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Understanding Linear Arithmetic Constraints for Efficient Computing

Learn how Linear Arithmetic Constraints represent sets of unbounded integers and their application in automata. Explore constraints, complexities, variable elimination, BDDs, and automata representation.

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Understanding Linear Arithmetic Constraints for Efficient Computing

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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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