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package util.graphOperations.dominance;
import util.objects.graphs.DirectedGraph;
import util.objects.setDataStructures.ISet;
/**
* Class enabling to compute LCA queries in constant time over the DFS tree of a given graph
* use a O(n+m) time preprocessing
*
* @author Jean-Guillaume Fages
*/
public class LCAGraphManager {
//***********************************************************************************
// VARIABLES
//***********************************************************************************
//
private int root;
private DirectedGraph graph;
private int nbNodes, nbActives;
//
private int[] father;
private int[] nodeOfDfsNumber;
private int[] dfsNumberOfNode;
//
private int[] I, L, h, A, htmp;
ISet[] successors;
//***********************************************************************************
// CONSTRUCTORS
//***********************************************************************************
public LCAGraphManager(int nb) {
nbNodes = nb;
successors = new ISet[nbNodes];
father = new int[nbNodes];
nodeOfDfsNumber = new int[nbNodes];
dfsNumberOfNode = new int[nbNodes];
htmp = new int[nbNodes];
A = new int[nbNodes];
I = new int[nbNodes];
L = new int[nbNodes];
h = new int[nbNodes];
}
public void preprocess(int r, DirectedGraph g) {
root = r;
graph = g;
initParams();
proceedFirstDFS();
performLCAPreprocessing();
}
//***********************************************************************************
// QUERIES
//***********************************************************************************
public int getLCA(int x, int y) {
return nodeOfDfsNumber[getDFS_LCA(dfsNumberOfNode[x], dfsNumberOfNode[y])];
}
//***********************************************************************************
// INITIALIZATION
//***********************************************************************************
private void initParams() {
nbActives = graph.getNodes().getSize();
for (int i = 0; i < nbNodes; i++) {
successors[i] = graph.getSuccOf(i);
dfsNumberOfNode[i] = -1;
father[i] = -1;
A[i] = -1;
}
}
/**
* perform a dfs in graph to label nodes
*/
private void proceedFirstDFS() {
int i = root;
int k = 0;
father[k] = k;
dfsNumberOfNode[root] = k;
nodeOfDfsNumber[k] = root;
int j;
boolean finished = false;
boolean first = true;
k++;
while (!finished) {
if (first) {
j = successors[i].getFirstElement();
first = false;
} else {
j = successors[i].getNextElement();
}
if (j < 0) {
if (i == root) {
finished = true;
break;
}
first = false;
i = nodeOfDfsNumber[father[dfsNumberOfNode[i]]];
} else {
if (dfsNumberOfNode[j] == -1) {
father[k] = dfsNumberOfNode[i];
dfsNumberOfNode[j] = k;
nodeOfDfsNumber[k] = j;
i = j;
first = true;
k++;
}
}
}
if (k != nbActives) {
throw new UnsupportedOperationException("LCApreprocess did not reach all nodes");
}
for (; k < nbNodes; k++) {
father[k] = -1;
}
}
//***********************************************************************************
// LCA attention : numerotation 1 - n (et non 0 - n-1) pour les manipulations de bit
//***********************************************************************************
/**
* LCA PREPROCESSING O(m+n) by Chris Lewis
* 1. Do a depth first traversal of the general tree to assign depth-first search numbers to the nodes. For each node set a pointer to its parent.
* 2. Using a linear time bottom up algorithm, compute I(v) for each node. For each k such that I(v) = k for some v, set L(k) to point to the head of the run containing k.
* (=>I(v) pointe sur le dernier element, L(I(v)) vers le premier)
* o The head of a run is identified when computing I values. v is identified as the head of its run if the I value of v's parent is not I(v).
* o After this step, the head of a run containing an arbitrary node v can be located in constant time. First compute I(v) then look up the value L(I(v)).
* 3. Given a complete binary tree with node-depth ceiling(log n)-1, map each node v in the general tree to I(v) in the binary tree (Fig 4).
* 4. For each node v in the general tree create on O(log n) bit number A v. Bit A v(i) is set to 1 if and only if node v has an ancestor
* in the general tree that maps to height i in the binary tree. i.e. iff v has an ancestor u such that h(I(u)) = i.
