/* * Copyright (C) 2011 The Android Open Source Project * * Licensed under the Apache License, Version 2.0 (the "License"); you may not * use this file except in compliance with the License. You may obtain a copy of * the License at * * http://www.apache.org/licenses/LICENSE-2.0 * * Unless required by applicable law or agreed to in writing, software * distributed under the License is distributed on an "AS IS" BASIS, WITHOUT * WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the * License for the specific language governing permissions and limitations under * the License. */ package com.android.inputmethod.latin.makedict; import java.util.ArrayList; import java.util.Arrays; import java.util.Collections; import java.util.HashMap; import java.util.Iterator; import java.util.LinkedList; /** * A dictionary that can fusion heads and tails of words for more compression. */ public class FusionDictionary implements Iterable { /** * A node of the dictionary, containing several CharGroups. * * A node is but an ordered array of CharGroups, which essentially contain all the * real information. * This class also contains fields to cache size and address, to help with binary * generation. */ public static class Node { ArrayList mData; // To help with binary generation int mCachedSize; int mCachedAddress; public Node() { mData = new ArrayList(); mCachedSize = Integer.MIN_VALUE; mCachedAddress = Integer.MIN_VALUE; } public Node(ArrayList data) { mData = data; mCachedSize = Integer.MIN_VALUE; mCachedAddress = Integer.MIN_VALUE; } } /** * A string with a frequency. * * This represents an "attribute", that is either a bigram or a shortcut. */ public static class WeightedString { final String mWord; int mFrequency; public WeightedString(String word, int frequency) { mWord = word; mFrequency = frequency; } @Override public int hashCode() { return Arrays.hashCode(new Object[] { mWord, mFrequency }); } @Override public boolean equals(Object o) { if (o == this) return true; if (!(o instanceof WeightedString)) return false; WeightedString w = (WeightedString)o; return mWord.equals(w.mWord) && mFrequency == w.mFrequency; } } /** * A group of characters, with a frequency, shortcut targets, bigrams, and children. * * This is the central class of the in-memory representation. A CharGroup is what can * be seen as a traditional "trie node", except it can hold several characters at the * same time. A CharGroup essentially represents one or several characters in the middle * of the trie trie; as such, it can be a terminal, and it can have children. * In this in-memory representation, whether the CharGroup is a terminal or not is represented * in the frequency, where NOT_A_TERMINAL (= -1) means this is not a terminal and any other * value is the frequency of this terminal. A terminal may have non-null shortcuts and/or * bigrams, but a non-terminal may not. Moreover, children, if present, are null. */ public static class CharGroup { public static final int NOT_A_TERMINAL = -1; final int mChars[]; ArrayList mShortcutTargets; ArrayList mBigrams; int mFrequency; // NOT_A_TERMINAL == mFrequency indicates this is not a terminal. Node mChildren; // The two following members to help with binary generation int mCachedSize; int mCachedAddress; public CharGroup(final int[] chars, final ArrayList shortcutTargets, final ArrayList bigrams, final int frequency) { mChars = chars; mFrequency = frequency; mShortcutTargets = shortcutTargets; mBigrams = bigrams; mChildren = null; } public CharGroup(final int[] chars, final ArrayList shortcutTargets, final ArrayList bigrams, final int frequency, final Node children) { mChars = chars; mFrequency = frequency; mShortcutTargets = shortcutTargets; mBigrams = bigrams; mChildren = children; } public void addChild(CharGroup n) { if (null == mChildren) { mChildren = new Node(); } mChildren.mData.add(n); } public boolean isTerminal() { return NOT_A_TERMINAL != mFrequency; } public boolean hasSeveralChars() { assert(mChars.length > 0); return 1 < mChars.length; } /** * Adds a word to the bigram list. Updates the frequency if the word already * exists. */ public void addBigram(final String word, final int frequency) { if (mBigrams == null) { mBigrams = new ArrayList(); } WeightedString bigram = getBigram(word); if (bigram != null) { bigram.mFrequency = frequency; } else { bigram = new WeightedString(word, frequency); mBigrams.add(bigram); } } /** * Gets the shortcut target for the given word. Returns null if the word is not in the * shortcut list. */ public WeightedString getShortcut(final String word) { if (mShortcutTargets != null) { final int size = mShortcutTargets.size(); for (int i = 0; i < size; ++i) { WeightedString shortcut = mShortcutTargets.get(i); if (shortcut.mWord.equals(word)) { return shortcut; } } } return null; } /** * Gets the bigram for the given word. * Returns null if the word is not in the bigrams list. */ public WeightedString getBigram(final String word) { if (mBigrams != null) { final int size = mBigrams.size(); for (int i = 0; i < size; ++i) { WeightedString bigram = mBigrams.get(i); if (bigram.mWord.equals(word)) { return bigram; } } } return null; } /** * Updates the CharGroup with the given properties. Adds the shortcut and bigram lists to * the existing ones if any. Note: unigram, bigram, and shortcut frequencies are only * updated if they are higher than the existing ones. */ public void update(int frequency, ArrayList shortcutTargets, ArrayList bigrams) { if (frequency > mFrequency) { mFrequency = frequency; } if (shortcutTargets != null) { if (mShortcutTargets == null) { mShortcutTargets = shortcutTargets; } else { final int size = shortcutTargets.size(); for (int i = 0; i < size; ++i) { final WeightedString shortcut = shortcutTargets.get(i); final WeightedString existingShortcut = getShortcut(shortcut.mWord); if (existingShortcut == null) { mShortcutTargets.add(shortcut); } else if (existingShortcut.mFrequency < shortcut.mFrequency) { existingShortcut.mFrequency = shortcut.mFrequency; } } } } if (bigrams != null) { if (mBigrams == null) { mBigrams = bigrams; } else { final int size = bigrams.size(); for (int i = 0; i < size; ++i) { final WeightedString bigram = bigrams.get(i); final WeightedString existingBigram = getBigram(bigram.mWord); if (existingBigram == null) { mBigrams.add(bigram); } else if (existingBigram.mFrequency < bigram.mFrequency) { existingBigram.mFrequency = bigram.mFrequency; } } } } } } /** * Options global to the dictionary. * * There are no options at the moment, so this class is empty. */ public static class DictionaryOptions { public final boolean mGermanUmlautProcessing; public final boolean mFrenchLigatureProcessing; public final HashMap mAttributes; public DictionaryOptions(final HashMap attributes, final boolean germanUmlautProcessing, final boolean frenchLigatureProcessing) { mAttributes = attributes; mGermanUmlautProcessing = germanUmlautProcessing; mFrenchLigatureProcessing = frenchLigatureProcessing; } } public final DictionaryOptions mOptions; public final Node mRoot; public FusionDictionary(final Node root, final DictionaryOptions options) { mRoot = root; mOptions = options; } public void addOptionAttribute(final String key, final String value) { mOptions.mAttributes.put(key, value); } /** * Helper method to convert a String to an int array. */ static private int[] getCodePoints(String word) { final int wordLength = word.length(); int[] array = new int[word.codePointCount(0, wordLength)]; for (int i = 0; i < wordLength; i = word.offsetByCodePoints(i, 1)) { array[i] = word.codePointAt(i); } return array; } /** * Helper method to add all words in a list as 0-frequency entries * * These words are added when shortcuts targets or bigrams are not found in the dictionary * yet. The same words may be added later with an actual frequency - this is handled by * the private version of add(). */ private void addNeutralWords(final ArrayList words) { if (null != words) { for (WeightedString word : words) { final CharGroup t = findWordInTree(mRoot, word.mWord); if (null == t) { add(getCodePoints(word.mWord), 0, null, null); } } } } /** * Helper method to add a word as a string. * * This method adds a word to the dictionary with the given frequency. Optional * lists of bigrams and shortcuts can be passed here. For each word inside, * they will be added to the dictionary as necessary. * * @param word the word to add. * @param frequency the frequency of the word, in the range [0..255]. * @param shortcutTargets a list of shortcut targets for this word, or null. * @param bigrams a list of bigrams, or null. */ public void add(final String word, final int frequency, final ArrayList shortcutTargets, final ArrayList bigrams) { if (null != bigrams) { addNeutralWords(bigrams); } add(getCodePoints(word), frequency, shortcutTargets, bigrams); } /** * Sanity check for a node. * * This method checks that all CharGroups in a node are ordered as expected. * If they are, nothing happens. If they aren't, an exception is thrown. */ private void checkStack(Node node) { ArrayList stack = node.mData; int lastValue = -1; for (int i = 0; i < stack.size(); ++i) { int currentValue = stack.get(i).mChars[0]; if (currentValue <= lastValue) throw new RuntimeException("Invalid stack"); else lastValue = currentValue; } } /** * Helper method to add a new bigram to the dictionary. * * @param word1 the previous word of the