public CaseIdentification_ResultType SplitTree_CaseIdentification(SplitTree ST, List<int> sigma, int idx, out TreeEdge e, out Vertex u, out SplitTree tree)
{
//nodeOrder is the inverse permutation of sigma
//Here idx is the index of x.
e = new TreeEdge();
u = null;
#region Line 1
MarkerVertex.ResetPerfectFlags();//new N(x) will be generated. All old perfect flags outdated.
tree = SmallestSpanningTree(ST, sigma[idx]);//N(x) is implied in neighbor flag of each vertex
#endregion
//remember tree is rooted in tree.root,and it perhaps has a valid parent!
//Also, check visited flag when traversing, as only those with visited flag true are included in subtree.
#region Line 2
int nodeCount = 0;
Queue<Node> processingQueue = new Queue<Node>();
Action<Node> testAllLeave = (n) =>
{
bool? allLeave = null;
(n as Node).ForEachChild((v) =>
{
if (!(v is Leaf))
{
allLeave = false;
return IterationFlag.Break;
}
allLeave = true;
return IterationFlag.Continue;
}, subtree: false);//TODO XXX here subtree should be true, according to Algorithm 5
if (!(allLeave == null || !allLeave.Value))//has all leave
processingQueue.Enqueue(n);
};
foreach (var n in tree.vertices)
if (n is Node)
{
++nodeCount;
testAllLeave(n as Node);
}
if (nodeCount > 1)//since there're not only 1 node, a node with all leaf children will certainly not be root
{
while (processingQueue.Count != 0)
{
var node = processingQueue.Dequeue();
//cast the spell of Remark 5.5
List<MarkerVertex> P, M, E;
node.ComputeStates(out P, out M, out E, exclude: node.rootMarkerVertex);
//since node has all leaf children, when we exclude the rootMarkerVertex, the time complexity is bound to |Children|
//also, M should be empty. Thus case 2 automatically satisfied. <- XXX not really, cannot assert here
//Debug.Assert(M.Count == 0, "A node with all leaf children has a mixed marker vertex!!");
var Pset = new System.Collections.Generic.HashSet<MarkerVertex>(P);
var inCount = 0;
var neighborCount = 0;
node.ForEachNeighbor(node.rootMarkerVertex, (q) =>
{
++neighborCount;
if (Pset.Contains(q))
++inCount;
return IterationFlag.Continue;
});
//P == N_Gu(q) iff P.Count == inCount == neighborCount
if (P.Count == inCount && inCount == neighborCount && M.Count == 0)//now we say that rootMarkerVertex'es opposite vertex is perfect
{
//(node.rootMarkerVertex.opposite as GLTVertex).
node.rootMarkerVertex.opposite.MarkAsPerfect();
tree.RemoveSubTree(node);
if (node.parent is Node)
testAllLeave(node.parent as Node);
}
}
}
#endregion
#region Line 3
u = tree.root;
bool unique = true;
Debug.Assert(u != null, "the root of subtree is null, after removing pendant perfect subtrees");
while (true)
{
Node child = null;
if (u is Leaf)
{
child = (u as Leaf).opposite.node;
if (child == null || !child.visited)//now the leaf root u is the unique vertex in T'
{
//this case is not discussed in paper... Thanks Emeric, problem solved.
return CaseIdentification_ResultType.SingleLeaf;
}
}
else
{
(u as Node).ForEachChild((v) =>
{
if (v is Node)
{
if (child == null)
{
unique = true;
child = v as Node;
}
else
{
unique = false;
return IterationFlag.Break;
}
}
return IterationFlag.Continue;
}, subtree: true);
}
//here unique indicates whether u has a unique node child
if (unique == false)
break;//there are at least two non-leaf children of the current root; break out and return the subtree, now.
else
if (child == null)//u contains no non-leaf child
{
//now we can say u is the unique node in T'
break;
}
else// the unique child node is located
{
if (u is Leaf)
{
//the root is a leaf, which indicates that root \in N(x), thus perfect.
(u as Leaf).visited = false;
tree.vertices.Remove(u as Leaf);
u = child;//prune it directly.
