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Decision Procedures for Presburger Arithmetic. Presented by Constantinos Bartzis. Presburger formulas. numeral ::= 0 | 1 | 2… var ::= x | y | z … relop ::= < | ≤ | = |  | > term ::= numeral | term + term | -term | numeral  term | var

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decision procedures for presburger arithmetic

Decision Procedures forPresburger Arithmetic

Presented by Constantinos Bartzis

presburger formulas
Presburger formulas
  • numeral ::= 0 | 1 | 2…
  • var ::= x | y | z …
  • relop ::= < | ≤ | = |  | >
  • term ::= numeral | term + term | -term | numeral  term | var
  • formula ::= term relop term | formula formula | formula formula | formula | var. formula | var. formula

numeral  term isn't really multiplication; it's short-hand for term + term + … + term

decision procedures
Decision Procedures
  • The aim is to produce an algorithm for determining whether or not a Presburger formula is valid with respect to the standard interpretation in arithmetic.
  • Such an algorithm is a decision procedure if it is guaranteed to correctly return “True” or “False” for all closedformulas.
  • Will discuss algorithms for determining truth of formulas of Presburger arithmetic:
    • Fourier-Motzkin variable elimination (FMVE)
    • Omega Test
    • Cooper's algorithm
    • Automata based
quantifier elimination
Quantifier Elimination
  • All the methods we'll look at are quantifier eliminationprocedures.
  • If a formula with no free variables has no quantifiers, then it is easy to determine its truth value,

e.g., 10 > 11  3+4 < 5  3 - 6

  • Quantifier elimination works by taking input Pwith nquantifiers and turning it into equivalent formula P’ with mquantifiers, and where m < n.
  • So, eventually P P’ … Qand Qhas no quantifiers.
  • Qwill be trivially true or false, and that's the decision
normalization
Normalization
  • Methods require input formulas to be normalized
    • e.g., collect coefficients, use only < and ≤
  • Methods eliminate innermost existentialquantifiers. Universal quantifiers are normalized with

(x. P(x))  (x. P(x))

  • In FMVE, the sub-formula under the innermost existential quantifier must be a conjunction of relations.
  • This means the inner formula must be converted to disjunctive normal form(DNF):

(c11c12 … c1n1)  ...  (cm1cm2 … cmnm)

normalization cont
Normalization (cont.)
  • The formula under  is in DNF. Next, the  must be moved inwards
  • First over disjuncts, using

(x. P Q)  (x. P)  (x. Q)

  • Must then ensure every conjunct under the quantifier mentions the bound variable. Use

(x. P(x) Q)  (x. P(x)) Q

  • For example:

(x. 3 < x x +2y ≤ 6 y < 0) 

(x. 3 < x x +2y ≤ 6) y < 0

fourier motzkin theorems
Fourier-Motzkin theorems
  • The following simple facts are the basis for a very simple quantifier elimination procedure.
  • Over R (or Q), with a,b > 0:

(x. c ≤ax bx ≤d) bc ≤ad

(x. c < ax bx ≤d) bc < ad

(x. c ≤ax bx < d) bc < ad

(x. c < ax bx < d) bc < ad

  • In all four, the right hand side is implied by the left because of transitivity
    • e.g., (x < y y ≤z)  x < z
fourier motzkin theorems cont
Fourier-Motzkin theorems (cont.)
  • For the other direction:

(bc < ad)  (x.c < ax bx ≤d)

take xto be d/b: c < a( d/b ), and b( d/b ) ≤d.

  • For (bc < ad)  (x.c < ax bx < d)

take xto be (bc+ad)/2ab:

c < a(bc+ad)/2ab  2bc < bc+ad bc < ad

  • Similarly for the other bound
extending to a full procedure
Extending to a full procedure
  • So far: a quantifier elimination procedure for formulas where the scope of each quantifier is 1 upper bound and 1 lower bound.
  • The method needs to extend to cover cases with multiple constraints.
  • No lower bound, many upper bounds:

(x: b1x < d1b2x < d2 … bnx < dn)

True!(take min(di/bi) as witness for x)

  • No upper bound, many lower bounds: obviously analogous.
combining many constraints
Combining many constraints
  • Example:

