Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
A Study of Spilling and Coalescing in Register Allocation as Two Separate Phases Ph.D. Defense Florent Bouchez
[email protected] Compsys LIP UMR CNRS — Inria — ENS Lyon — UCBL France
April 30th 2009
1 / 39 BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Compilation
foo.c
foo.bin
Our interest Register allocation, one of the latest steps.
2 / 39 ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
French folklore: Les Shadoks
3 / 39 MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Shadoks have a very small brain
4 / 39 BU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Shadoks have a very small brain
4 / 39 BU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Shadoks have a very small brain
4 / 39 BU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Shadoks have a very small brain
4 / 39 BU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Shadoks have a very small brain
4 / 39 BU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Shadoks have a very small brain
4 / 39 BU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The shadoks at the library
5 / 39 BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
What is register allocation? a ← 3425 n←0
Assign variables to memory locations Registers:
,
,
a 6= 1 ?
, ...
Memory: infinite
n ←n+1 a even ?
Rules of the game two interfering variables
à different registers
a ← a/2
a←3×a+1
not enough registers
à spill to memory print n BU ZO
6 / 39
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
What is register allocation?
Assign variables to memory locations Registers:
,
,
, ...
Memory: infinite
Rules of the game two interfering variables
à different registers
Plus constraints: register constraints pre-colored variables register pairing,
aliasing,. . .
not enough registers
à spill to memory 6 / 39 BU ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chaitin et al. model k j Live-ranges Live-in: k j g := mem[j+12] h := k-1 f := g+h e := mem[j+8] m := mem[j+16] b := mem[f] c := e+8 d := c k := m+4 j := b Live-out: d k j
g h f e m b c d k j 7 / 39 BU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chaitin et al. model Interference graph k j
k
g
Live-ranges g
h
h
j
f
e
e
f
m m
b
b c
c d
d k j
7 / 39 BU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chaitin et al. model Interference graph k j
k
g
Live-ranges g
h
h
j
f
e
e
f
m m
b
b c
c d
d k j
7 / 39 BU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Coloring the interference graph: the greedy way Register allocation is modeled as coloring the interference graph of the program.
Problem Graph-k-coloring is NP-complete (for k ≥ 3), and any interference graph can arise in programs. (Chaitin et al.’s proof) à register allocation is NP-complete in this model.
8 / 39 ZO GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Coloring the interference graph: the greedy way Register allocation is modeled as coloring the interference graph of the program.
Problem Graph-k-coloring is NP-complete (for k ≥ 3), and any interference graph can arise in programs. (Chaitin et al.’s proof) à register allocation is NP-complete in this model. A greedy coloring heuristic is used: Chaitin et al.’s greedy scheme. Greedy scheme
If coloring fails, usually spill.
8 / 39 ZO GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Coloring the interference graph: the greedy way Register allocation is modeled as coloring the interference graph of the program.
Problem Graph-k-coloring is NP-complete (for k ≥ 3), and any interference graph can arise in programs. (Chaitin et al.’s proof) à register allocation is NP-complete in this model. A greedy coloring heuristic is used: Chaitin et al.’s greedy scheme. Greedy scheme
If coloring fails, usually spill. Disadvantages: The scheme might fail even when there is a solution. A variable is supposed to be in exactly one register. 8 / 39 ZO GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Coloring the interference graph: the greedy way Register allocation is modeled as coloring the interference graph of the program.
Problem Graph-k-coloring is NP-complete (for k ≥ 3), and any interference graph can arise in programs. (Chaitin et al.’s proof) à register allocation is NP-complete in this model. A greedy coloring heuristic is used: Chaitin et al.’s greedy scheme. Greedy scheme
If coloring fails, usually spill. Disadvantages: The scheme might fail even when there is a solution. à need to spill more than necessary A variable is supposed to be in exactly one register. à restriction on the coloring ZO GA
8 / 39
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Outline 1
Introduction Les Shadoks Register allocation
2
Difficulty of register allocation Strict SSA graphs are chordal Questions with register allocation in two phases
3
The coalescing problem Introduction to coalescing Incremental coalescing
9 / 39 ZO BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Outline 1
Introduction Les Shadoks Register allocation
2
Difficulty of register allocation Strict SSA graphs are chordal Questions with register allocation in two phases
3
The coalescing problem Introduction to coalescing Incremental coalescing
10 / 39 ZO ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
SSA interference graphs are chordal In 2002, Darte realized that the interference graph of programs under strict SSA are chordal. (The same discovery was made independently by Brisk and Hack in 2004-2005.)
