parixit prasad december 4, 2013 parixit prasad | csa - iisc 1 deciding presburger arithmetic using...

Parixit Prasad | CSA - IISC

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PARIXIT PRASAD

December 4, 2013

Deciding Presburger Arithmetic Using Automata

Department of Computer Science and AutomatonIndian Institute of Science, Bangalore

December 4, 2013


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Structure of presentation

December 4, 2013

Recall Finite Automata Presburger Arithmetic

Sets of Integers recognized by a Finite Automaton

Presburger Formulas to Finite Automata Automata Associated with equalities Automata Associated with inequalities

Closure Properties and Automata for arbitrary formulas

Deciding satisfiability of a formula


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Finite Automata

December 4, 2013

A Finite Automaton is a tuple (Q,∑,δ,q0,F).› Q = Finite set of states› ∑ = Finite set of input symbols› δ (Qx∑xQ)› q0 Q (the start state)› F Q (Finite set of final states)

A run of the automaton on the word ω ∑* is a word ρ Q* , such that:

› ρ[1] = q0 › If ρ[i] = p and ρ[i+1] = q , then (p, ω[i],q) δ

A successful run ρ on ω is when ρ[n+1] F , where n=|ω|

A string is accepted by the automaton when there is a successful run on that string.

Class of languages accepted by finite automaton is closed under union, intersection and complementation.


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Presburger Arithmetic

December 4, 2013

A Basic term in presburger arithmetic consists of first order variable, constants 0 & 1 and sums of basic terms.

› x+x+x+y+1+1 is a basic term which can also be written as 3x+y+2

The Atomic formulas are equalities and inequalities between basic terms.

› x-2y = -2 is an atomic formula with equality.› x-2y ≤ -2 is an atomic formula with inequality.

The Formulas are first-order formulas build on atomic formulas using the connectives Λ(conjunction), V(disjunction), ¬(negation), x (existential quantification), x (universal quantification).

› x, y.(x=2y V x=2y+1) is a formula.› Τ is the empty conjunction (valid formula).› is the empty disjunction (unsatisfiable formula).


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The Free variables of a formula φ are defined as:› FV(φ) = variables which are not quantified in φ› FV(φ1 V φ2) = FV(φ1) U FV(φ2)› FV(x.φ) = FV(φ) \ {x}

The interpretation domain of the formulas is the set of natural

numbers N, with 0, 1, +, =, ≤ having their usual meaning.

A solution of the formula φ is an assignment of x1,x2,…,xn in N which satisfies φ, where FV(φ)={x1,x2,…,xn}

› ( x↦2, y↦1 ) is one solution of x+3 = 4y+1› ( x↦4 ) is one solution of y.(x=2y)

Presburger Arithmetic (Contd.)

December 4, 2013


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Sets of Integers recognized by Finite Automaton

December 4, 2013

Every Natural number can be written as a word in base-k (k≥1)

over the alphabet set {0,1,…,(k-1)}

› 29 in base-2 is written as

Since we want to read the word from LSB to MSB, we write the

word from right to left, with LSB in left and MSB in right.

› 29 in base-2 is written as


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Integers recognized by Finite Automaton(Contd.)

December 4, 2013

There is a mapping from {0,1,…,(k-1)}* to N denoted by ~k which

maps each word in base-k to its corresponding Natural number.

› 101110 maps to 29

n-tuples of natural number can be represented in base-k as a single

word over the alphabet set {0,1,…,(k-1)}n

› {0,1,…,(k-1)}3 set is {0,1,…,(k-1)} x {0,1,…,(k-1)} x {0,1,…,(k-1)}

› (13,29,6) (base-10) can be written in base-2 over the alphabet

set {0,1}3

as:

1 0 1 1 0

1 0 1 1 1

0 1 1 0 0

1 0 1 1 0 0

1 0 1 1 1 0

0 1 1 0 0 0

1 0 1 1 0 0 0

1 0 1 1 1 0 0

0 1 1 0 0 0 0


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Presburger Formula to Finite Automata

December 4, 2013

1. Build automaton for the atomic formulas in the given formula› Automata Associated with equalities› Automata Associated with inequalities

2. Use these automaton to create the automata for the given formula


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Automata associated with equality

December 4, 2013

For every atomic formula

a1x1 + a2x2 + … + anxn = b (where a1, a2,…, an, b Z)

we construct the automaton by repeating the following inference

rule untilno more states or transitions can be added:

where, i {0,1} clearly, {0,1}n


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Automata associated with equality (Contd.)

