603 lines
24 KiB
Java
603 lines
24 KiB
Java
/*
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* Copyright (C) 2011 The Android Open Source Project
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*
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* Licensed under the Apache License, Version 2.0 (the "License"); you may not
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* use this file except in compliance with the License. You may obtain a copy of
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* the License at
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*
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* http://www.apache.org/licenses/LICENSE-2.0
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*
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* Unless required by applicable law or agreed to in writing, software
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* distributed under the License is distributed on an "AS IS" BASIS, WITHOUT
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* WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the
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* License for the specific language governing permissions and limitations under
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* the License.
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*/
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package com.android.inputmethod.latin;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.Collections;
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import java.util.Iterator;
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import java.util.LinkedList;
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import java.util.List;
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/**
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* A dictionary that can fusion heads and tails of words for more compression.
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*/
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public class FusionDictionary implements Iterable<Word> {
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/**
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* A node of the dictionary, containing several CharGroups.
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*
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* A node is but an ordered array of CharGroups, which essentially contain all the
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* real information.
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* This class also contains fields to cache size and address, to help with binary
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* generation.
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*/
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public static class Node {
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ArrayList<CharGroup> mData;
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// To help with binary generation
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int mCachedSize;
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int mCachedAddress;
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public Node() {
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mData = new ArrayList<CharGroup>();
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mCachedSize = Integer.MIN_VALUE;
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mCachedAddress = Integer.MIN_VALUE;
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}
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public Node(ArrayList<CharGroup> data) {
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mData = data;
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mCachedSize = Integer.MIN_VALUE;
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mCachedAddress = Integer.MIN_VALUE;
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}
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}
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/**
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* A string with a frequency.
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*
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* This represents an "attribute", that is either a bigram or a shortcut.
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*/
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public static class WeightedString {
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final String mWord;
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final int mFrequency;
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public WeightedString(String word, int frequency) {
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mWord = word;
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mFrequency = frequency;
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}
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}
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/**
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* A group of characters, with a frequency, shortcuts, bigrams, and children.
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*
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* This is the central class of the in-memory representation. A CharGroup is what can
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* be seen as a traditional "trie node", except it can hold several characters at the
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* same time. A CharGroup essentially represents one or several characters in the middle
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* of the trie trie; as such, it can be a terminal, and it can have children.
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* In this in-memory representation, whether the CharGroup is a terminal or not is represented
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* in the frequency, where NOT_A_TERMINAL (= -1) means this is not a terminal and any other
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* value is the frequency of this terminal. A terminal may have non-null shortcuts and/or
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* bigrams, but a non-terminal may not. Moreover, children, if present, are null.
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*/
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public static class CharGroup {
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public static final int NOT_A_TERMINAL = -1;
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final int mChars[];
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final ArrayList<WeightedString> mBigrams;
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final int mFrequency; // NOT_A_TERMINAL == mFrequency indicates this is not a terminal.
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Node mChildren;
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// The two following members to help with binary generation
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int mCachedSize;
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int mCachedAddress;
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public CharGroup(final int[] chars,
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final ArrayList<WeightedString> bigrams, final int frequency) {
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mChars = chars;
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mFrequency = frequency;
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mBigrams = bigrams;
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mChildren = null;
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}
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public CharGroup(final int[] chars,
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final ArrayList<WeightedString> bigrams, final int frequency, final Node children) {
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mChars = chars;
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mFrequency = frequency;
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mBigrams = bigrams;
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mChildren = children;
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}
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public void addChild(CharGroup n) {
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if (null == mChildren) {
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mChildren = new Node();
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}
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mChildren.mData.add(n);
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}
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public boolean isTerminal() {
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return NOT_A_TERMINAL != mFrequency;
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}
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public boolean hasSeveralChars() {
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assert(mChars.length > 0);
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return 1 < mChars.length;
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}
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}
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/**
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* Options global to the dictionary.
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*
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* There are no options at the moment, so this class is empty.
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*/
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public static class DictionaryOptions {
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}
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public final DictionaryOptions mOptions;
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public final Node mRoot;
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public FusionDictionary() {
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mOptions = new DictionaryOptions();
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mRoot = new Node();
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}
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public FusionDictionary(final Node root, final DictionaryOptions options) {
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mRoot = root;
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mOptions = options;
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}
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/**
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* Helper method to convert a String to an int array.
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*/
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static private int[] getCodePoints(String word) {
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final int wordLength = word.length();
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int[] array = new int[word.codePointCount(0, wordLength)];
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for (int i = 0; i < wordLength; ++i) {
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array[i] = word.codePointAt(i);
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}
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return array;
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}
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/**
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* Helper method to add a word as a string.