à Starting point of this thesis
Theorem The interference graph of a program under strict SSA form is chordal.
11 / 39 ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
SSA interference graphs are chordal In 2002, Darte realized that the interference graph of programs under strict SSA are chordal. (The same discovery was made independently by Brisk and Hack in 2004-2005.)
à Starting point of this thesis
Theorem The interference graph of a program under strict SSA form is chordal.
11 / 39 ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chordal graphs Definition (Chordal graph) A graph is chordal iff every cycle of size ≥ 4 has a chord.
a
b
c
12 / 39 MEU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chordal graphs Definition (Chordal graph) A graph is chordal iff every cycle of size ≥ 4 has a chord. a
c
b
12 / 39 MEU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chordal graphs Definition (Chordal graph) A graph is chordal iff every cycle of size ≥ 4 has a chord. a b e c d
12 / 39 MEU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chordal graphs Definition (Chordal graph) A graph is chordal iff every cycle of size ≥ 4 has a chord. a b e c d
12 / 39 MEU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chordal graphs Definition (Chordal graph) A graph is chordal iff every cycle of size ≥ 4 has a chord. a b e c d
12 / 39 MEU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chordal graphs Definition (Chordal graph) A graph is chordal iff every cycle of size ≥ 4 has a chord. a b e c d
12 / 39 MEU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chordal graphs Definition (Chordal graph) A graph is chordal iff every cycle of size ≥ 4 has a chord. a b e c d
12 / 39 MEU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chordal graphs Definition (Chordal graph) A graph is chordal iff every cycle of size ≥ 4 has a chord. a b e c d
Chordal graphs are perfect and easy to color. 12 / 39 MEU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Static Single Assignment SSA : exactly one textual definition per variable
13 / 39 MEU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Static Single Assignment SSA : exactly one textual definition per variable
Example (Straight code converted to SSA form) a ← ... . . . ... ← a . . . a ← ... . . . ... ← a
a1 ← . . . . . . . . . ← a1 . . . a2 ← . . . . . . . . . ← a2
13 / 39 MEU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Static Single Assignment SSA : exactly one textual definition per variable
Example (Conditional code converted to SSA form) if (. . . )
if (. . . ) a←1
a←2 ... ← a
a1 ← 1
a2 ← 2
a3 ← φ(a1 , a2 ) . . . ← a3 13 / 39 MEU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Static Single Assignment SSA : exactly one textual definition per variable strictness : SSA where the definition always dominates its uses
Example (strict SSA or SSA with dominance property) if (. . . )
if (. . . ) a1 ← 1 . . . ← a1
...
a1 ← 1
...
a2 ← φ(a1 , ⊥) . . . ← a2 13 / 39 MEU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Proof that strict SSA interference graphs are chordal Theorem The interference graph of a program under strict SSA form is chordal.
Proof. a b e c d
14 / 39 MEU ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Proof that strict SSA interference graphs are chordal Theorem The interference graph of a program under strict SSA form is chordal.
Proof. a
dominance property
b e c d
14 / 39 MEU ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Proof that strict SSA interference graphs are chordal Theorem The interference graph of a program under strict SSA form is chordal.
Proof. a
dominance property
b e
a and b interfere ⇒ def (a) dominates def (b) (or the converse)
c d
14 / 39 MEU ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Proof that strict SSA interference graphs are chordal Theorem The interference graph of a program under strict SSA form is chordal.
Proof. a
dominance property
b e
a and b interfere ⇒ def (a) dominates def (b) (or the converse) direct each edge with dominance
c d
14 / 39 MEU ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Proof that strict SSA interference graphs are chordal Theorem The interference graph of a program under strict SSA form is chordal.
Proof. a
dominance property
b e
a and b interfere ⇒ def (a) dominates def (b) (or the converse) direct each edge with dominance def (d) is dominated by def (c) and def (e)
c d
14 / 39 MEU ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Proof that strict SSA interference graphs are chordal Theorem The interference graph of a program under strict SSA form is chordal.
