1. The Partitioning Algorithm for Detecting Congruent 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 We have already seen Local & Superlocal Value Numbering
Finds redundancy, constants, & identities in a block
Dominator Value Numbering
Extends scope to “almost” global (no back edges)
Uses dominance information to handle join points in CFG
Global Common Subexpression Elimination (GCSE)
Applying data-flow analysis to the problem
Lazy Code Motion
Extends ideas behind AVAIL to partial redundancies
Today’s lecture
The Partitioning Method
Using a well-known algorithm for a new purpose
Classic example of an optimistic algorithm
COMP 512, Fall 2003

3. Partitioning Method Key idea
Congruence is related to redundancy, but not the same
Using the idea
Start with SSA form (we need the name space)
Partition the set of all expressions into congruence classes
Find largest possible sets
Each class needs one (or more) representative computation
Other uses can be replaced with representer
Operations x and y are congruent iff they have the
same operator & their operands are congruent
Alpern, Wegman, & Zadeck
POPL 1988
COMP 512, Fall 2003

4. Partitioning Method The algorithm
Based on Hopcroft’s algorithm for DFA minimization
A widely-used, often re-invented algorithm
1. Partition operations into initial classes by opcode
2. Worklist  all classes
3. While Worklist is not empty
A. Remove a class c from Worklist
B. For each class s that uses some x  c
i. Split s into s1 and s2 around uses of c
ii. Put smaller of s1 and s2 onto Worklist
4. Pick a representer for each class & rewrite the code to replace other uses of the class with representer
COMP 512, Fall 2003

5. } Partitioning Method All Operations Some Initial Values *
m0  a + b
n0  a + b
p0  c + d
r0  c + d
q0  a + b
r1  c + d
e0  b + 18
s0  a + b
u0  e0 + f
e1  a + 17
t0  c + d
u1  e1 + f
e2  (e0,e1)
u2  (u0,u1)
v0  a + b
w0  c + d
x0  e2 + f
r2  (r0,r1)
y0  a + b
z0  c + d
a  1
b  2
c  3
d  4
f  5
All Operations
a  1
b  2
c  3
d  4
f  5
m0  a + b
n0  a + b A
p0  c + d
r0  c + d B
r2  (r0,r1)
y0  a + b
z0  c + d G
q0  a + b
r1  c + d C
e0  b + 18
s0  a + b
u0  e0 + f D
e1  a + 17
t0  c + d
u1  e1 + f E
e3  (e0,e1)
u2  (u0,u1)
v0  a + b
w0  c + d
x0  e2 + f F
COMP 512, Fall 2003 *

6. Partitioning Method Compute Initial Partition * Worklist
m0  a + b
n0  a + b
p0  c + d
r0  c + d
q0  a + b
r1  c + d
e0  b + 18
s0  a + b
u0  e0 + f
e1  a + 17
t0  c + d
u1  e1 + f
e2  (e0,e1)
u2  (u0,u1)
v0  a + b
w0  c + d
x0  e2 + f
r2  (r0,r1)
y0  a + b
z0  c + d
a  1
b  2
c  3
d  4
f  5
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
m0  a + b
n0  a + b
p0  c + d
r0  c + d
q0  a + b
r1  c + d
e0  b + 18
s0  a + b
u0  e0 + f
e1  a + 17
t0  c + d
u1  e1 + f
v0  a + b
w0  c + d
x0  e2 + f
y0  a + b
z0  c + d 7
Worklist
1, 2, 3, 4, 5, 6, 7
Compute
Initial
Partition
COMP 512, Fall 2003 *

7. Partitioning Method Split on a * Worklist 2, 3, 4, 5, 6, 71
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
m0  a + b
n0  a + b
p0  c + d
r0  c + d
q0  a + b
r1  c + d
e0  b + 18
s0  a + b
u0  e0 + f
e1  a + 17
t0  c + d
u1  e1 + f
v0  a + b
w0  c + d
x0  e2 + f
y0  a + b
z0  c + d 7
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
e1  a + 17
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
e0  b + 18
u0  e0 + f
t0  c + d
u1  e1 + f
w0  c + d
x0  e2 + f
z0  c + d 8
Split
on a
Worklist
2, 3, 4, 5, 6, 7
COMP 512, Fall 2003 *

8. Partitioning Method Split on b * Worklist 3, 4, 5, 6, 7, 9, 102
c  3 3
d  4 4
f  5 5
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
e1  a + 17
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
e0  b + 18
u0  e0 + f
t0  c + d
u1  e1 + f
w0  c + d
x0  e2 + f
z0  c + d 8
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
u0  e0 + f
t0  c + d
u1  e1 + f
w0  c + d
x0  e2 + f
z0  c + d 8
e1  a + 17 9
e0  b + 18 10
Split
on b
Worklist
3, 4, 5, 6, 7, 9, 10
COMP 512, Fall 2003 *

