1. Global Redundancy Elimination: Computing Available Expressions COMP 512 Rice University Houston, Texas Fall 2003
Copyright 2003, Keith D. Cooper & Linda Torczon, all rights reserved.
Students enrolled in Comp 512 at Rice University have explicit permission to make copies of these materials for their personal use.

2. Review So far, we have seen Local Value Numbering
Finds redundancy, constants, & identities in a block
Superlocal Value Numbering
Extends local value numbering to EBBs
Used SSA-like name space to simplify bookkeeping
Dominator Value Numbering
Extends scope to “almost” global (no back edges)
Uses dominance information to handle join points in CFG
Today’s Lecture
Global Common Subexpression Elimination (GCSE)
Applying global data-flow analysis to the problem
Today’s lecture: computing AVAIL
COMP 512, Fall 2003

3. The trick lies in finding these redundant subexpressions
The Idea
The evaluation of an expression e at point p is redundant
if and only if every path from the procedure’s entry to p
contains an evaluation of e and the value(s) of e’s
consitutent subexpressions do not change between
those earlier evaluations and p
Evaluating e at p always produces the same value as
those earlier evaluations
From the example in the last lecture:
u  e + f D E
u2  (u0,u1)
x  e + f F
e+f is redundant
The trick lies in finding these redundant subexpressions
COMP 512, Fall 2003

4. Using Available Expressions for GCSE
The goal
Find common subexpressions whose range spans basic blocks, and eliminate unnecessary re-evaluations
The mechanism
Pose the problem as a system of simultaneous equations
over the CFG of the code
Solve the equations to produce a set for each CFG node that
contains the names of every expression available on entry
Use these sets, AVAIL(n), as the basis for redundancy elimination
COMP 512, Fall 2003

5. Using Available Expressions for GCSE
The goal
Find common subexpressions whose range spans basic blocks, and eliminate unnecessary re-evaluations
Safety
x+y  AVAIL(n) proves that earlier value of x+y is the same
Transformation must provide a name for each such value
Several schemes for this mapping
Profitability
Don’t add any evaluations
Add some copy operations
Copies are inexpensive
Many copies coalesce away
Copies can shrink or stretch live ranges
COMP 512, Fall 2003 *

6. Computing Available Expressions
For each block b
Let AVAIL(b) be the set of expressions available on entry to b
Let EXPRKILL(b) be the set of expression killed in b
Let DEEXPR(b) be the set of downward exposed expressions
x  DEEXPR(b)  x defined in b & not subsequently killed in b
Now, AVAIL(b) can be defined as:
AVAIL(b) = xpred(b) (DEEXPR(x)  (AVAIL(x)  EXPRKILL(x) ))
AVAIL(n0) = Ø
where preds(b) is the set of b’s predecessors in the CFG
This system of simultaneous equations forms a data-flow problem
Solve it with a data-flow algorithm
Entry node in CFG is n0
COMP 512, Fall 2003

7. Using Available Expressions for GCSE
The Big Picture
 block b, compute AVAIL(b)
Assign unique global names to expressions in AVAIL(b)
 block b, value number b starting with AVAIL(b)
To compute AVAIL(b) :
 block b, compute DEEXPR(b) and EXPRKILL(b)
COMP 512, Fall 2003

8. Computing Available Expressions
First step is to compute DEEXPR & EXPRKILL
assume a block b with operations o1, o2, …, ok
VARKILL  Ø
DEEXPR(b)  Ø
for i = k to 1
assume oi is “x  y + z”
add x to VARKILL
if (y  VARKILL) and (z  VARKILL) then
add “y + z” to DEEXPR(b)
EXPRKILL(b)  Ø
For each expression e
for each variable v  e
if v  VARKILL(b) then
EXPRKILL(b)  EXPRKILL(b)  {e }
Many data-flow problems have initial information that costs less to compute
Backward through block }
O(k) steps
O(N) steps
N is # of operations *
COMP 512, Fall 2003

