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Arithmetic Constraints and Automata

Arithmetic Constraints and Automata. Linear Arithmetic Constraints. Can be used to represent sets of valuations of unbounded integers Linear integer arithmetic formulas can be stored as a set of polyhedra where each c kl is a linear equality or inequality constraint and each

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Arithmetic Constraints and Automata

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  1. Arithmetic Constraints and Automata

  2. Linear Arithmetic Constraints • Can be used to represent sets of valuations of unbounded integers • Linear integer arithmetic formulas can be stored as a set of polyhedra where each ckl is a linear equality or inequality constraint and each is a polyhedron

  3. Linear Arithmetic Constraints • Disjunction complexity: linear • Conjunction complexity: quadratic • Negation complexity: can be exponential • Because of the disjunctive representation • Satisfiability and Equivalence checking complexity: can be exponential • Uses existential variable elimination • Post and precondition computation complexity: can be exponential • Uses existential variable elimination • Existential variable elimination can be done by extending Fourier-Motzkin variable elimination to integers

  4. Fourier-Motzkin Variable Elimination • Given two constraints   bz and az   we have a  abz  b • We can eliminate z as: z . a  abz  b if and only if a  b • Every upper and lower bound pair can generate a separate constraint, the number of constraints can double for each eliminated variable real shadow

  5. Consider the constraints: y . 0  3y – x  7  1 x – 2y  5 We get the following bounds for y: 2x  6y 6y  2x + 14 3x - 15  6y 6y  3x - 3 When we combine 2 lower bounds with 2 upper bounds we get four constraints: 0  14 , 3  x , x  29 , 0  12 Result is: 3  x  29

  6. Integers are More Complicated • If z is integer z . a  abz  b if a + (a - 1)(b - 1) b • Remaining solutions can be characterized using periodicity constraints in the following form: z .  + i = bz real shadow

  7. y x – 5  2y 2y  x – 1 x  3y 3y  x + 7 3 29 x dark shadow real shadow

  8. What About Using BDDs for Encoding Arithmetic Constraints? • Arithmetic constraints on bounded integer variables can be represented using BDDs • Use a binary encoding • represent integer x as x0x1x2... xk • where x0, x1, x2, ... , xk are binary variables • You have to be careful about the variable ordering!

  9. Arithmetic Constraints vs. BDDs • Constraint based verification can be more efficient than BDDs for integers with large domains • BDD-based verification is more robust • Constraint based approach does not scale well when there are boolean or enumerated variables in the specification • Constraint based verification can be used to automatically verify infinite state systems • cannot be done using BDDs • Price of infinity • Verification becomes undecidable and fixpoints are not guaranteed to converge

  10. Fixpoints May Not Converge • Integer variables can increase without a bound • state space is infinite • Verification is undecidable for systems with unbounded integer variables • Must use approximation

  11. Widening • Assuming that i1 and i2 are conjunctions of atomic constraints (i.e., polyhedra), then i1 i2 is defined as: all the constraints in i1 which are also satisfied by i2 Example: i1=0count  count2 i2=0count  count3 i1 i2 =0count • Replace i2 with i1 i2 in c2 • This generates an upper approximation for the least fixpoint computation This constraint is not satisfied by i2 so we drop it

  12. Automata Representation for Arithmetic Constraints[Bartzis, Bultan CIAA’02, IJFCS ’02] • Given an atomic linear arithmetic constraint in one of the following two forms we construct an FA which accepts all the solutions to the given constraint • By combining such automata one can handle full Presburger arithmetic

  13. Basic Construction 0 1 2 • We first construct a basic state machine which • Reads one bit of each variable at each step, starting from the least significant bits • and executes bitwise binary addition and stores the carry in each step in its state 0 1 0 0 / / 0 1 0 1 / 1 1 0 / 0 0 1 1 1 / / 0 1 Example x + 2y 0 1 / 0 1 1 / 1 1 1 / 0 010 + 2  001 0 0 / 1 0 1 0 0 / / 0 1 1 0 0 Number of states:

  14. Automaton Construction • Equality With 0 • All transitions writing 1 go to a sink state • State labeled 0 is the only accepting state • For disequations (), state labeled 0 is the only rejecting state • Inequality (<0) • States with negative carries are accepting • No sink state • Non-zero Constant Term c • Same as before, but now -c is the initial state • If there is no such state, create one (and possibly some intermediate states which can increase the size by |c|)

  15. Conjunction and Disjunction 0 0 1 0,1,1 0 1 0,1 1 0 Automaton for x-y<1 1 0 1 0 -1 0 1 0 1 0 0 1 0,1,1 0 0 0,1 0 0 0,1 0,-1 0 0 1 1 1 0 1 1 1 0,1 0 1 Automaton for x-y<1  2x-y>0 1 0 0 1 0,1 Automaton for 2x-y>0 0 1 0,1 0 1 0,1 -1,-1 -1,0 -1 0 0 1 0 0 0 0 0 1 0 1 0 1 0 0 1 1,1 0 1 1,1 1 0 0 1 1 1,0,1 -2,-1 -2,0 -2,1 -2 0 1 1 0 1 1 • Conjunction and disjunction is handled by generating the product automaton

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