9. Partitioning Method Split on c * Worklist 4, 5, 6, 7, 9, 10, 11
e1  a + 17 9
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
u0  e0 + f
t0  c + d
u1  e1 + f
w0  c + d
x0  e2 + f
z0  c + d 8
e0  b + 18 10
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
e0  b + 18 10
u0  e0 + f
u1  e1 + f
x0  e2 + f 11
e1  a + 17 9
Split
on c
Worklist
4, 5, 6, 7, 9, 10, 11
COMP 512, Fall 2003 *

10. Partitioning Method Split on d, f1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
e0  b + 18 10
u0  e0 + f
u1  e1 + f
x0  e2 + f 11
e1  a + 17 9
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
e1  a + 17 9
e0  b + 18 10
u0  e0 + f
u1  e1 + f
x0  e2 + f 11
Neither d nor f refine the partitions
Split
on d, f
Worklist
6, 7, 9, 10, 11
COMP 512, Fall 2003 *

11. 11 is already on the Worklist
Partitioning Method
a  1 1
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
e1  a + 17 9
e0  b + 18 10
x0  e2 + f 11
u0  e0 + f
u1  e1 + f 12
11 is already on the Worklist
b  2 2
c  3 3
Split
on e2,u2,r2
d  4 4
f  5 5
e1  a + 17 9
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
u0  e0 + f
u1  e1 + f
x0  e2 + f 11
e0  b + 18 10
Worklist
7, 9, 10, 11
COMP 512, Fall 2003 *

12. Partitioning Method Split on m0, ...1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
e1  a + 17 9
e0  b + 18 10
x0  e2 + f 11
u0  e0 + f
u1  e1 + f 12
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
e1  a + 17 9
e0  b + 18 10
x0  e2 + f 11
u0  e0 + f
u1  e1 + f 12
Does not refine the partitions ...
Split
on m0, ...
Worklist
9, 10, 11
COMP 512, Fall 2003 *

13. Partitioning Method Split on e1 * Worklist 10, 11, 12, 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
e1  a + 17 9
e0  b + 18 10
x0  e2 + f 11
u1  e1 + f 12
u0  e0 + f 15
u2  (u0,u1)
r2  (r0,r1) 14
b  2 2
c  3 3
Split
on e1
d  4 4
f  5 5
e1  a + 17 9
e2  (e0,e1)
u2  (u0,u1)
r2  (r0,r1) 6
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
x0  e2 + f 11
x0  e2 + f 11
u0  e0 + f
u1  e1 + f 12
Worklist
10, 11, 12, 6
e0  b + 18 10
COMP 512, Fall 2003 *

14. Partitioning Method Split on e0 Does not refine the partitions ... *1
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
e1  a + 17 9
e0  b + 18 10
x0  e2 + f 11
u1  e1 + f 12
u0  e0 + f 14
u2  (u0,u1)
r2  (r0,r1)
Does not refine the partitions ...
b  2 2
c  3 3
Split
on e0
d  4 4
f  5 5
e1  a + 17 9
e2  (e0,e1) 6
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
u2  (u0,u1)
r2  (r0,r1) 14
x0  e2 + f 11
u1  e1 + f 12
u0  e0 + f 15
Worklist
11, 12, 6
e0  b + 18 10
COMP 512, Fall 2003 *

15. Partitioning Method Split on x0 Does not refine the partitions ... *1
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
e1  a + 17 9
e0  b + 18 10
x0  e2 + f 11
u1  e1 + f 12
u0  e0 + f 15
u2  (u0,u1)
r2  (r0,r1) 14
Does not refine the partitions ...
b  2 2
c  3 3
Split
on x0
d  4 4
f  5 5
e1  a + 17 9
e2  (e0,e1) 6
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
u2  (u0,u1)
r2  (r0,r1) 14
x0  e2 + f 11
u1  e1 + f 12
u0  e0 + f 15
Worklist
12, 6
e0  b + 18 10
COMP 512, Fall 2003 *

16. Partitioning Method Split on u1 * Worklist 6, 14 COMP 512, Fall 2003
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
e1  a + 17 9
e0  b + 18 10
x0  e2 + f 11
u1  e1 + f 12
u0  e0 + f 15
u2  (u0,u1) 14
r2  (r0,r1) 16
b  2 2
c  3 3
Split
on u1
d  4 4
f  5 5
e1  a + 17 9
e2  (e0,e1) 6
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
u2  (u0,u1)
r2  (r0,r1) 14
x0  e2 + f 11
u1  e1 + f 12
u0  e0 + f 15
Worklist
6, 14
e0  b + 18 10
COMP 512, Fall 2003 *

17. Partitioning Method Split on e2 Does not refine the partitions ... *1
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
e1  a + 17 9
e0  b + 18 10
x0  e2 + f 11
u1  e1 + f 12
u0  e0 + f 15
u2  (u0,u1) 14
r2  (r0,r1) 16
Does not refine the partitions ...
b  2 2
c  3 3
Split
on e2
d  4 4
f  5 5
e1  a + 17 9
e2  (e0,e1) 6
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
u2  (u0,u1) 14
r2  (r0,r1) 16
x0  e2 + f 11
u1  e1 + f 12
u0  e0 + f 15
Worklist 14
e0  b + 18 10
COMP 512, Fall 2003 *