9. Computing Available Expressions
The worklist iterative algorithm
Worklist  { all blocks, bi }
while (Worklist  Ø)
remove a block b from Worklist
recompute AVAIL(b ) as
AVAIL(bi) = xpred(b) (DEEXPR(x)  (AVAIL(x)  EXPRKILL(x) ))
if AVAIL(b ) changed then
Worklist  Worklist  successors(b ) }
How do we know these things?
Today, trust me
Finds fixed point solution to equation for AVAIL
That solution is unique
Identical to “meet over all paths” solution *
COMP 512, Fall 2003

10. Back to Our Example  AVAIL sets in blue A B G C D E F { a+b } { a+b }
m  a + b
n  a + b A
p  c + d
r  c + d B
y  a + b
z  c + d G
q  a + b C
e  b + 18
s  a + b
u  e + f D
e  a + 17
t  c + d E
v  a + b
w  c + d
x  e + f F
{ a+b }
{ a+b }
{ a+b,c+d }
{ a+b,c+d }
{ a+b,c+d,e+f }
{ a+b,c+d }
COMP 512, Fall 2003

11. Remember the Big Picture
 block b, compute AVAIL(b)
Assign unique global names to expressions in AVAIL(b)
 block b, value number b starting with AVAIL(b)
We’ve done step 1.
COMP 512, Fall 2003

12. Global CSE (replacement step)
Managing the name space
Need a unique name  e  AVAIL(b)
1. Can generate them as replacements are done (Fortran H)
2. Can compute a static mapping (Common strategy)
3. Can encode value numbers into names (Briggs 94)
Strategy
Works well, but requires 2 passes (or a lot of walking around IR)
Fast, but limits replacement to textually identical expressions
Requires more analysis (VN), but yields more CSEs
Assume, w.l.o.g., solution 2
COMP 512, Fall 2003

13. Global CSE (replacement step)
Compute a static mapping from expression to name
After analysis & before transformation
 b, e  AVAIL(b), assign e a global name by hashing on e
COMP 512, Fall 2003 *

14. Assigning unique names to global CSEs
Back to Our Example 
Assigning unique names to global CSEs
a+b  t1
c+d  t2
e+f  t3
m  a + b
n  a + b A
p  c + d
r  c + d B
y  a + b
z  c + d G
q  a + b C
e  b + 18
s  a + b
u  e + f D
e  a + 17
t  c + d E
v  a + b
w  c + d
x  e + f F
{ a+b }
{ a+b }
{ a+b,c+d }
{ a+b,c+d }
{ a+b,c+d,e+f }
{ a+b,c+d }
COMP 512, Fall 2003

15. Remember the Big Picture
 block b, compute AVAIL(b)
Assign unique global names to expressions in AVAIL(b)
 block b, value number b starting with AVAIL(b)
We’ve done steps 1 & 2.
COMP 512, Fall 2003

16. Global CSE (replacement step)
Compute a static mapping from expression to name
After analysis & before transformation
 b, e  AVAIL(b), assign e a global name by hashing on e
During transformation step
Evaluation of e  insert copy name(e)  e
Reference to e  replace e with name(e)
The major problem with this approach
Inserts extraneous copies
At all definitions and uses of any e  AVAIL(b),  b
Those extra copies are dead and easy to remove
The useful ones often coalesce away
Common strategy:
Insert copies that might be useful
Let DCE sort them out
Simplifies design & implementation
COMP 512, Fall 2003 *

17. An Aside on Dead Code Elimination
What does “dead” mean?
Useless code — result is never used
Unreachable code — code that cannot execute
Both are lumped together as “dead”
To perform DCE
Must have a global mechanism to recognize usefulness
Must have a global mechanism to eliminate unneeded stores
Must have a global mechanism to simplify control-flow predicates
All of these will come later in the course
COMP 512, Fall 2003

18. Value Numbering
To perform replacement, we can value numbering each block b
Initialize hash table with AVAIL(b)
Replace an expression in AVAIL(b) means copy from its name
At each evaluation of a global name, copy new value to its name
Otherwise, value number as in last two lectures
Net Result
Catches local redundancies with value numbering
Catches nonlocal redundancies because of AVAIL sets
Not quite same effect, but close
Local redundancies found by value
Global redundancies found by spelling
COMP 512, Fall 2003

