1 //===- llvm/Analysis/ET-Forest.h - ET-Forest implementation -----*- C++ -*-===//
3 // The LLVM Compiler Infrastructure
5 // This file was written by Daniel Berlin from code written by Pavel Nejedy, and
6 // is distributed under the University of Illinois Open Source License. See
7 // LICENSE.TXT for details.
9 //===----------------------------------------------------------------------===//
11 // This file defines the following classes:
12 // 1. ETNode: A node in the ET forest.
13 // 2. ETOccurrence: An occurrence of the node in the splay tree
14 // storing the DFS path information.
16 // The ET-forest structure is described in:
17 // D. D. Sleator and R. E. Tarjan. A data structure for dynamic trees.
18 // J. G'omput. System Sci., 26(3):362 381, 1983.
20 // Basically, the ET-Forest is storing the dominator tree (ETNode),
21 // and a splay tree containing the depth first path information for
22 // those nodes (ETOccurrence). This enables us to answer queries
23 // about domination (DominatedBySlow), and ancestry (NCA) in
24 // logarithmic time, and perform updates to the information in
27 //===----------------------------------------------------------------------===//
29 #ifndef LLVM_ANALYSIS_ETFOREST_H
30 #define LLVM_ANALYSIS_ETFOREST_H
38 /// ETOccurrence - An occurrence for a node in the et tree
40 /// The et occurrence tree is really storing the sequences you get from
41 /// doing a DFS over the ETNode's. It is stored as a modified splay
43 /// ET occurrences can occur at multiple places in the ordering depending
44 /// on how many ET nodes have it as their father. To handle
45 /// this, they are separate from the nodes.
49 ETOccurrence(ETNode *n): OccFor(n), Parent(NULL), Left(NULL), Right(NULL),
50 Depth(0), Min(0), MinOccurrence(this) {};
52 void setParent(ETOccurrence *n) {
53 assert(n != this && "Trying to set parent to ourselves");
57 // Add D to our current depth
58 void setDepthAdd(int d) {
63 // Reset our depth to D
64 void setDepth(int d) {
70 void setLeft(ETOccurrence *n) {
71 assert(n != this && "Trying to set our left to ourselves");
78 void setRight(ETOccurrence *n) {
79 assert(n != this && "Trying to set our right to ourselves");
85 // Splay us to the root of the tree
88 // Recompute the minimum occurrence for this occurrence.
89 void recomputeMin(void) {
90 ETOccurrence *themin = Left;
92 // The min may be our Right, too.
93 if (!themin || (Right && themin->Min > Right->Min))
96 if (themin && themin->Min < 0) {
97 Min = themin->Min + Depth;
98 MinOccurrence = themin->MinOccurrence;
101 MinOccurrence = this;
110 // Parent in the splay tree
111 ETOccurrence *Parent;
113 // Left Son in the splay tree
116 // Right Son in the splay tree
119 // Depth of the node is the sum of the depth on the path to the
123 // Subtree occurrence's minimum depth
126 // Subtree occurrence with minimum depth
127 ETOccurrence *MinOccurrence;
133 ETNode(void *d) : data(d), Father(NULL), Left(NULL),
134 Right(NULL), Son(NULL), ParentOcc(NULL) {
135 RightmostOcc = new ETOccurrence(this);
138 // This does *not* maintain the tree structure.
139 // If you want to remove a node from the forest structure, use
140 // removeFromForest()
145 void removeFromForest() {
146 // Split us away from all our sons.
150 // And then split us away from our father.
155 // Split us away from our parents and children, so that we can be
156 // reparented. NB: setFather WILL NOT DO WHAT YOU WANT IF YOU DO NOT
160 // Set our parent node to the passed in node
161 void setFather(ETNode *);
163 // Nearest Common Ancestor of two et nodes.
164 ETNode *NCA(ETNode *);
166 // Return true if we are below the passed in node in the forest.
167 bool Below(ETNode *);
169 Given a dominator tree, we can determine whether one thing
170 dominates another in constant time by using two DFS numbers:
172 1. The number for when we visit a node on the way down the tree
173 2. The number for when we visit a node on the way back up the tree
175 You can view these as bounds for the range of dfs numbers the
176 nodes in the subtree of the dominator tree rooted at that node
179 The dominator tree is always a simple acyclic tree, so there are
180 only three possible relations two nodes in the dominator tree have
183 1. Node A is above Node B (and thus, Node A dominates node B)
192 In the above case, DFS_Number_In of A will be <= DFS_Number_In of
193 B, and DFS_Number_Out of A will be >= DFS_Number_Out of B. This is
194 because we must hit A in the dominator tree *before* B on the walk
195 down, and we will hit A *after* B on the walk back up
197 2. Node A is below node B (and thus, node B dominates node B)
205 In the above case, DFS_Number_In of A will be >= DFS_Number_In of
206 B, and DFS_Number_Out of A will be <= DFS_Number_Out of B.
208 This is because we must hit A in the dominator tree *after* B on
209 the walk down, and we will hit A *before* B on the walk back up
211 3. Node A and B are siblings (and thus, neither dominates the other)
219 In the above case, DFS_Number_In of A will *always* be <=
220 DFS_Number_In of B, and DFS_Number_Out of A will *always* be <=
221 DFS_Number_Out of B. This is because we will always finish the dfs
222 walk of one of the subtrees before the other, and thus, the dfs
223 numbers for one subtree can't intersect with the range of dfs
224 numbers for the other subtree. If you swap A and B's position in
225 the dominator tree, the comparison changes direction, but the point
226 is that both comparisons will always go the same way if there is no
227 dominance relationship.
229 Thus, it is sufficient to write
231 A_Dominates_B(node A, node B) {
232 return DFS_Number_In(A) <= DFS_Number_In(B) &&
233 DFS_Number_Out(A) >= DFS_Number_Out(B);
236 A_Dominated_by_B(node A, node B) {
237 return DFS_Number_In(A) >= DFS_Number_In(A) &&
238 DFS_Number_Out(A) <= DFS_Number_Out(B);
241 bool DominatedBy(ETNode *other) const {
242 return this->DFSNumIn >= other->DFSNumIn &&
243 this->DFSNumOut <= other->DFSNumOut;
246 // This method is slower, but doesn't require the DFS numbers to
248 bool DominatedBySlow(ETNode *other) {
249 return this->Below(other);
252 void assignDFSNumber(int &num) {
256 Son->assignDFSNumber(num);
257 for (ETNode *son = Son->Right; son != Son; son = son->Right)
258 son->assignDFSNumber(num);
263 bool hasFather() const {
264 return Father != NULL;
267 // Do not let people play around with fathers.
268 const ETNode *getFather() const {
272 template <typename T>
274 return static_cast<T*>(data);
277 unsigned getDFSNumIn() const {
281 unsigned getDFSNumOut() const {
286 // Data represented by the node
290 unsigned DFSNumIn, DFSNumOut;
295 // Brothers. Node, this ends up being a circularly linked list.
296 // Thus, if you want to get all the brothers, you need to stop when
297 // you hit node == this again.
298 ETNode *Left, *Right;
303 // Rightmost occurrence for this node
304 ETOccurrence *RightmostOcc;
306 // Parent occurrence for this node
307 ETOccurrence *ParentOcc;
309 } // end llvm namespace