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      1 /* GENERATED SOURCE. DO NOT MODIFY. */
      2 //  2016 and later: Unicode, Inc. and others.
      3 // License & terms of use: http://www.unicode.org/copyright.html#License
      4 /*
      5  *******************************************************************************
      6  * Copyright (C) 2014, International Business Machines Corporation and         *
      7  * others. All Rights Reserved.                                                *
      8  *******************************************************************************
      9  */
     10 package android.icu.text;
     11 
     12 import java.io.IOException;
     13 import java.text.CharacterIterator;
     14 
     15 import android.icu.lang.UCharacter;
     16 import android.icu.lang.UProperty;
     17 import android.icu.lang.UScript;
     18 
     19 class LaoBreakEngine extends DictionaryBreakEngine {
     20 
     21     // Constants for LaoBreakIterator
     22     // How many words in a row are "good enough"?
     23     private static final byte LAO_LOOKAHEAD = 3;
     24     // Will not combine a non-word with a preceding dictionary word longer than this
     25     private static final byte LAO_ROOT_COMBINE_THRESHOLD = 3;
     26     // Will not combine a non-word that shares at least this much prefix with a
     27     // dictionary word with a preceding word
     28     private static final byte LAO_PREFIX_COMBINE_THRESHOLD = 3;
     29     // Minimum word size
     30     private static final byte LAO_MIN_WORD = 2;
     31 
     32     private DictionaryMatcher fDictionary;
     33     private static UnicodeSet fLaoWordSet;
     34     private static UnicodeSet fEndWordSet;
     35     private static UnicodeSet fBeginWordSet;
     36     private static UnicodeSet fMarkSet;
     37 
     38     static {
     39         // Initialize UnicodeSets
     40         fLaoWordSet = new UnicodeSet();
     41         fMarkSet = new UnicodeSet();
     42         fBeginWordSet = new UnicodeSet();
     43 
     44         fLaoWordSet.applyPattern("[[:Laoo:]&[:LineBreak=SA:]]");
     45         fLaoWordSet.compact();
     46 
     47         fMarkSet.applyPattern("[[:Laoo:]&[:LineBreak=SA:]&[:M:]]");
     48         fMarkSet.add(0x0020);
     49         fEndWordSet = new UnicodeSet(fLaoWordSet);
     50         fEndWordSet.remove(0x0EC0, 0x0EC4); // prefix vowels
     51         fBeginWordSet.add(0x0E81, 0x0EAE); // basic consonants (including holes for corresponding Thai characters)
     52         fBeginWordSet.add(0x0EDC, 0x0EDD); // digraph consonants (no Thai equivalent)
     53         fBeginWordSet.add(0x0EC0, 0x0EC4); // prefix vowels
     54 
     55         // Compact for caching
     56         fMarkSet.compact();
     57         fEndWordSet.compact();
     58         fBeginWordSet.compact();
     59 
     60         // Freeze the static UnicodeSet
     61         fLaoWordSet.freeze();
     62         fMarkSet.freeze();
     63         fEndWordSet.freeze();
     64         fBeginWordSet.freeze();
     65     }
     66 
     67     public LaoBreakEngine() throws IOException {
     68         super(BreakIterator.KIND_WORD, BreakIterator.KIND_LINE);
     69         setCharacters(fLaoWordSet);
     70         // Initialize dictionary
     71         fDictionary = DictionaryData.loadDictionaryFor("Laoo");
     72     }
     73 
     74     public boolean equals(Object obj) {
     75         // Normally is a singleton, but it's possible to have duplicates
     76         //   during initialization. All are equivalent.
