345 lines
15 KiB
Java
345 lines
15 KiB
Java
/*
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The contents of this file are subject to the THDL Open Community License
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Version 1.0 (the "License"); you may not use this file except in compliance
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with the License. You may obtain a copy of the License on the THDL web site
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(http://www.thdl.org/).
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Software distributed under the License is distributed on an "AS IS" basis,
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WITHOUT WARRANTY OF ANY KIND, either express or implied. See the
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License for the specific terms governing rights and limitations under the
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License.
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The Initial Developer of this software is the Tibetan and Himalayan Digital
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Library (THDL). Portions created by the THDL are Copyright 2003 THDL.
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All Rights Reserved.
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Contributor(s): ______________________________________.
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*/
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package org.thdl.tib.text.ttt;
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import java.util.ArrayList;
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/** A list of non-empty list of {@link TStackListList
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* TStackListLists} representing all the ways you could break up a
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* tsheg bar of ACIP into stacks (i.e., grapheme clusters).
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*
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* @author David Chandler */
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class TParseTree {
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/** a fast, non-thread-safe, random-access list implementation: */
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private ArrayList al = new ArrayList();
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/** Creates an empty list. */
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public TParseTree() { }
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/** Returns the ith list of stack lists in this parse tree. */
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public TStackListList get(int i) { return (TStackListList)al.get(i); }
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/** Adds p to the end of this list. */
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public void add(TStackListList p)
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throws IllegalArgumentException
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{
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if (p.isEmpty())
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throw new IllegalArgumentException("p is empty");
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al.add(p);
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}
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/** Returns the number of TStackListLists in this list. See
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* also {@link #numberOfParses()}, which gives a different
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* interpretation of the size of this tree. */
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public int size() { return al.size(); }
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/** Returns the number of different parses one could make from
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* this parse tree. Returns zero if this list is empty. */
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public int numberOfParses() {
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if (al.isEmpty()) return 0;
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int k = 1;
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int sz = size();
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for (int i = 0; i < sz; i++) {
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k *= get(i).size();
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}
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return k;
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}
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/** Returns the number of {@link TPair pairs} that are in a
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* parse of this tree. */
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public int numberOfPairs() {
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if (al.isEmpty()) return 0;
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int k = 1;
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int sz = size();
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for (int i = 0; i < sz; i++) {
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// get(i).get(0) is the same size as get(i).get(1),
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// get(i).get(2), ...
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k += get(i).get(0).size();
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}
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return k;
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}
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/** Returns an iterator that will iterate over the {@link
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* #numberOfParses} available. */
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public ParseIterator getParseIterator() {
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return new ParseIterator(al);
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}
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/** Returns a list containing the legal parses of this parse tree.
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* By "legal", we mean a sequence of stacks that is
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* legal by the rules of Tibetan tsheg bar syntax (sometimes
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* called spelling). This will return the {G-YA} parse of {GYA}
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* as well as the {GYA} parse, so watch yourself. */
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public TStackListList getLegalParses() {
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TStackListList sll = new TStackListList(2); // save memory
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ParseIterator pi = getParseIterator();
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while (pi.hasNext()) {
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TStackList sl = pi.next();
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if (sl.isLegalTshegBar(false).isLegal) {
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sll.add(sl);
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}
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}
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return sll;
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}
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/** Returns a list (never null) containing the parses of this
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* parse tree that are not clearly illegal. */
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public TStackListList getNonIllegalParses() {
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TStackListList sll = new TStackListList(2); // save memory
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ParseIterator pi = getParseIterator();
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while (pi.hasNext()) {
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TStackList sl = pi.next();
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if (!sl.isClearlyIllegal()) {
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sll.add(sl);
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}
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}
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return sll;
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}
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/** Returns the best parse, if there is a unique parse that is
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* clearly preferred to other parses. Basically, if there's a
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* unique legal parse, you get it. If there's not, but there is
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* a unique non-illegal parse, you get it. If there's not a
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* unique answer, null is returned. */
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// {TZANDRA} is not solved by this, DLC NOW. Solve PADMA PROBLEM!
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// DLC by using this we can get rid of single-sanskrit-gc, eh?