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*
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* This method adds a word to the dictionary with the given frequency. Optional
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* lists of bigrams and shortcuts can be passed here. For each word inside,
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* they will be added to the dictionary as necessary.
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*
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* @param word the word to add.
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* @param frequency the frequency of the word, in the range [0..255].
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* @param bigrams a list of bigrams, or null.
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*/
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public void add(String word, int frequency, ArrayList<WeightedString> bigrams) {
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if (null != bigrams) {
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for (WeightedString bigram : bigrams) {
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final CharGroup t = findWordInTree(mRoot, bigram.mWord);
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if (null == t) {
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add(getCodePoints(bigram.mWord), 0, null);
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}
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}
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}
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add(getCodePoints(word), frequency, bigrams);
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}
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/**
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* Sanity check for a node.
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*
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* This method checks that all CharGroups in a node are ordered as expected.
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* If they are, nothing happens. If they aren't, an exception is thrown.
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*/
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private void checkStack(Node node) {
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ArrayList<CharGroup> stack = node.mData;
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int lastValue = -1;
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for (int i = 0; i < stack.size(); ++i) {
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int currentValue = stack.get(i).mChars[0];
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if (currentValue <= lastValue)
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throw new RuntimeException("Invalid stack");
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else
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lastValue = currentValue;
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}
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}
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/**
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* Add a word to this dictionary.
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*
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* The bigrams, if any, have to be in the dictionary already. If they aren't,
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* an exception is thrown.
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*
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* @param word the word, as an int array.
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* @param frequency the frequency of the word, in the range [0..255].
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* @param bigrams an optional list of bigrams for this word (null if none).
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*/
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private void add(int[] word, int frequency, ArrayList<WeightedString> bigrams) {
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assert(frequency >= 0 && frequency <= 255);
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Node currentNode = mRoot;
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int charIndex = 0;
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CharGroup currentGroup = null;
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int differentCharIndex = 0; // Set by the loop to the index of the char that differs
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int nodeIndex = findIndexOfChar(mRoot, word[charIndex]);
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while (CHARACTER_NOT_FOUND != nodeIndex) {
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currentGroup = currentNode.mData.get(nodeIndex);
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differentCharIndex = compareArrays(currentGroup.mChars, word, charIndex);
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if (ARRAYS_ARE_EQUAL != differentCharIndex
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&& differentCharIndex < currentGroup.mChars.length) break;
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if (null == currentGroup.mChildren) break;
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charIndex += currentGroup.mChars.length;
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if (charIndex >= word.length) break;
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currentNode = currentGroup.mChildren;
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nodeIndex = findIndexOfChar(currentNode, word[charIndex]);
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}
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if (-1 == nodeIndex) {
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// No node at this point to accept the word. Create one.
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final int insertionIndex = findInsertionIndex(currentNode, word[charIndex]);
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final CharGroup newGroup = new CharGroup(
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Arrays.copyOfRange(word, charIndex, word.length), bigrams, frequency);
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currentNode.mData.add(insertionIndex, newGroup);
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checkStack(currentNode);
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} else {
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// There is a word with a common prefix.
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if (differentCharIndex == currentGroup.mChars.length) {
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if (charIndex + differentCharIndex >= word.length) {
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// The new word is a prefix of an existing word, but the node on which it
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// should end already exists as is.
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if (currentGroup.mFrequency > 0) {
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throw new RuntimeException("Such a word already exists in the dictionary : "
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+ new String(word, 0, word.length));
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} else {
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final CharGroup newNode = new CharGroup(currentGroup.mChars,
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bigrams, frequency, currentGroup.mChildren);
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currentNode.mData.set(nodeIndex, newNode);
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checkStack(currentNode);
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}
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} else {
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// The new word matches the full old word and extends past it.
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// We only have to create a new node and add it to the end of this.
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final CharGroup newNode = new CharGroup(
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Arrays.copyOfRange(word, charIndex + differentCharIndex, word.length),
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bigrams, frequency);
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currentGroup.mChildren = new Node();
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currentGroup.mChildren.mData.add(newNode);
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}
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} else {
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if (0 == differentCharIndex) {
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// Exact same word. Check the frequency is 0 or -1, and update.
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if (0 != frequency) {
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if (0 < currentGroup.mFrequency) {
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throw new RuntimeException("This word already exists with frequency "
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+ currentGroup.mFrequency + " : "
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+ new String(word, 0, word.length));
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}
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final CharGroup newGroup = new CharGroup(word,
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currentGroup.mBigrams, frequency, currentGroup.mChildren);
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currentNode.mData.set(nodeIndex, newGroup);
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}
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} else {
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// Partial prefix match only. We have to replace the current node with a node
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// containing the current prefix and create two new ones for the tails.
