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I recall concluding that a single, fixed weight would actually be sufficient to make it significantly better than XZ, without needing tweaking for each block. Cheers, / h+ -- -- The early bird gets the worm, but the second mouse gets the cheese. I think you missed his point there. These are clipping rectangles - = so you don't need to clip to the bits inside a rectangle which is already = being clipped to. So for your examples, the first becomes two rectangles = thus: +---+||| +---+|||| +---+|| +---+ And the second becomes one thus: +--------+|||||||||| +--------+ No further rectangles are required. -----Original Message----- From: gdalgorithms-list-admin@. [mailto:gdalgorithms-list-admin@.] On Behalf Of = Stuart Harrison Sent: 20 January 2006 11:49 To: gdalgorithms-list@. Subject: RE: [Algorithms] rectangle clipping (non-noob question) You're missing whole batches of possible overlap issues there (I assume you're only dealing with horizontal & vertical quad edges otherwise it gets really messy), consider (excuse ASCII-art): +---+||| +-----+|||| +-----+|| +---+ The above example must be split to at least 4 quads if you don't want any overlap (top, middle, bottom for the vertical quad). And how about. +--------+||| +--+|||||| +--+||| +--------+ That requires at least 5 quads to avoid overlap (top, centre-left, centre-right, bottom for the big quad). I seem to remember doing something very similar to this for a graduate window manager application; I worked out there were something like 19 different possible arrangements of two rectangles (including not overlapping). I can't remember the precise algorithm, but it dealt with two rect-lists (a list being 1-or-more rectangles) and created a third 'merged' rect-list from the clipping of one list against the other (by piecemeal NxM clipping). Each rect-list was defined as having no overlap to start with (so you didn't need inter-list clipping, only intra-list clipping). You're=20missing=20whole=20batches=20of=20possible=20overlap=20issues=20th= ere=20(I=20assume you're=20only=20dealing=20with=20horizontal=20&=20vertical=20quad=20edges=20= otherwise=20it gets=20really=20messy),=20consider=20(excuse=20ASCII-art): +---+|=20=20=20||=20+-----+|=20|=20=20=20=20=20||=20+-----+|=20=20=20| +---+ The=20above=20example=20must=20be=20split=20to=20at=20least=204=20quads=20= if=20you=20don't=20want any=20overlap=20(top,=20middle,=20bottom=20for=20the=20vertical=20quad).=20= =20And=20how about. +--------+|=20=20=20=20=20=20=20=20||=20=20+--+=20=20||=20=20|=20=20|=20=20||=20=20+--+=20=20||=20=20=20=20=20=20=20=20| +--------+ That=20requires=20at=20least=205=20quads=20to=20avoid=20overlap=20(top,=20= centre-left, centre-right,=20bottom=20for=20the=20big=20quad). I=20seem=20to=20remember=20doing=20something=20very=20similar=20to=20this=20= for=20a=20graduate window=20manager=20application;=20I=20worked=20out=20there=20were=20someth= ing=20like=2019 different=20possible=20arrangements=20of=20two=20rectangles=20(including=20= not overlapping). I=20can't=20remember=20the=20precise=20algorithm,=20but=20it=20dealt=20wit= h=20two=20rect-lists (a=20list=20being=201-or-more=20rectangles)=20and=20created=20a=20third=20= 'merged' rect-list=20from=20the=20clipping=20of=20one=20list=20against=20the=20othe= r=20(by=20piecemeal NxM=20clipping).=20=20Each=20rect-list=20was=20defined=20as=20having=20no=20= overlap=20to=20start with=20(so=20you=20didn't=20need=20inter-list=20clipping,=20only=20intra-l= ist=20clipping). Jon Watte wrote: 'What I've experimented with is measuring the distance along U and V for each line of edges across the tile, and proportioning the U and V=20 according to these distances (rather than just the X/Z projected [.]' I experimented with this as well, but didn't get satisfactory results; it just didn't look good enough from *every* angle.=20 'or two rounds of relaxation on the values, it actually reaches a happy' That's interesting, what energy function are you minimising here? Cheers, Willem.
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