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public TStackList getBestParse() {
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TStackListList up = getUniqueParse(false);
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if (up.size() == 1)
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return up.get(0);
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up = getNonIllegalParses();
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int sz = up.size();
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if (sz == 1) {
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return up.get(0);
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} else if (sz > 1) {
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// {PADMA}, for example. Our technique is to go from the
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// left and stack as much as we can. So {PA}{D}{MA} is
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// inferior to {PA}{D+MA}, and {PA}{D+MA}{D}{MA} is
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// inferior to {PA}{D+MA}{D+MA}. We do not look for the
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// minimum number of glyphs, though -- {PA}{N+D}{B+H+R}
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// and {PA}{N}{D+B+H+R} tie by that score, but the former
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// is the clear winner.
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// We give a warning about these, optionally, so that
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// users can produce output that even a dumb ACIP reader
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// can understand. See getWarning("All", ..).
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// if j is in this list, then up.get(j) is still a
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// potential winner.
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ArrayList candidates = new ArrayList(sz);
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for (int i = 0; i < sz; i++)
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candidates.add(new Integer(i));
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boolean keepGoing = true;
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int stackNumber = 0;
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boolean someoneHasThisStack = true;
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while (someoneHasThisStack && candidates.size() > 1) {
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// maybe none of the candidates have stackNumber+1
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// stacks. If none do, we'll quit.
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someoneHasThisStack = false;
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int maxGlyphsInThisStack = 0;
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for (int k = 0; k < candidates.size(); k++) {
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TStackList sl = up.get(((Integer)candidates.get(k)).intValue());
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if (sl.size() > stackNumber) {
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int ng;
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if ((ng = sl.get(stackNumber).size()) > maxGlyphsInThisStack)
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maxGlyphsInThisStack = ng;
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someoneHasThisStack = true;
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}
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}
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// Remove all candidates that aren't keeping up.
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if (someoneHasThisStack) {
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for (int k = 0; k < candidates.size(); k++) {
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TStackList sl = up.get(((Integer)candidates.get(k)).intValue());
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if (sl.size() > stackNumber) {
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if (sl.get(stackNumber).size() != maxGlyphsInThisStack)
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candidates.remove(k--);
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} else throw new Error("impossible!");
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}
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}
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++stackNumber;
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}
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if (candidates.size() == 1)
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return up.get(((Integer)candidates.get(0)).intValue());
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else
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return null;
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}
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return null;
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}
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/** Returns a list containing the unique legal parse of this parse
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* tree if there is a unique legal parse. Returns an empty list
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* if there are no legal parses. Returns a list containing all
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* legal parses if there two or more equally good parses. By
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* "legal", we mean a sequence of stacks that is legal
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* by the rules of Tibetan tsheg bar syntax (sometimes called
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* spelling).
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* @param noPrefixTests true if you want to pretend that every
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* stack can take every prefix, which is not the case in
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* reality */
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public TStackListList getUniqueParse(boolean noPrefixTests) {
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// For Sanskrit+Tibetan:
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TStackListList allNonillegalParses = new TStackListList(2); // save memory
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// For Tibetan only:
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TStackListList allStrictlyLegalParses = new TStackListList(2); // save memory
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TStackListList legalParsesWithVowelOnRoot = new TStackListList(1);
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ParseIterator pi = getParseIterator();
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while (pi.hasNext()) {
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TStackList sl = pi.next();
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BoolTriple bt = sl.isLegalTshegBar(noPrefixTests);
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if (bt.isLegal) {
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if (bt.isLegalAndHasAVowelOnRoot)
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legalParsesWithVowelOnRoot.add(sl);
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if (!bt.isLegalButSanskrit)
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allStrictlyLegalParses.add(sl);
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allNonillegalParses.add(sl);
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}
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}
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if (legalParsesWithVowelOnRoot.size() == 1)
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return legalParsesWithVowelOnRoot;
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else {
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if (allStrictlyLegalParses.size() == 1)
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return allStrictlyLegalParses;
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if (allStrictlyLegalParses.size() > 2)
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throw new Error("can this happen?");
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if (legalParsesWithVowelOnRoot.size() == 2) {
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if (legalParsesWithVowelOnRoot.get(0).size()
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!= 1 + legalParsesWithVowelOnRoot.get(1).size()) {
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// MARDA is MAR+DA or MA-R-DA -- both are legal if
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// noPrefixTests.
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return new TStackListList();
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} else {
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// G-YA vs. GYA.