}
else
{
List<MarkerVertex> P, M, E;
(u as Node).ComputeStates(out P, out M, out E, exclude: child.rootMarkerVertex.opposite as MarkerVertex, excludeRootMarkerVertex: false);
bool perfect = false;
if (M.Count == 0)//condition 2 is satisfied
{
var Pset = new System.Collections.Generic.HashSet<MarkerVertex>(P);
var inCount = 0;
var neighborCount = 0;
(u as Node).ForEachNeighbor(child.rootMarkerVertex.opposite as MarkerVertex, (q) =>
{
++neighborCount;
if (Pset.Contains(q))
++inCount;
return IterationFlag.Continue;
});
//P == N_Gu(q) iff P.Count == inCount == neighborCount
if (P.Count == inCount && inCount == neighborCount)//now we say that child's rootMarkerVertex is perfect
{
child.rootMarkerVertex.MarkAsPerfect();
perfect = true;
}
}
if (perfect)
{
tree.RemoveSubTree(u as Node, exclude: child);
u = child;
}
else
{
//the tree edge uv is fully mixed. thus the subtree T' is fully mixed.
unique = false;
break;
}
}
}
tree.root = u as GLTVertex;//let the tree root follow u after each iteration of the processing loop
}
#endregion
#region Line 4
if (!unique)
return CaseIdentification_ResultType.FullyMixedSubTree;
else
{
#region if (u is DegenerateNode)
if (u is DegenerateNode)
{
#region Lemma 5.6
var uDeg = u as DegenerateNode;
MarkerVertex q = null;
List<MarkerVertex> P, M, E;
//Note, we computed P(u), thus if q exists, its perfect flag is available
(uDeg).ComputeStates(out P, out M, out E);
//Debug.Assert(M.Count == 0, "Algorithm 5 Line 4, lemma 5.6 requires P(u) == NE(u)!");
if (M.Count == 0)
{
if (P.Count == 1 && uDeg.isStar && uDeg.center == P[0])//case 3
{
uDeg.ForEachMarkerVertex((v) =>
{
if (v != uDeg.center)
{
q = v;
return IterationFlag.Break;
}
return IterationFlag.Continue;
});
}
else if (P.Count == 2 && uDeg.isStar && P.Contains(uDeg.center))//case 4
{
if (uDeg.center == P[0])
q = P[1];
else q = P[0];
}
else
{
int count = P.Count + E.Count;//faster than the commented line below
//uDeg.ForEachMarkerVertex((v) => { ++count; return IterationFlag.Continue; });
if (count == P.Count)//case 1
{
if (uDeg.isClique)
q = P[0];
else q = uDeg.center;
}
else if (P.Count == count - 1 && (uDeg.isClique || E[0] == uDeg.center))//case 2
{
//E[0]: V(u) \ P(u)
q = E[0];
}
}
}
#endregion
if (q != null)
{
e.u = q;
e.v = q.opposite;
q.opposite.MarkAsPerfect();
return CaseIdentification_ResultType.TreeEdge;
}
else
{
return CaseIdentification_ResultType.HybridNode;
}
}
#endregion
#region else: prime node
else//prime node
{
var uPrime = u as PrimeNode;
MarkerVertex q = uPrime.lastMarkerVertex;//last leaf vertex in \sigma[G(u)]
MarkerVertex qPrime = (u as PrimeNode).universalMarkerVetex;//the universal marker (if any)
//Again cast the spell of Remark 5.5
List<MarkerVertex> P, M, E;
//Note, here we didn't exclude anything. Which means that, if a PP/PE edge is returned, perfect flags will be available.