19. After replacement & local value numbering
Back to Our Example
m  a + b
t1  m
n  t1 A
p  c + d
t2  p
r  t2 B
y  t1
z  t2 G
q  t1
r  c + d
t2  r C
e  b + 18
s  t1
u  e + f
t3  u D
e  a + 17
t  t2 E
v  t1
w  t2
x  t3 F
After replacement & local value numbering
COMP 512, Fall 2003

20. Back to Our Example A B G C D E F
m  a + b
t1  m
n  t1 A
p  c + d
t2  p
r  t2 B
y  t1
z  t2 G
q  t1
r  c + d
t2  r C
e  b + 18
s  t1
u  e + f
t3  u D
e  a + 17
t  t2 E
v  t1
w  t2
x  t3 F
In practice, most of these copies will be folded into subsequent uses…
We leave copy folding to another pass where it can be done with appropriate tools (interference graph) m m p r u m r
COMP 512, Fall 2003

21. Some Copies Serve a Critical Purpose
In the example, all the copies coalesce away.
Sometimes, the copies are needed.
Copies into t1 create a common name along two paths
Makes the replacement possible
Later uses of w or x may preclude their sharing storage
w  a + b
t1  w
x  a + b
t1  x
y  t1
w  a + b
x  a + b
y  a + b 
Cannot write
“w or x”
COMP 512, Fall 2003

22. Example does not highlight value identity versus lexical identity
Back to Our Example
m  a + b
n  a + b A
p  c + d
r  c + d B
y  a + b
z  c + d G
q  a + b C
e  b + 18
s  a + b
u  e + f D
e  a + 17
t  c + d E
v  a + b
w  c + d
x  e + f F
N.B.:
SVN subsumes LVN
DVN subsumes SVN
GRE & xVN are not directly comparable
LVN
GRE,SVN
LVN
GRE,SVN
GRE,SVN
GRE, DVN
GRE, DVN
GRE
GRE, DVN
GRE
Example does not highlight value identity versus lexical identity
COMP 512, Fall 2003

23. Next Class Interative data-flow analysis: Does it halt?
Does it produce the desired answer?
How fast does it converge?
Implementation strategies
COMP 512, Fall 2003

24. And that’s the end of my story ….
Extra Slides
COMP 512, Fall 2003

25. Data-flow Analysis Definition
Data-flow analysis is a collection of techniques for compile-time reasoning about the run-time flow of values
Almost always involves building a graph
Problems are trivial on a basic block
Global problems  control-flow graph (or derivative)
Whole program problems  call graph (or derivative)
Usually formulated as a set of simultaneous equations
Sets attached to nodes and edges
Lattice (or semilattice) to describe values
We solved AVAIL with an iterative fixed-point algorithm
Desired result is usually meet over all paths solution
“What is true on every path from the entry?”
“Can this happen on any path from the entry?”
Related to the safety of optimization
COMP 512, Fall 2003

26. Data-flow Analysis Limitations
1. Precision – “up to symbolic execution”
Assume all paths are taken
2. Solution – cannot afford to compute MOP solution
Large class of problems where MOP = MFP= LFP
Not all problems of interest are in this class
3. Arrays – treated naively in classical analysis
Represent whole array with a single fact
4. Pointers – difficult (and expensive) to analyze
Imprecision rapidly adds up
Need to ask the right questions
Summary
For scalar values, we can quickly solve simple problems
Good news:
Simple problems can carry us pretty far *
COMP 512, Fall 2003

27. Data-flow Analysis Semilattice
A semilattice is a set L and a meet operation  such that,
 a, b, & c  L :
1. a  a = a
2. a  b = b  a
3. a  (b  c) = (a  b)  c
 imposes an order on L,  a, b, & c  L :
1. a ≥ b  a  b = b
2. a > b  a ≥ b and a ≠ b
A semilattice has a bottom element, denoted 
1.  a  L,   a = 
2.  a  L, a ≥ 
COMP 512, Fall 2003