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return new TStackListList(legalParsesWithVowelOnRoot.get(1));
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}
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}
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if (allNonillegalParses.size() == 2) {
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if (allNonillegalParses.get(0).size() != 1 + allNonillegalParses.get(1).size()) {
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// BDREN, e.g., if noPrefixTests:
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return new TStackListList();
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}
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return new TStackListList(allNonillegalParses.get(1));
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}
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return allNonillegalParses;
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}
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}
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/** Returns a human-readable representation. */
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public String toString() {
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return al.toString();
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}
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/** Returns true if and only if either x is an TParseTree
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* object representing the same TPairLists in the same order
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* or x is a String that is equals to the result of {@link
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* #toString()}. */
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public boolean equals(Object x) {
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if (x instanceof TParseTree) {
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return al.equals(((TParseTree)x).al);
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} else if (x instanceof String) {
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return toString().equals(x);
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}
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return false;
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}
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/** Returns null if this parse tree is perfectly legal and valid.
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* Returns a warning for users otherwise. If and only if
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* warningLevel is "All", then even unambiguous ACIP like PADMA,
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* which could be improved by being written as PAD+MA, will cause
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* a warning.
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* @param warningLevel "All" if you're paranoid, "Most" to see
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* warnings about lacking vowels on final stacks, "Some" to see
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* warnings about lacking vowels on non-final stacks and also
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* warnings about when prefix rules affect you, "None" if you
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* like to see IllegalArgumentExceptions.
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* @param pl the pair list from which this parse tree originated
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* @param originalACIP the original ACIP, or null if you want
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* this parse tree to make a best guess. */
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public String getWarning(String warningLevel,
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TPairList pl,
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String originalACIP) {
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if (warningLevel != "Some"
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&& warningLevel != "Most"
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&& warningLevel != "All")
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throw new IllegalArgumentException("warning level bad: is it interned?");
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TStackList bestParse = getBestParse();
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{
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TStackListList noPrefixTestsUniqueParse = getUniqueParse(true);
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if (noPrefixTestsUniqueParse.size() == 1
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&& !noPrefixTestsUniqueParse.get(0).equals(bestParse)) {
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if (warningLevel != "Some")
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return "Warning: We're going with " + bestParse + ", but only because our knowledge of prefix rules says that " + noPrefixTestsUniqueParse.get(0) + " is not a legal Tibetan tsheg bar (\"syllable\")";
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}
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}
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TStackListList up = getUniqueParse(false);
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if (null == up || up.size() != 1) {
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// DLC FIXME: code duplication
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boolean isLastStack[] = new boolean[1];
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TStackListList nip = getNonIllegalParses();
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if (nip.size() != 1) {
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if (null == bestParse) {
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return "Warning: There's not even a unique, non-illegal parse for ACIP {" + ((null != originalACIP) ? originalACIP : recoverACIP()) + "}";
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} else {
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if (bestParse.hasStackWithoutVowel(pl, isLastStack)) {
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if (isLastStack[0]) {
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if (warningLevel == "All")
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return "Warning: The last stack does not have a vowel in the ACIP {" + ((null != originalACIP) ? originalACIP : recoverACIP()) + "}; this may indicate a typo, because Sanskrit, which this is (because it's not legal Tibetan), should have a vowel after each stack.";
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} else {
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throw new Error("Can't happen now that we stack greedily");
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}
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}
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if ("All" == warningLevel) {
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return "Warning: Though the ACIP {" + ((null != originalACIP) ? originalACIP : recoverACIP()) + "} is unambiguous, it would be more computer-friendly if + signs were used to stack things because there are two (or more) ways to interpret this ACIP if you're not careful.";
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}
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}
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} else {
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if (nip.get(0).hasStackWithoutVowel(pl, isLastStack)) {
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if (isLastStack[0]) {
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if (warningLevel == "All")
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return "Warning: The last stack does not have a vowel in the ACIP {" + ((null != originalACIP) ? originalACIP : recoverACIP()) + "}; this may indicate a typo, because Sanskrit, which this is (because it's not legal Tibetan), should have a vowel after each stack.";
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} else {
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throw new Error("Can't happen now that we stack greedily");
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}
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}
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}
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}
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return null;
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}
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/** Returns something akin to the ACIP input (okay, maybe 1-2-3-4
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* instead of 1234, and maybe AUTPA instead of AUT-PA)
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* corresponding to this parse tree. */
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public String recoverACIP() {
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ParseIterator pi = getParseIterator();
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if (pi.hasNext()) {
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return pi.next().recoverACIP();
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}
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return null;
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}
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/** Returns a hashCode appropriate for use with our {@link
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* #equals(Object)} method. */
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public int hashCode() { return al.hashCode(); }
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/** Returns true if and only if this parse tree is empty. */
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public boolean isEmpty() { return al.isEmpty(); }
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}
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