/**
* Sets the enum to operate in linear fashion, as we have found a looping transition at position:
* we set an upper bound and act like a TermRangeQuery for this portion of the term space.
*/
private void setLinear(int position) {
assert linear == false;
int state = runAutomaton.getInitialState();
int maxInterval = 0xff;
for (int i = 0; i < position; i++) {
state = runAutomaton.step(state, seekBytesRef.bytes[i] & 0xff);
assert state >= 0 : "state=" + state;
}
for (int i = 0; i < allTransitions[state].length; i++) {
Transition t = allTransitions[state][i];
if (t.getMin() <= (seekBytesRef.bytes[position] & 0xff)
&& (seekBytesRef.bytes[position] & 0xff) <= t.getMax()) {
maxInterval = t.getMax();
break;
}
}
// 0xff terms don't get the optimization... not worth the trouble.
if (maxInterval != 0xff) maxInterval++;
int length = position + 1; /* position + maxTransition */
if (linearUpperBound.bytes.length < length) linearUpperBound.bytes = new byte[length];
System.arraycopy(seekBytesRef.bytes, 0, linearUpperBound.bytes, 0, position);
linearUpperBound.bytes[position] = (byte) maxInterval;
linearUpperBound.length = length;
linear = true;
}
/**
* Returns the next String in lexicographic order that will not put the machine into a reject
* state.
*
* <p>This method traverses the DFA from the given position in the String, starting at the given
* state.
*
* <p>If this cannot satisfy the machine, returns false. This method will walk the minimal path,
* in lexicographic order, as long as possible.
*
* <p>If this method returns false, then there might still be more solutions, it is necessary to
* backtrack to find out.
*
* @param state current non-reject state
* @param position useful portion of the string
* @return true if more possible solutions exist for the DFA from this position
*/
private boolean nextString(int state, int position) {
/*
* the next lexicographic character must be greater than the existing
* character, if it exists.
*/
int c = 0;
if (position < seekBytesRef.length) {
c = seekBytesRef.bytes[position] & 0xff;
// if the next byte is 0xff and is not part of the useful portion,
// then by definition it puts us in a reject state, and therefore this
// path is dead. there cannot be any higher transitions. backtrack.
if (c++ == 0xff) return false;
}
seekBytesRef.length = position;
visited[state] = curGen;
Transition transitions[] = allTransitions[state];
// find the minimal path (lexicographic order) that is >= c
for (int i = 0; i < transitions.length; i++) {
Transition transition = transitions[i];
if (transition.getMax() >= c) {
int nextChar = Math.max(c, transition.getMin());
// append either the next sequential char, or the minimum transition
seekBytesRef.grow(seekBytesRef.length + 1);
seekBytesRef.length++;
seekBytesRef.bytes[seekBytesRef.length - 1] = (byte) nextChar;
state = transition.getDest().getNumber();
/*
* as long as is possible, continue down the minimal path in
* lexicographic order. if a loop or accept state is encountered, stop.
*/
while (visited[state] != curGen && !runAutomaton.isAccept(state)) {
visited[state] = curGen;
/*
* Note: we work with a DFA with no transitions to dead states.
* so the below is ok, if it is not an accept state,
* then there MUST be at least one transition.
*/
transition = allTransitions[state][0];
state = transition.getDest().getNumber();
// append the minimum transition
seekBytesRef.grow(seekBytesRef.length + 1);
seekBytesRef.length++;
seekBytesRef.bytes[seekBytesRef.length - 1] = (byte) transition.getMin();
// we found a loop, record it for faster enumeration
if (!finite && !linear && visited[state] == curGen) {
setLinear(seekBytesRef.length - 1);
}
}
return true;
}
}
return false;
}