Proof. a
dominance property
b e
a and b interfere ⇒ def (a) dominates def (b) (or the converse) direct each edge with dominance def (d) is dominated by def (c) and def (e)
c d
c and e are live at def (d)
14 / 39 MEU ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Strict SSA programs are easy to color Chordal graphs are perfect graphs, hence easy to color. We proved more:
Theorem Chordal graphs are colorable using Chaitin et al. greedy scheme. They are greedy-k-colorable. General program: NP-complete
strict SSA program: greedy-k-colorable
15 / 39 MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Strict SSA programs are easy to color Chordal graphs are perfect graphs, hence easy to color. We proved more:
Theorem Chordal graphs are colorable using Chaitin et al. greedy scheme. They are greedy-k-colorable. General program: NP-complete
strict SSA program: greedy-k-colorable
Under strict SSA, Maxlive, the maximum number of simultaneously live variables, is the coloring indicator: Maxlive ≤ R
15 / 39 MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Register allocation in two phases Using Maxlive, it seems possible to use a very simple register allocation scheme: 1
spill variables until Maxlive ≤ R
2
transform program into strict SSA form
3
allocate variables using R registers
4
go out of colored SSA form
16 / 39 BU GA GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Register allocation in two phases Using Maxlive, it seems possible to use a very simple register allocation scheme: 1
spill variables until Maxlive ≤ R
2
transform program into strict SSA form
3
allocate variables using R registers
4
go out of colored SSA form
Questions SSA seems to transform an NP-complete problem into polynomial one. . . Where is the complexity now? What else is simplified?
16 / 39 BU GA GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Questions raised by register allocation in two phases
SSA = chordal (2002)
Coalescing (out-of-SSA)
Chaitin’s proof
spilling simplified by SSA? 17 / 39 BU GA BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Questions raised by register allocation in two phases
critical flow-edge splitting
SSA = chordal (2002)
revisit Chaitin’s Chaitin’s proof proof
swap problem
Coalescing (out-of-SSA)
# simult. defs
spilling simplified by SSA? 17 / 39 BU GA BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chaitin’s proof revisited
Pro
of
it es spl edg e l l a iab tic var cri
g
tin
y xit
le mp
co NP-complete
Chaitin et al.
No
Yes
18 / 39 BU GA ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chaitin’s proof revisited
Pro
of
it es spl edg e l l a iab tic var cri
g
tin
y xit
le mp
co NP-complete
Chaitin et al.
NP-complete
Thm. 3.1
No
Yes
18 / 39 BU GA ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chaitin’s proof revisited
Pro
of
ng ion itti truct es l g p ed es ins cal abl wap i i t r i va S cr
Chaitin et al.
co NP-complete NP-complete
Thm. 3.1 Sec. 3.2.1 (SSA)
polynomial
*
No * on basic block borders only
y xit
le mp
Yes
18 / 39 BU GA ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chaitin’s proof revisited
Pro
of
ng ion itti truct es l g p ed es ins cal abl wap i i t r i va S cr
Chaitin et al.
co NP-complete NP-complete
Thm. 3.1 Sec. 3.2.1 (SSA)
*
Thm. 3.2
*
No * on basic block borders only
y xit
le mp
polynomial NP-complete
Yes
18 / 39 BU GA ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chaitin’s proof revisited
Pro
of
Chaitin et al.
ng ion . itti truct es l g def p ty d s s . e e lt l in exi l l u a b p p c ia a im ti var Sw com cri 2s NP-complete NP-complete
Thm. 3.1 Sec. 3.2.1 (SSA)
*
Thm. 3.2
*
polynomial NP-complete polynomial
Thm. 3.4 (Col. prop.)
NP-complete
Thm. 3.3 No * on basic block borders only
Yes
18 / 39 BU GA ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Questions raised by register allocation in two phases
critical flow-edge splitting
SSA = chordal (2002)
revisit Chaitin’s Chaitin’s proof proof
swap problem
Coalescing (out-of-SSA)
# simult. defs
spilling simplified by SSA? 19 / 39 BU GA MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Questions raised by register allocation in two phases
π-motion
critical flow-edge splitting
SSA = chordal (2002)
revisit Chaitin’s Chaitin’s proof proof
swap problem
Coalescing (out-of-SSA)
# simult. defs
spilling simplified by SSA? 19 / 39 BU GA MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Spilling under SSA is still a difficult problem In this thesis, we studied the complexity of spill everywhere under SSA to know whether it would simplify the problem or not.
