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.