Control Dependence. Stanford University CS243 Winter 2006 Wei Li 1

Transcription

1 Control Dependence Wei Li 1

2 Control Dependence Introduction Formal Definition Optimal Control Dependence Computation 2

3 xample START a b c d CFG ND b is control dependent on a c is control dependent on a a and d are control dependent on START 3

4 Applications of CD Dead code elimination Scheduling (hyper-block formation) Predication 4

5 Simple Dead Code limination Mark inherently live statement live Store to memory, print, For each variable in these live statements, mark its definition statement live. For each live statement, mark it live the node that it is control dependent on. Remove everything that is not marked. 5

6 xample if (x > 0) { printf( greater than zero ); } The printf statement is inherently live. You also need to mark the if (x>0) live because the print statement is control dependent on the if. 6

7 Control Dependence Introduction Formal Definition Optimal Control Dependence Computation 7

8 Post-dominator Relation If X appears on every path from START to Y, then X dominates Y. If X appears on every path from Y to ND, then X postdominates Y. Postdominator Tree ND is the root Any node Y other than ND has ipdom(y) ) as its parent Parent, child, ancestor, descendant 8

9 Control Dependence Relation There are two possible definitions. Node w is control dependent on edge (u v)) if w postdominates v If w u, w does not postdominate u Node w is control dependent on node u if there exists an edge u v w postdominates v If w u, w does not postdominate u 9

10 xample S CFG?? a Pdom Tree b a b c d e c e S a d b c Control Dep Relation 10

11 xample c d S a b e CFG e S b Pdom Tree d a c a b c d e S a b c Control Dep Relation 11

12 xample c d S a b e CFG e S b Pdom Tree d a c a b c d e S a b c Control Dep Relation 12

13 Control Dependence Queries CD(e): set of nodes control dependent on edge e CONDS(v): set of edges that node v is control dependent on 13

14 Dominance Frontier Reverse control flow graph (RCFG) Let X and Y be nodes in CFG. X in DF(Y) in CFG iff Y is control dependent on X in RCFG. DF(Y) in CFG = conds(y) ) in RCFG, where conds(y) ) is the set of nodes that Y is control dependent on. 14

15 Worst-case Size of CDR S CFG a b c d a S a b c d d a c b Control Dependence Relation 15

16 Worst-case Size of CDR S CFG a b c d a S a b c d d a c b Control Dependence Relation Size = O(n 2 ) 16

17 Control Dependence Introduction Formal Definition Optimal Linear Control Dependence Computation 17

18 APT APT: Augmented Postdominator Tree which can be built in O( ) space and time which can be used to answer CD and CONDS queries in time proportional to output size Optimal control dependence computation Solution: reduced control computation to a graph problem called Roman Chariots Problem 18

19 Key Idea (I): xploit Structure How to avoid building the entire control dependence relation (O(n 2 ))? Nodes that are control dependent on an edge e form a simple path in the postdominator tree In a tree, a simple path is uniquely specified by its endpoints. Pdom tree + endpoints of each control dependence path can be built in O( ) space and time 19

20 xample CFG S a b c d e e S b d a c Pdom Tree S a b c a S a b c path?? Path Array b d e Control Dep Relation c 20

21 xample CFG S a b c d e e S b d a c Pdom Tree S a b c a S a b c path [a,e] [c,b] Path Array b d e Control Dep Relation c 21

22 CD Queries How can we use the compact representation of the CDR (Control Dependence Relation) to answer queries for CD and CONDS sets in time proportional to output size? 22

23 Roman Chariots Problem Roma Milano Bologna Napoli Verona Pompeii Venezia Corleone Route # I II III path [Milano,Roma] [Pompeii,Bologna] [Venezia,Roma] CD(n): which cities are served by chariot n? CONDS(w): which chariots serve city w? 23

24 CD(n): which cities are served by chariot n? Look up entry for chariot n in Route Array (say [x,y[ x,y]) Traverse nodes in tree T, starting at x and ending at y Output all nodes encountered in traversal Query time is proportional to output size 24

25 CONDS(w): which chariots serve city w? For each chariot c in Route Array do Let route of c be [x,y[ x,y]; If w is an ancestor of x and w is a descendant of y then Output c; Can we avoid examining all routes? 25

