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Expressiveness and Complexity of Crosscut Languages

Karl Lieberherr, Jeffrey Palm and Ravi Sundaram

Northeastern UniversityFOAL 2005 presentation

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Goal

• Crosscut Languages are important in AOP• Study them abstractly using expressions

on graphs: lower bounds and upper bounds• Assumption: know entire program or class

graph• Of interest to: AOSD language designers

and tool builders and mathematicians.

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Regular Expressions on Graphs• Questions: Given G and reg. exp. r:

– Is there a path in G satisfying r? (SAT)– Do all paths in G that satisfy r contain n in G?

(ALWAYS)– Is there a path in G that starts with e and satisfies r?

(FIRST)• Questions: Given G and reg. exps r1 and r2:

– Is the set of paths in G satisfying r1 a subset of the set of paths satisfying r2? (IMPL)

• What has this to do with AOSD?

ALL PROBLEMSARE POLYNOMIAL

Generalizes regular expressions on strings:sentences must be node paths in graphs. Work by Tarjan and Mendelzon.

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Enhanced Regular Expressions

• ERE = regular expressions (primitive, concatenation, union, star) with – complement/negation– nodes and edges

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Canonical Crosscut Language

Enhanced Regular Expressions

AspectJ Pointcuts Traversal Strategies subset of XPath

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Some PARC-Northeastern History about Crosscut Languages:

Enhanced Regular Expressions (ERE)>From lamping@parc.xerox.com Thu Aug 31 13:33:57 1995 >To lieber@ccs.neu.edu (cc to Gregor, Crista and Jens Palsberg et al.)Subject: Re: Boolean and Regular We seem to be converging, but I still think that enhanced regular

expressions can express all of the operators. Here is the enhanced regular expression language from a while back:

Atomic expressions: A The empty traversal at class Alnk A link of type lnk ("any" is a special case of any link type)

For combining expressions, the usual regular expression crowd: . concatenation \cap intersection \cup union * repetition not negation

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Same message continued:Demeter in ERE

[A,B] A.any*.B through edges any*.lnk.any* bypassing edges not(any*.lnk.any*) through vertices any*.A.any* bypassing vertices not(any*.A.any*) d1 join d2 [d1].[d2] d1 merge d2 [d1] \cup [d2] d1 intersect d2 [d1] \cap [d2] not d1 not([d1])

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Enhanced Regular Expressions on Graphs

• Questions: Given G and reg. exp. r:– Is there a path in G satisfying r? (SAT)– Do all paths in G that satisfy r contain n in G?

(ALWAYS)– Is there a path in G that starts with e and satisfies r?

(FIRST) • Questions: Given G and reg. exps. r1 and r2:

– Is the set of paths in G satisfying r1 a subset of the set of paths satisfying r2? (IMPL)

• Ok, related to Demeter but how does AspectJ come in?

ALL PROBLEMSBECOME NP-COMPLETE

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My responseFrom lieber Thu Aug 31 13:51:57 1995From: Karl Lieberherr <lieber>To: lamping@parc.xerox.com, lieber@ccs.neu.eduSubject: Re: Boolean and RegularCc: Gregor@parc.xerox.com, boaz@ccs.neu.edu, crista@ccs.neu.edu, huersch@ccs.neu.edu, ivbaev@ccs.neu.edu, palsberg@theory.lcs.mit.edu, salil@ccs.neu.edu, seiter@ccs.neu.edu, yangl@ccs.neu.edu

Hi John:

yes, we agree. The operators of what I called Boolean algebra operatorsare just as well counted as regular expression operators.

I like your integration; have to think more about how expressive it is.

-- Karl CLAIM: ERE are the foundation for crosscut languages

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Using ERE for AspectJ

AspectJk (a primitive)cflow(k)&&||!

EREmain any* kmain any* k any* \cap\cup!

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We continue the study of crosscut languages

• and show that AspectJ pointcuts are equivalent to Demeter strategies and vice versa if you abstract from the unimportant details.

• we show the correspondence by direct translations in both directions.

