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A PolynomialTime Algorithm for Global Value Numbering

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RKS Algorithm (SAS 1999) Polynomial, Incomplete, Improvement on AWZ ... AWZ Algorithm: functions are uninterpreted. fails to discover second assertion. RKS ... – PowerPoint PPT presentation

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Title: A PolynomialTime Algorithm for Global Value Numbering


1
A Polynomial-Time Algorithm for Global Value
Numbering
  • SAS 2004
  • Sumit Gulwani George C. Necula

2
Global Value Numbering
  • Goal Discover equivalent expressions in
    procedures
  • Applications
  • Compiler optimizations
  • Copy propagation, Constant propagation, Common
    sub-expression elimination, Induction variable
    elimination etc.
  • Program verification
  • Discover loop invariants, verify program
    assertions
  • Discover equivalent computations across programs
  • Translation validation, Plagiarism detection
    tools

3
Global Value Numbering
  • Challenge Undecidable Problem
  • Simplification Assumptions
  • Operators are uninterpreted (will not discover u
    x)
  • Conditionals are non-deterministic (will not
    discover v x)
  • Will discover x y

4
Non-trivial Example

x b y b z F(b)
x a y a z F(a)
assert(x y) assert(z F(y))
5
Related Work
  • Algorithms that work on SSA form of the program
  • AWZ Algorithm (POPL 1988)
  • Polynomial, Incomplete
  • RKS Algorithm (SAS 1999)
  • Polynomial, Incomplete, Improvement on AWZ
  • Dataflow analysis or Abstract interpretation
    based algorithm
  • Kildalls Algorithm (POPL 1973)
  • Exponential, Complete
  • Our Algorithm (this paper)
  • Polynomial, Complete

6
Non-trivial Example
x ?(a,b) y ?(a,b) z ?(F(a),F(b)) F(y)
F(?(a,b))

x b y b z F(b)
x a y a z F(a)
assert(x y) assert(z F(y))
  • AWZ Algorithm ? functions are uninterpreted
  • fails to discover second assertion
  • RKS Algorithm uses rewrite rules for
    normalization
  • Does not discover all assertions in little more
    involved examples.
  • Rewrite rules not applied exhaustively (exp
    applications o.w.)
  • Rules are pessimistic in handling loops

7
Outline
  • Strong Equivalence DAG (SED)
  • The Assignment Operation
  • The Join Operation
  • Pruning an SED
  • Fixed Point Computation

8
Representing Equivalences
a 1 b 2 x F(1,2)
a,1 b,2 x, F(1,2)
9
Representing Equivalences
a 1 b 2 x F(1,2)
a,1 b,2 x, F(1,2), F(a,2), F(1,b),
F(a,b)
Such an explicit representation can be
exponential.
10
Strong Equivalence DAG (SED)
  • A data structure for representing equivalences.
  • Nodes n ltSet of variables, Typegt
  • Type c, ?, F(n1,n2)
  • For every variable x, exactly one node ltV,tgt s.t.
    x 2 V
  • called Node(x)
  • Terms(n) set of equivalent expressions
  • Terms(ltV, ?gt) V
  • Terms(ltV, cgt) V c
  • Terms(ltV, F(n1,n2)gt) V
  • F(e1,e2) e1 2 Terms(n1),
    e2 2 Terms(n2)

11
SED Example
e, F
d,c, F
a, 2
b, ?
  • Terms(n1) a, 2
  • Terms(n2) b
  • Terms(n3) c, d, F(a,b), F(2,b)
  • Terms(n4) e, F(c,b), F(d,b), F(F(a,b),b),
    F(F(2,b),b)
  • Note that e F(d,b) since F(d,b) 2
    Terms(Node(e))

12
Abstract Interpretation based algorithm
G
Assignment Node
x e
G Assignment(G,x e)
G
Conditional Node


G2 G
G1 G
G1
G2
Join Node
G Join(G1, G2)
13
Example
x 1 y 1 z F(1,1)
x 2 y 2 z F(2,2)
L1
L2
L3
u F(x,y)
L4
Assert(u z)
14
Outline
  • Strong equivalence DAG (SED)
  • The assignment operation
  • The join operation
  • Pruning an SED
  • Fixed point computation

15
The Assignment Operation
  • Assignment(G, x e)
  • It is the strongest postcondition of equivalences
    represented by G w.r.t the assignment x e
  • Delete label x from Node(x) in G
  • Let nltV,tgt be the node in G s.t. e 2 Terms(n)
  • (Add such a node to G if it does not already
    exists)
  • Add x to node n.

16
Example The Assignment Operation
F
u, F
G0 Assignment(G, u F(z,x))
17
Outline
  • Strong Equivalence DAG (SED)
  • The Assignment Operation
  • The Join Operation
  • Pruning an SED

18
The Join Operation
  • Join(G1, G2)
  • Product Construction of G1 and G2
  • If nltV1,t1gt 2 G1 and mltV2,t2gt 2 G2, then
  • (n,m) ltV1 Å V2, t1 t t2gt 2 Join(G1,G2)
  • Definition of t1 t t2
  • c t c c
  • F(n1,n2) t F(m1,m2) F ((n1,m1),(n2,m2))
  • t1 t t2 ?, otherwise

19
Example The Join Operation
y1, F
y2, F
F
F
y6,?
y7,?
y4,y5 ?
y3,?
G1
G2
G Join(G1,G2)
G
20
Outline
  • Strong equivalence DAG (SED)
  • The assignment operation
  • The join operation
  • Pruning an SED
  • Fixed point computation

21
Motivation The Prune Operation
  • If GJoin(G1,G2), then Size(G) can be Size(G1)
    Size(G2)
  • There are programs with n joins such that size of
    the SEDs after joins is exponential in n.

Discovering equivalences among all expressions
Discovering equivalences among program expressions
vs.
For the latter, it is sufficient to discover
equivalences among all terms of size at most t at
each program point (where t variables size
of program). Thus, SEDs can be pruned to have a
small size (k t)
22
The Prune Operation
  • Prune(G,k)
  • For each node ltV,tgt such that V ? , check if it
    is a root of some DAG with all leaves labelled
    with at least one variable of size k.
  • If not, then delete all the nodes that are
    reachable from only ltV,tgt

23
Example The Prune Operation
G
24
Outline
  • Strong equivalence DAG (SED)
  • The assignment operation
  • The join operation
  • Pruning an SED
  • Fixed point computation

25
Fixed Point Computation and Complexity
  • The lattice of sets of such equivalences has
    height at most k
  • Complexity
  • Dominated by the cost of join operations
  • Each join operation O(k2 N)
  • This requires doing pruning while computing join
  • of join operations O(j k)
  • Total cost O(k3 N j)
  • k of variables
  • N size of program
  • j of join points in program

26
Conclusion
  • Discovering all
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