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swapped angles to match anchor points
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jsh
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EDIT: The incorrect A rhomb substitution has been fixed. Apologies to anoyone who had started working on a solution.

Consider the following substitutions, where the substituted rhomb(us) is scaled up and shown in red:

EDIT: Note the anchor points (circle) created where a corner of each new rhomb touches the substituted rhomb. Also note the substitution's anchor point (black circle).

A rhomb (144°, 36°)

rhomb a ... becomes ... rhomb a subsitution 1

B rhomb (108°72°, 72°108°)

rhomb b ... becomes ... rhomb b substitution 1

Each substitution is made up of some combination of A rhombs and B rhombs, on which the same substitutions can be performed.

A tiling can be created by repeating the substitutions n times.

After n=2 substitutions of A rhomb:

EDIT: The substitution is oriented by aligning the rhomb's anchor point with the substitution's anchor point

rhomb a substitution 2

In the fewest bytes of code (given input n>=0) draw the tiling after n substitutions, where n=0 is A rhomb


Notes:

  • The red lines and anchor points shouldn't be included, they only illustrate the substitution.
  • In each substitution the tiling is scaled up so the rhombs have a constant size.
  • Any rotation is optional.
  • Graphical choices (colour, dimensions, cropping etc.) are optional if not exploited for loopholes.

EDIT: The incorrect A rhomb substitution has been fixed. Apologies to anoyone who had started working on a solution.

Consider the following substitutions, where the substituted rhomb(us) is scaled up and shown in red:

EDIT: Note the anchor points (circle) created where a corner of each new rhomb touches the substituted rhomb. Also note the substitution's anchor point (black circle).

A rhomb (144°, 36°)

rhomb a ... becomes ... rhomb a subsitution 1

B rhomb (108°, 72°)

rhomb b ... becomes ... rhomb b substitution 1

Each substitution is made up of some combination of A rhombs and B rhombs, on which the same substitutions can be performed.

A tiling can be created by repeating the substitutions n times.

After n=2 substitutions of A rhomb:

EDIT: The substitution is oriented by aligning the rhomb's anchor point with the substitution's anchor point

rhomb a substitution 2

In the fewest bytes of code (given input n>=0) draw the tiling after n substitutions, where n=0 is A rhomb


Notes:

  • The red lines and anchor points shouldn't be included, they only illustrate the substitution.
  • In each substitution the tiling is scaled up so the rhombs have a constant size.
  • Any rotation is optional.
  • Graphical choices (colour, dimensions, cropping etc.) are optional if not exploited for loopholes.

EDIT: The incorrect A rhomb substitution has been fixed. Apologies to anoyone who had started working on a solution.

Consider the following substitutions, where the substituted rhomb(us) is scaled up and shown in red:

EDIT: Note the anchor points (circle) created where a corner of each new rhomb touches the substituted rhomb. Also note the substitution's anchor point (black circle).

A rhomb (144°, 36°)

rhomb a ... becomes ... rhomb a subsitution 1

B rhomb (72°, 108°)

rhomb b ... becomes ... rhomb b substitution 1

Each substitution is made up of some combination of A rhombs and B rhombs, on which the same substitutions can be performed.

A tiling can be created by repeating the substitutions n times.

After n=2 substitutions of A rhomb:

EDIT: The substitution is oriented by aligning the rhomb's anchor point with the substitution's anchor point

rhomb a substitution 2

In the fewest bytes of code (given input n>=0) draw the tiling after n substitutions, where n=0 is A rhomb


Notes:

  • The red lines and anchor points shouldn't be included, they only illustrate the substitution.
  • In each substitution the tiling is scaled up so the rhombs have a constant size.
  • Any rotation is optional.
  • Graphical choices (colour, dimensions, cropping etc.) are optional if not exploited for loopholes.
explained anchor points
Source Link
jsh
  • 919
  • 7
  • 22

EDIT: The incorrect A rhomb substitution has been fixed. Apologies to anoyone who has started working on a solution. The incorrect A rhomb substitution has been fixed. Apologies to anoyone who had started working on a solution.

Consider the following substitutions, where the substituted rhomb(us) is scaled up and shown in red:

EDIT: Note the anchor points (circle) created where a corner of each new rhomb touches the substituted rhomb. Also note the substitution's anchor point (black circle).

A rhomb (144°, 36°)

rhomb a ... becomes ... rhomb a subsitution 1

B rhomb (108°, 72°)

rhomb b ... becomes ... rhomb b substitution 1

Each substitution is made up of some combination of A rhombs and B rhombs, on which the same substitutions can be performed.

A tiling can be created by repeating the substitutions n times.

After n=2 substitutions of A rhomb:

EDIT: The substitution is oriented by aligning the rhomb's anchor point with the substitution's anchor point

rhomb a substitution 2

In the fewest bytes of code (given input n>=0) draw the tiling after n substitutions, where n=0 is A rhomb


Notes:

  • Substitutions are tiled without leaving gaps or overlapping.
  • The red lines and anchor points shouldn't be included, they only illustrate the substitution.
  • In each substitution the tiling is scaled up so the rhombs have a constant size.
  • Any rotation is optional.
  • Graphical choices (colour, dimensions, cropping etc.) are optional if not exploited for loopholes.

EDIT: The incorrect A rhomb substitution has been fixed. Apologies to anoyone who has started working on a solution.

