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With inspiration from other libraries
// © 2016 and later: Unicode, Inc. and others.
// License & terms of use: http://www.unicode.org/copyright.html#License
/*
*******************************************************************************
* Copyright (C) 2014, International Business Machines Corporation and *
* others. All Rights Reserved. *
*******************************************************************************
*/
package com.ibm.icu.text;
import java.io.IOException;
import java.text.CharacterIterator;
import com.ibm.icu.lang.UCharacter;
import com.ibm.icu.lang.UProperty;
import com.ibm.icu.lang.UScript;
class LaoBreakEngine extends DictionaryBreakEngine {
// Constants for LaoBreakIterator
// How many words in a row are "good enough"?
private static final byte LAO_LOOKAHEAD = 3;
// Will not combine a non-word with a preceding dictionary word longer than this
private static final byte LAO_ROOT_COMBINE_THRESHOLD = 3;
// Will not combine a non-word that shares at least this much prefix with a
// dictionary word with a preceding word
private static final byte LAO_PREFIX_COMBINE_THRESHOLD = 3;
// Minimum word size
private static final byte LAO_MIN_WORD = 2;
private DictionaryMatcher fDictionary;
private static UnicodeSet fLaoWordSet;
private static UnicodeSet fEndWordSet;
private static UnicodeSet fBeginWordSet;
private static UnicodeSet fMarkSet;
static {
// Initialize UnicodeSets
fLaoWordSet = new UnicodeSet();
fMarkSet = new UnicodeSet();
fBeginWordSet = new UnicodeSet();
fLaoWordSet.applyPattern("[[:Laoo:]&[:LineBreak=SA:]]");
fLaoWordSet.compact();
fMarkSet.applyPattern("[[:Laoo:]&[:LineBreak=SA:]&[:M:]]");
fMarkSet.add(0x0020);
fEndWordSet = new UnicodeSet(fLaoWordSet);
fEndWordSet.remove(0x0EC0, 0x0EC4); // prefix vowels
fBeginWordSet.add(0x0E81, 0x0EAE); // basic consonants (including holes for corresponding Thai characters)
fBeginWordSet.add(0x0EDC, 0x0EDD); // digraph consonants (no Thai equivalent)
fBeginWordSet.add(0x0EC0, 0x0EC4); // prefix vowels
// Compact for caching
fMarkSet.compact();
fEndWordSet.compact();
fBeginWordSet.compact();
// Freeze the static UnicodeSet
fLaoWordSet.freeze();
fMarkSet.freeze();
fEndWordSet.freeze();
fBeginWordSet.freeze();
}
public LaoBreakEngine() throws IOException {
super(BreakIterator.KIND_WORD, BreakIterator.KIND_LINE);
setCharacters(fLaoWordSet);
// Initialize dictionary
fDictionary = DictionaryData.loadDictionaryFor("Laoo");
}
public boolean equals(Object obj) {
// Normally is a singleton, but it's possible to have duplicates
// during initialization. All are equivalent.