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Node newChildren = new Node();
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final CharGroup newOldWord = new CharGroup(
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Arrays.copyOfRange(currentGroup.mChars, differentCharIndex,
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currentGroup.mChars.length),
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currentGroup.mBigrams, currentGroup.mFrequency, currentGroup.mChildren);
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newChildren.mData.add(newOldWord);
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final CharGroup newParent;
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if (charIndex + differentCharIndex >= word.length) {
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newParent = new CharGroup(
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Arrays.copyOfRange(currentGroup.mChars, 0, differentCharIndex),
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bigrams, frequency, newChildren);
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} else {
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newParent = new CharGroup(
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Arrays.copyOfRange(currentGroup.mChars, 0, differentCharIndex),
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null, -1, newChildren);
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final CharGroup newWord = new CharGroup(
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Arrays.copyOfRange(word, charIndex + differentCharIndex,
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word.length), bigrams, frequency);
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final int addIndex = word[charIndex + differentCharIndex]
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> currentGroup.mChars[differentCharIndex] ? 1 : 0;
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newChildren.mData.add(addIndex, newWord);
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}
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currentNode.mData.set(nodeIndex, newParent);
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}
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checkStack(currentNode);
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}
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}
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}
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/**
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* Custom comparison of two int arrays taken to contain character codes.
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*
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* This method compares the two arrays passed as an argument in a lexicographic way,
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* with an offset in the dst string.
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* This method does NOT test for the first character. It is taken to be equal.
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* I repeat: this method starts the comparison at 1 <> dstOffset + 1.
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* The index where the strings differ is returned. ARRAYS_ARE_EQUAL = 0 is returned if the
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* strings are equal. This works BECAUSE we don't look at the first character.
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*
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* @param src the left-hand side string of the comparison.
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* @param dst the right-hand side string of the comparison.
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* @param dstOffset the offset in the right-hand side string.
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* @return the index at which the strings differ, or ARRAYS_ARE_EQUAL = 0 if they don't.
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*/
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private static int ARRAYS_ARE_EQUAL = 0;
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private static int compareArrays(final int[] src, final int[] dst, int dstOffset) {
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// We do NOT test the first char, because we come from a method that already
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// tested it.
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for (int i = 1; i < src.length; ++i) {
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if (dstOffset + i >= dst.length) return i;
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if (src[i] != dst[dstOffset + i]) return i;
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}
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if (dst.length > src.length) return src.length;
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return ARRAYS_ARE_EQUAL;
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}
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/**
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* Helper class that compares and sorts two chargroups according to their
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* first element only. I repeat: ONLY the first element is considered, the rest
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* is ignored.
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* This comparator imposes orderings that are inconsistent with equals.
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*/
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static private class CharGroupComparator implements java.util.Comparator {
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public int compare(Object o1, Object o2) {
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final CharGroup c1 = (CharGroup)o1;
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final CharGroup c2 = (CharGroup)o2;
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if (c1.mChars[0] == c2.mChars[0]) return 0;
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return c1.mChars[0] < c2.mChars[0] ? -1 : 1;
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}
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public boolean equals(Object o) {
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return o instanceof CharGroupComparator;
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}
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}
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final static private CharGroupComparator CHARGROUP_COMPARATOR = new CharGroupComparator();
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/**
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* Finds the insertion index of a character within a node.
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*/
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private static int findInsertionIndex(final Node node, int character) {
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final List data = node.mData;
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final CharGroup reference = new CharGroup(new int[] { character }, null, 0);
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int result = Collections.binarySearch(data, reference, CHARGROUP_COMPARATOR);
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return result >= 0 ? result : -result - 1;
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}
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/**
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* Find the index of a char in a node, if it exists.
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*
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* @param node the node to search in.
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* @param character the character to search for.
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* @return the position of the character if it's there, or CHARACTER_NOT_FOUND = -1 else.
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*/
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private static int CHARACTER_NOT_FOUND = -1;
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private static int findIndexOfChar(final Node node, int character) {
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final int insertionIndex = findInsertionIndex(node, character);
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if (node.mData.size() <= insertionIndex) return CHARACTER_NOT_FOUND;
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return character == node.mData.get(insertionIndex).mChars[0] ? insertionIndex
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: CHARACTER_NOT_FOUND;
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}
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/**
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* Helper method to find a word in a given branch.
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*/
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public static CharGroup findWordInTree(Node node, final String s) {
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int index = 0;
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final StringBuilder checker = new StringBuilder();
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CharGroup currentGroup;
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do {
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int indexOfGroup = findIndexOfChar(node, s.codePointAt(index));
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if (CHARACTER_NOT_FOUND == indexOfGroup) return null;
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currentGroup = node.mData.get(indexOfGroup);
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checker.append(new String(currentGroup.mChars, 0, currentGroup.mChars.length));
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index += currentGroup.mChars.length;
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if (index < s.length()) {
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node = currentGroup.mChildren;
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}
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} while (null != node && index < s.length());
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if (!s.equals(checker.toString())) return null;
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return currentGroup;
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}
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/**
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* Recursively count the number of character groups in a given branch of the trie.