28. Data-flow Analysis How does this relate to data-flow analysis?
Choose a semilattice to represent the facts
Attach a meaning to each a  L
Each a  L is a distinct set of known facts
With each node n, associate a function fn : L  L
fn models behavior of code in block corresponding to n
Let F be the set of all functions that the code might generate
Example — AVAIL
Semilattice is (2E,), where E is the set of all expressions &  is 
Set are bigger than |variables|,  is Ø
For a node n, fn has the form fn(x) = Dn  (x Nn)
Where Dn is DEF(n) and Nn is NKILL(n)
COMP 512, Fall 2003

29. Concrete Example: Available Expressions
E = {a+b,c+d,e+f,a+17,b+18}
2E is the set of all subsets of E
m  a + b
n  a + b A
p  c + d
r  c + d B
y  a + b
z  c + d G
q  a + b C
e  b + 18
s  a + b
u  e + f D
e  a + 17
t  c + d E
v  a + b
w  c + d
x  e + f F
2E = [ {a+b,c+d,e+f,a+17,b+18}, {a+b,c+d,e+f,a+17}, {a+b,c+d,e+f,b+18}, {a+b,c+d,a+17,b+18}, {a+b,e+f,a+17,b+18}, {c+d,e+f,a+17,b+18}, {a+b,c+d,e+f}, {a+b,c+d,b+18}, {a+b,c+d,a+17}, {a+b,e+f,a+17}, {a+b,e+f,b+18},{a+b,a+17,b+18}, {c+d,e+f,a+17}, {c+d,e+f,b+18}, {c+d,a+17,b+18},{e+f,a+17,b+18}, {a+b,c+d},{a+b,e+f},{a+b,a+17}, {a+b,b+18},{c+d,e+f},{c+d,a+17}, {c+d,b+18},{e+f,a+17},{e+f,b+18}, {a+17,b+18},{a+b}, {c+d}, {e+f}, {a+17}, {b+18}, {} ]
COMP 512, Fall 2003

30. Concrete Example: Available Expressions
The Lattice
{ }
{a+b} {c+d} {e+f} {a+17} {b+18}
{a+b,c+d} {a+b,a+17} {c+d,e+f} {c+d,b+18} {e+f,b+18}
{a+b,e+f} {a+b,b+18} {c+d,a+17} {e+f,a+17} {a+17,b+18}
{a+b,c+d,e+f} {a+b,c+d,b+18} {a+b,c+d,a+17} {a+b,e+f,a+17} {a+b,e+f,b+18} {a+b,a+17,b+18} {c+d,e+f,a+17} {c+d,e+f,b+18} {c+d,a+17,b+18} {e+f,a+17,b+18},
{a+b,c+d,e+f,a+17} {a+b,c+d,e+f,b+18} {a+b,c+d,a+17,b+18} {a+b,e+f,a+17,b+18} {c+d,e+f,a+17,b+18}
{a+b,c+d,e+f,a+17,b+18},
Comparability
(transitive)
meet
COMP 512, Fall 2003 *

31.  Lattice Theory This stuff is somewhat dry
Everybody stand up and stretch 
COMP 512, Fall 2003

32. Data-flow Analysis
What does this have to do with the iterative algorithm?
We can use a lattice-theoretic formulation to prove
Termination – it halts on an instance of AVAIL
Correctness – it produces the desired result for AVAIL
Complexity – it runs pretty quickly (d(CFG)+3 passes)
Worklist  { all blocks, bi }
while (Worklist  Ø)
remove a block bi from Worklist
recompute AVAIL(bi ) as
AVAIL(b) = xpred(b) (DEF(x)  (AVAIL(x)  NKILL(x) ))
if AVAIL(bi ) changed then
Worklist  Worklist  successors(bi )
COMP 512, Fall 2003

33. Data-flow Analysis Termination
If every fn  F is monotone, i.e., f(xy) ≤ f(x)  f(y), and
If the lattice is bounded, i.e., every descending chain is finite
Chain is sequence x1, x2, …, xn where xi  L, 1 ≤ i ≤ n
xi > xi+1, 1 ≤ i < n  chain is descending
Then
The iterative algorithm must halt on an instance of the problem
Set at each block can only change a finite number of times 
Any finite semilattice is bounded
Some infinite semilattices are bounded
COMP 512, Fall 2003