*/
private void performLCAPreprocessing() {
// step 1 : DFS already done
// step 2
ISet nei;
for (int i = nbActives - 1; i >= 0; i--) {
h[i] = BitOperations.getFirstExp(i + 1);
if (h[i] == -1) {
throw new UnsupportedOperationException();
}
htmp[i] = h[i];
I[i] = i;
L[i] = i;
int sucInRun = -1;
nei = graph.getSuccOf(nodeOfDfsNumber[i]);
int s;
for (int k = nei.getFirstElement(); k >= 0; k = nei.getNextElement()) {
s = dfsNumberOfNode[k];
if (i != s && father[s] == i && htmp[s] > htmp[i]) {
htmp[i] = htmp[s];
sucInRun = s;
}
}
if (sucInRun != -1) {
I[i] = I[sucInRun];
L[I[i]] = i;
}
}
// step 3 : mapping to a binary tree : the binary tree is virtual here so there is nothing to do
//step 4
// A[0] = I[0];
int exp = BitOperations.getFirstExp(I[0] + 1);
if (exp == -1) {
throw new UnsupportedOperationException();
}
A[0] = BitOperations.pow(2, exp);
for (int i = 0; i < nbActives; i++) {
A[i] = A[father[i]];
if (I[i] != I[father[i]]) {
exp = BitOperations.getFirstExp(I[i] + 1);
if (exp == -1) {
throw new UnsupportedOperationException();
}
A[i] += BitOperations.pow(2, exp);
}
}
}
/**
* Get the lowest common ancestor of two nodes in O(1) time
* Query by Chris Lewis
* 1. Find the lowest common ancestor b in the binary tree of nodes I(x) and I(y).
* 2. Find the smallest position j ≥ h(b) such that both numbers A x and A y have 1-bits in position j.
* This gives j = h(I(z)).
* 3. Find node x', the closest node to x on the same run as z:
* a. Find the position l of the right-most 1 bit in A x
* b. If l = j, then set x' = x {x and z are on the same run in the general graph} and go to step 4.
* c. Find the position k of the left-most 1-bit in A x that is to the right of position j.
* Form the number consisting of the bits of I(x) to the left of the position k,
* followed by a 1-bit in position k, followed by all zeros. {That number will be I(w)}
* Look up node L(I(w)), which must be node w. Set node x' to be the parent of node w in the general tree.
* 4. Find node y', the closest node to y on the same run as z using the approach described in step 3.
* 5. If x' < y' then set z to x' else set z to y'
*
* @param x node dfs number
* @param y node dfs number
* @return the dfs number of the lowest common ancestor of a and b in the dfs tree
*/
private int getDFS_LCA(int x, int y) {
// trivial cases
if (x == y || x == father[y]) {
return x;
}
if (y == father[x]) {
return y;
}
// step 1
int b = BitOperations.binaryLCA(I[x] + 1, I[y] + 1);
// step 2
int hb = BitOperations.getFirstExp(b);
if (hb == -1) {
throw new UnsupportedOperationException();
}
int j = BitOperations.getFirstExpInBothXYfromI(A[x], A[y], hb);
if (j == -1) {
throw new UnsupportedOperationException();
}
// step 3 & 4
int xPrim = closestFrom(x, j);
int yPrim = closestFrom(y, j);
// step 5
if (xPrim < yPrim) {
return xPrim;
}
return yPrim;
}
private int closestFrom(int x, int j) {
// 3.a
int l = BitOperations.getFirstExp(A[x]);
if (l == -1) {
throw new UnsupportedOperationException();
}
if (l == j) { // 3.b
return x;
} else { // 3.c
int k = BitOperations.getMaxExpBefore(A[x], j);
int IW = I[x] + 1;
if (k != -1) { // there is at least one 1-bit at the right of j
IW = BitOperations.replaceBy1and0sFrom(IW, k);
} else {
throw new UnsupportedOperationException();
}
IW--;
return father[L[IW]];
}
}
//***********************************************************************************
// ACCESSORS
//***********************************************************************************
/**
* Get the parent of x in the dfs tree
*
* @param x the focused node
* @return the parent of x in the dfs tree
*/
public int getParentOf(int x) {
return nodeOfDfsNumber[father[dfsNumberOfNode[x]]];
}
/**
* @return an array containing the nodes sorted according to the dfs order
*/
public int[] getNodeOfDfsNumber() {
return nodeOfDfsNumber;
}
/**
* @return an array containing the dfs numbers of nodes
*/
public int[] getDfsNumberOfNode() {
return dfsNumberOfNode;
}
}