20 / 39 BU BU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Spilling under SSA is still a difficult problem In this thesis, we studied the complexity of spill everywhere under SSA to know whether it would simplify the problem or not. Short answer: No
20 / 39 BU BU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Spilling under SSA is still a difficult problem In this thesis, we studied the complexity of spill everywhere under SSA to know whether it would simplify the problem or not. Short answer: No Longer answer: weighted Chordal graph = SSA case Interval graph = basic block
Ω0 ≤ C
Ω0 ≤ R
Ω0 ≤ Ω − 1 X3C
no yes
dyn. prog.
no
furthest use
yes
ILP
polynomial
NP-complete
dyn. prog. new result
20 / 39 BU BU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Spilling under SSA is still a difficult problem In this thesis, we studied the complexity of spill everywhere under SSA to know whether it would simplify the problem or not. Short answer: No Longer answer: weighted Chordal graph = SSA case Interval graph = basic block
Ω0 ≤ C
Ω0 ≤ R
Ω0 ≤ Ω − 1 X3C
no yes
dyn. prog.
no
furthest use
yes
ILP
polynomial
dyn. prog. new result
NP-complete
Previous studies by Belady, Yannakakis&Gavril, and Liberatore. 20 / 39 BU BU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Spilling under SSA is still a difficult problem In this thesis, we studied the complexity of spill everywhere under SSA to know whether it would simplify the problem or not. Short answer: No
The problem of “holes.” On RISC architectures, spilled variables leave small intervals behind.
a ← ... . . . ... ← a . . . ... ← a
a1 ← . . . store a1 . . . spill a
a2 ← load . . . ← a2 . . . a3 ← load . . . ← a3 20 / 39 BU BU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Spilling under SSA is still a difficult problem In this thesis, we studied the complexity of spill everywhere under SSA to know whether it would simplify the problem or not. Short answer: No With “holes,” it becomes harder even on a basic block: weight Ω0 ≤ C Ω0 ≤ R Ω0 ≤ Ω − C no ? h=1 yes stable set no stable set h≥2 yes dyn. prog. no h not bounded yes dyn. prog. polynomial
NP-complete
Ω0 ≤ Ω − 1
set cover
new result 20 / 39 BU BU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Questions raised by register allocation in two phases
π-motion
critical flow-edge splitting
SSA = chordal (2002)
revisit Chaitin’s Chaitin’s proof proof
swap problem
Coalescing (out-of-SSA)
# simult. defs
spilling spilling NOT simplified by by SSA? SSA 21 / 39 BU BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Questions raised by register allocation in two phases
π-motion
critical flow-edge splitting
SSA = chordal (2002)
revisit Chaitin’s Chaitin’s proof proof
swap problem
edge splitting π-motion
Coalescing (out-of-SSA)
# simult. defs
spilling spilling NOT simplified by by SSA? SSA
complexity
heuristics
21 / 39 BU BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Outline 1
Introduction Les Shadoks Register allocation
2
Difficulty of register allocation Strict SSA graphs are chordal Questions with register allocation in two phases
3
The coalescing problem Introduction to coalescing Incremental coalescing
22 / 39 BU BU ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
What is coalescing? Goal of coalescing Removing the register-to-register copies [move a ← b]
Numerous move due to: live-range splitting to
avoid spilling
23 / 39 BU BU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
What is coalescing? Goal of coalescing Removing the register-to-register copies [move a ← b]
Numerous move due to: live-range splitting to
avoid spilling register constraints
a ← ... b ← ... c ← f (a, b)
a ← ... b ← ... move R0 ,a move R1 ,b call f move c,R0
23 / 39 BU BU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
What is coalescing? Goal of coalescing Removing the register-to-register copies [move a ← b]
Numerous move due to:
if (. . . )
live-range splitting to
avoid spilling register constraints
a1 ← 1
a2 ← 2
SSA destruction
a3 ← φ(a1 , a2 ) · · · ← a3 23 / 39 BU BU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
What is coalescing? Goal of coalescing Removing the register-to-register copies [move a ← b]
Numerous move due to:
if (. . . )
live-range splitting to
avoid spilling register constraints SSA destruction
a1 ← 1 move a3 ← a1
a2 ← 2 move a3 ← a2
a3 ← φ(a1 , a2 ) · · · ← a3 23 / 39 BU BU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
What is coalescing? Goal of coalescing Removing the register-to-register copies [move a ← b]
Numerous move due to:
if (. . . )
live-range splitting to
avoid spilling register constraints SSA destruction
Not a new problem, but never studied not intermixed with spilling.
a1 ← 1 move a3 ← a1
a2 ← 2 move a3 ← a2
a3 ← φ(a1 , a2 ) · · · ← a3 23 / 39 BU BU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Traditional modeling of the coalescing problem Given an instruction [move a ← b] Fact I Giving the same color to both a and b saves the instruction. Fact II Merging nodes a and b forces them to have the same color.