34. Not distributive  answer may not be unique
Data-flow Analysis
Correctness
Does the iterative algorithm compute the desired answer?
Admissible Function Spaces
1.  f  F,  x,y  L, f (xy) = f (x)  f (y)
2.  fi  F such that  x  L, fi(x) = x
3. f,g  F   h  F such that h(x ) = f (g(x))
4.  x  L,  a finite subset H  F such that x = f  H f ()
If F meets these four conditions, then the problem (L,F,) has a unique fixed point solution
 LFP = MFP = MOP
 order of evaluation does not matter
Not distributive  answer may not be unique
COMP 512, Fall 2003 *

35. Data-flow Analysis Complexity
Sets stabilize in two passes around a loop
Each pass does O(E ) meets & O(N ) other operations
Data-flow Analysis
Complexity
For a problem with an admissible function space & a bounded semilattice,
If the functions all meet the rapid condition, i.e.,
f,g  F,  x  L, f (g()) ≥ g()  f (x)  x
then, a round-robin, reverse-postorder iterative algorithm
will halt in d(G)+3 passes over a graph G
d(G) is the loop-connectedness of the graph w.r.t a DFST
Maximal number of back edges in an acyclic path
Several studies suggest that, in practice, d(G) is small (<3)
For most CFGs, d(G) is independent of the specific DFST
COMP 512, Fall 2003 *

36. Data-flow analysis What does this mean? Reverse postorder
Number the nodes in a postorder traversal
Reverse the order
Round-robin iterative algorithm
Visit all the nodes in a consistent order (RPO)
Do it again until the sets stop changing
So, these conditions are easily met
Admissible framework, rapid function space
Round-robin, reverse-postorder, iterative algorithm
The analysis runs in (effectively) linear time
COMP 512, Fall 2003

37. Data-flow Analysis How do we use these results?
Prove that data-flow framework is admissible & rapid
Its just algebra
Most (but not all) global data-flow problems are rapid
This is a property of F
Code up the iterative algorithm
World’s simplest data-flow algorithm
Other versions (worklist) have similar behavior
This lets us ignore most of the other data-flow algorithms in 512
COMP 512, Fall 2003

38. EXTRA SLIDES START HERE
COMP 512, Fall 2003

39. Global CSE (replacement step)
Managing the name space
Need a unique name  e  AVAIL(b)
1. Can generate them as replacements are done (Fortran H)
2. Can compute a static mapping
3. Can encode value numbers into names (Briggs 94)
Strategy
1. This works; it is the classic method
2. Fast, but limits replacement to textually identical expressions
3. Requires more analysis (VN), but yields more CSEs
Assume, w.l.o.g., solution 2
COMP 512, Fall 2003

40. Computing Available Expressions
The Big Picture
1. Build a control-flow graph
2. Gather the initial (local) data — DEF(b) & NKILL(b)
3. Propagate information around the graph, evaluating the equation
4. Post-process the information to make it useful (if needed)
All data-flow problems are solved, essentially, this way
Next lecture:
Iterative computation of AVAIL information
From Chapter 8 of EaC
COMP 512, Fall 2003

41. Example A B G C D E F m  a + b n  a + b p  c + d r  c + d
y  a + b
z  c + d G
q  a + b C
e  b + 18
s  a + b
u  e + f D
e  a + 17
t  c + d E
v  a + b
w  c + d
x  e + f F
COMP 512, Fall 2003

42. Example does not highlight value identity versus lexical identity
Back to Our Example 
AVAIL sets in blue
m  a + b
n  a + b A
p  c + d
r  c + d B
y  a + b
z  c + d G
q  a + b C
e  b + 18
s  a + b
u  e + f D
e  a + 17
t  c + d E
v  a + b
w  c + d
x  e + f F
LVN
{ a+b }
{ a+b }
GRE,SVN
LVN
{ a+b,c+d }
{ a+b,c+d }
GRE,SVN
GRE,SVN
{ a+b,c+d,e+f }
GRE, DVN
GRE, DVN
GRE
{ a+b,c+d }
GRE, DVN
GRE
Example does not highlight value identity versus lexical identity
COMP 512, Fall 2003