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Examples first

• Show two programs and their graph abstractions

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class Example { // AspectJ program public static void main(String[] s) {x1(); nx1();} static void x1() { x2(); nx2(); } static void x2() { x3(); nx3(); } static void x3() { target(); } static void nx1() { x2(); nx2(); } static void nx2() { x3(); nx3(); } static void nx3() { target(); } static void target() {}}aspect Aspect { pointcut p1(): cflow(call (void x1())) || cflow(call (void nx2())) || cflow(call (void x3())); pointcut p2() : cflow(call (void nx1())) || cflow(call (void x2())); pointcut p3() : cflow(call (void x1())); pointcut p4() : cflow(call (void nx3())); pointcut all(): p1() && p2() && p3() && p4(); before(): all() && !within(Aspect) { System.out.println(thisJoinPoint); }}

x1

main

nx1

x2 nx2

x3 nx3

targetmain x1 x2 x3 target nx3 target…

Meta graph=Call graph

Instance treeCall tree

Selected by all()

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class Main { // Java Program with DJ X1 x1; Nx1 nx1; public static void main(String[] s) { ClassGraph cg = new ClassGraph(); Main m = new Main(); cg.traverse(m, // m is the complete tree with 8 leaves "intersect(" + // union is expressed by concatenation of edges "{source: Main -> X1 X1 -> target: Target " + "source: Main ->Nx2 Nx2 -> target: Target " + "Main -> X3 X3 -> Target}," + "{source: Main -> Nx1 Nx1 -> target: Target " + "Main -> X2 X2 -> Target}," + "{source:Main -> X1 X1 -> target: Target}," + "{source:Main -> Nx3 Nx3 -> target: Target})", new Visitor(){ public void start (){System.out.println(" start traversal");} public void finish (){System.out.println(" finish traversal");} void before (Target host){System.out.print(host + ' ');} void before (Nx3 host) {System.out.print(host + ' ');} void before (X2 host) {System.out.print(host + ' ');} void before (X1 host) {System.out.print(host + ' ');} });}}class X1 { X2 x2; Nx2 nx2; } class Nx1 { X2 x2; Nx2 nx2; }class X2 { X3 x3; Nx3 nx3; } class Nx2 { X3 x3; Nx3 nx3; }class X3 { Target t; } class Nx3 { Target t; }class Target {}

X1

Main

Nx1

X2 Nx2

X3 Nx3

TargetMain X1 X2 X3 Target Nx3 Target…

Meta graph=Class graph

Instance treeObject tree

Selected by strategy

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Crosscut Language SAJ

S ::= a set of nodesk | set of nodes having label kflow(S) | set of nodes reachable from SS | S | unionS & S | intersection!S complement

base language

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Crosscut language SD

D ::= a set of paths[A,B] | paths from A to BD . D | concatenation of pathsD | D | union of pathsD & D | intersection of paths!D complement of paths

base language

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Instance Tree

J is called an instance tree of graph G, if J is a tree, Root(J)=Start(G) and for each edge e=(u,v) in E(J), there is an edge e’ = (u’, v’) in G so that Label(u)=Label(u’) and Label(v)=Label(v’). J is a rooted tree with edges directed away from the root. (think of Label = Class)

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Expressions on GraphsExpressions on Instances

• Questions: Given graph G and r: Exists J sat G:– Is there a path in J satisfying r?– For a given node m in G: Do all paths in J that satisfy r

contain a node n in J with Label(n) = m?– Is there a path in J that starts with e and satisfies r?

• Questions: Given G and r1 and r2: Exists J sat G:– Is the set of paths in J satisfying r1 a subset of the set

of paths satisfying r2?

push down to instances

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Expressions on instances

• Questions: Given graph G, r and instance J satisfying G:– Determine the set of edges e from Root(J) so

that there is a path in J starting with e and satisfying r?

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Connections between SAJ and SD

SAJ• selects set of nodes

in tree (but there is a unique path from root to each node)

• set expression flavor

SD• selects set of paths in

tree• regular expression

flavor

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Equivalence of node sets and path sets

In a rooted tree, such as an instance tree, there is a one-to-one correspondence between nodes, and, paths from the root, because there is a unique path from the root to each node.