18. Partitioning Method Split on u2 Does not refine the partitions ... *1
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
a  1 1
b  2 2
c  3 3
d  4 4
f  5 5
e2  (e0,e1) 6
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
e1  a + 17 9
e0  b + 18 10
x0  e2 + f 11
u1  e1 + f 12
u0  e0 + f 15
u2  (u0,u1) 14
r2  (r0,r1) 16
Does not refine the partitions ...
b  2 2
c  3 3
Split
on u2
d  4 4
f  5 5
e1  a + 17 9
e2  (e0,e1) 6
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
u2  (u0,u1) 14
r2  (r0,r1) 16
x0  e2 + f 11
u1  e1 + f 12
u0  e0 + f 15
Worklist is empty
e0  b + 18 10
COMP 512, Fall 2003 *

19. Partitioning Method Now what? We have proved a set of theorems
Must rewrite code to use them
Knowledge alone won’t do it !
a  1 1
m0  a + b
n0  a + b
q0  a + b
s0  a + b
v0  a + b
y0  a + b 7
b  2 2
c  3 3
d  4 4
f  5 5
e1  a + 17 9
Use the partition to rebuild the code
Sets with > 1 member  reuse
Sets with 1 member  no reuse
Let’s look at the code
Opportunities are a+b & c+d
Can we make this work?
e2  (e0,e1) 6
p0  c + d
r0  c + d
r1  c + d
t0  c + d
w0  c + d
z0  c + d 8
u2  (u0,u1) 14
r2  (r0,r1) 16
x0  e2 + f 11
u1  e1 + f 12
u0  e0 + f 15
e0  b + 18 10
COMP 512, Fall 2003

20. Partitioning Method A B G C D E F Fixing a+b is easy
m0  a + b
n0  a + b A
p0  c + d
r0  c + d B
r2  (r0,r1)
y0  a + b
z0  c + d G
q0  a + b
r1  c + d C
e0  b + 18
s0  a + b
u0  e + f D
e1  a + 17
t0  c + d
u1  e + f E
e3  (e0,e1)
u2  (u0,u1)
v0  a + b
w0  c + d
x0  e + f F
Fixing a+b is easy
Need to choose the right instance to keep
Rest are redundant
COMP 512, Fall 2003

21. Partitioning Method A B G C D E F
m0  a + b
n0  m0 A
p0  c + d
r0  c + d B
r2  (r0,r1)
y0  m0
z0  c + d G
q0  m0
r1  c + d C
e0  b + 18
s0  m0
u0  e + f D
e1  a + 17
t0  c + d
u1  e + f E
e3  (e0,e1)
u2  (u0,u1)
v0  m0
w0  c + d
x0  e + f F
Notice that when we move to a global scope for optimization, we suddenly need to separate analysis & transformation.
Fixing c+d is harder
Need some rationale for replacement scheme
COMP 512, Fall 2003 *

22. Partitioning Method A B G C D E F
m0  a + b
n0  a + b A
p0  c + d
r0  c + d B
r2  (r0,r1)
y0  a + b
z0  c + d G
q0  a + b
r1  c + d C
e0  b + 18
s0  a + b
u0  e + f D
e1  a + 17
t0  c + d
u1  e + f E
e3  (e0,e1)
u2  (u0,u1)
v0  a + b
w0  c + d
x0  e + f F
AWZ suggest removing any computation that is dominated by another computation in the same partition
For a+b, the def of m0 dominates all other instances
For c+d no single instance will do
Some cases cannot be caught this way
Simpson suggested an alternative approach.
COMP 512, Fall 2003 *

23. Partitioning Method Simpson’s suggestion
Encode congruences into the name space
Same idea Briggs used in reassociation
Use LCM to remove redundancies
Simplification
In SSA name space, there are no KILLs
Knocks terms out of some equations
Makes initial information cheaper to compute
Larger name space, but O(n) rather than O(n2) initialization
Consistently outdoes dominator-based replacement
COMP 512, Fall 2003

24. } Partitioning Method Is this optimistic or pessimistic?
Earlier view said
Initialize to top  optimistic
Initialize to bottom  pessimistic
This algorithm does neither
AWZ builds a set of congruence relations
Assumes maximal size sets
Knocks out infeasible elements
Over-estimates set of congruences
Finds largest set of self-consistent congruences
 Must let it run to completion, or sets might contain lies }
inherently
optimistic
COMP 512, Fall 2003 *

25. Partitioning Method Strengths
Global algorithm for finding redundancies
Handles arbitrary control flow (loops)
Finds largest congruence classes consistent with definition
Relies on well understood technology
Weaknesses
Cannot put 2xa and a+a in same class
Not obvious how to handle identities or constants (Click)
Slower than DVNT (1(LVN):4(DVNT):10(AWZ))
COMP 512, Fall 2003