December 4, 2013

The resulting automata would be (Q, ∑, δ, qb, F)

› Q = {qi’s computed by the inference rule}

› ∑ = {0,1}n

› δ (Qx∑xQ) computed by the reference rule

› qb is the initial state

› F =


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Automata associated with equality (Example)

December 4, 2013

Consider the equation : x-2y = -2

› We have b=-2 , so the initial state q-2Q

Let c=b

Compute the solutions θ=(xθ1, yθ2) of the equation: x-2y =2 c

› The solutions are {(x , y ), (x , y )} for c=-2

Compute the new states for each θ and the transition to them from

c

› For θ=(x , y ), d = = = -1

• add q-1Q

• add (q-2,,q-1)δ

Similarly calculate the state and transition for other θ’s

Repeat for each new state until no more new states can be added.


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Automaton for: x-2y=-2

December 4, 2013

q-2

q-1

q0

00

01

10

11

00

q1

01

10

11


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Getting solutions from the automaton

December 4, 2013

For every successful run of the automaton, we get one solution.

For example, for the successful run ρ = q-2 q-1 q-1 q0 q1 q0

q-2 q-1 q-1 q0 q1 q0

› x = 01101 = 22

› y = 00110 = 12

› This is one solution to the equation: x-2y=-2


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Observations on the automaton

December 4, 2013

For an automaton for: a1x2+a2x2+…+anxn=b

1. The absolute value of the states are bounded by:

max(|b|,

2. Any successful run starting from qc gives a solution to:

a1x2+a2x2+…+anxn=c


Automata associated with inequality

December 4, 2013

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Every inequality can be written in the form of:

a1x1 + a2x2 + … + anxn ≤ b (where a1, a2,…, an, b Z)

we construct the automaton by repeating the following inference

rule untilno more states or transitions can be added:

where, i {0,1} clearly, {0,1}n

we apply this formula for all possible combinations of


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Automata associated with inequality (Contd.)

December 4, 2013

The resulting automata would be (Q, ∑, δ, qb, F)

› Q = {qi’s computed by the inference rule}

› ∑ = {0,1}n

› δ (Qx∑xQ) computed by the reference rule

› qb is the initial state

› F = {qc|c≥0}


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Automata associated with inequality (Example)

December 4, 2013

Consider the equation : x-2y ≤ -2

› We have b=-2 , so the initial state q-2Q

Let c=b

Take θ=(xθ1, yθ2), where θ1, θ2 {0,1}

› θ {(x , y ), (x , y ) , (x , y ) , (x , y )}

Compute the new states for each θ and the transition to them from

c

› For θ=(x , y ), d = = = -1

• add q-1Q

• add (q-2,,q-1)δ

Similarly calculate the state and transition for other θ’s

Repeat for each new state until no more new states can be added.


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Automaton for: x-2y≤-2

December 4, 2013

q-2

q-1

q0

00

01

,10

,11

00

q1

01

,10

,

10

00

01

,11

00

,11011

0


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Getting solutions from the automaton

December 4, 2013

For every successful run of the automaton, we get one solution.

For example, for the successful run ρ = q-2 q-1 q-1 q0 q1 q0

q-2 q-1 q-1 q0 q1 q0

› x = 11001 = 19

› y = 10110 = 13

› This is one solution to the equation: x-2y≤-2


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Closure Properties and automata for arbitrary formula

December 4, 2013

Let Aφ(x1,x2,…,xn) be an automaton over alphabet {0,1}n which

accepts the solutions of the formula φ.

› Then FV(φ) {x1,x2,…,xn}

› xi accepting or not accepting a word in the automaton will not

depend on the ith track.

Exercise:

How to compute Aφ(x1,x2,…,xn,y)? Given Aφ(x1,x2,…,xn) and y .


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Automata for arbitrary formulas (Contd.)

December 4, 2013

Given 2 Automata A() and A()

› A can be computed by the intersection of the two given

automaton

› A can be computed by the union of the two given automaton

› We have to make sure that contains FV(φ1) U FV(φ2), which can

be done easily.

Given Automata Aφcan construct A¬φ by taking complement of Aφ

A can be computed from Aφ by projection, i.e. in Aφ simply forgetting

about the track corresponding to x in the transition labels.

Since is logically equivalent to ¬, we can easily compute A Aφ


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An example of Projection

December 4, 2013

Automaton for: x-2y = -2)

q-2

q-1

q0

0¿

0¿

1¿

1¿

0¿

q1

0¿

1¿

1¿


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Deciding satisfiability of a formula

December 4, 2013

We can apply induction on the Presburger formula φ to compute an

automaton φ which accepts the set of solutions of φ.

By computing φ , deciding satisfiability of φ reduces to deciding the

emptiness of the language of φ, which can be done in linear time

w.r.t number of states.


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Thank you

December 4, 2013

Questions…

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