/*
* 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;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
/**
* A dictionary that can fusion heads and tails of words for more compression.
*/
public class FusionDictionary implements Iterable<Word> {
/**
* 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<CharGroup> mData;
// To help with binary generation
int mCachedSize;
int mCachedAddress;
public Node() {
mData = new ArrayList<CharGroup>();
mCachedSize = Integer.MIN_VALUE;
mCachedAddress = Integer.MIN_VALUE;
}
public Node(ArrayList<CharGroup> 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;
final int mFrequency;
public WeightedString(String word, int frequency) {
mWord = word;
mFrequency = frequency;
}
}
/**
* A group of characters, with a frequency, shortcuts, 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[];
final ArrayList<WeightedString> mBigrams;
final 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<WeightedString> bigrams, final int frequency) {
mChars = chars;
mFrequency = frequency;
mBigrams = bigrams;
mChildren = null;
}
public CharGroup(final int[] chars,
final ArrayList<WeightedString> bigrams, final int frequency, final Node children) {
mChars = chars;
mFrequency = frequency;
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;
}
}
/**
* Options global to the dictionary.
*
* There are no options at the moment, so this class is empty.
*/
public static class DictionaryOptions {
}
public final DictionaryOptions mOptions;
public final Node mRoot;
public FusionDictionary() {
mOptions = new DictionaryOptions();
mRoot = new Node();
}
public FusionDictionary(final Node root, final DictionaryOptions options) {
mRoot = root;
mOptions = options;
}
/**
* 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) {
array[i] = word.codePointAt(i);
}
return array;
}
/**
* 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 bigrams a list of bigrams, or null.
*/
public void add(String word, int frequency, ArrayList<WeightedString> bigrams) {
if (null != bigrams) {
for (WeightedString bigram : bigrams) {
final CharGroup t = findWordInTree(mRoot, bigram.mWord);
if (null == t) {
add(getCodePoints(bigram.mWord), 0, null);
}
}
}
add(getCodePoints(word), frequency, 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<CharGroup> 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;
}
}
/**
* Add a word to this dictionary.
*
* The 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 bigrams an optional list of bigrams for this word (null if none).
*/
private void add(int[] word, int frequency, ArrayList<WeightedString> 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), 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.
if (currentGroup.mFrequency > 0) {
throw new RuntimeException("Such a word already exists in the dictionary : "
+ new String(word, 0, word.length));
} else {
final CharGroup newNode = new CharGroup(currentGroup.mChars,
bigrams, frequency, currentGroup.mChildren);
currentNode.mData.set(nodeIndex, newNode);
checkStack(currentNode);
}
} 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),
bigrams, frequency);
currentGroup.mChildren = new Node();
currentGroup.mChildren.mData.add(newNode);
}
} else {
if (0 == differentCharIndex) {
// Exact same word. Check the frequency is 0 or -1, and update.
if (0 != frequency) {
if (0 < currentGroup.mFrequency) {
throw new RuntimeException("This word already exists with frequency "
+ currentGroup.mFrequency + " : "
+ new String(word, 0, word.length));
}
final CharGroup newGroup = new CharGroup(word,
currentGroup.mBigrams, frequency, currentGroup.mChildren);
currentNode.mData.set(nodeIndex, newGroup);
}
} 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.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),
bigrams, frequency, newChildren);
} else {
newParent = new CharGroup(
Arrays.copyOfRange(currentGroup.mChars, 0, differentCharIndex),
null, -1, newChildren);
final CharGroup newWord = new CharGroup(
Arrays.copyOfRange(word, charIndex + differentCharIndex,
word.length), 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(Object o1, Object o2) {
final CharGroup c1 = (CharGroup)o1;
final CharGroup c2 = (CharGroup)o2;
if (c1.mChars[0] == c2.mChars[0]) return 0;
return c1.mChars[0] < c2.mChars[0] ? -1 : 1;
}
public boolean equals(Object o) {
return o instanceof CharGroupComparator;
}
}
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 List data = node.mData;
final CharGroup reference = new CharGroup(new int[] { character }, 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;
}
/**
* 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<String, ArrayList<Node>> repository =
// new HashMap<String, ArrayList<Node>>();
// mergeTailsInner(repository, root);
//
// MakedictLog.i("Number of different pseudohashes : " + repository.size());
// int size = 0;
// for (ArrayList<Node> 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<String, ArrayList<Node>> mergeTailsInner(
// final HashMap<String, ArrayList<Node>> map, final Node node) {
// final ArrayList<CharGroup> 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<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);
}
}