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<div class="section">
<div class="titlepage"><div><div><h2 class="title" style="clear: both">
<a name="a-clustering-example-for-levels-0-and-1"></a>A clustering example for levels 0 and 1</h2></div></div></div>
<p>
The basic shaping operations affect clusters in a predictable
manner when using level 0 or level 1:
</p>
<div class="itemizedlist"><ul class="itemizedlist" style="list-style-type: disc; ">
<li class="listitem"><p>
When two or more clusters <span class="emphasis"><em>merge</em></span>, the
resulting merged cluster takes as its cluster value the
<span class="emphasis"><em>minimum</em></span> of the incoming cluster values.
</p></li>
<li class="listitem"><p>
When a cluster <span class="emphasis"><em>decomposes</em></span>, all of the
resulting child clusters inherit as their cluster value the
cluster value of the parent cluster.
</p></li>
<li class="listitem"><p>
When a character is <span class="emphasis"><em>reordered</em></span>, the
reordered character and all clusters that the character
moves past as part of the reordering are merged into one cluster.
</p></li>
</ul></div>
<p>
The functionality, guarantees, and benefits of level 0 and level
1 behavior can be seen with some examples. First, let us examine
what happens with cluster values when shaping involves cluster
merging with ligatures and decomposition.
</p>
<p>
Let's say we start with the following character sequence (top row) and
initial cluster values (bottom row):
</p>
<pre class="programlisting">
A,B,C,D,E
0,1,2,3,4
</pre>
<p>
During shaping, HarfBuzz maps these characters to glyphs from
the font. For simplicity, let us assume that each character maps
to the corresponding, identical-looking glyph:
</p>
<pre class="programlisting">
A,B,C,D,E
0,1,2,3,4
</pre>
<p>
Now if, for example, <code class="literal">B</code> and <code class="literal">C</code>
form a ligature, then the clusters to which they belong
"merge". This merged cluster takes for its cluster
value the minimum of all the cluster values of the clusters that
went in to the ligature. In this case, we get:
</p>
<pre class="programlisting">
A,BC,D,E
0,1 ,3,4
</pre>
<p>
because 1 is the minimum of the set {1,2}, which were the
cluster values of <code class="literal">B</code> and
<code class="literal">C</code>.
</p>
<p>
Next, let us say that the <code class="literal">BC</code> ligature glyph
decomposes into three components, and <code class="literal">D</code> also
decomposes into two components. Whenever a cluster decomposes,
its components each inherit the cluster value of their parent:
</p>
<pre class="programlisting">
A,BC0,BC1,BC2,D0,D1,E
0,1 ,1 ,1 ,3 ,3 ,4
</pre>
<p>
Next, if <code class="literal">BC2</code> and <code class="literal">D0</code> form a
ligature, then their clusters (cluster values 1 and 3) merge into
<code class="literal">min(1,3) = 1</code>:
</p>
<pre class="programlisting">
A,BC0,BC1,BC2D0,D1,E
0,1 ,1 ,1 ,1 ,4
</pre>
<p>
Note that the entirety of cluster 3 merges into cluster 1, not
just the <code class="literal">D0</code> glyph. This reflects the fact
that the cluster <span class="emphasis"><em>must</em></span> be treated as an
indivisible unit.
</p>
<p>
At this point, cluster 1 means: the character sequence
<code class="literal">BCD</code> is represented by glyphs
<code class="literal">BC0,BC1,BC2D0,D1</code> and cannot be broken down any
further.
</p>
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