Clusters
In shaping text, a cluster is a sequence of
code points that needs to be treated as a single, indivisible unit.
When you add text to a HB buffer, each character is associated with
a cluster value. This is an arbitrary number as
far as HB is concerned.
Most clients will use UTF-8, UTF-16, or UTF-32 indices, but the
actual number does not matter. Moreover, it is not required for the
cluster values to be monotonically increasing, but pretty much all
of HB's tests are performed on monotonically increasing cluster
numbers. Nevertheless, there is no such assumption in the code
itself. With that in mind, let's examine what happens with cluster
values during shaping under each cluster-level.
HarfBuzz provides three levels of clustering
support. Level 0 is the default behavior and reproduces the behavior
of the old HarfBuzz library. Level 1 tweaks this behavior slightly
to produce better results, so level 1 clustering is recommended for
code that is not required to implement backward compatibility with
the old HarfBuzz.
Level 2 differs significantly in how it treats cluster values.
Levels 0 and 1 both process ligatures and glyph decomposition by
merging clusters; level 2 does not.
The conceptual model for what the cluster values mean, in levels 0
and 1, is this:
the sequence of cluster values will always remain monotone
each value represents a single cluster
each cluster contains one or more glyphs and one or more
characters
Assuming that initial cluster numbers were monotonically increasing
and distinct, then all adjacent glyphs having the same cluster
number belong to the same cluster, and all characters belong to the
cluster that has the highest number not larger than their initial
cluster number. This will become clearer with an example.
A clustering example for levels 0 and 1
Let's say we start with the following character sequence and cluster
values:
A,B,C,D,E
0,1,2,3,4
We then map the characters to glyphs. For simplicity, let's assume
that each character maps to the corresponding, identical-looking
glyph:
A,B,C,D,E
0,1,2,3,4
Now if, for example, B and C
ligate, then the clusters to which they belong "merge".
This merged cluster takes for its cluster number the minimum of all
the cluster numbers of the clusters that went in. In this case, we
get:
A,BC,D,E
0,1 ,3,4
Now let's assume that the BC glyph decomposes
into three components, and D also decomposes into
two. The components each inherit the cluster value of their parent:
A,BC0,BC1,BC2,D0,D1,E
0,1 ,1 ,1 ,3 ,3 ,4
Now if BC2 and D0 ligate, then
their clusters (numbers 1 and 3) merge into
min(1,3) = 1:
A,BC0,BC1,BC2D0,D1,E
0,1 ,1 ,1 ,1 ,4
At this point, cluster 1 means: the character sequence
BCD is represented by glyphs
BC0,BC1,BC2D0,D1 and cannot be broken down any
further.
Reordering in levels 0 and 1
Another common operation in the more complex shapers is when things
reorder. In those cases, to maintain monotone clusters, HB merges
the clusters of everything in the reordering sequence. For example,
let's again start with the character sequence:
A,B,C,D,E
0,1,2,3,4
If D is reordered before B,
then the B, C, and
D clusters merge, and we get:
A,D,B,C,E
0,1,1,1,4
This is clearly not ideal, but it is the only sensible way to
maintain monotone indices and retain the true relationship between
glyphs and characters.
The distinction between levels 0 and 1
So, the above is pretty much what cluster levels 0 and 1 do. The
only difference between the two is this: in level 0, at the very
beginning of the shaping process, we also merge clusters between
base characters and all Unicode marks (combining or not) following
them. E.g.:
A,acute,B
0,1 ,2
will become:
A,acute,B
0,0 ,2
This is the default behavior. We do it because Windows did it and
old HarfBuzz did it, so this remained the default. But this behavior
makes it impossible to color diacritic marks differently from their
base characters. That's why in level 1 we do not perform this
initial merging step.
For clients, level 0 is more convenient if they rely on HarfBuzz
clusters for cursor positioning. But that's wrong anyway: cursor
positions should be determined based on Unicode grapheme boundaries,
NOT shaping clusters. As such, level 1 clusters are preferred.
One last note about levels 0 and 1. We currently don't allow a
MultipleSubst lookup to replace a glyph with zero
glyphs (i.e., to delete a glyph). But in some other situations,
glyphs can be deleted. In those cases, if the glyph being deleted is
the last glyph of its cluster, we make sure to merge the cluster
with a neighboring cluster.
This is, primarily, to make sure that the starting cluster of the
text always has the cluster index pointing to the start of the text
for the run; more than one client currently relies on this
guarantee.
Incidentally, Apple's CoreText does something else to maintain the
same promise: it inserts a glyph with id 65535 at the beginning of
the glyph string if the glyph corresponding to the first character
in the run was deleted. HarfBuzz might do something similar in the
future.
Level 2
Level 2 is a different beast from levels 0 and 1. It is simple to
describe, but hard to make sense of. It simply doesn't do any
cluster merging whatsoever. When things ligate or otherwise multiple
glyphs turn into one, the cluster value of the first glyph is
retained.
Here are a few examples of why processing cluster values produced at
this level might be tricky:
Ligatures with combining marks
Imagine capital letters are bases and lower case letters are
combining marks. With an input sequence like this:
A,a,B,b,C,c
0,1,2,3,4,5
if A,B,C ligate, then here are the cluster
values one would get under the various levels:
level 0:
ABC,a,b,c
0 ,0,0,0
level 1:
ABC,a,b,c
0 ,0,0,5
level 2:
ABC,a,b,c
0 ,1,3,5
Making sense of the last example is the hardest for a client,
because there is nothing in the cluster values to suggest that
B and C ligated with
A.
Reordering
Another tricky case is when things reorder. Under level 2:
A,B,C,D,E
0,1,2,3,4
Now imagine D moves before
B:
A,D,B,C,E
0,3,1,2,4
Now, if D ligates with B, we
get:
A,DB,C,E
0,3 ,2,4
In a different scenario, A and
B could have ligated
before D reordered; that
would have resulted in:
AB,D,C,E
0 ,3,2,4
There's no way to differentiate between these two scenarios based
on the cluster numbers alone.
Another problem happens with ligatures under level 2 if the
direction of the text is forced to opposite of its natural
direction (e.g. left-to-right Arabic). But that's too much of a
corner case to worry about.