Consider the following substitutions, where the substituted rhomb(us) is scaled up and shown in red:

A rhomb (144°, 36°)

rhomb a ... becomes ... rhomb a subsitution 1

B rhomb (108°, 72°)

rhomb b ... becomes ... rhomb b substitution 1

Each substitution is made up of some combination of A rhombs and B rhombs, on which the same substitutions can be performed.

A tiling can be created by repeating the substitutions n times.

After n=2 substitutions of A rhomb:

rhomb a substitution 2

In the fewest bytes of code (given input n>=0) draw the tiling after n substitutions, where n=0 is A rhomb


Notes:

  • Substitutions are tiled without leaving gaps or overlapping.
  • The red lines shouldn't be included, they only illustrate the substitution.
  • In each substitution the tiling is scaled up so the rhombs have a constant size.
  • Any rotation is optional.
  • Graphical choices (colour, dimensions, cropping etc.) are optional if not exploited for loopholes.

EDIT: The incorrect A rhomb substitution has been fixed. Apologies to anoyone who had started working on a solution.

Consider the following substitutions, where the substituted rhomb(us) is scaled up and shown in red:

EDIT: Note the anchor points (circle) created where a corner of each new rhomb touches the substituted rhomb. Also note the substitution's anchor point (black circle).

A rhomb (144°, 36°)

rhomb a ... becomes ... rhomb a subsitution 1

B rhomb (108°, 72°)

rhomb b ... becomes ... rhomb b substitution 1

Each substitution is made up of some combination of A rhombs and B rhombs, on which the same substitutions can be performed.

A tiling can be created by repeating the substitutions n times.

After n=2 substitutions of A rhomb:

EDIT: The substitution is oriented by aligning the rhomb's anchor point with the substitution's anchor point

rhomb a substitution 2

In the fewest bytes of code (given input n>=0) draw the tiling after n substitutions, where n=0 is A rhomb


Notes:

  • The red lines and anchor points shouldn't be included, they only illustrate the substitution.
  • In each substitution the tiling is scaled up so the rhombs have a constant size.
  • Any rotation is optional.
  • Graphical choices (colour, dimensions, cropping etc.) are optional if not exploited for loopholes.
fixed a rhomb substitution
Source Link
jsh
  • 919
  • 7
  • 22

EDIT: The incorrect A rhomb substitution has been fixed. Apologies to anoyone who has started working on a solution.

Consider the following substitutions, where the substituted rhomb(us) is scaled up and shown in red:

A rhomb (144°, 36°)

rhomb a ... becomes ... rhomb a substitution 1rhomb a subsitution 1

B rhomb (108°, 72°)

rhomb b    ... becomes ... rhomb b substitution 1 rhomb b substitution 1

Each substitution is made up of some combination of A rhombs and B rhombs, on which the same substitutions can be performed.

A tiling can be created by repeating the substitutions n times.

After n=2 substitutions of A rhomb:

rhomb a substitution 2rhomb a substitution 2

In the fewest bytes of code (given input n>=0) draw the tiling after n substitutions, where n=0 is A rhomb


Notes:

  • Substitutions are tiled without leaving gaps or overlapping.
  • The red lines shouldn't be included, they only illustrate the substitution.
  • In each substitution the tiling is scaled up so the rhombs have a constant size.
  • Any rotation is optional.
  • Graphical choices (colour, dimensions, cropping etc.) are optional if not exploited for loopholes.

Consider the following substitutions, where the substituted rhomb(us) is scaled up and shown in red:

A rhomb (144°, 36°)

rhomb a ... becomes ... rhomb a substitution 1

B rhomb (108°, 72°)

rhomb b  ... becomes ... rhomb b substitution 1

Each substitution is made up of some combination of A rhombs and B rhombs, on which the same substitutions can be performed.

A tiling can be created by repeating the substitutions n times.

After n=2 substitutions of A rhomb:

rhomb a substitution 2

In the fewest bytes of code (given input n>=0) draw the tiling after n substitutions, where n=0 is A rhomb


Notes:

  • Substitutions are tiled without leaving gaps or overlapping.
  • The red lines shouldn't be included, they only illustrate the substitution.
  • In each substitution the tiling is scaled up so the rhombs have a constant size.
  • Any rotation is optional.
  • Graphical choices (colour, dimensions, cropping etc.) are optional if not exploited for loopholes.

EDIT: The incorrect A rhomb substitution has been fixed. Apologies to anoyone who has started working on a solution.

Consider the following substitutions, where the substituted rhomb(us) is scaled up and shown in red:

A rhomb (144°, 36°)

rhomb a ... becomes ... rhomb a subsitution 1

B rhomb (108°, 72°)

rhomb b  ... becomes ... rhomb b substitution 1

Each substitution is made up of some combination of A rhombs and B rhombs, on which the same substitutions can be performed.

A tiling can be created by repeating the substitutions n times.

After n=2 substitutions of A rhomb:

rhomb a substitution 2

In the fewest bytes of code (given input n>=0) draw the tiling after n substitutions, where n=0 is A rhomb


Notes:

  • Substitutions are tiled without leaving gaps or overlapping.
  • The red lines shouldn't be included, they only illustrate the substitution.
  • In each substitution the tiling is scaled up so the rhombs have a constant size.
  • Any rotation is optional.
  • Graphical choices (colour, dimensions, cropping etc.) are optional if not exploited for loopholes.
Tweeted twitter.com/StackCodeGolf/status/728078133975384064
added rhomb angles
Source Link
jsh
  • 919
  • 7
  • 22
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Source Link
jsh
  • 919
  • 7
  • 22
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