uPrime.ComputeStates(out P, out M, out E);
//first, for q
if (M.Count == 0 || (M.Count == 1 && M[0] == q))//condition 2
{
var Pset = new System.Collections.Generic.HashSet<MarkerVertex>(P);
var inCount = 0;
var neighborCount = 0;
uPrime.ForEachNeighbor(q, (r) =>
{
++neighborCount;
if (Pset.Contains(r))
++inCount;
return IterationFlag.Continue;
});
//P == N_Gu(q) iff P.Count == inCount == neighborCount, or P.Count == inCount+1 == neighborCount+1 and q in P
if ((P.Count == inCount && inCount == neighborCount)
||
(P.Count == inCount + 1 && P.Count == neighborCount + 1 && Pset.Contains(q)))
{
e.u = q;
e.v = q.opposite;
q.opposite.MarkAsPerfect();
return CaseIdentification_ResultType.TreeEdge;
}
}
//then for q'
if (qPrime != null)
{
if (M.Count == 0 || (M.Count == 1 && M[0] == qPrime))//condition 2
{
var Pset = new System.Collections.Generic.HashSet<MarkerVertex>(P);
var inCount = 0;
var neighborCount = 0;
uPrime.ForEachNeighbor(qPrime, (r) =>
{
++neighborCount;
if (Pset.Contains(r))
++inCount;
return IterationFlag.Continue;
});
//P == N_Gu(q) iff P.Count == inCount == neighborCount, or P.Count == inCount+1 == neighborCount+1 and q in P
if ((P.Count == inCount && inCount == neighborCount)
||
(P.Count == inCount + 1 && P.Count == neighborCount + 1 && Pset.Contains(qPrime)))
{
e.u = qPrime;
e.v = qPrime.opposite;
qPrime.opposite.MarkAsPerfect();
return CaseIdentification_ResultType.TreeEdge;
}
}
}
//else, just return u
return CaseIdentification_ResultType.HybridNode;
}
#endregion
}
#endregion
}
//Proposition 4.17
internal void SplitEdgeToStar(TreeEdge e, int xId)
{
//Assume e is P-E
GLTVertex oParent = e.u.GetGLTVertex();
GLTVertex oChild = e.v.GetGLTVertex();
bool uIsParent = true;
//test and swap
if (oChild.parent != oParent)
{
var tmp = oParent;
oParent = oChild;
oChild = tmp;
uIsParent = false;
}
////Fixing multiple-star forking problem
//var oChildDeg = oChild as DegenerateNode;
//if (oChildDeg != null && oChildDeg.isStar && oChildDeg.center == oChildDeg.rootMarkerVertex)
//{
// //note that if child's center is not root, we cannot merge.
// AttachToDegenerateNode(oChildDeg, xId);
// return;
//}
MarkerVertex v1 = new MarkerVertex();
MarkerVertex v2 = new MarkerVertex();
MarkerVertex v3 = new MarkerVertex();
//center is E (e.v) => v2
var deg = new DegenerateNode()
{
active = false,
center = v2,
parent = oParent,
rootMarkerVertex = uIsParent ? v1 : v2,
visited = false,
Vu = new List<MarkerVertex> { v1, v2, v3 }
};
if (uIsParent)
v1.node = deg;
else v2.node = deg;
deg.parent = oParent;
if (oParent is PrimeNode == false)
{
oChild.parent = deg;
}
else
{
//(oParent as PrimeNode).ForEachChild(c =>
// {
// if (c.unionFind_parent == oChild)
// Console.WriteLine("!!!");
// return IterationFlag.Continue;
// }, false);
//oChild.parent = deg;
var newChild = oChild.Clone();//since oChild is potentially pointed by others' unionFind_parent, we have to clone.
if (newChild is Leaf)
{
LeafMapper[(newChild as Leaf).id] = newChild as Leaf;
//a new leaf is cloned. thus e.u / e.v is cloned. check and update.
if (e.u is Leaf)
e.u = newChild;
else
e.v = newChild;
}
newChild.parent = deg;
}
//linking e.u & e.v to v1 & v2
if (e.u is MarkerVertex)
(e.u as MarkerVertex).opposite = v1;
else
(e.u as Leaf).opposite = v1;
if (e.v is MarkerVertex)
(e.v as MarkerVertex).opposite = v2;
else
(e.v as Leaf).opposite = v2;
v1.opposite = e.u;
v2.opposite = e.v;
Leaf x = new Leaf
{
id = xId,
opposite = v3,
};
v3.opposite = x;
x.parent = deg;
//vertices.Add(deg);
AddLeaf(x);
}
