Sorry wrong link, I'm sure there is a component on the site somewhere that does what you need (I think) ;)
On 5/9/06, Mick G <[EMAIL PROTECTED]> wrote:
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 > > > > _______________________________________________ > > Flashcoders@chattyfig.figleaf.com > > To change your subscription options or search the archive: > > http://chattyfig.figleaf.com/mailman/listinfo/flashcoders > > > > Brought to you by Fig Leaf Software > > Premier Authorized Adobe Consulting and Training > > http://www.figleaf.com > > http://training.figleaf.com > > > _______________________________________________ > Flashcoders@chattyfig.figleaf.com > To change your subscription options or search the archive: > http://chattyfig.figleaf.com/mailman/listinfo/flashcoders > > Brought to you by Fig Leaf Software > Premier Authorized Adobe Consulting and Training > http://www.figleaf.com > http://training.figleaf.com >
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