(Yet another) decision procedure for Equality Logic

Transcription

1 (Yet another) decision procedure for Equality Logic Ofer Strichman and Orly Meir echnion echnion 1

2 Equality Logic E : (x 1 = x 2 Æ (x 2 x 3 Ç x 1 x 3 )) Domain: x 1,x 2,x 3 2 N he satisfiability problem: is there an assignment to x 1,x 2,x 3 that satisfies E? Q: When is Equality Logic useful?... echnion 2

3 Equality Logic E : (x 1 = x 2 Æ (x 2 x 3 Ç x 1 x 3 )) A: Mainly when combined with Uninterpreted Functions f(x,y), g(z), Mainly used in proving equivalences, but not only. echnion 3

4 Basic notions E : x = y Æ y = z Æ z x (non-polar) Equality Graph: y x z Gives an abstract view of E echnion 4

5 From Equality to Propositional Logic Bryant & Velev CAV 00 the Sparse method x 1 E : x 1 = x 2 Æ x 2 = x 3 Æ x 1 x 3 B : e 1,2 Æ e 2,3 Æ :e 1,3 e 1,3 e 1,2 x 2 e 2,3 x 3 Encode all edges with Boolean variables his is an abstraction ransitivity of equality is lost! Must add transitivity constraints! echnion 5

6 From Equality to Propositional Logic Bryant & Velev CAV 00 the Sparse method x 1 E : x 1 = x 2 Æ x 2 = x 3 Æ x 1 x 3 B : e 1,2 Æ e 2,3 Æ :e 1,3 e 1,3 e 1,2 x 2 e 2,3 ransitivity Constraints: For each cycle of size n, forbid a true assignment to n-1 edges S = (e 1,2 Æ e 2,3! e 1,3 ) Æ (e 1,2 Æ e 1,3! e 2,3 ) Æ (e 1,3 Æ e 2,3! e 1,2 ) Check: B Æ S x 3 echnion 6

7 From Equality to Propositional Logic Bryant & Velev CAV 00 the Sparse method hm-1: It is sufficient to constrain simple cycles only e 1 e 2 e 3 e4 F e 6 e 5 echnion 7

8 From Equality to Propositional Logic Bryant & Velev CAV 00 the Sparse method hm-2: It is sufficient to constrain chord-free simple cycles e 2 F e 1 e 5 F e 3 e 4 echnion 8

9 From Equality to Propositional Logic Bryant & Velev CAV 00 the Sparse method Still, there can be an exponential number of chordfree simple cycles. Solution: make the graph chordal! echnion 9

10 From Equality to Propositional Logic Bryant & Velev CAV 00 the Sparse method Dfn: A graph is chordal iff every cycle of size 4 or more has a chord. How to make a graph chordal? eliminate vertices one at a time, and connect their neighbors. echnion 10

11 From Equality to Propositional Logic Bryant & Velev CAV 00 the Sparse method In a chordal graph, it is sufficient to constrain only triangles. Contradiction! F Polynomial # of edges and constraints. # constraints = 3 #triangles echnion 11

12 An improvement Reduced ransitivity Constraints (RC) So far we did not consider the polarity of the edges. E : x = y Æ y = z Æ z x Assuming E is in Negation Normal Form (polar) Equality Graph: y = = x z echnion 12

13 Monotonicity of NNF hm-3: NNF formulas are monotonically satisfied (in CNF this is simply the pure literal rule) Let be in NNF and satisfiable. hm-3 implies: Let ² Derive from by switching the value of a mis-assigned pure literal in Now ² echnion 13

14 An improvement Reduced ransitivity Constraints (RC) Claim: in the following graph R = e 3 Æ e 2! e 1 is sufficient z e 1 = e 3 Allowing e.g. e 1 =e 2 =, e 3 = F x = e 2 y his is only true because of monotonicity of NNF (an extension of the pure literal rule) echnion 14

15 Basic notions y x z Equality Path: a path made of equalities. we write x =*z Disequality Path: a path made of equalities and exactly one disequality. We write x *y Contradictory Cycle: two nodes x and y, s.t. x=*y and x * y form a contradictory cycle echnion 17

16 Basic notions hm-4: Every contradictory cycle is either simple or contains a simple contradictory cycle echnion 18

17 Definitions Dfn: A contradictory Cycle C is constrained under if does not allow this assignment C = F echnion 19

18 Main theorem If R constrains all simple contradictory cycles, and For every assignment S, S ² S! S ² R then From the Sparse method E is satisfiable iff B Æ R is satisfiable he Equality Formula echnion 21

19 Proof of the main theorem ( ) E is satisfiable BÆ S is satisfiable BÆ R is satisfiable ( ) Proof strategy: Let R be a satisfying assignment to B Æ R We will construct S that satisfies B Æ S From this we will conclude that E is satisfiable Skip proof echnion 22