(x.c ≤ax b1x ≤d1b2x ≤d2) b1c ≤ad1b2c ≤ad2

  • From left to right, result just depends on transitivity.
  • From right to left, take xto be min(d1/b1, d2/b2).
  • In general, with many constraints, combine all possible lower-upper bound pairs.
  • Proof that this is possible is by induction on number of constraints.
combining many constraints1
Combining many constraints
  • The core elimination formula is
  • With nconstraints initially, evenly divided between upper and lower bounds, this formula generates n2/4 new constraints.
fmve example
FMVE example

x. 20+x ≤ 0 y. 3y +x ≤ 10  20 ≤y - x

(re-arrange)

x. 20+x ≤ 0 y. 20+x ≤y  3y ≤ 10 - x

(eliminate y)

x. 20+x ≤ 0  60+3x ≤ 10 - x

(re-arrange)

x. 20+x ≤ 0  4x +50 ≤ 0

(normalize universal)

x. 20+x ≤ 0  0 < 4x +50

(re-arrange)

 x. -50 < 4x x ≤ -20

(eliminate x)

 (-50 < -80)  T

complexity
Complexity
  • As before, when eliminating an existential over nconstraints we may introduce n2/4 new constraints.
  • With kquantiers to eliminate, we might end with n2k/4kconstraints.
  • If dealing with alternating quantifiers, repeated conversions to DNF may really hurt.
expressivity over integers
Expressivity over Integers
  • Can do divisibility by specific numerals:

2|e x. 2x = e

for example:

x. 0 < x < 30  (2|x  3|x  5|x)

  • Can do integer division and modulus, as long as divisor is constant. Use one of the following results (similar for division)

P(x mod d)  q,r. (x = qd +r ) (0 ≤r < d d < r ≤ 0) P(r )

P(x mod d)  q,r. (x = qd +r )  (0 ≤r < d d < r ≤ 0) P(r )

  • Any formula involving modulus or integer division by a constant can be translated to one without.
expressivity over integers1
Expressivity over Integers
  • Any procedure for Z trivially extends to be one for N (or any mixture of N and Z) too: add extra constraints stating that variables are  0
  • Relations < and ≤ can be converted into one another:

x ≤y x < y +1

x < y x +1 ≤y

  • Decision procedures normalize to one of these relations.
fourier motzkin for integers
Fourier-Motzkin for Integers?
  • Central theorem is false. E.g.,

(xZ. 3 ≤ 2x 2x ≤ 3)  6 ≤ 6

  • But one direction still works (thanks to transitivity):

(x. c ≤ax bx ≤d) bc ≤ad

  • We can compute consequences of existentially quantified formulas

/

fourier motzkin for integers1
Fourier-Motzkin for Integers?
  • We know (x. c ≤ax bx ≤d) bc ≤ad
  • Thus an incomplete procedure for universal formulas over Z:
  • Compute negation: (x. P(x)) (x. P(x))
  • Compute consequences:

if (x. P(x)) then (x. P(x)) and (x. P(x))  T

  • Repeat for all quantified variables.
  • This is Phase 1 of the Omega Test
omega phase 1 example
Omega Phase 1 - Example

x,yZ. 0 < x y < x y +1 < 2x

(normalize)

 x,y. 1 ≤x y +1 ≤x  2x ≤y +1

x,y. 1 ≤x y +1 ≤x  2x ≤y +1

(eliminate y)

x. 1 ≤x  2x ≤x

(normalize)

x. 1 ≤x x ≤ 0

(eliminate x)

 1 ≤ 0 

omega phase 1 and the interactive theorem provers
Omega Phase 1 and the Interactive Theorem Provers
  • The Omega Test's Phase 1 is used by systems like Coq, HOL4, HOL Light and Isabelle to decide arithmetic problems.
  • Against:
    • it's incomplete
    • conversion to DNF
    • quadratic increase in numbers of constraints
  • For:
    • it's easy to implement
    • it's easy to adapt the procedures to create proofs that can be checked by other tools
some shadows
Some Shadows
  • Given x. (i ci≤aix) (j bjx ≤dj)
  • The formula

i,j bjci≤aidj

is known as the real shadow.

  • If all of the aior all of the bjare equal to 1, then the real shadow is exact.
  • If the shadow is exact, then the formula can be used as an equivalence.
exact shadows
Exact Shadows
  • When a = 1 or b = 1, the core theorem

(x. c ≤ax bx ≤d) bc ≤ad is valid because

    •  transitivity still holds
    •  take x = dif b = 1 or x = cif a = 1
  • Omega Test's inventor, Bill Pugh, claims many problems in his domain (compiler optimization) have exact shadows.
  • Experience suggests the same is true in other domains too, such as interactive theorem-proving.
  • When shadows are exact, can pretend problem is over R rather than Z and life is easy.
dark shadows
Dark Shadows
  • The formula i,j (ai-1)(bj-1) ≤ aidj - bjci

is known as the dark shadow.

