Expressiveness and Complexity of Crosscut Languages

Expressiveness and Complexity of Crosscut Languages Karl Lieberherr, Jeffrey Palm and Ravi Sundaram Northeastern University FOAL 2005 presentation

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.

ALL PROBLEMS ARE POLYNOMIAL 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? Generalizes regular expressions on strings: sentences must be node paths in graphs. Work by Tarjan and Mendelzon.

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

Canonical Crosscut Language

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 A lnk 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

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])

Enhanced Regular Expressions on Graphs ALL PROBLEMS BECOME NP-COMPLETE • 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?

My response From lieber Thu Aug 31 13:51:57 1995 From: Karl Lieberherr <lieber> To: lamping@parc.xerox.com, lieber@ccs.neu.edu Subject: Re: Boolean and Regular Cc: 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 operators are 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

AspectJ k (a primitive) cflow(k) && || ! ERE main any* k main any* k any* \cap \cup ! Using ERE for AspectJ

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.

Examples first • Show two programs and their graph abstractions

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); } } Meta graph= Call graph main nx1 x1 x2 nx2 x3 nx3 main x1 x2 x3 target nx3 target … target Instance tree Call tree Selected by all()

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 {} Meta graph= Class graph Main Nx1 X1 X2 Nx2 X3 Nx3 Main X1 X2 X3 Target Nx3 Target … Target Instance tree Object tree Selected by strategy

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)

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

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?

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 Connections between SAJ and SD

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.

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.

SD T([A,B]) T(D1.D2) T(D1 | D2) T(D1 & D2) !D SAJ flow(A) & B flow(T(D1)) & T(D2) T(D1) | T(D2) T(D1) & T(D2) !T(D) Proof: T: SD to SAJ

SAJ T(k) SD Start(G) Proof

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); } } Meta graph main nx1 x1 x2 nx2 x3 nx3 main x1 x2 x3 target nx3 target … target Instance tree Selected by all() APPROXIMATION

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.

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.

Same results for 5 problems • We don’t know yet how to unify all the proofs. • So we prove the results separately.

Results (Problem)

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

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.

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.

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.

Crosscut Language SAJ/SD S ::= a set of nodes k | set of nodes having label k nodes(D) | set of nodes selected by D in SD S | S | union S & S | intersection !Scomplement SAJ/SD seems interesting. Have both capabilities of AspectJ pointcuts and Demeter traversals. This is basically what Gregor Kiczales suggested a few years ago:he called it traversal(D), instead of nodes(D).

Mathematics for AOP

SAT: is there a path in G satisfying r?

SAT: is there a path in G satisfying r? results identical for class graphs

Abbreviations

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.

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.

translate row to column N: no, unless P=NP; NN: no Polynomial Translations ( any mistakes?)

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).

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.

Expressiveness and Complexity of Crosscut Languages