24 / 39 BU ZO GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Traditional modeling of the coalescing problem Given an instruction [move a ← b] Fact I Giving the same color to both a and b saves the instruction. Fact II Merging nodes a and b forces them to have the same color.
a
d
d
Merge a and b b
ab
c
c
24 / 39 BU ZO GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Traditional modeling of the coalescing problem Given an instruction [move a ← b] Fact I Giving the same color to both a and b saves the instruction. Fact II Merging nodes a and b forces them to have the same color. Idea Express this as an “affinity” between a and b in the interference graph to drive the algorithm. a
d
d
Merge a and b b
ab
c
c
24 / 39 BU ZO GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Traditional modeling of the coalescing problem Given an instruction [move a ← b] Fact I Giving the same color to both a and b saves the instruction. Fact II Merging nodes a and b forces them to have the same color. Idea Express this as an “affinity” between a and b in the interference graph to drive the algorithm. a
d
d
Merge a and b b
ab
c
c
We work on graphs instead of programs. 24 / 39 BU ZO GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Different coalescing problems Aggressive
Conservative
Incremental
Optimistic
25 / 39 BU ZO BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Different coalescing problems Aggressive
Conservative
Coalesce as many affinities as possible.
Incremental
Optimistic
25 / 39 BU ZO BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Different coalescing problems Aggressive s3
Conservative v
s1
v
s3 w
u
w
u
s2 edge removed
s1
s2
affinity not coalesced
multiway-cut
Incremental
Optimistic
25 / 39 BU ZO BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Different coalescing problems Aggressive s3
Conservative v
s1
v
s3 w
u
w
u
s2 edge removed
s1
s2
affinity not coalesced
Coalesce as many affinities as possible but remains k-colorable.
multiway-cut
Incremental
Optimistic
25 / 39 BU ZO BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Different coalescing problems Aggressive
Conservative xhd,bi
s3 u s1
v
v
s3
w u s2 edge removed
d
w s1
s2
affinity not coalesced
e c
b a
yhc,di
e xhc,di
c
b xhb,ai
yhc,ei xhc,ei
yha,ci yhb,ai
multiway-cut
d
yhd,bi
a
xha,ci
graph k-coloring
Incremental
Optimistic
25 / 39 BU ZO BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Different coalescing problems Aggressive
Conservative xhd,bi
s3 u s1
v
v
s3
w u s2 edge removed
d
w s1
s2
affinity not coalesced
e c
b a
yhc,di
e xhc,di
c
b xhb,ai
yhc,ei xhc,ei
yha,ci yhb,ai
multiway-cut
d
yhd,bi
a
xha,ci
graph k-coloring
Incremental
Optimistic Perform aggressive coalescing, then de-coalescing to get k-colorable.
25 / 39 BU ZO BU
Introduction
Difficulty of register allocation
v2
The coalescing problem
Conclusion
Different coalescing problems Aggressive
Conservative
d
xhd,bi
s3 u s1
v
v
s3
w u s2 edge removed
d
w s1
s2
affinity not coalesced
e
xhc,di
c
xhb,ai
a
e
cv
b
c
b
yhc,di
yhd,bi
xhc,ei
yha,ci yhb,ai
multiway-cut
yhc,ei
xha,ci
a
graph k-coloring
Incremental
Optimistic v2
v1 cv
c0v cv
v3
v3
3-vertex-cover 25 / 39 BU ZO BU
Introduction
Difficulty of register allocation
v2
The coalescing problem
Conclusion
Different coalescing problems Aggressive
Conservative
d
xhd,bi
s3 u s1
v
v
s3
w u s2 edge removed
d
w s1
s2
affinity not coalesced
e
xhc,di
c
xhb,ai
a
e
cv
b
c
b
yhc,di
yhd,bi
xhc,ei
yha,ci yhb,ai
multiway-cut
yhc,ei
xha,ci
a
graph k-coloring
Incremental
Optimistic v2
Coalesce one affinity while staying k-colorable.
