|  | Koch snowflakes | | |
| | | conrad |  |
|  Posted: Sun Sep 07, 2008 1:50 am Post subject: Koch snowflakes |  |
If I generate a Koch snowflake of
an order higher than 1, then it conflicts
with my recursive algorithm because:
i) A Koch Snowflake of order 1 involves
generating three triangles from the sides
of the initial triangle.
ii) All Koch Snowflakes of order >= 2
involves the creation of only two
triangles. The third is not needed
because the third side becomes
non-existent.
How can I rectify this?
I initially thought: keep it general
in the sense that I should not
treat the order = 1 case as special.
In this way, I can check for an
order >= 2 Koch Snowflake by
determining which point generated ends
up within the area of the polygon.
I'm only posting this because I'm
curious if there is a different approach
that someone would have taken.
--
conrad |
| |
| | | Gene | |
| | Posted: Sun Sep 07, 2008 3:13 am Post subject: Re: Koch snowflakes | |
| |  | |
On Sep 6, 9:50 pm, conrad <con...@lawyer.com> wrote:
| Quote: | If I generate a Koch snowflake of
an order higher than 1, then it conflicts
with my recursive algorithm because:
i) A Koch Snowflake of order 1 involves
generating three triangles from the sides
of the initial triangle.
ii) All Koch Snowflakes of order >= 2
involves the creation of only two
triangles. The third is not needed
because the third side becomes
non-existent.
How can I rectify this?
I initially thought: keep it general
in the sense that I should not
treat the order = 1 case as special.
In this way, I can check for an
order >= 2 Koch Snowflake by
determining which point generated ends
up within the area of the polygon.
I'm only posting this because I'm
curious if there is a different approach
that someone would have taken.
|
I'm not sure what you're saying. Here's an alogorithm to return a
list of all the star vertices:
function koch (k)
let s = sqrt(3)/2
if k = 1 return [ [0, 1], [-s -1/2], [s, -1/2], [0, 1] ]
rtn = [];
for each adjacent pair [a, b] in koch(k - 1)
let v = b - a
c = a + 1/3 * v
d = a + 2/3 * v
e = s * perp(d - c)
append [ a, c, e, d, b ] onto rtn
return rtn
Note that
perp([x, y]) = [-y, x] |
| |
| | | CBFalconer | |
| | Posted: Sun Sep 07, 2008 6:23 am Post subject: Re: Koch snowflakes | |
| | | |
conrad wrote:
| Quote: |
If I generate a Koch snowflake of an order higher than 1, then it
conflicts with my recursive algorithm because:
i) A Koch Snowflake of order 1 involves generating three
triangles from the sides of the initial triangle.
ii) All Koch Snowflakes of order >= 2 involves the creation of
only two triangles. The third is not needed because the
third side becomes non-existent.
How can I rectify this? I initially thought: keep it general in
the sense that I should not treat the order = 1 case as special.
In this way, I can check for an order >= 2 Koch Snowflake by
determining which point generated ends up within the area of the
polygon.
I'm only posting this because I'm curious if there is a different
approach that someone would have taken.
|
What is a Koch snowflake? What is a Koch snowflake order? What is
your recursive algorithm?
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section. |
| |
| | | jellybean stonerfish | |
| | Posted: Sun Sep 07, 2008 6:50 am Post subject: Re: Koch snowflakes | |
On Sun, 07 Sep 2008 04:23:54 -0400, CBFalconer wrote:
| Quote: |
What is a Koch snowflake?
Stop pretending your stupid. |
| Quote: | What is a Koch snowflake order?
Stop pretending your stupid. |
| Quote: | What is your recursive algorithm?
|
Good guestion. |
| |
| | | Martin Eisenberg | |
| | Posted: Sun Sep 07, 2008 8:40 am Post subject: Re: Koch snowflakes | |
conrad wrote:
| Quote: | If I generate a Koch snowflake of
an order higher than 1, then it conflicts
with my recursive algorithm because:
i) A Koch Snowflake of order 1 involves
generating three triangles from the sides
of the initial triangle.
ii) All Koch Snowflakes of order >= 2
involves the creation of only two
triangles. The third is not needed
because the third side becomes
non-existent.
How can I rectify this?
|
If I understand well:
Don't *count* the edges, just *walk* them.