context * @param word2 the next word of the context * @param frequency the bigram frequency */ public void setBigram(final String word1, final String word2, final int frequency) { CharGroup charGroup = findWordInTree(mRoot, word1); if (charGroup != null) { final CharGroup charGroup2 = findWordInTree(mRoot, word2); if (charGroup2 == null) { // TODO: refactor with the identical code in addNeutralWords add(getCodePoints(word2), 0, null, null); } charGroup.addBigram(word2, frequency); } else { throw new RuntimeException("First word of bigram not found"); } } /** * Add a word to this dictionary. * * The shortcuts and bigrams, if any, have to be in the dictionary already. If they aren't, * an exception is thrown. * * @param word the word, as an int array. * @param frequency the frequency of the word, in the range [0..255]. * @param shortcutTargets an optional list of shortcut targets for this word (null if none). * @param bigrams an optional list of bigrams for this word (null if none). */ private void add(final int[] word, final int frequency, final ArrayList shortcutTargets, final ArrayList bigrams) { assert(frequency >= 0 && frequency <= 255); Node currentNode = mRoot; int charIndex = 0; CharGroup currentGroup = null; int differentCharIndex = 0; // Set by the loop to the index of the char that differs int nodeIndex = findIndexOfChar(mRoot, word[charIndex]); while (CHARACTER_NOT_FOUND != nodeIndex) { currentGroup = currentNode.mData.get(nodeIndex); differentCharIndex = compareArrays(currentGroup.mChars, word, charIndex); if (ARRAYS_ARE_EQUAL != differentCharIndex && differentCharIndex < currentGroup.mChars.length) break; if (null == currentGroup.mChildren) break; charIndex += currentGroup.mChars.length; if (charIndex >= word.length) break; currentNode = currentGroup.mChildren; nodeIndex = findIndexOfChar(currentNode, word[charIndex]); } if (-1 == nodeIndex) { // No node at this point to accept the word. Create one. final int insertionIndex = findInsertionIndex(currentNode, word[charIndex]); final CharGroup newGroup = new CharGroup( Arrays.copyOfRange(word, charIndex, word.length), shortcutTargets, bigrams, frequency); currentNode.mData.add(insertionIndex, newGroup); checkStack(currentNode); } else { // There is a word with a common prefix. if (differentCharIndex == currentGroup.mChars.length) { if (charIndex + differentCharIndex >= word.length) { // The new word is a prefix of an existing word, but the node on which it // should end already exists as is. Since the old CharNode was not a terminal, // make it one by filling in its frequency and other attributes currentGroup.update(frequency, shortcutTargets, bigrams); } else { // The new word matches the full old word and extends past it. // We only have to create a new node and add it to the end of this. final CharGroup newNode = new CharGroup( Arrays.copyOfRange(word, charIndex + differentCharIndex, word.length), shortcutTargets, bigrams, frequency); currentGroup.mChildren = new Node(); currentGroup.mChildren.mData.add(newNode); } } else { if (0 == differentCharIndex) { // Exact same word. Update the frequency if higher. This will also add the // new bigrams to the existing bigram list if it already exists. currentGroup.update(frequency, shortcutTargets, bigrams); } else { // Partial prefix match only. We have to replace the current node with a node // containing the current prefix and create two new ones for the tails. Node newChildren = new Node(); final CharGroup newOldWord = new CharGroup( Arrays.copyOfRange(currentGroup.mChars, differentCharIndex, currentGroup.mChars.length), currentGroup.mShortcutTargets, currentGroup.mBigrams, currentGroup.mFrequency, currentGroup.mChildren); newChildren.mData.add(newOldWord); final CharGroup newParent; if (charIndex + differentCharIndex >= word.length) { newParent = new CharGroup( Arrays.copyOfRange(currentGroup.mChars, 0, differentCharIndex), shortcutTargets, bigrams, frequency, newChildren); } else { newParent = new CharGroup( Arrays.copyOfRange(currentGroup.mChars, 0, differentCharIndex), null, null, -1, newChildren); final CharGroup newWord = new CharGroup( Arrays.copyOfRange(word, charIndex + differentCharIndex, word.length), shortcutTargets, bigrams, frequency); final int addIndex = word[charIndex + differentCharIndex] > currentGroup.mChars[differentCharIndex] ? 