v1 cv
c0v cv
v3
v3
3-vertex-cover 25 / 39 BU ZO BU
Introduction
Difficulty of register allocation
v2
The coalescing problem
Conclusion
Different coalescing problems Aggressive
Conservative
d
xhd,bi
v
s3
s3
w u s2 edge removed
u s1
v
d
w s2
s1
affinity not coalesced
e
xhc,di
c
xhb,ai
a
e
cv
b
c
b
yhc,di
yhd,bi
xhc,ei
yha,ci yhb,ai
multiway-cut
yhc,ei
xha,ci
a
graph k-coloring
Optimistic
Incremental yi,1
ai,1
v2
bi,1 yi,2
ai,2
yi,3
ai,3
v1
ci,1 F T
cv c0v
R yi,4
ai,4
bi,2
ci,2
3-SAT
cv
v3
v3
3-vertex-cover 25 / 39 BU ZO BU
Introduction
Difficulty of register allocation
v2
The coalescing problem
Conclusion
Different coalescing problems Aggressive
Conservative
d
xhd,bi
v
s3
s3
w u s2 edge removed
u s1
v
d
w s2
s1
affinity not coalesced
e
xhc,di
c
xhb,ai
a
e
cv
b
c
b
yhc,di
yhd,bi
xhc,ei
yha,ci yhb,ai
multiway-cut
yhc,ei
xha,ci
a
graph k-coloring
Optimistic
Incremental yi,1
ai,1
v2
bi,1 yi,2
ai,2
yi,3
ai,3
v1
ci,1 F T
cv c0v
R yi,4
ai,4
bi,2
ci,2
3-SAT
NP-complete for general graphs. Polynomial for chordal graphs.
cv
v3
v3
3-vertex-cover 25 / 39 BU ZO BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
The problem of incremental coalescing The goal of incremental is to perform conservative coalescing by coalescing affinities one by one.
Problem (Incremental coalescing) Given a k-colorable graph G and two nodes x and y , is it possible to color G such that x and y have the same color?
Theorem The incremental coalescing problem is NP-complete.
26 / 39 BU ZO ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). First, equivalence between graph-3-coloring and 4-SAT. T
3 nodes for True, False, and X
F
X
27 / 39 BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). First, equivalence between graph-3-coloring and 4-SAT. T
F x
3 nodes for True, False, and X
X
2 nodes for each variable: v and v¯
x¯
27 / 39 BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). First, equivalence between graph-3-coloring and 4-SAT. T
F x
3 nodes for True, False, and X
X
x¯
2 nodes for each variable: v and v¯
y y¯ 27 / 39 BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). First, equivalence between graph-3-coloring and 4-SAT. T
F x
3 nodes for True, False, and X
X
x¯
2 nodes for each variable: v and v¯
y y¯ z¯
z 27 / 39
BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). First, equivalence between graph-3-coloring and 4-SAT. T
F x
3 nodes for True, False, and X
X
x¯
2 nodes for each variable: v and v¯
y
w ¯ w
y¯ z¯
z 27 / 39
BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). First, equivalence between graph-3-coloring and 4-SAT. T
F
v¯ 3 nodes for True, False, and X 2 nodes for each variable: v and v¯
x X
v
x¯ y
w ¯ w
y¯ z¯
z 27 / 39
BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). First, equivalence between graph-3-coloring and 4-SAT. T
F
v¯ 3 nodes for True, False, and X 2 nodes for each variable: v and v¯ . . . and a widget to forbid every
variable of a clause to be false
x X
v
x¯ y
w ¯ w
y¯ z¯
z 27 / 39
BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). x F
y T z¯
X
w Need a widget that is 3-colorable only if not all 4 variables are false. 27 / 39 BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). x F
y T z¯
X
w Need a widget that is 3-colorable only if not all 4 variables are false. 27 / 39 BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). x F
y T z¯
X
w If all 4 variables are false, not 3-colorable. 27 / 39 BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). x F
y T z¯
X
w If at least one variable is true 27 / 39 BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). x F
y T z¯
X
w If at least one variable is true 27 / 39 BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). x F
y T z¯
X
w If at least one variable is true, 3-colorable. 27 / 39 BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). Now, transform 3-SAT instance into 4-SAT by adding x0 to every clause: (y ∨ z¯ ∨ w ) ∧ · · · ∧ (z ∨ y¯ ∨ u) becomes (x0 ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (x0 ∨ z ∨ y¯ ∨ u) Clearly, x0 =True satisfies the equation (i.e., the graph is 3-colorable).
