Brief explanation of the hyphenation algorithm herein.
Raph Levien
4 Aug 1998
The hyphenation algorithm is basically the same as Knuth's TeX
algorithm. However, the implementation is quite a bit faster.
The hyphenation files from TeX can almost be used directly. There
is a preprocessing step, however. If you don't do the preprocessing
step, you'll get bad hyphenations (i.e. a silent failure).
Start with a file such as hyphen.us. This is the TeX ushyph1.tex
file, with the exception dictionary encoded using the same rules as
the main portion of the file. Any line beginning with % is a comment.
Each other line should contain exactly one rule.
Then, do the preprocessing - "perl substrings.pl hyphen.us". The
resulting file is hyphen.mashed. It's in Perl, and it's fairly slow
(it uses brute force algorithms; about 17 seconds on a P100), but it
could probably be redone in C with clever algorithms. This would be
valuable, for example, if it was handle user-supplied exception
dictionaries by integrating them into the rule table.
Once the rules are preprocessed, loading them is quite quick -
about 200ms on a P100. It then hyphenates at about 40,000 words per
second on a P100. I haven't benchmarked it against other
implementations (both TeX and groff contain essentially the same
algorithm), but expect that it runs quite a bit faster than any of
them.
Knuth's algorithm
This section contains a brief explanation of Knuth's algorithm, in
case you missed it from the TeX books. We'll use the semi-word
"example" as our running example.
Since the beginning and end of a word are special, the algorithm is
actually run over the prepared word (prep_word in the source)
".example.". Knuths algorithm basically just does pattern matches from
the rule set, then applies the matches. The patterns in this case that
match are "xa", "xam", "mp", and "pl". These are actually stored as
"x1a", "xam3", "4m1p", and "1p2l2". Whenever numbers appear between
the letters, they are added in. If two (or more) patterns have numbers
in the same place, the highest number wins. Here's the example:
. e x a m p l e .
x1a
x a m3
4m1p
1p2l2
-----------------
. e x1a4m3p2l2e .
Finally, hyphens are placed wherever odd numbers appear. They are,
however, suppressed after the first letter and before the last letter
of the word (TeX actually suppresses them before the next-to-last, as
well). So, it's "ex-am-ple", which is correct.
Knuth uses a trie to implement this. I.e. he stores each rule in a
trie structure. For each position in the word, he searches the trie,
searching for a match. Most patterns are short, so efficiency should
be quite good.
Theory of the algorithm
The algorithm works as a slightly modified finite state machine.
There are two kinds of transitions: those that consume one letter of
input (which work just like your regular finite state machine), and
"fallback" transitions, which don't consume any input. If no
transition matching the next letter is found, the fallback is used.
One way of looking at this is a form of compression of the transition
tables - i.e. it behaves the same as a completely vanilla state
machine in which the actual transition table of a node is made up of
the union of transition tables of the node itself, plus its fallbacks.
Each state is represented by a string. Thus, if the current state
is "am" and the next letter is "p", then the next state is "amp".
Fallback transitions go to states which chop off one or (sometimes)
more letters from the beginning. For example, if none of the
transitions from "amp" match the next letter, then it will fall back
to "mp". Similarly, if none of the transitions from "mp" match the
next letter, it will fall back to "m".
Each state is also associated with a (possibly null) "match"
string. This represents the union of all patterns which are
right-justified substrings of the match string. I.e. the pattern "mp"
is a right-justified substring of the state "amp", so it's numbers get
added in. The actual calculation of this union is done by the
Perl preprocessing script, but could probably be done in C just about
as easily.
Because each state transition either consumes one input character
or shortens the state string by one character, the total number of
state transitions is linear in the length of the word.