1 : 0; newChildren.mData.add(addIndex, newWord); } currentNode.mData.set(nodeIndex, newParent); } checkStack(currentNode); } } } /** * Custom comparison of two int arrays taken to contain character codes. * * This method compares the two arrays passed as an argument in a lexicographic way, * with an offset in the dst string. * This method does NOT test for the first character. It is taken to be equal. * I repeat: this method starts the comparison at 1 <> dstOffset + 1. * The index where the strings differ is returned. ARRAYS_ARE_EQUAL = 0 is returned if the * strings are equal. This works BECAUSE we don't look at the first character. * * @param src the left-hand side string of the comparison. * @param dst the right-hand side string of the comparison. * @param dstOffset the offset in the right-hand side string. * @return the index at which the strings differ, or ARRAYS_ARE_EQUAL = 0 if they don't. */ private static int ARRAYS_ARE_EQUAL = 0; private static int compareArrays(final int[] src, final int[] dst, int dstOffset) { // We do NOT test the first char, because we come from a method that already // tested it. for (int i = 1; i < src.length; ++i) { if (dstOffset + i >= dst.length) return i; if (src[i] != dst[dstOffset + i]) return i; } if (dst.length > src.length) return src.length; return ARRAYS_ARE_EQUAL; } /** * Helper class that compares and sorts two chargroups according to their * first element only. I repeat: ONLY the first element is considered, the rest * is ignored. * This comparator imposes orderings that are inconsistent with equals. */ static private class CharGroupComparator implements java.util.Comparator { public int compare(CharGroup c1, CharGroup c2) { if (c1.mChars[0] == c2.mChars[0]) return 0; return c1.mChars[0] < c2.mChars[0] ? -1 : 1; } } final static private CharGroupComparator CHARGROUP_COMPARATOR = new CharGroupComparator(); /** * Finds the insertion index of a character within a node. */ private static int findInsertionIndex(final Node node, int character) { final ArrayList data = node.mData; final CharGroup reference = new CharGroup(new int[] { character }, null, null, 0); int result = Collections.binarySearch(data, reference, CHARGROUP_COMPARATOR); return result >= 0 ? result : -result - 1; } /** * Find the index of a char in a node, if it exists. * * @param node the node to search in. * @param character the character to search for. * @return the position of the character if it's there, or CHARACTER_NOT_FOUND = -1 else. */ private static int CHARACTER_NOT_FOUND = -1; private static int findIndexOfChar(final Node node, int character) { final int insertionIndex = findInsertionIndex(node, character); if (node.mData.size() <= insertionIndex) return CHARACTER_NOT_FOUND; return character == node.mData.get(insertionIndex).mChars[0] ? insertionIndex : CHARACTER_NOT_FOUND; } /** * Helper method to find a word in a given branch. */ public static CharGroup findWordInTree(Node node, final String s) { int index = 0; final StringBuilder checker = new StringBuilder(); CharGroup currentGroup; do { int indexOfGroup = findIndexOfChar(node, s.codePointAt(index)); if (CHARACTER_NOT_FOUND == indexOfGroup) return null; currentGroup = node.mData.get(indexOfGroup); checker.append(new String(currentGroup.mChars, 0, currentGroup.mChars.length)); index += currentGroup.mChars.length; if (index < s.length()) { node = currentGroup.mChildren; } } while (null != node && index < s.length()); if (!s.equals(checker.toString())) return null; return currentGroup; } /** * Helper method to find out whether a word is in the dict or not. */ public boolean hasWord(final String s) { if (null == s || "".equals(s)) { throw new RuntimeException("Can't search for a null or empty string"); } return null != findWordInTree(mRoot, s); } /** * Recursively count the number of character groups in a given branch of the trie. * * @param node the parent node. * @return the number of char groups in all the branch under this node. */ public static int countCharGroups(final Node node) { final int nodeSize = node.mData.size(); int size = nodeSize; for (int i = nodeSize - 1; i >= 0; --i) { CharGroup group = node.mData.get(i); if (null != group.mChildren) size += countCharGroups(group.mChildren); } return size; } /** * Recursively count the number of nodes in a given branch of the trie. * * @param node the node to count. * @result the number of nodes in this branch. */ public static int countNodes(final Node node) { int size = 1; for (int i = node.mData.size() - 1; i >= 0; --i) { CharGroup group = node.mData.get(i); if (null != group.mChildren) size += countNodes(group.mChildren); } return size; } // Historically, the tails of the words were going to be merged to save space. // However, that would prevent the code to search for a specific address in log(n) // time so this was abandoned. // The code is still of interest as it does add some compression to any dictionary // that has no need for attributes. Implementations that does not read attributes should be // able to read a dictionary with merged tails. // Also, the