Better staggering of accidental placements. (issue 7101045)

[k-ohara5a5a]

Obfuscated, but so was the old code.

I'll leave here the notes I used to decipher it, and maybe come back
later to clean up the code.


https://codereview.appspot.com/7101045/diff/1/lily/accidental-placement.cc
File lily/accidental-placement.cc (right):

https://codereview.appspot.com/7101045/diff/1/lily/accidental-placement.cc#newcode123
lily/accidental-placement.cc:123: Real ape_priority
(Accidental_placement_entry const *a)
Evil name.  The return values are not used as a priority.  Formerly,
"ape_highest_extent()" would have been appropriate, because it was not
even then a priority.  Rather, the highest ape was placed first, then
the lowest, then the second highest...

https://codereview.appspot.com/7101045/diff/1/lily/accidental-placement.cc#newcode129
lily/accidental-placement.cc:129: bool ape_less
(Accidental_placement_entry *const &a,
Bad name.  The apes were formerly sorted by "priority" (really height)
so this should have been ape_lower().  Now, they are sorted by number of
accidentals in the ape, then by "priority" (really height) to break
ties.

https://codereview.appspot.com/7101045/diff/1/lily/accidental-placement.cc#newcode186
lily/accidental-placement.cc:186: vector<Accidental_placement_entry *>
asc = *apes;
accidental sorted-columns (asc)

https://codereview.appspot.com/7101045/diff/1/lily/accidental-placement.cc#newcode189
lily/accidental-placement.cc:189: // we do the staggering below based on
size
'size' must mean number of accidentals in the ape, but you don't stagger
based on 'size', you prioritize based on 'size', then within each groups
of apes with the same 'size', you stagger the based on their height.

https://codereview.appspot.com/7101045/diff/1/lily/accidental-placement.cc#newcode190
lily/accidental-placement.cc:190: // this ensures that if a placement
has 4 entries, it will
An accidental-placement-entry (ape) contains the accidentals that get
placed together (same pitch names in different octaves) so it would
rarely have 4 entries -- more likely just 2 accidentals.

https://codereview.appspot.com/7101045/diff/1/lily/accidental-placement.cc#newcode191
lily/accidental-placement.cc:191: // always be closer to the NoteColumn
than a placement with 1
Not always closer, merely placed before.  In one of the regression
tests, a single accidental had room to slide closer to a couple
previously-placed octave-pairs of accidentals.

https://codereview.appspot.com/7101045/diff/1/lily/accidental-placement.cc#newcode193
lily/accidental-placement.cc:193: // while still preserving octave
alignment
Nothing in here deals with octaves.  The grouping seems to be done in
scheme, so here we can only promise to keep the apes together.

https://codereview.appspot.com/7101045/diff/1/lily/accidental-placement.cc#newcode194
lily/accidental-placement.cc:194:
vector<vector<Accidental_placement_entry *> > ascs;
accidental sorted-column sets (ascs)

https://codereview.appspot.com/7101045/diff/1/lily/accidental-placement.cc#newcode196
lily/accidental-placement.cc:196: vsize sz = INT_MAX;
Number of accidentals in a group to be placed as a column (sz)

https://codereview.appspot.com/7101045/diff/1/lily/accidental-placement.cc#newcode197
lily/accidental-placement.cc:197: for (vsize i = 0; i < asc.size ();
i++)
The "accidental sorted-columns" (asc) is sorted in an order not directly
related to the order of placement.  We need to break it up in to subsets
(ascs) that contain columns with similar numbers of accidentals in them.
 Each sub-set is 'staggered' but the subset containing groups-of-2
accidentals placed before the subset with single accidentals.

https://codereview.appspot.com/7101045/diff/1/lily/accidental-placement.cc#newcode210
lily/accidental-placement.cc:210: int parity = 1;
Not really "parity", but which end, highest-or-lowest, of the sub-lists
of accidentals we pick from.

Apparently, when we finish with the accidentals in octaves, we want to
pick the highest of the remaining accidentals to place next.

https://codereview.appspot.com/7101045/