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Tremendously better EWTS->Unicode and EWTS->TMW conversion, though still not tested end-to-end and without perfect unit tests. See EWTSTest.RUN_FAILING_TESTS, for example, to find imperfection.

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This commit is contained in:
dchandler 2005-07-06 02:19:38 +00:00
parent affb9e4b5e
commit 0b3a636f63
20 changed files with 797 additions and 350 deletions
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52
source/org/thdl/tib/text/ttt/TParseTree.java
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@ -105,26 +105,33 @@ class TParseTree {
ParseIterator pi = getParseIterator();
while (pi.hasNext()) {
TStackList sl = pi.next();
if (!sl.isClearlyIllegal()) {
BoolTriple bt = sl.isLegalTshegBar(false);
if (!sl.isClearlyIllegal(bt.candidateType)) {
sll.add(sl);
}
}
return sll;
}
private static final boolean debug = false;
/** Returns the best parse, if there is a unique parse that is
* clearly preferred to other parses. Basically, if there's a
* unique legal parse, you get it. If there's not, but there is
* a unique non-illegal parse, you get it. If there's not a
* unique answer, null is returned. */
public TStackList getBestParse() {
if (debug) System.out.println("getBestParse: parse tree is " + toString());
TStackListList up = getUniqueParse(false);
if (up.size() == 1)
if (up.size() == 1) {
if (debug) System.out.println("getBestParse: unique parse");
return up.get(0);
}
up = getNonIllegalParses();
int sz = up.size();
if (sz == 1) {
if (debug) System.out.println("getBestParse: sole non-illegal parse");
return up.get(0);
} else if (sz > 1) {
// TODO(DLC)[EWTS->Tibetan]: does this still happen? If so, when?
@ -132,12 +139,14 @@ class TParseTree {
// System.out.println("SHO NUFF, >1 non-illegal parses still happens");
// {PADMA}, for example. Our technique is to go from the
// left and stack as much as we can. So {PA}{D}{MA} is
// inferior to {PA}{D+MA}, and {PA}{D+MA}{D}{MA} is
// inferior to {PA}{D+MA}{D+MA}. We do not look for the
// minimum number of glyphs, though -- {PA}{N+D}{B+H+R}
// and {PA}{N}{D+B+H+R} tie by that score, but the former
// is the clear winner.
// left and stack as much as we can (when
// !traits.stackingMustBeExplicit() only!
// TODO(DLC)[EWTS->Tibetan]: fix these comments). So
// {PA}{D}{MA} is inferior to {PA}{D+MA}, and
// {PA}{D+MA}{D}{MA} is inferior to {PA}{D+MA}{D+MA}. We
// do not look for the minimum number of glyphs, though --
// {PA}{N+D}{B+H+R} and {PA}{N}{D+B+H+R} tie by that
// score, but the former is the clear winner.
// We give a warning about these, optionally, so that
// users can produce output that even a dumb ACIP reader
@ -177,11 +186,27 @@ class TParseTree {
}
++stackNumber;
}
if (candidates.size() == 1)
if (candidates.size() == 1) {
if (debug) System.out.println("getBestParse: one candidate");
return up.get(((Integer)candidates.get(0)).intValue());
else
} else {
if (debug) {
System.out.println("getBestParse: no parse, num candidates="
+ candidates.size());
for (int i = 0; i < candidates.size(); i++) {
System.out.println("candidate " + i + " is "
+ up.get(((Integer)candidates.get(i)).intValue()));
if (i + 1 < candidates.size()) {
boolean eq = (up.get(((Integer)candidates.get(i)).intValue()).equals(up.get(((Integer)candidates.get(i + 1)).intValue())));
System.out.println("This candidate and the next are"
+ (eq ? "" : " not") + " equal.");
}
}
}
return null;
}
}
if (debug) System.out.println("getBestParse: no non-illegal parses");
return null;
}
@ -480,9 +505,10 @@ n+t+s
middle = pl.get(1).getLeft();
right = pl.get(2).getLeft();
if (pl.get(0).getRight() == null
&& !pl.get(1).endsACIPStack()
&& pl.get(2).endsACIPStack()
&& !pl.get(1).endsStack()
&& pl.get(2).endsStack()
&& null != left && null != right) {
// TODO(DLC)[EWTS->Tibetan]: This is ACIP-specific.
if (("D".equals(left) && "G".equals(middle) && "R".equals(right))
|| ("D".equals(left) && "G".equals(middle) && "Y".equals(right))) {
if (pl.size() == 3) {
@ -503,7 +529,7 @@ n+t+s
String left, right;
left = pl.get(0).getLeft();
right = pl.get(1).getLeft();
if (pl.get(0).getRight() == null && pl.get(1).endsACIPStack()
if (pl.get(0).getRight() == null && pl.get(1).endsStack()
&& null != left && null != right) {
if (("D".equals(left) && "B".equals(right))
|| ("B".equals(left) && "D".equals(right))