27 / 39 BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). Now, transform 3-SAT instance into 4-SAT by adding x0 to every clause: (y ∨ z¯ ∨ w ) ∧ · · · ∧ (z ∨ y¯ ∨ u) becomes (x0 ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (x0 ∨ z ∨ y¯ ∨ u) Clearly, x0 =True satisfies the equation (i.e., the graph is 3-colorable). Now, ask x0 and False to be coalesced. . . 27 / 39 BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing is NP-complete in the general case Reduction from 3-SAT. (Similar to reduction of graph-3-coloring from 3-SAT).
Example on (x ∨ y ∨ z¯ ∨ w ) ∧ · · · ∧ (¯ x ∨ z ∨ y¯ ∨ u). To conclude:
3-SAT is true
⇐⇒
4-SAT is true with x0 = False
⇐⇒
graph is 3-colorable with x0 in red/False
⇐⇒
incremental coalescing of x0 with False is possible
27 / 39 BU ZO MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Incremental coalescing for chordal graphs Problem (Incremental coalescing for chordal graphs) Given a k-colorable chordal graph G and two nodes x and y . Is it possible to color G such that x and y have the same color?
This problem is polynomial!
Moreover, if the answer is yes, it is possible to modify G so that x and y are merged and G stays chordal. The same question with greedy-k-colorable graphs is still open. 28 / 39 BU MEU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Example on a chordal graph Let us consider a 3-colorable chordal graph.
Demo: graph in ubigraph
29 / 39 BU MEU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chordal graphs as subtrees of a tree
30 / 39 BU MEU ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chordal graphs as subtrees of a tree
30 / 39 BU MEU ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Chordal graphs as subtrees of a tree
30 / 39 BU MEU ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
y x
31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
x
y 31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
x
y 31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
x
y 31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
x
y 31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
x
y 31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
x
y 31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
x
y 31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
x
y 31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
x
y 31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
x
y 31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
x
y 31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Finding a path on the interval graph Once the branches are pruned, an interval graph remains.
x
y 31 / 39 BU MEU MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
A simple algorithmic strategy for chordal incremental coalescing
Building the representation of a chordal graph as subtrees of a tree is painful. We have devised an algorithmic strategy that works directly on the graph, using the same ideas as in Chaitin et al.’s greedy coloring algorithm.
32 / 39 ZO GA GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Demonstration of coalescing Demonstration of conservative coalescing on graph #311 of the “Coalescing Challenge.” (Appel&George)
33 / 39
Demo: conservative coalescing
ZO GA BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Questions raised by register allocation in two phases not published
critical flow-edge splitting
π-motion
SSA = chordal (2002)
edge splitting π-motion
WDDD’06 revisit Chaitin’s Chaitin’s proof proof
Coalescing (out-of-SSA)
LCPC’06 (ext) swap problem
# simult. defs
LCTES’07 spilling spilling NOT simplified by by SSA? SSA
CGO’07 (best paper)
CASES’08
complexity
heuristics
34 / 39 ZO GA ZO
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Conclusion: practical work going on
This work was part of a collaboration with STMicroelectronics in Grenoble. Practical experiments were made during this thesis (coalescing). But a lot more are going on, including: including Hack’s spilling in LAO experimenting new coalescing strategies in LAO experimenting the permutation motion to avoid splitting edges
35 / 39 ZO GA MEU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Conclusion Why are those results important for register allocation? The idea of separating register allocation in two phases has been around for a while: spilling is the most important issue splitting helps (Briggs, Cooper) but copies are hard to remove greedy coloring is not a good indicator for spilling
Assuming that some copies are always better than a spill: Maxlive is an “oracle” for spilling spilling can be optimized first so that Maxlive ≤ R good coalescing strategies: do not worry about splitting anymore
36 / 39 ZO BU GA
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
Remaining problems to work on continuing the work on permutation motion improving the spilling as a first phase having a better “oracle” that handles registers constraints
(Pereira&Palsberg) register allocation in two phases for JIT Maxlive indicator can be used for better instruction scheduling
Bonus: what will be written left to the pumping shadoks in next slide?
