Handy TILE component...
http://chq.emehmedovic.com/?id=2
On 5/9/06, Bernard Poulin <[EMAIL PROTECTED]> wrote:
Wow!
I just tried your algorithm with my previous example numbers and it does
output the correct square size (100) - also, internally it has the right
number of columns/lines: e.g. 3x4 (p=3, q=4) As for performance, it took
6
iterations: Since the output was 3x4, the number of iterations was (3 + 4
-
1) = 6. (it started at 1x1 and did 6 iterations: something like: 1x1,
1x2,
2x2, 2x3, 3x3, 3x4)
For N = 100, the number of iteration depends on the ratio: it could be
anywhere from 19 (10x10-1) to 100 iterations (worst case happens if the
output is a single line or a single column). So that would make N
iterations (in worst cases) and ~2*SQRT(N)-1 best case.
I took the liberty of optimizing the method a little bit and renaming a
few
internal variables to my personal liking. ;-) -- I did not do any "real
life" testing on this. But it seems to work on paper.
/**
* computes the largest N square size (and layout) that can fit an area
(width,height).
*
* @return an Object containing 'squareSize' and the number of 'cols' and
'rows'.
*
* 98% of the credits goes to Danny Kodicek
*/
function computeLargestSquareSizeAndLayout(width:Number, height:Number,
Nsquares:Number): Object
{
var cols:Number = 1;
var rows:Number = 1;
var swidth:Number = width;
var sheight:Number = height;
var next_swidth:Number = width/(cols+1);
var next_sheight:Number = height/(rows+1);
while (true)
{
if (cols*rows >= Nsquares)
{
var squareSize:Number = Math.min(swidth, sheight);
return { squareSize:squareSize, cols:cols, rows:rows };
}
if (next_swidth > next_sheight)
{
cols++;
swidth = next_swidth;
next_swidth = width/(cols+1);
}
else
{
rows++;
sheight = next_sheight;
next_sheight = height/(rows+1);
}
}
}
B.
2006/5/8, Danny Kodicek <[EMAIL PROTECTED]>:
>
> >Danny: I do not understand your algorithm - could you shed some more
> (high-level) light on what it is doing?
>
> Sure. The idea is that the optimal size will always be an exact fraction
> of
> either the width or the height. So what we do is drop down by multiples
of
> these until we get to the first size that will contain N or more
squares.
> At
> any particular width, we keep track of the two 'next-smallest' widths
and
> drop down to the largest of these. Run through the algorithm with a few
> sample numbers and it should make sense.
>
> It may be that there's a more algebraic approach. The problem is,
though,
> that there is no simple relationship between floor(x), floor(y) and
> floor(xy), which would be needed to come up with any truly useful
> solution.
> In the end, it turns into quite a complex optimisation problem, and
given
> that the brute force algorithm is actually pretty fast, it hardly seems
> worth the effort :)
>
> Danny
>
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