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/*
 * Copyright 1999-2004 The Apache Software Foundation.
 *
 * 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
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 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
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 * See the License for the specific language governing permissions and
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package com.itextpdf.text.pdf.hyphenation;

import java.io.Serializable;
import java.util.Enumeration;
import java.util.Stack;

/**
 * 

Ternary Search Tree.

* *

A ternary search tree is a hybrid between a binary tree and * a digital search tree (trie). Keys are limited to strings. * A data value of type char is stored in each leaf node. * It can be used as an index (or pointer) to the data. * Branches that only contain one key are compressed to one node * by storing a pointer to the trailer substring of the key. * This class is intended to serve as base class or helper class * to implement Dictionary collections or the like. Ternary trees * have some nice properties as the following: the tree can be * traversed in sorted order, partial matches (wildcard) can be * implemented, retrieval of all keys within a given distance * from the target, etc. The storage requirements are higher than * a binary tree but a lot less than a trie. Performance is * comparable with a hash table, sometimes it outperforms a hash * function (most of the time can determine a miss faster than a hash).

* *

The main purpose of this java port is to serve as a base for * implementing TeX's hyphenation algorithm (see The TeXBook, * appendix H). Each language requires from 5000 to 15000 hyphenation * patterns which will be keys in this tree. The strings patterns * are usually small (from 2 to 5 characters), but each char in the * tree is stored in a node. Thus memory usage is the main concern. * We will sacrifice 'elegance' to keep memory requirements to the * minimum. Using java's char type as pointer (yes, I know pointer * it is a forbidden word in java) we can keep the size of the node * to be just 8 bytes (3 pointers and the data char). This gives * room for about 65000 nodes. In my tests the English patterns * took 7694 nodes and the German patterns 10055 nodes, * so I think we are safe.

* *

All said, this is a map with strings as keys and char as value. * Pretty limited!. It can be extended to a general map by * using the string representation of an object and using the * char value as an index to an array that contains the object * values.

* * @author [email protected] */ public class TernaryTree implements Cloneable, Serializable { /** * We use 4 arrays to represent a node. I guess I should have created * a proper node class, but somehow Knuth's pascal code made me forget * we now have a portable language with virtual memory management and * automatic garbage collection! And now is kind of late, furthermore, * if it ain't broken, don't fix it. */ private static final long serialVersionUID = 5313366505322983510L; /** * Pointer to low branch and to rest of the key when it is * stored directly in this node, we don't have unions in java! */ protected char[] lo; /** * Pointer to high branch. */ protected char[] hi; /** * Pointer to equal branch and to data when this node is a string terminator. */ protected char[] eq; /** *

The character stored in this node: splitchar. * Two special values are reserved:

*
  • 0x0000 as string terminator
  • *
  • 0xFFFF to indicate that the branch starting at * this node is compressed
*

This shouldn't be a problem if we give the usual semantics to * strings since 0xFFFF is guaranteed not to be an Unicode character.

