On Sat, Aug 18, 2012 at 5:34 AM, Keith OHara <k-ohara5...@oco.net> wrote:
> On Fri, 17 Aug 2012 10:16:25 -0700, <mts...@gmail.com> wrote: > > >> http://codereview.appspot.com/**5626052/diff/106004/lily/axis-** >> group-interface.cc#newcode780<http://codereview.appspot.com/5626052/diff/106004/lily/axis-group-interface.cc#newcode780> >> lily/axis-group-interface.cc:**780: while (dirty); >> On 2012/08/17 08:12:56, Keith wrote: >> >>> I am beginning to understand the new code. Would you consider again using >>> distance() instead of the repeated calls to intersects() in the >>> while(dirty) >>> loop ? It might look simpler now that you've been away from the code >>> for awhile. >>> >> > floors[j] = padding - forest[j][UP]~distance~pair[**DOWN] >>> ceilings[j] = forest[j][DOWN]~distance~pair[**UP] >>> >> > You know the solution will be to move an amount equal to one of the >>> floors[j], >>> and you want to find the smallest that fits, so make >>> trials = sort (floors) >>> and find the smallest among trials that satisfies >>> trials[n] > 0 >>> && for all j (trials[n] > floors[j] >>> || trials[n] < ceilings[j] ) >>> >> > I am not adverse to using distance if possible - my trouble case is the >> following. >> >> AAAAAAAAA >> >> BBBBBBBBBBBBB >> >> Image that object A above is placed first and LilyPond is now >> considering whether it should leave B where it is or place it above A. >> The distance function would dictate that it needs to be shifted above A >> because its DOWN skyline is below A's UP. >> > > We need not let the distance function between A[UP] and B[DOWN] dictate, > because having positive distance between A[DOWN] and B[UP] is another > solution. > If you check out the dev/jneem-skylines (which is a simplified but not (yet) feature-compatible version of Mike's branch), you'll see that I'm using this method under the name 'bool Skyline_pair::intersects.' Cheers, Joe
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