edited yesterday

answered yesterday

Arnauld

69k584291

edited yesterday

answered yesterday

Arnauld

69k584291

answered yesterday

Arnauld

69k584291

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 '│'

edited 19 hours ago

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 '│'

edited 19 hours ago

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 '│'

edited 19 hours ago

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 '│'

edited 19 hours ago

answered yesterday

nwellnhof

5,9881122

edited 19 hours ago

answered yesterday

nwellnhof

5,9881122

answered yesterday

nwellnhof

5,9881122

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

edited yesterday

answered yesterday

ngn

6,24312459

1

To make the output more beautiful, highlight all occurences of 1 (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

edited yesterday

answered yesterday

ngn

6,24312459

1

To make the output more beautiful, highlight all occurences of 1 (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

edited yesterday

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

edited yesterday

answered yesterday

ngn

6,24312459

edited yesterday

answered yesterday

ngn

6,24312459

answered yesterday

ngn

6,24312459

answered yesterday

ngn

6,24312459

1

To make the output more beautiful, highlight all occurences of 1 (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 of 1 (if supported by your browser)
– user202729
yesterday

3

@user202729 done - in the footer so it don't affect the byte count
– ngn
yesterday

To make the output more beautiful, highlight all occurences of 1 (if supported by your browser)
– user202729
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]

Peano curve

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]

sequence

edited 3 hours ago

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]

Peano curve

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]

sequence

edited 3 hours ago

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]

Peano curve

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]

sequence

edited 3 hours ago

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]

Peano curve

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]

sequence

edited 3 hours ago

answered yesterday

DavidC

23.6k243101

edited 3 hours ago

answered yesterday

DavidC

23.6k243101

answered yesterday

DavidC

23.6k243101

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 |

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

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

enter image description here

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

enter image description here

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

enter image description here

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

enter image description here

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

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)

edited yesterday

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)

edited yesterday

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)

edited yesterday

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)

edited yesterday

answered yesterday

Arnauld

69k584291

edited yesterday

answered yesterday

Arnauld

69k584291

answered yesterday

Arnauld

69k584291

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 |

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

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

add a comment |

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Drawing the Peano curve

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8 Answers 8

Mathematica, score 60 - 100 - 100 = -140

GFA Basic 3.51 (Atari ST), 156 134 124 bytes

Example output

Perl 6, 117 bytes

Explanation

K (ngn/k), 37 27 26 bytes

Wolfram Language 83 36 bytes, (possibly -48 bytes with bonus)

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BBC BASIC, 142 ASCII characters (130 bytes tokenised)

HTML+SVG+JS, 224 213 bytes

Logo, 89 bytes

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8 Answers 8

8 Answers 8

Mathematica, score 60 - 100 - 100 = -140

Mathematica, score 60 - 100 - 100 = -140

Mathematica, score 60 - 100 - 100 = -140

Mathematica, score 60 - 100 - 100 = -140

GFA Basic 3.51 (Atari ST), 156 134 124 bytes

Example output

GFA Basic 3.51 (Atari ST), 156 134 124 bytes

Example output

GFA Basic 3.51 (Atari ST), 156 134 124 bytes

Example output

GFA Basic 3.51 (Atari ST), 156 134 124 bytes

Example output

Perl 6, 117 bytes

Explanation

Perl 6, 117 bytes

Explanation

Perl 6, 117 bytes

Explanation

Perl 6, 117 bytes

Explanation

K (ngn/k), 37 27 26 bytes

K (ngn/k), 37 27 26 bytes

K (ngn/k), 37 27 26 bytes

K (ngn/k), 37 27 26 bytes

Wolfram Language 83 36 bytes, (possibly -48 bytes with bonus)

Bonus

Wolfram Language 83 36 bytes, (possibly -48 bytes with bonus)

Bonus

Wolfram Language 83 36 bytes, (possibly -48 bytes with bonus)

Bonus

Wolfram Language 83 36 bytes, (possibly -48 bytes with bonus)

8 Answers
8

8 Answers
8

8 Answers
8