We say a set of paths P is equivalent to a set of nodes N if for each n in N there is a path p in P that starts at the root and ends at n and similarly for each p in P it is the case that p starts at the root and ends in a node n in N.

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Theorem 1

• A selector expression in SD (SAJ) can be transformed into an expression in SAJ (SD) in polynomial-time, such that for all meta graphs and instance trees the set of paths (nodes) selected by the SD (SAJ) selector is equivalent to the set of nodes (paths) selected by the SAJ (SD) selector.

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Proof: T: SD to SAJ

SDT([A,B])T(D1.D2)T(D1 | D2)T(D1 & D2)!D

SAJflow(A) & Bflow(T(D1)) & T(D2)T(D1) | T(D2)T(D1) & T(D2)!T(D)

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Proof

SAJT(k)

SDStart(G)

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class Example { // AspectJ program public static void main(String[] s) {x1(); nx1();} static void x1() {if (false) x2(); nx2(); } static void x2() { if (false) x3(); nx3(); } static void x3() { if (false) target(); } static void nx1() {if (false) x2(); nx2(); } static void nx2() {if (false) x3(); nx3(); } static void nx3() {if (false) target(); } static void target() {}}aspect Aspect { pointcut p1(): cflow(call (void x1())) || cflow(call (void nx2())) || cflow(call (void x3())); pointcut p2() : cflow(call (void nx1())) || cflow(call (void x2())); pointcut p3() : cflow(call (void x1())); pointcut p4() : cflow(call (void nx3())); pointcut all(): p1() && p2() && p3() && p4(); before(): all() && !within(Aspect) { System.out.println(thisJoinPoint); }}

x1

main

nx1

x2 nx2

x3 nx3

targetmain x1 x2 x3 target nx3 target…

Meta graph

Instance tree

Selected by all()

APPROXIMATION

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Computational Properties

• Select-Sat: Given a selector p and a meta graph G, is there an instance tree for G for which p selects a non-empty set of nodes.

• X/Y/Z– X is a problem, e.g., Select-Sat– Z is a language, e.g. SAJ or SD– Y is one of -,&,! representing a version of Z.

• X/-/Z base language of Z. • X/&/Z is base language of Z plus intersection. • X/!/Z is base language of Z plus negation.

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Approximation and Computational Properties

• Not Select-Sat: Given a selector p and a meta graph G, for all instance trees for G selector p selects an empty set of nodes, i.e. p is useless.

• If Not Select-Sat(p,G)/*/SAJ holds then also for the original Java program the selector p (pointcut) is useless.

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Same results for 5 problems

• We don’t know yet how to unify all the proofs.

• So we prove the results separately.

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Results (Problem)

Problem SD SAJ

- P P

& NP-complete NP-complete

! NP-complete NP-complete

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Results (Problem)

• Results(Select-Sat)• Results(Not Select-Impl)• Results(Select-First)• Results(Not Select-Always)• Results(Not Select-Never)

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Select-Sat

• Select-Sat/&/SAJ is NP-complete• This is unexpected because we have only

primitive pointcuts (e.g., call), cflow, union and intersection. Looks like Satisfiability of a monotone Boolean expression which is polynomial.

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An idea by Gregor

• add a new primitive pointcut to AspectJ: traversal(D).

• cflow(call (void class(traversal({A->B})). foo())) && this(B)– in the cflow of a call to void foo() of a class

between A and B and the currently executing object is of class B.

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Combining SAJ and SD

• Extend SD with [A,*]: all nodes reachable from A

• Replace in SAJ: flow(S) by nodes(D)• Can simulate flow(S): use [X,*] for each X

in S and take the union.

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Crosscut Language SAJ/SD

S ::= a set of nodesk | set of nodes having label knodes(D) | set of nodes selected by D in SDS | S | unionS & S | intersection!S complement

SAJ/SD seems interesting. Have both capabilities of AspectJ pointcutsand Demeter traversals.This is basically what Gregor Kiczales suggested a few years ago:he called ittraversal(D), instead of nodes(D).