  • If all aior all bjare one, then this is the same as the real shadow (or exact).
  • The real shadow provides a test for unsatisfiability.
  • The dark shadow tests for satisfiability, because

(a-1)(b-1) ≤ad - bc  (x. c ≤ax bx ≤d)

  • This is the Phase 2 of the Omega Test
omega test phases 1 2
Omega Test Phases 1 & 2
  • Problem is x. P(x)
  • If input is exact for one elements of x, then eliminate them

x. P(x) x’. P’(x’)

  • Otherwise, calculate real shadow R:

x. P(x) R

so, if R , then input formula is .

  • Otherwise, calculate dark shadow D:

D x. P(x)

so, if D = T, then input formula is T.

omega phase 2 example
Omega Phase 2 - Example

(a-1)(b-1) ≤ad - bc  (x. c ≤ax bx ≤d)

x,y. 3x +2y ≤ 18  3y ≤ 4x  3x ≤ 2y +1

3y ≤ 4x 3x ≤ 2y +1 3y ≤ 4x 3x ≤ 18 - 2y

6 ≤ 8y + 4 - 9y 6 ≤ 72 - 8y - 9y

y ≤ -2 17y ≤ 66

y ≤ 3

  • This gives a suitable value for y, and by back-substitution, finds

x = -1, y = -2 as a possible solution.

splinters
Splinters
  • Purely existential formulas are often proved false by their real shadow; or proved true by their dark shadow
  • But in “rare” cases, the main theorem is needed. Let mbe the maximum of all the djs. Then

splinter

dark shadow

splinters1
Splinters
  • A splinter doesrepresent a smaller problem than the original because the extra equality allows xto be eliminated immediately.
  • When quantifiers alternate, and there is no exact shadow, the main theorem is used as an equivalence, and splinters can't be avoided.
  • Splinters must also be checked if neither real nor dark shadows decide an input formula.
eliminating equalities
Eliminating Equalities
  • In an expression

x. … cx = e  …

the existential can be eliminated.

  • First, multiply all terms involving xso that they have a common coefficient.
  • Formula becomes

x. …c’x … c’x = e’  …c’x…

  • This is equivalent to

…e’… c’|e’ …e’…

eliminating divisibilities
Eliminating Divisibilities

x. … c | dx + e  …

  • Note: d < c(otherwise, replace d with d mod c).
  • Introduce temporary new existential variable:

x,y. … cy = dx + e  …

  • Rearrange:

x,y. … dx = cy -e  …

  • Use equality elimination to derive

y. … d | cy -e  …

  • Because d < c, this process must terminate with elimination of divisibility term.
implementation normalization
Implementation - Normalization
  • Omega Test's bigdisadvantage is that it requires the formula under quantifier to be in DNF
  • Consider

x. (x  10 x  11  9 < x ≤ 12) x = 12

  • Negate, remove , <:

x. (x ≤ 9  11 ≤x)  (x ≤ 10  12 ≤x)  10 ≤x x ≤ 12  (x ≤ 11  13 ≤x)

  • Evaluate 8 (= 23) DNF terms.
  • Clever preparation of input formulas can make orders of magnitude difference
implementation normalization1
Implementation - Normalization
  • The propositional tautology

(p (q q’))  (p q p q’)

justifies the following procedure:

  • If Pis an atomic formula, then when processing P Q, assume Pis true while processing Q:
    • If a sub-formula Q0 of Qis such that P Q0, then replace Q0 in Qby T.
    • If a sub-formula Q0 of Qis such that P Q0, then replace Q0 in

Qby .

  • Similarly, (p (q q’))  (p q p q’) for disjunctions.
contextual rewriting example
Contextual Rewriting example
  • Over  :

0 ≤x + y + 4  (0 ≤x + y + 6  0 ≤ 2x + 3y + 6)

is equivalent to 0 ≤x + y + 4

  • Whereas

0 ≤x + y + 4  0 ≤ -x -y -6  0 ≤ 2x + 3y + 6

is equivalent to 

  • Over  :

