Decision Procedures. Jochen Hoenicke. Software Engineering Albert-Ludwigs-University Freiburg. Summer 2012

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1 Decision Procedures Jochen Hoenicke Software Engineering Albert-Ludwigs-University Freiburg Summer 2012 Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

2 Quantifier-free Rationals

3 Conjunctive Quantifier-free Fragment In the next lectures, we consider conjunctive quantifier-free Σ-formulae, ie, conjunctions of Σ-literals (Σ-atoms or negations of Σ-atoms) Remark 1: From this an algorithm for arbitrary quantifier-free formulae can be built For given arbitrary quantifier-free Σ-formula F, convert it into DNF Σ-formula F 1 F k where each F i conjunctive F is T -satisfiable iff at least one F i is T -satisfiable Remark 2: One can also combine a decision procedure for conjunctive fragment with DPLL Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

4 Conjunctive Quantifier-free Fragment of Rationals For T Q a formula in the conjunctive fragment looks like this: a 11 x 1 + a 12 x a 1n x n b 1 a 21 x 1 + a 22 x a 2n x n b 2 a m1 x 1 + a m2 x a mn x n b m as vectors: A x b Note: x = b can be expressed as x b x b (x b) can be expressed as x < b x < b requires some additional handling (later) Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

5 Dutertre de Moura Algorithm Presented 2006 by B Dutertre and L de Moura Based on Simplex algorithm Simpler; it doesn t optimize Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

6 Nonbasic and Basic Variables The set of variables in the formula is called N (set of non-basic variables) Additionally we introduce basic variables B, one variable for each linear term in the formula: y i := a i1 x 1 + a i2 x a in x n The basic variables depend on the non-basic variables Note: The naming is counter-intuitive Unfortunately it is the standard naming for Simplex algorithm We need to find a solution for y 1 b 1,, y m b m Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

7 Computing Basic from Non-basic Variables The basic variables can be computed by a simple Matrix computation: y 1 a 11 a 1n x 1 = y m a m1 a mn x n One can also use tableaux notation: x 1 x n y 1 a 11 a 1n y m a m1 a mn We start by setting all non-basic to 0 and computing the basic variables, denoted as β 0 (x) := 0 The valuation β s assigns values for the variables at step s Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

8 Configuration A configuration at step s of the algorithm consists of a partition of the variables into non-basic and basic variables N s B s = {x 1,, x n, y 1, y m }, a tableaux A (a m n matrix) where the columns correspond to non-basic and rows correspond to basic variables, and a valuation β s, that assigns β s (x i ) = 0 for x i N s, β s (y i ) = b i for y i N s, β s (z i ) = z j N s a ij β(z j ) for z i B s (Here z stands for either an x or a y variable) The initial configuration is: N 0 = {x 1,, x n }, B 0 = {y 1,, y m }, A 0 = A, β 0 (x i ) = 0 In later steps variables from N and B are swapped Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

9 Pivoting aka Exchanging Basic and Non-basic Variables Suppose β s is not a solution for y 1 b 1,, y m b m Let y i be a variable whose value β s (y i ) > b i Consider the row in the matrix: y i = a i1 z 1 + a i2 z a in z n Idea: Choose a z j, then solve z j in the above equation Thus, z j becomes non-basic variable, y i becomes basic Then decrease β(y i ) to b i This will either decrease z j (if a ij > 0) or increase z j (if a ij < 0, z j must be a x-variable) Solving z j in the above equation gives: z j = a i1 a ij z 1 + a i2 a ij z a in z n + 1 y i a ij a ij Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

10 Result of Pivoting After pivoting y i and z j the matrix looks as follows: y 1 = (a 11 a 1j a i1 a ij )z a 1j a ij y i + + (a 1n a 1j a in a ij )z n z j = a i1 a ij z a ij y i + + a in a ij z n y m = (a m1 a mj a i1 a ij )z a mj a ij y i + + (a mn a mj a in a ij )z n Now, set β s+1 (y i ) to b i and recompute basic variables Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

11 Detecting Conflicts We may arrive at a configuration like: y i = 0 x a ij1 y j1 + + a ijk y jk + 0 x n where the non-basic y variables are set to their bound: β s (y j1 ) = b j1,, β s (y jk ) = b jk, coefficients of x variables are zero, coefficients a ij1,, a ijk β s (y i ) > b i 0, and Then, we have a conflict: The formula is not satisfiable y j1 b j1 y jk b jk y i > b i Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

12 Example Consider the formula F : x 1 + x 2 4 x 1 x 2 1 We have two non-basic variables N = {x 1, x 2 } Define basic variables B = {y 1, y 2 }: y 1 = x 1 x 2, y 1 4 y 2 = x 1 x 2, y 2 1 We write the equation as a tableaux: x 1 x 2 y y Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

13 Example (cont) Tableaux: x 1 x 2 y y Values: x 1 = x 2 = 0 y 1 = 0 > 4 (!) y 2 = 0 1 Pivot y 1 against x 1 : x 1 = y 1 x 2 New Tableaux: y 1 x 2 x y Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

14 Example (cont) Tableaux: y 1 x 2 x y Values: y 1 = 4, x 2 = 0 x 1 = 4 y 2 = 4 > 1 (!) y 2 cannot be pivoted with y 1, since 1 negative Pivot y 2 and x 2 : New Tableaux: y 1 y 2 x x Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

15 Example (cont) Tableaux: y 1 y 2 x x Values: y 1 = 4, y 2 = 1 x 1 = 25 x 2 = 15 We found a satisfying interpretation for: F : x 1 + x 2 4 x 1 x 2 1 Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