Martin
--
Quidquid latine scriptum est, altum videtur. |
| |
| | | Richard Heathfield | |
| | Posted: Sun Sep 07, 2008 10:27 am Post subject: Re: Koch snowflakes | |
CBFalconer said:
<snip>
| Quote: | What is a Koch snowflake?
|
Is this a joke?
| Quote: | What is a Koch snowflake order?
|
An equilateral triangle is a Koch snowflake of order 0.
To obtain a Koch snowflake S of order N, given a Koch snowflake K of order
(N - 1):
for each straight line segment on K (as it existed before any mods):
(a) identify C, the central third (33.3etc%) of the line segment
(b) consider C to be the base of an equilateral triangle "pointing"
away from the centre of the figure
(c) complete the equilateral triangle by drawing two new line segments
(d) remove C from the figure
endfor
| Quote: | What is your recursive algorithm?
|
The Web contains many examples.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999 |
| |
| | | Willem | |
| | Posted: Sun Sep 07, 2008 11:11 am Post subject: Re: Koch snowflakes | |
conrad wrote:
) If I generate a Koch snowflake of
) an order higher than 1, then it conflicts
) with my recursive algorithm because:
) i) A Koch Snowflake of order 1 involves
) generating three triangles from the sides
) of the initial triangle.
) ii) All Koch Snowflakes of order >= 2
) involves the creation of only two
) triangles. The third is not needed
) because the third side becomes
) non-existent.
Doesn't a Koch Snowflake work by changing each
edge of the current shape into four edges ?
They just happen to be at 60-degree angles,
and you just happen to start with a triangle.
How are you using triangles to do it ?
As I see it, that would miss out a lot of sides.
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT |
| |
| | | conrad | |
| | Posted: Sun Sep 07, 2008 12:48 pm Post subject: Re: Koch snowflakes | |
| | | |
On Sep 7, 6:11 am, Willem <wil...@stack.nl> wrote:
| Quote: | conrad wrote:
) If I generate a Koch snowflake of
) an order higher than 1, then it conflicts
) with my recursive algorithm because:
) i) A Koch Snowflake of order 1 involves
) generating three triangles from the sides
) of the initial triangle.
) ii) All Koch Snowflakes of order >= 2
) involves the creation of only two
) triangles. The third is not needed
) because the third side becomes
) non-existent.
Doesn't a Koch Snowflake work by changing each
edge of the current shape into four edges ?
They just happen to be at 60-degree angles,
and you just happen to start with a triangle.
How are you using triangles to do it ?
As I see it, that would miss out a lot of sides.
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
|
Yeah, I only realized this after I had started to write the algorithm
for it.
The book that I was doing the problem from did not have a complete
explanation of a Koch Snowflake and so I filled in some gaps
with my own ideas. However, after having found a complete explanation
I understand the problem.
And understanding the problem is half the battle.
--
conrad |
| |
| | | CBFalconer | |
| | Posted: Sun Sep 07, 2008 8:23 pm Post subject: Re: Koch snowflakes | |
| | | |
Richard Heathfield wrote:
| Quote: | CBFalconer said:
snip
What is a Koch snowflake?
Is this a joke?
|
No. I never heard of such.
| Quote: |
What is a Koch snowflake order?
An equilateral triangle is a Koch snowflake of order 0.
|
Is it the only one?
| Quote: |
To obtain a Koch snowflake S of order N, given a Koch snowflake K of order
(N - 1):
for each straight line segment on K (as it existed before any mods):
(a) identify C, the central third (33.3etc%) of the line segment
(b) consider C to be the base of an equilateral triangle "pointing"
away from the centre of the figure
(c) complete the equilateral triangle by drawing two new line segments
(d) remove C from the figure
endfor
What is your recursive algorithm?
The Web contains many examples.
|
Thanks. Two seconds work with a triangle and piece of paper gives
me the idea. What are they used for?
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section. |
| |
| | | Richard Heathfield | |
| | Posted: Mon Sep 08, 2008 3:42 am Post subject: Re: Koch snowflakes | |
CBFalconer said:
| Quote: | Richard Heathfield wrote:
CBFalconer said:
snip
What is a Koch snowflake?
Is this a joke?
No. I never heard of such.
|
Sierpinksi carpet? Menger sponge? Mandelbrot set? Julia set?
| Quote: | What is a Koch snowflake order?
An equilateral triangle is a Koch snowflake of order 0.
Is it the only one?
|
Yes. Nevertheless, you could use some other figure to get a pseudoKoch
snowflake. For example, Eric Haines has produced a "sphereflake".
<snip>
| Quote: | What are they used for?
|
What are Mandelbrot sets used for?
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999 |
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