return obj instanceof LaoBreakEngine;
}
public int hashCode() {
return getClass().hashCode();
}
public boolean handles(int c, int breakType) {
if (breakType == BreakIterator.KIND_WORD || breakType == BreakIterator.KIND_LINE) {
int script = UCharacter.getIntPropertyValue(c, UProperty.SCRIPT);
return (script == UScript.LAO);
}
return false;
}
public int divideUpDictionaryRange(CharacterIterator fIter, int rangeStart, int rangeEnd,
DequeI foundBreaks) {
if ((rangeEnd - rangeStart) < LAO_MIN_WORD) {
return 0; // Not enough characters for word
}
int wordsFound = 0;
int wordLength;
int current;
PossibleWord words[] = new PossibleWord[LAO_LOOKAHEAD];
for (int i = 0; i < LAO_LOOKAHEAD; i++) {
words[i] = new PossibleWord();
}
int uc;
fIter.setIndex(rangeStart);
while ((current = fIter.getIndex()) < rangeEnd) {
wordLength = 0;
//Look for candidate words at the current position
int candidates = words[wordsFound%LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd);
// If we found exactly one, use that
if (candidates == 1) {
wordLength = words[wordsFound%LAO_LOOKAHEAD].acceptMarked(fIter);
wordsFound += 1;
}
// If there was more than one, see which one can take us forward the most words
else if (candidates > 1) {
boolean foundBest = false;
// If we're already at the end of the range, we're done
if (fIter.getIndex() < rangeEnd) {
do {
int wordsMatched = 1;
if (words[(wordsFound+1)%LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd) > 0) {
if (wordsMatched < 2) {
// Followed by another dictionary word; mark first word as a good candidate
words[wordsFound%LAO_LOOKAHEAD].markCurrent();
wordsMatched = 2;
}
// If we're already at the end of the range, we're done
if (fIter.getIndex() >= rangeEnd) {
break;
}
// See if any of the possible second words is followed by a third word
do {
// If we find a third word, stop right away
if (words[(wordsFound+2)%LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd) > 0) {
words[wordsFound%LAO_LOOKAHEAD].markCurrent();
foundBest = true;
break;
}
} while (words[(wordsFound+1)%LAO_LOOKAHEAD].backUp(fIter));
}
} while (words[wordsFound%LAO_LOOKAHEAD].backUp(fIter) && !foundBest);
}
wordLength = words[wordsFound%LAO_LOOKAHEAD].acceptMarked(fIter);
wordsFound += 1;
}
// We come here after having either found a word or not. We look ahead to the
// next word. If it's not a dictionary word, we will combine it with the word we
// just found (if there is one), but only if the preceding word does not exceed
// the threshold.
// The text iterator should now be positioned at the end of the word we found.
if (fIter.getIndex() < rangeEnd && wordLength < LAO_ROOT_COMBINE_THRESHOLD) {
// If it is a dictionary word, do nothing. If it isn't, then if there is
// no preceding word, or the non-word shares less than the minimum threshold
// of characters with a dictionary word, then scan to resynchronize
if (words[wordsFound%LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd) <= 0 &&
(wordLength == 0 ||
words[wordsFound%LAO_LOOKAHEAD].longestPrefix() < LAO_PREFIX_COMBINE_THRESHOLD)) {
// Look for a plausible word boundary
int remaining = rangeEnd - (current + wordLength);
int pc = fIter.current();
int chars = 0;
for (;;) {
fIter.next();
uc = fIter.current();
chars += 1;
if (--remaining <= 0) {
break;
}
if (fEndWordSet.contains(pc) && fBeginWordSet.contains(uc)) {
// Maybe. See if it's in the dictionary.
int candidate = words[(wordsFound + 1) %LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd);
fIter.setIndex(current + wordLength + chars);
if (candidate > 0) {
break;
}
}
pc = uc;
}
// Bump the word count if there wasn't already one
if (wordLength <= 0) {
wordsFound += 1;
}
// Update the length with the passed-over characters
wordLength += chars;
} else {
// Backup to where we were for next iteration
fIter.setIndex(current+wordLength);
}
}
// Never stop before a combining mark.
int currPos;
while ((currPos = fIter.getIndex()) < rangeEnd && fMarkSet.contains(fIter.current())) {
fIter.next();
wordLength += fIter.getIndex() - currPos;
}
// Look ahead for possible suffixes if a dictionary word does not follow.
// We do this in code rather than using a rule so that the heuristic
// resynch continues to function. For example, one of the suffix characters
// could be a typo in the middle of a word.
// NOT CURRENTLY APPLICABLE TO LAO
// Did we find a word on this iteration? If so, push it on the break stack
if (wordLength > 0) {
foundBreaks.push(Integer.valueOf(current + wordLength));
}
}
// Don't return a break for the end of the dictionary range if there is one there
if (foundBreaks.peek() >= rangeEnd) {
foundBreaks.pop();
wordsFound -= 1;
}
return wordsFound;
}
}