16 Example Now, consider the formula F : x 1 + x 2 4 x 1 x 2 1 x 2 1 We have two non-basic variables N = {x 1, x 2 } Define basic variables B = {y 1, y 2, y 3 }: y 1 = x 1 x 2, y 1 4 y 2 = x 1 x 2, y 2 1 y 3 = x 2, y 3 1 We write the equation as tableaux: x 1 x 2 y y y Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

17 Example (cont) The first two steps are identical: pivot y 1 resp y 2 and x 1 resp x 2 y 1 y 2 x x y Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

18 Example (cont) Tableaux: y 1 y 2 x x y Values: y 1 = 4, y 2 = 1 x 1 = 25 x 2 = 15 y 3 = 15 > 1! Now, y 3 cannot pivot, since all coefficients in that row are negative Conflict is x 1 x 2 4 x 1 x 2 1 x 2 > 1 Formula F is unsatisfiable Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

19 Termination To guarantee termination we need a fixed pivot selection rule The following rule works: When choosing the basic variable (row) to pivot: Choose the y-variable with the smallest index, whose value exceeds the bound If there is no such variable, return satisfiable When choosing the non-basic variable (column) to pivot with: if possible, take a x-variable Otherwise, take the y-variable with the smallest index, such that the corresponding coefficient in the matrix is positive If there is no such variable, return unsatisfiable Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

20 Termination Proof Assume we have an infinite computation of the algorithm Let y j be the variable with the largest index, that is infinitely often pivoted Look at the step where y j is pivoted to a non-basic variable and where for k > j, y k is not pivoted any more The (ordered) tableaux at the point of pivoting looks like this: x x y y y j y y i 0 0 /0 /0 + ±/0 (+ denotes a positive coefficient, a negative coefficient) After pivoting the tableaux changes to: x x y y y i y y j 0 0 +/0 +/0 + /0 Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

21 Termination Proof (cont) After pivoting the tableaux changes to: x x y y y i y y j 0 0 +/0 +/0 + /0 k<j,y k N s a k b k + k>j,y k N s a k b k = β s (y j ) < b j, where a k 0 for k < j Now look at the step s where y j is pivoted back By the pivoting rule: β s (y k ) b k for all k < j For k > j, the non-basic/basic variables do not change Therefore, the value of y j can only get smaller β s (y j ) = This contradicts β s (y j ) > b j k<j,y k N s a k β s (y k ) + k>j,y k N s a k b k < b j Therefore, assumption was wrong and algorithm terminates Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

22 Strict Bounds With strict bounds the formula looks like this: a 11 x 1 + a 12 x a 1n x n b 1 a i1 x 1 + a i2 x a in x n b i a (i+1)1 x 1 + a (i+1)2 x a (i+1)n x n < b i+1 a m1 x 1 + a m2 x a mn x n < b m If the formula is satisfiable, then there is an ε > 0 with: a 11 x 1 + a 12 x a 1n x n b 1 a i1 x 1 + a i2 x a in x n b i a (i+1)1 x 1 + a (i+1)2 x a (i+1)n x n b i+1 ε a m1 x 1 + a m2 x a mn x n b m ε Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

23 Infinitesimal Numbers We compute with ε symbolically Our bounds are elements of Q ε := {a 1 + a 2 ε a 1, a 2 Q} The arithmetical operators and the ordering are defined as: (a 1 + a 2 ε) + (b 1 + b 2 ε) = (a 1 + b 1 ) + (a 2 + b 2 )ε a (b 1 + b 2 ε) = ab 1 + ab 2 ε a 1 + a 2 ε b 1 + b 2 ε iff a 1 < b 1 (a 1 = b 1 a 2 b 2 ) Note: Q ε is a two-dimensional vector space over Q Changes to the configuration: β gives values for variables in Q ε The tableaux does not contain ε It is still a Q m n matrix Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

24 Example F 1 : 3x 1 + 2x 2 < 5 2x 1 + 3x 2 < 1 x 1 + x 2 > 1 Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

25 Example F 1 Step 1: x 1 x 2 β b i β 0 0 y ε y ε y ε (!) Step 2: y 3 x 2 β b i β 1 ε 0 y ε 5 ε y ε 1 ε (!) x ε Step 3: y 3 y 2 β b i β 1 ε 1 ε y ε 5 ε x ε x ε β(y 1 ) = 4 + 6ε 5 ε (for 0 < ε 1/7) Solution (ε = 01): x 1 = 24, x 2 = 13 Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

26 Example F 2 : 3x 1 + 2x 2 < 5 2x 1 x 2 > 1 x 1 + 3x 2 > 4 Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

27 Example F 2 Step 1: x 1 x 2 β b i β 0 0 y ε y ε (!) y ε (!) Step 2: x 1 y 2 β b i β 0 1 ε y ε 5 ε x ε y ε 4 ε (!) Step 3: y 3 y 2 β b i β 4 ε 1 ε y ε 5 ε (!) x 2-2/7 1/ /7ε x 1-1/7-3/ /7ε Now 5 + 2ε > 5 ε but all coefficients in first row negative Unsatisfiable Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

28 Correctness of the Algorithm Theorem (Sound and Complete) Quantifier-free conjunctive Σ Q -formula F is T Q -satisfiable iff the Dutertre-de-Moura algorithm returns satisfiable Jochen Hoenicke (Software Engineering) Decision Procedures Summer / 28

Gomory Cuts. Chapter 5. Linear Arithmetic. Decision Procedures. An Algorithmic Point of View. Revision 1.0

Gomory Cuts. Chapter 5. Linear Arithmetic. Decision Procedures. An Algorithmic Point of View. Revision 1.0 Chapter 5 Linear Arithmetic Decision Procedures An Algorithmic Point of View D.Kroening O.Strichman Revision 1.0 Cutting planes Recall that in Branch & Bound we first solve a relaxed problem (i.e., no