following code does support frequencies, as in, it will only merges // tails that share the same frequency. Though it would result in the above loss of // performance while searching by address, it is still technically possible to merge // tails that contain attributes, but this code does not take that into account - it does // not compare attributes and will merge terminals with different attributes regardless. public void mergeTails() { MakedictLog.i("Do not merge tails"); return; // MakedictLog.i("Merging nodes. Number of nodes : " + countNodes(root)); // MakedictLog.i("Number of groups : " + countCharGroups(root)); // // final HashMap> repository = // new HashMap>(); // mergeTailsInner(repository, root); // // MakedictLog.i("Number of different pseudohashes : " + repository.size()); // int size = 0; // for (ArrayList a : repository.values()) { // size += a.size(); // } // MakedictLog.i("Number of nodes after merge : " + (1 + size)); // MakedictLog.i("Recursively seen nodes : " + countNodes(root)); } // The following methods are used by the deactivated mergeTails() // private static boolean isEqual(Node a, Node b) { // if (null == a && null == b) return true; // if (null == a || null == b) return false; // if (a.data.size() != b.data.size()) return false; // final int size = a.data.size(); // for (int i = size - 1; i >= 0; --i) { // CharGroup aGroup = a.data.get(i); // CharGroup bGroup = b.data.get(i); // if (aGroup.frequency != bGroup.frequency) return false; // if (aGroup.alternates == null && bGroup.alternates != null) return false; // if (aGroup.alternates != null && !aGroup.equals(bGroup.alternates)) return false; // if (!Arrays.equals(aGroup.chars, bGroup.chars)) return false; // if (!isEqual(aGroup.children, bGroup.children)) return false; // } // return true; // } // static private HashMap> mergeTailsInner( // final HashMap> map, final Node node) { // final ArrayList branches = node.data; // final int nodeSize = branches.size(); // for (int i = 0; i < nodeSize; ++i) { // CharGroup group = branches.get(i); // if (null != group.children) { // String pseudoHash = getPseudoHash(group.children); // ArrayList similarList = map.get(pseudoHash); // if (null == similarList) { // similarList = new ArrayList(); // map.put(pseudoHash, similarList); // } // boolean merged = false; // for (Node similar : similarList) { // if (isEqual(group.children, similar)) { // group.children = similar; // merged = true; // break; // } // } // if (!merged) { // similarList.add(group.children); // } // mergeTailsInner(map, group.children); // } // } // return map; // } // private static String getPseudoHash(final Node node) { // StringBuilder s = new StringBuilder(); // for (CharGroup g : node.data) { // s.append(g.frequency); // for (int ch : g.chars){ // s.append(Character.toChars(ch)); // } // } // return s.toString(); // } /** * Iterator to walk through a dictionary. * * This is purely for convenience. */ public static class DictionaryIterator implements Iterator { private static class Position { public Iterator pos; public int length; public Position(ArrayList groups) { pos = groups.iterator(); length = 0; } } final StringBuilder mCurrentString; final LinkedList mPositions; public DictionaryIterator(ArrayList root) { mCurrentString = new StringBuilder(); mPositions = new LinkedList(); final Position rootPos = new Position(root); mPositions.add(rootPos); } @Override public boolean hasNext() { for (Position p : mPositions) { if (p.pos.hasNext()) { return true; } } return false; } @Override public Word next() { Position currentPos = mPositions.getLast(); mCurrentString.setLength(mCurrentString.length() - currentPos.length); do { if (currentPos.pos.hasNext()) { final CharGroup currentGroup = currentPos.pos.next(); currentPos.length = currentGroup.mChars.length; for (int i : currentGroup.mChars) mCurrentString.append(Character.toChars(i)); if (null != currentGroup.mChildren) { currentPos = new Position(currentGroup.mChildren.mData); mPositions.addLast(currentPos); } if (currentGroup.mFrequency >= 0) return new Word(mCurrentString.toString(), currentGroup.mFrequency, currentGroup.mShortcutTargets, currentGroup.mBigrams); } else { mPositions.removeLast(); currentPos = mPositions.getLast(); mCurrentString.setLength(mCurrentString.length() - mPositions.getLast().length); } } while(true); } @Override public void remove() { throw new UnsupportedOperationException("Unsupported yet"); } } /** * Method to return an iterator. * * This method enables Java's enhanced for loop. With this you can have a FusionDictionary x * and say : for (Word w : x) {} */ @Override public Iterator iterator() { return new DictionaryIterator(mRoot.mData); } }