     77         return obj instanceof LaoBreakEngine;
     78     }
     79 
     80     public int hashCode() {
     81         return getClass().hashCode();
     82     }
     83 
     84     public boolean handles(int c, int breakType) {
     85         if (breakType == BreakIterator.KIND_WORD || breakType == BreakIterator.KIND_LINE) {
     86             int script = UCharacter.getIntPropertyValue(c, UProperty.SCRIPT);
     87             return (script == UScript.LAO);
     88         }
     89         return false;
     90     }
     91 
     92     public int divideUpDictionaryRange(CharacterIterator fIter, int rangeStart, int rangeEnd,
     93             DequeI foundBreaks) {
     94 
     95 
     96         if ((rangeEnd - rangeStart) < LAO_MIN_WORD) {
     97             return 0;  // Not enough characters for word
     98         }
     99         int wordsFound = 0;
    100         int wordLength;
    101         int current;
    102         PossibleWord words[] = new PossibleWord[LAO_LOOKAHEAD];
    103         for (int i = 0; i < LAO_LOOKAHEAD; i++) {
    104             words[i] = new PossibleWord();
    105         }
    106         int uc;
    107 
    108         fIter.setIndex(rangeStart);
    109         while ((current = fIter.getIndex()) < rangeEnd) {
    110             wordLength = 0;
    111 
    112             //Look for candidate words at the current position
    113             int candidates = words[wordsFound%LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd);
    114 
    115             // If we found exactly one, use that
    116             if (candidates == 1) {
    117                 wordLength = words[wordsFound%LAO_LOOKAHEAD].acceptMarked(fIter);
    118                 wordsFound += 1;
    119             }
    120 
    121             // If there was more than one, see which one can take us forward the most words
    122             else if (candidates > 1) {
    123                 boolean foundBest = false;
    124                 // If we're already at the end of the range, we're done
    125                 if (fIter.getIndex() < rangeEnd) {
    126                     do {
    127                         int wordsMatched = 1;
    128                         if (words[(wordsFound+1)%LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd) > 0) {
    129                             if (wordsMatched < 2) {
    130                                 // Followed by another dictionary word; mark first word as a good candidate
    131                                 words[wordsFound%LAO_LOOKAHEAD].markCurrent();
    132                                 wordsMatched = 2;
    133                             }
    134 
    135                             // If we're already at the end of the range, we're done
    136                             if (fIter.getIndex() >= rangeEnd) {
    137                                 break;
    138                             }
    139 
    140                             // See if any of the possible second words is followed by a third word
    141                             do {
    142                                 // If we find a third word, stop right away
    143                                 if (words[(wordsFound+2)%LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd) > 0) {
    144                                     words[wordsFound%LAO_LOOKAHEAD].markCurrent();
    145                                     foundBest = true;
    146                                     break;
    147                                 }
    148                             } while (words[(wordsFound+1)%LAO_LOOKAHEAD].backUp(fIter));
    149                         }
    150                     } while (words[wordsFound%LAO_LOOKAHEAD].backUp(fIter) && !foundBest);
    151                 }
    152                 wordLength = words[wordsFound%LAO_LOOKAHEAD].acceptMarked(fIter);
    153                 wordsFound += 1;
    154             }
    155 
    156             // We come here after having either found a word or not. We look ahead to the
    157             // next word. If it's not a dictionary word, we will combine it with the word we
    158             // just found (if there is one), but only if the preceding word does not exceed
    159             // the threshold.
    160             // The text iterator should now be positioned at the end of the word we found.
    161             if (fIter.getIndex() < rangeEnd && wordLength < LAO_ROOT_COMBINE_THRESHOLD) {
    162                 // If it is a dictionary word, do nothing. If it isn't, then if there is
    163                 // no preceding word, or the non-word shares less than the minimum threshold
    164                 // of characters with a dictionary word, then scan to resynchronize
    165                 if (words[wordsFound%LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd) <= 0 &&
    166                         (wordLength == 0 ||
    167                                 words[wordsFound%LAO_LOOKAHEAD].longestPrefix() < LAO_PREFIX_COMBINE_THRESHOLD)) {
    168                     // Look for a plausible word boundary
    169                     int remaining = rangeEnd - (current + wordLength);
    170                     int pc = fIter.current();
    171                     int chars = 0;
    172                     for (;;) {
    173                         fIter.next();
    174                         uc = fIter.current();
    175                         chars += 1;
    176                         if (--remaining <= 0) {
    177                             break;
    178                         }
    179                         if (fEndWordSet.contains(pc) && fBeginWordSet.contains(uc)) {
    180                             // Maybe. See if it's in the dictionary.
    181                             int candidate = words[(wordsFound + 1) %LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd);
    182                             fIter.setIndex(current + wordLength + chars);
    183                             if (candidate > 0) {
    184                                 break;
    185                             }
    186                         }
    187                         pc = uc;
    188                     }
    189 
    190                     // Bump the word count if there wasn't already one
    191                     if (wordLength <= 0) {
    192                         wordsFound += 1;
    193                     }
    194 
    195                     // Update the length with the passed-over characters
    196                     wordLength += chars;
    197                 } else {
    198                     // Backup to where we were for next iteration
    199                     fIter.setIndex(current+wordLength);
    200                 }
    201             }
    202 
    203             // Never stop before a combining mark.
    204             int currPos;
    205             while ((currPos = fIter.getIndex()) < rangeEnd && fMarkSet.contains(fIter.current())) {
    206                 fIter.next();
    207                 wordLength += fIter.getIndex() - currPos;
    208             }
    209 
    210             // Look ahead for possible suffixes if a dictionary word does not follow.
    211             // We do this in code rather than using a rule so that the heuristic
    212             // resynch continues to function. For example, one of the suffix characters
    213             // could be a typo in the middle of a word.
    214             // NOT CURRENTLY APPLICABLE TO LAO
    215 
    216             // Did we find a word on this iteration? If so, push it on the break stack
    217             if (wordLength > 0) {
    218                 foundBreaks.push(Integer.valueOf(current + wordLength));
    219             }
    220         }
    221 
    222         // Don't return a break for the end of the dictionary range if there is one there
    223         if (foundBreaks.peek() >= rangeEnd) {
    224             foundBreaks.pop();
    225             wordsFound -= 1;
    226         }
    227 
    228         return wordsFound;
    229     }
    230 
    231 }
    232