37 / 39 ZO BU BU
Introduction
Difficulty of register allocation
The coalescing problem
Conclusion
That’s all for today
Answer: ZO BU ZO 38 / 39 ZO BU ZO
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors. a
3
j
Stack
2
d0 1
c0
c
4
4
b
3
d
e
3
4
f 2
g0
1
3
i g
2 Return
Hi-degree node Low-degree node 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors. a
3
j 2
Stack
d0 1
c0
j
c
4
4
b
3
d
e
3
4
f 2
g0
2
i g
2 Return
Hi-degree node Low-degree node 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors. a
3
j 2
Stack
d0 1
c0 g j
c
4
4
b
3
d
e
3
3
f 2
g0
1
i g Return
Hi-degree node Low-degree node 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors. a
3
j 2
Stack
d0 1
c0 i g j
c
4
4
b
3
d
e
3
2
f 2
g0
i g Return
Hi-degree node Low-degree node 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors. a
3
j 2
Stack
d0 1
f i g j
c0
c
4
4
b
3
d
e
2
f 1
g0
i g Return
Hi-degree node Low-degree node 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors. a
3
j
Stack
2
d0 g0 f i g j
1
c0
c
4
4
b
3
d
e
1
g0
f
i g Return
Hi-degree node Low-degree node 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors. a
3
j
Stack
2
d0 e g0 f i g j
1
c0
c
4
4
b
2
d
e g0
f
i g Return
Hi-degree node Low-degree node 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors. a
3
j
Stack
2
d0 d e g0 f i g j
1
c0
c
3
3
b
d
e g0
f
i g Return
Hi-degree node Low-degree node 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors.
Stack
a d0 d e g0 f i g j
3
j d0
1
c0
c
2
2
b
d
e g0
f
i g Return
Hi-degree node Low-degree node 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors.
Stack
a c d0 d e g0 f i g j
2
j d0
1
c0
c
1
b
d
e g0
f
i g Return
Hi-degree node Low-degree node 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors.
Stack
a b c d0 d e g0 f i g j
1
j d0
1
c0
c
b
d
e g0
f
i g Return
Hi-degree node Low-degree node 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al.
Stack
Greedy-k-colorable-graphs: simplify nodes with < k neighbors.
a b c d0 d e g0 f i g j
a
j d0
c0
c
b
d
e g0
f
i g Return
Hi-degree node Low-degree node 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al.
Stack
Greedy-k-colorable-graphs: simplify nodes with < k neighbors. c0 a b c d0 d e g0 f i g j
a
j d0
c0
c
b
d
e g0
f
i g Return
Hi-degree node Low-degree node 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al.
Stack
Greedy-k-colorable-graphs: simplify nodes with < k neighbors. c0 a b c d0 d e g0 f i g j
a
j d0
c0
c
b
d
e g0
f
i g Return
39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al.
Stack
Greedy-k-colorable-graphs: simplify nodes with < k neighbors.
a b c d0 d e g0 f i g j
a
j d0
c0
c
b
d
e g0
f
i g Return
39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors.
Stack
a b c d0 d e g0 f i g j
j d0
c0
c
b
d
e g0
f
i g Return
39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors.
Stack
a c d0 d e g0 f i g j
j d0
c0
c
b
d
e g0
f
i g Return
39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors.
Stack
a d0 d e g0 f i g j
j d0
c0
c
b
d
e g0
f
i g Return
39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors.
Stack
a
j d0
d e g0 f i g j
c0
c
b
d
e g0
f
i g Return
39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors.
Stack
a
j d0
e g0 f i g j
c0
c
b
d
e g0
f
i g Return
39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors.
Stack
a
j d0
g0 f i g j
c0
c
b
d
e g0
f
i g Return
39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors. a
j
Stack
d0
f i g j
c0
c
b
d
e g0
f
i g Return
39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors. a
j
Stack
d0 c0
c
b
d
e g0
f
i g Return
i g j 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors. a
j
Stack
d0 c0
c
b
d
e g0
f
i g Return
g j 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors. a
j
Stack
d0 c0
c
b
d
e g0
f
i g Return
j 39 / 39 ZO BU MEU
Greedy coloring scheme of Chaitin et al. Greedy-k-colorable-graphs: simplify nodes with < k neighbors.
Stack
a
j d0
c0
c
b
d
e g0
f
i g Return
39 / 39 ZO BU MEU