*/ protected char[] sc; /** * This vector holds the trailing of the keys when the branch is compressed. */ protected CharVector kv; protected char root; protected char freenode; protected int length; // number of items in tree protected static final int BLOCK_SIZE = 2048; // allocation size for arrays TernaryTree() { init(); } protected void init() { root = 0; freenode = 1; length = 0; lo = new char[BLOCK_SIZE]; hi = new char[BLOCK_SIZE]; eq = new char[BLOCK_SIZE]; sc = new char[BLOCK_SIZE]; kv = new CharVector(); } /** * Branches are initially compressed, needing * one node per key plus the size of the string * key. They are decompressed as needed when * another key with same prefix * is inserted. This saves a lot of space, * specially for long keys. */ public void insert(String key, char val) { // make sure we have enough room in the arrays int len = key.length() + 1; // maximum number of nodes that may be generated if (freenode + len > eq.length) { redimNodeArrays(eq.length + BLOCK_SIZE); } char strkey[] = new char[len--]; key.getChars(0, len, strkey, 0); strkey[len] = 0; root = insert(root, strkey, 0, val); } public void insert(char[] key, int start, char val) { int len = strlen(key) + 1; if (freenode + len > eq.length) { redimNodeArrays(eq.length + BLOCK_SIZE); } root = insert(root, key, start, val); } /** * The actual insertion function, recursive version. */ private char insert(char p, char[] key, int start, char val) { int len = strlen(key, start); if (p == 0) { // this means there is no branch, this node will start a new branch. // Instead of doing that, we store the key somewhere else and create // only one node with a pointer to the key p = freenode++; eq[p] = val; // holds data length++; hi[p] = 0; if (len > 0) { sc[p] = 0xFFFF; // indicates branch is compressed lo[p] = (char)kv.alloc(len + 1); // use 'lo' to hold pointer to key strcpy(kv.getArray(), lo[p], key, start); } else { sc[p] = 0; lo[p] = 0; } return p; } if (sc[p] == 0xFFFF) { // branch is compressed: need to decompress // this will generate garbage in the external key array // but we can do some garbage collection later char pp = freenode++; lo[pp] = lo[p]; // previous pointer to key eq[pp] = eq[p]; // previous pointer to data lo[p] = 0; if (len > 0) { sc[p] = kv.get(lo[pp]); eq[p] = pp; lo[pp]++; if (kv.get(lo[pp]) == 0) { // key completely decompressed leaving garbage in key array lo[pp] = 0; sc[pp] = 0; hi[pp] = 0; } else { // we only got first char of key, rest is still there sc[pp] = 0xFFFF; } } else { // In this case we can save a node by swapping the new node // with the compressed node sc[pp] = 0xFFFF; hi[p] = pp; sc[p] = 0; eq[p] = val; length++; return p; } } char s = key[start]; if (s < sc[p]) { lo[p] = insert(lo[p], key, start, val); } else if (s == sc[p]) { if (s != 0) { eq[p] = insert(eq[p], key, start + 1, val); } else { // key already in tree, overwrite data eq[p] = val; } } else { hi[p] = insert(hi[p], key, start, val); } return p; } /** * Compares 2 null terminated char arrays */ public static int strcmp(char[] a, int startA, char[] b, int startB) { for (; a[startA] == b[startB]; startA++, startB++) { if (a[startA] == 0) { return 0; } } return a[startA] - b[startB]; } /** * Compares a string with null terminated char array */ public static int strcmp(String str, char[] a, int start) { int i, d, len = str.length(); for (i = 0; i < len; i++) { d = str.charAt(i) - a[start + i]; if (d != 0) { return d; } if (a[start + i] == 0) { return d; } } if (a[start + i] != 0) { return -a[start + i]; } return 0; } public static void strcpy(char[] dst, int di, char[] src, int si) { while (src[si] != 0) { dst[di++] = src[si++]; } dst[di] = 0; } public static int strlen(char[] a, int start) { int len = 0; for (int i = start; i < a.length && a[i] != 0; i++) { len++; } return len; } public static int strlen(char[] a) { return strlen(a, 0); } public int find(String key) { int len = key.length(); char strkey[] = new char[len + 1]; key.getChars(0, len, strkey, 0); strkey[len] = 0; return find(strkey, 0); } public int find(char[] key, int start) { int d; char p = root; int i = start; char c; while (p != 0) { if (sc[p] == 0xFFFF) { if (strcmp(key, i, kv.getArray(), lo[p]) == 0) { return eq[p]; } else { return -1; } } c = key[i]; d = c - sc[p]; if (d == 0) { if (c == 0) { return eq[p]; } i++; p = eq[p]; } else if (d < 0) { p = lo[p]; } else { p = hi[p]; } } return -1; } public boolean knows(String key) { return find(key) >= 0; } // redimension the arrays private void redimNodeArrays(int newsize) { int len = newsize < lo.length ? newsize : lo.length; char[] na = new char[newsize]; System.arraycopy(lo, 0, na, 0, len); lo = na; na = new char[newsize]; System.arraycopy(hi, 0, na, 0, len); hi = na; na = new char[newsize]; System.arraycopy(eq, 0, na, 0, len); eq = na; na = new char[newsize]; System.arraycopy(sc, 0, na, 0, len); sc = na; } public int size() { return length; } @Override public Object clone() { TernaryTree t = new TernaryTree(); t.lo = this.lo.clone(); t.hi = this.hi.clone(); t.eq = this.eq.clone(); t.sc = this.sc.clone(); t.kv = (CharVector)this.kv.clone(); t.root = this.root; t.freenode = this.freenode; t.length = this.length; return t; } /** * Recursively insert the median first and then the median of the * lower and upper halves, and so on in order to get a balanced * tree. The array of keys is assumed to be sorted in ascending * order. */ protected void insertBalanced(String[] k, char[] v, int offset, int n) { int m; if (n < 1) { return; } m = n >> 1; insert(k[m + offset], v[m + offset]); insertBalanced(k, v, offset, m); insertBalanced(k, v, offset + m + 1, n - m - 1); } /** * Balance the tree for best search performance */ public void balance() { // System.out.print("Before root splitchar = "); System.out.println(sc[root]); int i = 0, n = length; String[] k = new String[n]; char[] v = new char[n]; Iterator iter = new Iterator(); while (iter.hasMoreElements()) { v[i] = iter.getValue(); k[i++] = iter.nextElement(); } init(); insertBalanced(k, v, 0, n); // With uniform letter distribution sc[root] should be around 'm' // System.out.print("After root splitchar = "); System.out.println(sc[root]); } /** * Each node stores a character (splitchar) which is part of * some key(s). In a compressed branch (one that only contain * a single string key) the trailer of the key which is not * already in nodes is stored externally in the kv array. * As items are inserted, key substrings decrease. * Some substrings may completely disappear when the whole * branch is totally decompressed. * The tree is traversed to find the key substrings actually * used. In addition, duplicate substrings are removed using * a map (implemented with a TernaryTree!). * */ public void trimToSize() { // first balance the tree for best performance balance(); // redimension the node arrays redimNodeArrays(freenode); // ok, compact kv array CharVector kx = new CharVector(); kx.alloc(1); TernaryTree map = new TernaryTree(); compact(kx, map, root); kv = kx; kv.trimToSize(); } private void compact(CharVector kx, TernaryTree map, char p) { int k; if (p == 0) { return; } if (sc[p] == 0xFFFF) { k = map.find(kv.getArray(), lo[p]); if (k < 0) { k = kx.alloc(strlen(kv.getArray(), lo[p]) + 1); strcpy(kx.getArray(), k, kv.getArray(), lo[p]); map.insert(kx.getArray(), k, (char)k); } lo[p] = (char)k; } else { compact(kx, map, lo[p]); if (sc[p] != 0) { compact(kx, map, eq[p]); } compact(kx, map, hi[p]); } } public Enumeration keys() { return new Iterator(); } public class Iterator implements Enumeration { /** * current node index */ int cur; /** * current key */ String curkey; private class Item implements Cloneable { char parent; char child; public Item() { parent = 0; child = 0; } public Item(char p, char c) { parent = p; child = c; } @Override public Item clone() { return new Item(parent, child); } } /** * Node stack */ Stack ns; /** * key stack implemented with a StringBuffer */ StringBuffer ks; public Iterator() { cur = -1; ns = new Stack(); ks = new StringBuffer(); rewind(); } public void rewind() { ns.removeAllElements(); ks.setLength(0); cur = root; run(); } public String nextElement() { String res = curkey; cur = up(); run(); return res; } public char getValue() { if (cur >= 0) { return eq[cur]; } return 0; } public boolean hasMoreElements() { return cur != -1; } /** * traverse upwards */ private int up() { Item i = new Item(); int res = 0; if (ns.empty()) { return -1; } if (cur != 0 && sc[cur] == 0) { return lo[cur]; } boolean climb = true; while (climb) { i = ns.pop(); i.child++; switch (i.child) { case 1: if (sc[i.parent] != 0) { res = eq[i.parent]; ns.push(i.clone()); ks.append(sc[i.parent]); } else { i.child++; ns.push(i.clone()); res = hi[i.parent]; } climb = false; break; case 2: res = hi[i.parent]; ns.push(i.clone()); if (ks.length() > 0) { ks.setLength(ks.length() - 1); // pop } climb = false; break; default: if (ns.empty()) { return -1; } climb = true; break; } } return res; } /** * traverse the tree to find next key */ private int run() { if (cur == -1) { return -1; } boolean leaf = false; while (true) { // first go down on low branch until leaf or compressed branch while (cur != 0) { if (sc[cur] == 0xFFFF) { leaf = true; break; } ns.push(new Item((char)cur, '\u0000')); if (sc[cur] == 0) { leaf = true; break; } cur = lo[cur]; } if (leaf) { break; } // nothing found, go up one node and try again cur = up(); if (cur == -1) { return -1; } } // The current node should be a data node and // the key should be in the key stack (at least partially) StringBuffer buf = new StringBuffer(ks.toString()); if (sc[cur] == 0xFFFF) { int p = lo[cur]; while (kv.get(p) != 0) { buf.append(kv.get(p++)); } } curkey = buf.toString(); return 0; } } public void printStats() { System.out.println("Number of keys = " + Integer.toString(length)); System.out.println("Node count = " + Integer.toString(freenode)); // System.out.println("Array length = " + Integer.toString(eq.length)); System.out.println("Key Array length = " + Integer.toString(kv.length())); /* * for(int i=0; i




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