/// <summary>Legal as long as there are at least two items on the state's stack.</summary>
public virtual bool IsLegal(State state, IList <ParserConstraint> constraints)
{
// some of these quotes come directly from Zhang Clark 09
if (state.finished)
{
return(false);
}
if (state.stack.Size() <= 1)
{
return(false);
}
// at least one of the two nodes on top of stack must be non-temporary
if (ShiftReduceUtils.IsTemporary(state.stack.Peek()) && ShiftReduceUtils.IsTemporary(state.stack.Pop().Peek()))
{
return(false);
}
if (ShiftReduceUtils.IsTemporary(state.stack.Peek()))
{
if (side == BinaryTransition.Side.Left)
{
return(false);
}
if (!ShiftReduceUtils.IsEquivalentCategory(label, state.stack.Peek().Value()))
{
return(false);
}
}
if (ShiftReduceUtils.IsTemporary(state.stack.Pop().Peek()))
{
if (side == BinaryTransition.Side.Right)
{
return(false);
}
if (!ShiftReduceUtils.IsEquivalentCategory(label, state.stack.Pop().Peek().Value()))
{
return(false);
}
}
// don't allow binarized labels if it makes the state have a stack
// of size 1 and a queue of size 0
if (state.stack.Size() == 2 && IsBinarized() && state.EndOfQueue())
{
return(false);
}
// when the stack contains only two nodes, temporary resulting
// nodes from binary reduce must be left-headed
if (state.stack.Size() == 2 && IsBinarized() && side == BinaryTransition.Side.Right)
{
return(false);
}
// when the queue is empty and the stack contains more than two
// nodes, with the third node from the top being temporary, binary
// reduce can be applied only if the resulting node is non-temporary
if (state.EndOfQueue() && state.stack.Size() > 2 && ShiftReduceUtils.IsTemporary(state.stack.Pop().Pop().Peek()) && IsBinarized())
{
return(false);
}
// when the stack contains more than two nodes, with the third
// node from the top being temporary, temporary resulting nodes
// from binary reduce must be left-headed
if (state.stack.Size() > 2 && ShiftReduceUtils.IsTemporary(state.stack.Pop().Pop().Peek()) && IsBinarized() && side == BinaryTransition.Side.Right)
{
return(false);
}
if (constraints == null)
{
return(true);
}
Tree top = state.stack.Peek();
int leftTop = ShiftReduceUtils.LeftIndex(top);
int rightTop = ShiftReduceUtils.RightIndex(top);
Tree next = state.stack.Pop().Peek();
int leftNext = ShiftReduceUtils.LeftIndex(next);
// The binary transitions are affected by constraints in the
// following two circumstances. If a transition would cross the
// left boundary of a constraint, that is illegal. If the
// transition is exactly the right size for the constraint and
// would make a temporary node, that is also illegal.
foreach (ParserConstraint constraint in constraints)
{
if (leftTop == constraint.start)
{
// can't binary reduce away from a tree which doesn't match a constraint
if (rightTop == constraint.end - 1)
{
if (!ShiftReduceUtils.ConstraintMatchesTreeTop(top, constraint))
{
return(false);
}
else
{
continue;
}
}
else
{
if (rightTop >= constraint.end)
{
continue;
}
else
{
// can't binary reduce if it would make the tree cross the left boundary
return(false);
}
}
}
// top element is further left than the constraint, so
// there's no harm to be done by binary reduce
if (leftTop < constraint.start)
{
continue;
}
// top element is past the end of the constraint, so it must already be satisfied
if (leftTop >= constraint.end)
{
continue;
}
// now leftTop > constraint.start and < constraint.end, eg inside the constraint
// the next case is no good because it crosses the boundary
if (leftNext < constraint.start)
{
return(false);
}
if (leftNext > constraint.start)
{
continue;
}
// can't transition to a binarized node when there's a constraint that matches.
if (rightTop == constraint.end - 1 && IsBinarized())
{
return(false);
}
}
return(true);
}