0 ≤x + y + 4  0 ≤x + y + 1  0 ≤ 2x + 3y + 6

is equivalent to

0 ≤x + y + 4  0 ≤ 2x + 3y + 6

cooper s algorithm1
Cooper's Algorithm
  • Cooper's algorithm is a decision procedure for Presburger arithmetic.
  • A non-Fourier-Motzkin alternative
  • It is also a quantifier elimination procedure, which also works from the inside out, eliminating existentials.
  • Its bigadvantage is that it doesn't need to normalize input formulas to DNF.
  • Description is of simplest possible implementation; many tweaks are possible.
preprocessing
Preprocessing
  • To eliminate the quantifier in x. P(x):
  • Normalize so that only operators are <, and divisibility (c|e), and negations only occur around divisibility leaves.
  • Compute least common multiplec of all coefficients of x, and multiply all terms by appropriate numbers so that in every term the coefficient of x is c.
  • Now apply

(x. P(cx))  (x. P(x) c|x).

preprocessing example
Preprocessing Example

x,y  Z. 0 < y x < y x +1 < 2y

(normalize)

x,y. 0 < y x < y  2y < x +2

(transform y to 2y everywhere)

 x,y. 0 < 2y  2x < 2y  2y < x +2

(give y unit coefficient)

 x,y. 0 < y  2x < y y < x +2  2|y

two cases
Two cases
  • How might x. P(x) be true?
  • Either:
    • there is a least xmaking Ptrue; or
    • there is no least x: however small you go, there will be a smaller xthat still makes Ptrue
  • Construct two formulas corresponding to both cases.
case 1 infinitely many small solutions
Case 1:Infinitely many small solutions
  • Look at the atomic formulas in P, and think about their values when xhas been made arbitrarily small:
    • x < e: if xgoes as small as we like, this will be T
    • e < x: if xgoes small, this will be
    • c | x+e: unchanged
  • This constructs P-, a formula where xonly occurs in divisibility terms.
  • Say  is the l.c.m.of the constants involved in divisibility terms. Need just test P- on 1,…, .
p example
P- example
  • For y. 0 < y  2x < y y < x +2  2|y
    • 0 < ywill become  as ygets small
    • 2x < yalso becomes  as y gets small
    • y < x +2 will be T as y gets small
    • 2|ydoesn't change (it tests if yis even or not)
  • So in this case,

P- (y)  ( T  2|y) 

case 2 least solution
Case 2: Least solution
  • The case when there is a least xsatisfying P.
  • For there to be a least xsatisfying P, it must be the case that one of the terms e < xis T, and that if xwas any smaller the formula would become .
  • Let B = {b | b < x is a term of P}
  • Need just consider P(b+j), where b Band 1 ≤j ≤.
  • Final elimination formula is:
example continued
Example continued
  • For

y. 0 < y  2x < y y < x +2  2|y

  • least solutions, if they exist, will be at

y = 1, y = 2, y = 2x +1, or y = 2x +2

  • The divisibility constraint eliminates two of these.
  • Original formula is equivalent to:

(2x < 2  0 < x)  (0 < 2x +2 x < 0)

Which is unsatisfiable.

symbolic representation

0 1 1

0,0,1

0 0 1

0,1,1

0

1

1

0

Symbolic Representation

We use finite automata to represent the integer solutions (in binary) of atomic linear constraints.

Example: The constraint x1x20 has solutions:

(0,0), (1,0), (1,1), (2,0), (2,1), (2,2), (3,0), …

The corresponding automaton

fa construction

0

1

2

FA Construction
  • Based on the basic state machine (BSM)
  • A finite state transducer for computing linear integer expressions

0 1

0 0

/ /

0 1

0

1

/

1

1

0

/

0

0 1

1 1

/ /

0 1

Example BSM

for 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

equality with 0
Equality With 0
  • Construct the BSM
  • 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
  • Size
inequality 0
Inequality (<0)
  • Construct the BSM
  • States with negative carries are accepting
  • No sink state
  • All other types of inequalities can be derived from this
  • Size
non zero constant term c
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)
  • Size
  • Alternatively, we can stack log2c BSMs that compare with 0 or 1, depending on the corresponding bit of c
boolean connectives
Boolean Connectives
  • For  compute the intersection
  • For  compute the union
  • For compute the complement
conjunction example

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 example
existential quantification
Existential Quantification
  • Project the quantified variables away
  • The resulting FA is non-deterministic
  • Determinization results in exponential blowup of the FA size
  • Rarely occurs in practice
  • For universal quantification use negation
constraints on all integers
Constraints on All Integers
  • Use 2’s complement arithmetic
  • The BSM remains the same
  • Create two clones for each state of the BSM one accepting and one rejecting
  • For =0, accepting states contain loops that write 0
  • For <0, accepting states contain loops that write 1
  • Size
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