main/classes/core/src/com/ibm/icu/text/LaoBreakEngine.java - external/github.com/unicode-org/icu - Git at Google

 /*
  *******************************************************************************
  * Copyright (C) 2013, International Business Machines Corporation and         *
  * others. All Rights Reserved.                                                *
  *******************************************************************************
  */
 package com.ibm.icu.text;

 import java.io.IOException;
 import java.text.CharacterIterator;
 import java.util.Stack;

 import com.ibm.icu.lang.UCharacter;
 import com.ibm.icu.lang.UProperty;
 import com.ibm.icu.lang.UScript;

 class LaoBreakEngine implements LanguageBreakEngine {
     /* Helper class for improving readability of the Lao word break
      * algorithm.
      */
     static class PossibleWord {
         // List size, limited by the maximum number of words in the dictionary
         // that form a nested sequence.
         private final static int POSSIBLE_WORD_LIST_MAX = 20;
         //list of word candidate lengths, in increasing length order
         private int lengths[];
         private int count[];    // Count of candidates
         private int prefix;     // The longest match with a dictionary word
         private int offset;     // Offset in the text of these candidates
         private int mark;       // The preferred candidate's offset
         private int current;    // The candidate we're currently looking at

         // Default constructor
         public PossibleWord() {
             lengths = new int[POSSIBLE_WORD_LIST_MAX];
             count = new int[1]; // count needs to be an array of 1 so that it can be pass as reference
             offset = -1;
         }

         // Fill the list of candidates if needed, select the longest, and return the number found
         public int candidates(CharacterIterator fIter, DictionaryMatcher dict, int rangeEnd) {
             int start = fIter.getIndex();
             if (start != offset) {
                 offset = start;
                 prefix = dict.matches(fIter, rangeEnd - start, lengths, count, lengths.length);
                 // Dictionary leaves text after longest prefix, not longest word. Back up.
                 if (count[0] <= 0) {
                     fIter.setIndex(start);
                 }
             }
             if (count[0] > 0) {
                 fIter.setIndex(start + lengths[count[0]-1]);
             }
             current = count[0] - 1;
             mark = current;
             return count[0];
         }

         // Select the currently marked candidate, point after it in the text, and invalidate self
         public int acceptMarked(CharacterIterator fIter) {
             fIter.setIndex(offset + lengths[mark]);
             return lengths[mark];
         }

         // Backup from the current candidate to the next shorter one; return true if that exists
         // and point the text after it
         public boolean backUp(CharacterIterator fIter) {
             if (current > 0) {
                 fIter.setIndex(offset + lengths[--current]);
                 return true;
             }
             return false;
         }

         // Return the longest prefix this candidate location shares with a dictionary word
         public int longestPrefix() {
             return prefix;
         }

         // Mark the current candidate as the one we like
         public void markCurrent() {
             mark = current;
         }
     }

     // 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();
         fEndWordSet = new UnicodeSet();
         fBeginWordSet = new UnicodeSet();

         fLaoWordSet.applyPattern("[[:Laoo:]&[:LineBreak=SA:]]");
         fLaoWordSet.compact();

         fMarkSet.applyPattern("[[:Laoo:]&[:LineBreak=SA:]&[:M:]]");
         fMarkSet.add(0x0020);
         fEndWordSet = 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 {
         // Initialize dictionary
         fDictionary = DictionaryData.loadDictionaryFor("Laoo");
     }

     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 findBreaks(CharacterIterator fIter, int rangeStart, int rangeEnd, boolean reverse, int breakType,
             Stack<Integer> 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().intValue() >= rangeEnd) {
             foundBreaks.pop();
             wordsFound -= 1;
         }

         return wordsFound;
     }

 }
	/*
	*******************************************************************************
	* Copyright (C) 2013, International Business Machines Corporation and *
	* others. All Rights Reserved. *
	*******************************************************************************
	*/
	package com.ibm.icu.text;

	import java.io.IOException;
	import java.text.CharacterIterator;
	import java.util.Stack;

	import com.ibm.icu.lang.UCharacter;
	import com.ibm.icu.lang.UProperty;
	import com.ibm.icu.lang.UScript;

	class LaoBreakEngine implements LanguageBreakEngine {
	/* Helper class for improving readability of the Lao word break
	* algorithm.
	*/
	static class PossibleWord {
	// List size, limited by the maximum number of words in the dictionary
	// that form a nested sequence.
	private final static int POSSIBLE_WORD_LIST_MAX = 20;
	//list of word candidate lengths, in increasing length order
	private int lengths[];
	private int count[]; // Count of candidates
	private int prefix; // The longest match with a dictionary word
	private int offset; // Offset in the text of these candidates
	private int mark; // The preferred candidate's offset
	private int current; // The candidate we're currently looking at

	// Default constructor
	public PossibleWord() {
	lengths = new int[POSSIBLE_WORD_LIST_MAX];
	count = new int[1]; // count needs to be an array of 1 so that it can be pass as reference
	offset = -1;
	}

	// Fill the list of candidates if needed, select the longest, and return the number found
	public int candidates(CharacterIterator fIter, DictionaryMatcher dict, int rangeEnd) {
	int start = fIter.getIndex();
	if (start != offset) {
	offset = start;
	prefix = dict.matches(fIter, rangeEnd - start, lengths, count, lengths.length);
	// Dictionary leaves text after longest prefix, not longest word. Back up.
	if (count[0] <= 0) {
	fIter.setIndex(start);
	}
	}
	if (count[0] > 0) {
	fIter.setIndex(start + lengths[count[0]-1]);
	}
	current = count[0] - 1;
	mark = current;
	return count[0];
	}

	// Select the currently marked candidate, point after it in the text, and invalidate self
	public int acceptMarked(CharacterIterator fIter) {
	fIter.setIndex(offset + lengths[mark]);
	return lengths[mark];
	}

	// Backup from the current candidate to the next shorter one; return true if that exists
	// and point the text after it
	public boolean backUp(CharacterIterator fIter) {
	if (current > 0) {
	fIter.setIndex(offset + lengths[--current]);
	return true;
	}
	return false;
	}

	// Return the longest prefix this candidate location shares with a dictionary word
	public int longestPrefix() {
	return prefix;
	}

	// Mark the current candidate as the one we like
	public void markCurrent() {
	mark = current;
	}
	}

	// 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();
	fEndWordSet = new UnicodeSet();
	fBeginWordSet = new UnicodeSet();

	fLaoWordSet.applyPattern("[[:Laoo:]&[:LineBreak=SA:]]");
	fLaoWordSet.compact();

	fMarkSet.applyPattern("[[:Laoo:]&[:LineBreak=SA:]&[:M:]]");
	fMarkSet.add(0x0020);
	fEndWordSet = 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 {
	// Initialize dictionary
	fDictionary = DictionaryData.loadDictionaryFor("Laoo");
	}

	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 findBreaks(CharacterIterator fIter, int rangeStart, int rangeEnd, boolean reverse, int breakType,
	Stack<Integer> 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().intValue() >= rangeEnd) {
	foundBreaks.pop();
	wordsFound -= 1;
	}

	return wordsFound;
	}

	}