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Crosscut language SD

D ::= a set of paths[A,B] | paths from A to BD . D | concatenation of pathsD | D | union of pathsD & D | intersection of paths!D complement of paths

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Mathematics for AOP

string graph/instance tree

class graph/instance tree

reg. exp. Kleene Mendelzon (SIAM Comp. 95, no instance trees)

Palsberg/Xiao/Lieberherr (TOPLAS 95)

e. reg. exp Kleene PARC/Northeastern (summer 95)Palsberg/Patt-Shamir/ Lieberherr (96)

Palsberg/Patt-Shamir/Lieberherr (96)Palm/Sundaram/ Lieberherr (04)

strategy graph

? Patt-Shamir/ Orleans/Lieberherr (97,05)

Patt-Shamir/ Orleans/Lieberherr (97, 05)Wand/Lieberherr (01)

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SAT: is there a path in G satisfying r?

graph G/instance tree

class graph G/instance tree

reg. exp. r Mendelzon (SIAM J. Comp. 95, no instance trees): polynomial

Palsberg/Xiao/Lieberherr (TOPLAS 95): polynomial (special case)

e. reg. expr

PARC/Northeastern (summer 95)Palm/Sundaram/ Lieberherr (04): NP-complete

Palm/Sundaram/ Lieberherr (04): NP-complete

strategy graph r

Patt-Shamir/ Orleans/Lieberherr (97,05): polynomial

Patt-Shamir/ Orleans/Lieberherr (97, 05): polynomial

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SAT: is there a path in G satisfying r?

graph G/instance tree

class graph G/instance tree

reg. exp. r Mendelzon (SIAM J. Comp. 95, no instance trees): polynomial

Palsberg/Xiao/Lieberherr (TOPLAS 95): polynomial (special case)

e. reg. expr

PARC/Northeastern (summer 95)Palm/Sundaram/ Lieberherr (04): NP-complete

Palm/Sundaram/ Lieberherr (04): NP-complete

strategy graph r

Patt-Shamir/ Orleans/Lieberherr (97,05): polynomial

Patt-Shamir/ Orleans/Lieberherr (97, 05): polynomial

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SAT: is there a path in G satisfying r?

graph Greg. exp. poly.e. reg. exp. NPC (add negation)strat. graph poly.e. strat. graph NPC (add intersection/negation)SAJ (AspectJ) NPCSD (Demeter) NPCSAJ-base poly. (without intersection)SD-base poly. (without intersection)

results identical for class graphs

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Abbreviations

Language Abbreviationregular exp. REenhanced regular exp. EREstrategy graph SGenhanced strategy graph ESGSAJ (AspectJ) SAJSD (Demeter) SDSAJ-base SAJBSD-base SDB

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Polynomial Translations

• We want to know which languages are fundamental. We conjecture that all languages can be translated in polynomial time into ERE. Maybe we also need ESG?

• The translations must preserve the meaning: – same set of nodes or – same set of paths or – set of paths corresponding to a set of nodes or – set of nodes corresponding to a set of paths.

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Motivation for polynomial translations

• If a large number of languages can be translated efficiently to ERE, we only need an efficient implementation for ERE.

• Currently the AP Library uses SG with intersection. If we would add complement, the AP Library would use ESG.

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Polynomial Translations ( any mistakes?)

RE ERE SG ESG SAJ SD SAJB SDB

RE Y Y Y N NERE NN NSG YESG NN NSAJ Y Y Y N NSD Y Y Y N NSAJB Y Y Y Y YSDB Y Y Y Y NN

translate row to column N: no, unless P=NP; NN: no

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Discussion

• some results are trivial: an RE sentence is trivially an ERE sentence.

• an ERE sentence can not be translated in polynomial time to an RE sentence because negation cannot be simulated by union et al.

• An SAJ sentence cannot be translated to an SDB sentence in polynomial time because otherwise P=NP (consider SAT).

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Assignments

• We want to fill in all 64 entries and have a proof for them. This is a good opportunity for a beginning PhD student.

• Yuantai: please can you do the upper triangle.

• Jingsong: please can you do the lower triangle.