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*
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* @param node the parent node.
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* @return the number of char groups in all the branch under this node.
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*/
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public static int countCharGroups(final Node node) {
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final int nodeSize = node.mData.size();
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int size = nodeSize;
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for (int i = nodeSize - 1; i >= 0; --i) {
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CharGroup group = node.mData.get(i);
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if (null != group.mChildren)
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size += countCharGroups(group.mChildren);
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}
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return size;
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}
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/**
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* Recursively count the number of nodes in a given branch of the trie.
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*
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* @param node the node to count.
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* @result the number of nodes in this branch.
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*/
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public static int countNodes(final Node node) {
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int size = 1;
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for (int i = node.mData.size() - 1; i >= 0; --i) {
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CharGroup group = node.mData.get(i);
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if (null != group.mChildren)
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size += countNodes(group.mChildren);
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}
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return size;
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}
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// Historically, the tails of the words were going to be merged to save space.
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// However, that would prevent the code to search for a specific address in log(n)
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// time so this was abandoned.
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// The code is still of interest as it does add some compression to any dictionary
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// that has no need for attributes. Implementations that does not read attributes should be
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// able to read a dictionary with merged tails.
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// Also, the following code does support frequencies, as in, it will only merges
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// tails that share the same frequency. Though it would result in the above loss of
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// performance while searching by address, it is still technically possible to merge
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// tails that contain attributes, but this code does not take that into account - it does
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// not compare attributes and will merge terminals with different attributes regardless.
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public void mergeTails() {
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MakedictLog.i("Do not merge tails");
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return;
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// MakedictLog.i("Merging nodes. Number of nodes : " + countNodes(root));
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// MakedictLog.i("Number of groups : " + countCharGroups(root));
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//
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// final HashMap<String, ArrayList<Node>> repository =
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// new HashMap<String, ArrayList<Node>>();
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// mergeTailsInner(repository, root);
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//
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// MakedictLog.i("Number of different pseudohashes : " + repository.size());
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// int size = 0;
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// for (ArrayList<Node> a : repository.values()) {
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// size += a.size();
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// }
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// MakedictLog.i("Number of nodes after merge : " + (1 + size));
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// MakedictLog.i("Recursively seen nodes : " + countNodes(root));
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}
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// The following methods are used by the deactivated mergeTails()
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// private static boolean isEqual(Node a, Node b) {
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// if (null == a && null == b) return true;
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// if (null == a || null == b) return false;
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// if (a.data.size() != b.data.size()) return false;
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// final int size = a.data.size();
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// for (int i = size - 1; i >= 0; --i) {
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// CharGroup aGroup = a.data.get(i);
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// CharGroup bGroup = b.data.get(i);
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// if (aGroup.frequency != bGroup.frequency) return false;
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// if (aGroup.alternates == null && bGroup.alternates != null) return false;
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// if (aGroup.alternates != null && !aGroup.equals(bGroup.alternates)) return false;
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// if (!Arrays.equals(aGroup.chars, bGroup.chars)) return false;
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// if (!isEqual(aGroup.children, bGroup.children)) return false;
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// }
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// return true;
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// }
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// static private HashMap<String, ArrayList<Node>> mergeTailsInner(
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// final HashMap<String, ArrayList<Node>> map, final Node node) {
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// final ArrayList<CharGroup> branches = node.data;
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// final int nodeSize = branches.size();
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// for (int i = 0; i < nodeSize; ++i) {
|
|
// CharGroup group = branches.get(i);
|
|
// if (null != group.children) {
|
|
// String pseudoHash = getPseudoHash(group.children);
|
|
// ArrayList<Node> similarList = map.get(pseudoHash);
|
|
// if (null == similarList) {
|
|
// similarList = new ArrayList<Node>();
|
|
// 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<Word> {
|
|
|
|
private static class Position {
|
|
public Iterator<CharGroup> pos;
|
|
public int length;
|
|
public Position(ArrayList<CharGroup> groups) {
|
|
pos = groups.iterator();
|
|
length = 0;
|
|
}
|
|
}
|
|
final StringBuilder mCurrentString;
|
|
final LinkedList<Position> mPositions;
|
|
|
|
public DictionaryIterator(ArrayList<CharGroup> root) {
|
|
mCurrentString = new StringBuilder();
|
|
mPositions = new LinkedList<Position>();
|
|
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.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<Word> iterator() {
|
|
return new DictionaryIterator(mRoot.mData);
|
|
}
|
|
}
|