/**
* Assembly a reply from a decomp rule and the input. If the reassembly rule is goto, return null
* and give the gotoKey to use. Otherwise return the response.
*/
String assemble(Decomp d, String reply[], Key gotoKey) {
String lines[] = new String[3];
d.stepRule();
String rule = d.nextRule();
if (EString.match(rule, "goto *", lines)) {
// goto rule -- set gotoKey and return false.
gotoKey.copy(keys.getKey(lines[0]));
if (gotoKey.key() != null) return null;
System.out.println("Goto rule did not match key: " + lines[0]);
return null;
}
String work = "";
while (EString.match(rule, "* (#)*", lines)) {
// reassembly rule with number substitution
rule = lines[2]; // there might be more
int n = 0;
try {
n = Integer.parseInt(lines[1]) - 1;
} catch (NumberFormatException e) {
System.out.println("Number is wrong in reassembly rule " + lines[1]);
}
if (n < 0 || n >= reply.length) {
System.out.println("Substitution number is bad " + lines[1]);
return null;
}
reply[n] = post.translate(reply[n]);
work += lines[0] + " " + reply[n];
}
work += rule;
if (d.mem()) {
mem.save(work);
return null;
}
return work;
}
/**
* Decompose a string according to the given key. Try each decomposition rule in order. If it
* matches, assemble a reply and return it. If assembly fails, try another decomposition rule. If
* assembly is a goto rule, return null and give the key. If assembly succeeds, return the reply;
*/
String decompose(Key key, String s, Key gotoKey) {
String reply[] = new String[10];
for (int i = 0; i < key.decomp().size(); i++) {
Decomp d = (Decomp) key.decomp().elementAt(i);
String pat = d.pattern();
if (syns.matchDecomp(s, pat, reply)) {
String rep = assemble(d, reply, gotoKey);
if (rep != null) return rep;
if (gotoKey.key() != null) return null;
}
}
return null;
}
/**
* Process a sentence. (1) Make pre transformations. (2) Check for quit word. (3) Scan sentence
* for keys, build key stack. (4) Try decompositions for each key.
*/
String sentence(String s) {
s = pre.translate(s);
s = EString.pad(s);
if (quit.find(s)) {
finished = true;
return finl;
}
keys.buildKeyStack(keyStack, s);
for (int i = 0; i < keyStack.keyTop(); i++) {
Key gotoKey = new Key();
String reply = decompose(keyStack.key(i), s, gotoKey);
if (reply != null) return reply;
// If decomposition returned gotoKey, try it
while (gotoKey.key() != null) {
reply = decompose(gotoKey, s, gotoKey);
if (reply != null) return reply;
}
}
return null;
}
public boolean haskey(Key data) // tests if node contains a certain key
{
if (data.key().equals(left.key()) && left().present()) // if left key is data
{
return true;
} else if (middle != null
&& data.key().equals(middle.key())
&& middle().present()) // if middle key is data
{
return true;
} else if (right != null
&& data.key().equals(right.key())
&& right().present()) // if right key is data
{
return true;
}
return false;
}
/** Constructs a ChainableOrdering. */
public ChainableOrdering(Key<?> key, SortOrder sortOrder) {
this(key.key(), sortOrder);
}
public void insert(String newkey) // inserts new key into the node
{
keycount++;
if (keycount == 2) // if there are now 2 keys
{
Key temp = left;
Key temp2 = new Key(newkey);
if (temp.key().compareTo(temp2.key()) < 0) {
left = temp;
middle = temp2;
} else {
left = temp2;
middle = temp;
}
} else {
Key temp = left;
Key temp2 = middle;
Key temp3 = new Key(newkey);
if (temp.key().compareTo(temp2.key()) < 0 && temp2.key().compareTo(temp3.key()) < 0) {
right = temp3; // order of temp1, 2, 3
} else if (temp.key().compareTo(temp3.key()) < 0 && temp3.key().compareTo(temp2.key()) < 0) {
middle = temp3; // order temp1, 3, 2
right = temp2;
} else if (temp2.key().compareTo(temp.key()) < 0 && temp.key().compareTo(temp3.key()) < 0) {
left = temp2; // order temp2, 1, 3
middle = temp;
right = temp3;
} else if (temp2.key().compareTo(temp3.key()) < 0 && temp3.key().compareTo(temp.key()) < 0) {
left = temp2; // order temp2, 3, 1
middle = temp3;
right = temp;
} else if (temp3.key().compareTo(temp.key()) < 0 && temp.key().compareTo(temp2.key()) < 0) {
left = temp3; // order temp3, 1, 2
middle = temp;
right = temp2;
} else {
left = temp3; // order temp3, 2, 1
middle = temp2;
right = temp;
}
}
}
public Node nextnode(
Key data) // given data, decides which child's subtree data should be located in
{
if (keycount == 1) // if only 1 key
{
if (data.key().compareTo(left.key()) < 0) // go to leftmost child
{
return llnode;
} else // otherwise go to the second leftmost child
{
return lmnode;
}
} else if (keycount == 2) // if only 2 keys
{
if (data.key().compareTo(left.key()) < 0) // if belongs to the left of the first key
{
return llnode; // return leftmost node
} else if (data.key().compareTo(middle.key())
< 0) // if belongs to the left of the middle key and right of first key
{
return lmnode; // return second leftmost node
} else // must belong t right of second key
{
return mrnode; // second rightmost node
}
} else // 3 keys
{
if (data.key().compareTo(left.key()) < 0) // if belongs to left of first key
{
return llnode; // left most node
} else if (data.key().compareTo(middle.key())
< 0) // belongs right of first key left of second
{
return lmnode; // second leftmost node
} else if (data.key().compareTo(right.key()) < 0) // belongs right of second key left of third
{
return mrnode; // second rightmost node
} else // belongs to the right of the third key
{
return rrnode; // rightmost node
}
}
}