"Peter Popov" <pet### [at] vipbg> wrote:
>
> >So I wonder, if there are other, more advanced algorithms out there, which are
> >faster ...
>
> Use an oct-tree like POV does for bounding slabs.

in POV-SDL ? ...

I've thought about this, but I dunno how I could
do this, if I haven't some kind of "references"

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"Jan Walzer" <jan### [at] lzernet> schrieb im Newsbeitrag
news:3cbd317a$1@news.povray.org...
> yeah ... interesting technique ...

Thanx, I did it that way, because I never used trace() before :-)

> how well will this work with
> different sized spheres ?

Good, it actually does. Look again at p.b.images ;-) The size of the sphere
radius is a random number between half of square size and 0.

> would this mean that I have to
> mark some adjacent boxes as
> loaded ?

Not in my approach. But your idea should also work, but is more complicated.
You have to have
a bigger array which is a lot slower. The advance of your approach is that
there is no free space in form of a square around a sphere:

+--------+
|        |
|        |
|        |
|     O  |
+--------+

If I put a sphere into a square like above, no other sphere can be put into
this
square, even if there is enough space left. That means that you never will
fill the whole place totally. As long, as you have small amounts of spheres
(e.g. 50000) compared to the highest amount of spheres (e.g. 250000) that
fits in the array (e.g. 500x500), the result will be o.k.

So long,

Bonsai

P.S. If my englisch explanation is not clear enough, you can also mail me
directly to switch over to German ;-)

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"JRG" <jrg### [at] hotmailcom> wrote:
> Usually jittering is done in such way to make it impossible to have intersections.
> That would make it a O(0)... ;-)

BTW: I just read your text again...

O(0) is not possible for any real algorithm ...
O(1) is for normal cases the lowest you can get ...
and your algorithm is
O(n) when n is number of spheres ...

... just some nitpicking ...

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Jan Walzer <jan### [at] lzernet> wrote in message
news:3cbc5eab@news.povray.org...

I wouldn't consider myself an advanced user, but I do have a suggestion.

I would cover the viewing area in rows of boxes with a random x and z size
within a range. I would then randomly select the number of balls to be
placed in each box and place the balls as I go, checking for intersections
within this small box. I think that this would give the best pseudo-random
look in the least amount of time.

 -Shay

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In article <3cbd31c0$1@news.povray.org>, "Jan Walzer" <jan### [at] lzernet> 
wrote:

> in POV-SDL ? ...
> 
> I've thought about this, but I dunno how I could
> do this, if I haven't some kind of "references"

You can use unions to group objects, and you can put unions inside 
unions.

Assuming you have a line of 16 objects like this:

A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P

Split it into two groups, put each into a union and make a union of the 
unions. Split each sub-union the same way.
The resulting structure would be like:

union {
    union {
        union {
            union {A, B},
            union {C, D}
        },
        union {
            union {E, F},
            union {G, H}
        }
    },
    union {
        union {
            union {I, J},
            union {K, L}
        },
        union {
            union {M, N},
            union {O, P}
        }
    }
}

This basically makes POV do a binary search through the objects instead 
of brute-force testing all of them. I used a line for simplicity, but 
the technique could be used for 2D and 3D grids or other structures.

-- 
Christopher James Huff <chr### [at] maccom>
POV-Ray TAG e-mail: chr### [at] tagpovrayorg
TAG web site: http://tag.povray.org/

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Sorry, but I don't get, how this could
help me with my current problem. Maybe
because its a bit late already... Give
me the chance to sleep about this idea
,but I doubt that this union-structure
will help me very much with a good way
to test for intersections.

What I want to to is to compute all of
the positions in SDL before postioning
them. But to do this efficient I think
I need something like references to be
able to create list or tree structures
like they are common in other programm-
ing languages.

It could be able to create such things
with the help of some arrays and a lot
macros. But the overhead would be more
than neccesary and the efficiency gets
lost, so it would no longer be faster
than my current approach...

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"Jan Walzer" wrote:
>
> "JRG" <jrg### [at] hotmailcom> wrote:
> > Usually jittering is done in such way to make it impossible to have
intersections.
> > That would make it a O(0)... ;-)
>
> BTW: I just read your text again...
>
> O(0) is not possible for any real algorithm ...

I was referring to the intersection testing algorithm alone, not to the placing
algorithm as a whole... no test, no algorithm, no operations, nada de nada... hence
the O(0) which, anyway, was just a joke... :)

--
Jonathan.

Home: http://digilander.iol.it/jrgpov

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> So I wonder, if there are other, more advanced algorithms out there, which
are
> faster ...

Here's an idea I just came up with.

Do it just like you are. Except, as you place the spheres, keep the array
that holds their positions *sorted*. So, when you add a new value to the
array, use some sort of binary search to find the part of the array where
the new position should be kept. (To sort the array, do something like
sorting first in the X direction, and then in the Z direction.) Then insert
the new value there, pushing the rest of the values ahead.

Then, when you have a new position, you still have to check the array to
make sure it doesn't conflict with any of the positions already there.
However, you no longer have to check the whole array - you only have to
check the part of the array that's within twice the maximum possible radius
from the new position you're testing.

It might be difficult to implement, but it might really speed it up. The key
thing here is to search the array with a binary search or something else
that takes advantage of its sortedness.

- Slime
[ http://www.slimeland.com/ ]
[ http://www.slimeland.com/images/ ]

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On Wed, 17 Apr 2002 10:26:41 +0200, "Jan Walzer" <jan### [at] lzernet>
wrote:

>in POV-SDL ? ...

Supposedly :)

>I've thought about this, but I dunno how I could
>do this, if I haven't some kind of "references"

You can make a binary tree with a one-dimensional array. The extension
to oct-trees should be straightforward.


Peter Popov ICQ : 15002700
Personal e-mail : pet### [at] vipbg
TAG      e-mail : pet### [at] tagpovrayorg

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I have come up with an algorithm that, although it is still O(n^2), can
place 16,000 spheres in, if I remember correctly, 13 minutes in the POV SDL.
It gets faster, of course, when the spheres are less densely placed. In this
example, about one out of every ten spheres needed more than one placement
to find a place that didn't intersect with other spheres.

I'm rendering now, I can post the source or the steps to the algorithm when
it's done, if you like. Might be a while... with focal blur and media, it's
going at 2 pps along the horizon line; so expect at least another day.

- Slime
[ http://www.slimeland.com/ ]
[ http://www.slimeland.com/images/ ]

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