26 Key Idea (II): Caching At each node in the tree, keep a list of chariot # s# s whose bottom node is n. d c e S b a {?} {?} Chariot # I II route [a,e] [c,b] 26

27 Key Idea (II): Caching At each node in the tree, keep a list of chariot # s# s whose bottom node is n. d c e S b a {I} {II} Chariot # I II route [a,e] [c,b] 27

28 CONDS(w): which chariots serve city w? For each descendant d of w do For each route c = [x,y[ x,y] ] in list at d do If w is a descendant of y then Output c; Query time is proportional to # of descendants + size of all lists at d 28

29 Sorting Lists Sort each list by decreasing length c d a b S Chariot # I II III IV route [e,d] [e,c] [e,b] [e,a] e {IV,III,II,I} 29

30 CONDS(w): which chariots serve city w? For each descendant d of w do For each route c = [x,y[ x,y] ] in list at d do If w is a descendant of y then Output c; else break Query time is proportional to size of output + # of descendants 30

31 Caching at All Nodes on Route Sort each list by decreasing length e a b c d {IV,III,II,I} e a {IV} b {IV,III} c {IV,III,II} d {IV,III,II,I} {IV,III,II,I} 31

32 Space Time Tradeoff Chariot # stored only at bottom node of the route Space: O( V + A ) Query Time: O( V + Output ) Chariot # stored at all nodes on route Space: O( V * A ) Query Time: O( Output ) ) V is the set of tree nodes, and A is the Route Array. 32

33 Key idea (III): Caching Zones Divide tree into ZONS Nodes in Zone: Boundary nodes: lowest nodes in zone Interior nodes: all other nodes Query procedure: Visit only nodes below query node and in the same zone as query node 33

34 Caching Rule Boundary node: store all chariots serving node Interior node: store all chariots whose bottom node is that node Algorithm: bottom-up, greedy zone construction Query time: A v + Z v (α + 1) A v Space requirements A + V / α 34

35 Constructing Zones (I) Invariant: for any node v, Z v α A v + 1, where α is a design parameter. Query time for CONDS(v) = O( A v + Z v ) = O((α + 1) A v + 1) = O( A v ) 35

36 Constructing Zones (II) Build zones bottom-up, making them as large as possible without violating invariant V is a leaf node, then make v a boundary node V is an interior node then If (1 + u є children(v) Z u ) > α A v + 1 then make v a boundary node else make v an interior node 36

37 α = 1 (some caching) S 3 α Av f 6 e d 5 c b a g h {e d} {f c} {g b} {S a, h a} {} {} {S a, h a, g a, f c} {} Zones: {a,b,c,d,e}, {f,g,h{ f,g,h}, {}, {S} Boundary nodes: a, f,, S 37

38 α = >>(no caching) α Av + 1 d c b a S e f g {} {} h {e d} {f c} {g b} {} {S a, h a} {} Zones: {a,b,c,d,e,f,g,h}, {}, {S} Boundary nodes: a,, S 38

39 α = << (full caching) 1 d 1 c b a h S 1 {S a, h a} α Av g {S a, h a, g a} 1 f {S a, h a, g a, f c} 1 e {S a, h a, g b, f c, e d} 1 1 {S a, h a, g b, f c, e d} {S a, h a, g b, f c} {S a, h a, g b} {S a, h a} Zones: every node Boundary nodes: every node 39

40 APT (I) Postdominator tree with bidirectional edges dfs-number[v number[v]: integer Used for ancestorship determination in CONDS query Boundary?[v]: boolean True if v is a boundary node, false otherwise Used in CONDS query 40

41 APT (II) L[v]: list pf chariots # s/control# dependences Boundary node: all chariots serving v (all control dependences of v) Interior node: all chariots whose bottom node is v (all immediate control dependences of v) Used in CONDS query 41

42 Computational Complexity Query time: (α( + 1) * output-size Space: + V /α 42

43 Reference Optimal Control Dependence Computation and the Roman Chariots Problem, Keshav Pingali, Gianfranco Bilardi,, ACM Transactions on Programming Languages and Systems (TOPLAS), May pers/toplas1997.pdf 43