20 Definitions for the proof A Violating cycle under an assignment R e F F e 1 Either dashed or solid e 2 his assignment violates S but not necessarily R echnion 23

21 More definitions for the proof An edge e = (v i,v j ) is equal under an assignment iff there is an equality path between v i and v j all assigned under Denote: v 3 F v 1 v 2 echnion 24

22 More definitions for the proof An edge e = (v i,v j ) is disequal under an assignment iff there is a disequality path between v i and v j in which the solid edge is the only one assigned false by Denote: v 3 F v 1 v 2 echnion 25

23 Proof Observation 1: he combination is impossible if = R (recall: R ² R) v 3 F v 1 v 2 Observation 2: if (v 1,v 3 ) is solid, then echnion 26

24 ReConstructing S ype 1: It is not the case that ype 2: Otherwise it is not the case that v 3 v 3 F F F v 1 v 2 v 1 v 2 Assign S (e 23 ) = F Assign (e 13 ) = In all other cases S = R echnion 27

25 ReConstructing S Starting from R, repeat until convergence: (e ) := F in all ype 1 cycles (e F ) := in all ype 2 cycles All ype 1 and ype 2 triangles now satisfy S B is still satisfied (monotonicity of NNF) Left to prove: all contradictory cycles are still satisfied echnion 28

26 Proof Invariant: contradictory cycles are not violating throughout the reconstruction. v 3 F F v 1 v 2 contradicts the precondition to make this assignment echnion 29

27 Proof Invariant: contradictory cycles are not violating throughout the reconstruction. v 3 F F v 1 v 2 contradicts the precondition to make this assignment echnion 30

28 Applying RC How can we use the theorem without enumerating contradictory cycles? Answer: Consider the chordal graph. Constrain triangles if they are part of a (simple) contradictory cycle How? echnion 31

29 Focus on Bi-connected dashed components built on top of a solid edge Includes all contradictory cycles involving this edge echnion 33

30 Make the component chordal Chordal-ity guarantees: every cycle contains a simplicial vertex, i.e. a vertex that its neighbors are connected. echnion 34

31 he RC algorithm Constraints cache: e 2 Æ e 3! e 1 e 4 Æ e 7! e 2 e 5 Æ e 8! e echnion 35

32 Constrains all contradictory cycles Constraints cache: e 2 Æ e 3! e 1 e 4 Æ e 7! e 2 e 6 Æ e 3! e echnion 36

33 Constraining simple contradictory cycles he constraint e 3,6 Æ e 3,5 e 5,6 is not added x x 0 2 x 4 cache: e 5,6 Æ e 4,6 e 4,5 x 1 x 3 x 6 x 5 Open problem: constrain simple contradictory cycles in P time echnion 37

34 Constraining simple contradictory cycles the constraint Suppose the e 3,6 graph Æ e 3,5 has e 5,6 3 more is not edges added, though needed Here we will stop, although cache: e 5,6 Æ e 4,6 e 4,5 x 0 x 2 x 4 x 1 x 3 x 6 x 5 Open problem: constrain simple contradictory cycles in P time echnion 38

35 Results echnion 39

36 Example: Circuit ransformations Stage 1 Stage 2 A pipeline processes data in stages Data is processed in parallel as in an assembly line Formal Model: Stage 3 echnion 40

37 Example: Circuit ransformations he maximum clock frequency depends on the longest path between two latches Note that the output of g is used as input to k We want to speed up the design by postponing k to the third stage echnion 41

38 Validating Circuit ransformations =? echnion 42

39 Validating a compilation process arget program u 1 = x 1 + y 1 ; u 2 = x 2 + y 2 ; z = u 1 u 2 ; Compilation Source program z = (x 1 + y 1 ) (x 2 + y 2 ); Need to prove that: (u 1 = x 1 + y 1 u 2 = x 2 + y 2 z = u 1 u 2 ) $ z = (x 1 + y 1 ) (x 2 + y 2 ) arget Source echnion 43

40 Validating a compilation process arget program u 1 = x 1 + y 1 ; u 2 = x 2 + y 2 ; z = u 1 u 2 ; Compilation Source program z = (x 1 + y 1 ) (x 2 + y 2 ); Need to prove that: (u 1 = x 1 + y 1 u 2 = x 2 + y 2 z = u 1 u 2 ) $ z = (x 1 + y 1 ) (x 2 + y 2 ) f 1 f 2 g 1 g2 f 1 f 2 echnion 44

41 Validating a compilation process Instead, prove: under functional consistency: for every uninterpreted function f x = y! f(x) = f(y) Need to prove that: (u 1 = x 1 + y 1 u 2 = x 2 + y 2 z = u 1 u 2 ) $ z = (x 1 + y 1 ) (x 2 + y 2 ) Which translates to (via Ackermann s reduction): f 1 f 2 g 1 g2 f 1 f 2 echnion 45