Decidability and Complexity of Tree Share Formulashobor/Publications/2016/...Decidability and...

Decidability and Complexity of Tree Share Formulas

Decidability and Complexity of Tree ShareFormulas

Xuan Bach Le1 Aquinas Hobor1 Anthony W. Lin2

1 National University of Singapore 2 University of Oxford

December 14, 2016

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Introduction

tree(x , τ) ∧WRITE(τ)

tree(x , τ1) ∧ READ(τ1) tree(x , τ2) ∧ READ(τ2)

tree(x , τ) ∧WRITE(τ)

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Introduction

Shares

Shares are embedded into separation logic to reason aboutresource accounting:

addrτ1⊕τ2↦ val ⇔ addr

τ1↦ val ⋆ addrτ2↦ val

Allow resources to be split and shared in large scale:

tree(`, τ)def= (` = null ∧ emp) ∨ ∃`l , `r . (`

τ↦ (`l , `r) ⋆ tree(`l , τ) ⋆ tree(`r , τ))

tree(`, τ1 ⊕ τ2) ⇔ tree(`, τ1) ⋆ tree(`, τ2)

Share policies to reason about permissions for single writerand multiple readers:

WRITE(τ)READ(τ)

Write-Read

READ(τ)∃τ1, τ2. τ1 ⊕ τ2 = τ ∧

READ(τ1) ∧ READ(τ2)

Split-Read

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Introduction

Shares

WRITE(τ)READ(τ)

Write-Read

READ(τ)∃τ1, τ2. τ1 ⊕ τ2 = τ ∧

Split-Read

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Introduction

Shares

WRITE(τ)READ(τ)

Write-Read

READ(τ)∃τ1, τ2. τ1 ⊕ τ2 = τ ∧

Split-Read

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Introduction

Shares

Shares enable resource reasoning in concurrent programming

Rational numbers [Boyland (2003)]: disjointness problemmakes tree split equivalence false:¬(tree(`, τ1 ⊕ τ2)⇐ tree(`, τ1) ⋆ tree(`, τ2))Subsets of natural numbers [Parkinson (2005)]

Finite sets: recursion depth is finiteInfinite sets: intersections may not be in the model

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Introduction

Shares

Rational numbers [Boyland (2003)]: disjointness problemmakes tree split equivalence false:¬(tree(`, τ1 ⊕ τ2)⇐ tree(`, τ1) ⋆ tree(`, τ2))

Subsets of natural numbers [Parkinson (2005)]

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Introduction

Shares

Rational numbers [Boyland (2003)]: disjointness problemmakes tree split equivalence false:¬(tree(`, τ1 ⊕ τ2)⇐ tree(`, τ1) ⋆ tree(`, τ2))Subsets of natural numbers [Parkinson (2005)]

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Introduction

Tree Share Definition

A tree share τ ∈ T is a boolean binary tree equipped with thereduction rules R1 and R2 (their inverses are E1,E2 resp.):

τdef= ○ ∣ ● ∣ τ τ R1 ∶ ● ●↦ ● R2 ∶ ○ ○↦ ○

The tree domain T contains canonical trees which areirreducible with respect to the reduction rules.

● ● ○ ○ ○

Ri↦ ● ○ ○Ri↦ ● ○

○ is the empty tree, and ● the full tree.

READ(τ) def= τ ≠ ○ WRITE(τ) def= τ = ●

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Introduction

τdef= ○ ∣ ● ∣ τ τ R1 ∶ ● ●↦ ● R2 ∶ ○ ○↦ ○

● ● ○ ○ ○

Ri↦ ● ○ ○Ri↦ ● ○

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Introduction

τdef= ○ ∣ ● ∣ τ τ R1 ∶ ● ●↦ ● R2 ∶ ○ ○↦ ○

● ● ○ ○ ○

Ri↦ ● ○ ○Ri↦ ● ○

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Introduction

τdef= ○ ∣ ● ∣ τ τ R1 ∶ ● ●↦ ● R2 ∶ ○ ○↦ ○

● ● ○ ○ ○

Ri↦ ● ○ ○Ri↦ ● ○

READ(τ) def= τ ≠ ○ WRITE(τ) def= τ = ●5 / 30

Introduction

Tree Share Operators

The complement 2:

●1 ○2 ○3¬↦○1 ●2 ●3

The Boolean function union ⊔ and intersection ⊓ operator:

● ○ ●⊔○ ● ○

Ei↦●1 ●2 ○3 ●4

⊔○1 ●2 ○3 ○4

∨↦●1 ●2 ○3 ●4

Ri↦ ● ○ ●

● ○ ●⊓○ ● ○

Ei↦●1 ●2 ○3 ●4

⊓○1 ●2 ○3 ○4

∧↦○1 ●2 ○3 ○4

Ri↦○ ● ○

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Introduction

The complement 2:

●1 ○2 ○3¬↦○1 ●2 ●3

● ○ ●⊔○ ● ○

Ei↦●1 ●2 ○3 ●4

⊔○1 ●2 ○3 ○4

∨↦●1 ●2 ○3 ●4

Ri↦ ● ○ ●

● ○ ●⊓○ ● ○

Ei↦●1 ●2 ○3 ●4

⊓○1 ●2 ○3 ○4

∧↦○1 ●2 ○3 ○4

Ri↦○ ● ○

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Introduction

The complement 2:

●1 ○2 ○3¬↦○1 ●2 ●3

● ○ ●⊔○ ● ○

Ei↦●1 ●2 ○3 ●4

⊔○1 ●2 ○3 ○4

∨↦●1 ●2 ○3 ●4

Ri↦ ● ○ ●

● ○ ●⊓○ ● ○

Ei↦●1 ●2 ○3 ●4

⊓○1 ●2 ○3 ○4

∧↦○1 ●2 ○3 ○4

Ri↦○ ● ○

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Introduction

The complement 2:

●1 ○2 ○3¬↦○1 ●2 ●3

● ○ ●⊔○ ● ○

Ei↦●1 ●2 ○3 ●4

⊔○1 ●2 ○3 ○4

∨↦●1 ●2 ○3 ●4

Ri↦ ● ○ ●

● ○ ●⊓○ ● ○

Ei↦●1 ●2 ○3 ●4

⊓○1 ●2 ○3 ○4

∧↦○1 ●2 ○3 ○4

Ri↦○ ● ○

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Introduction

The complement 2:

●1 ○2 ○3¬↦○1 ●2 ●3

● ○ ●⊔○ ● ○

Ei↦●1 ●2 ○3 ●4

⊔○1 ●2 ○3 ○4

∨↦●1 ●2 ○3 ●4

Ri↦ ● ○ ●

● ○ ●⊓○ ● ○

Ei↦●1 ●2 ○3 ●4

⊓○1 ●2 ○3 ○4

∧↦○1 ●2 ○3 ○4

Ri↦○ ● ○

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Introduction

Properties of ⊔, ⊓ and 2

M = (⊔,⊓,2, ●, ○) forms a Boolean Algebra [Dockins et al.(2009)]:

B1a. (τ1 ⊓ τ2) ⊓ τ3 = τ1 ⊓ (τ2 ⊓ τ3) B1b. (τ1 ⊔ τ2) ⊔ τ3 = τ1 ⊔ (τ2 ⊔ τ3) (associativity)B2a. τ1 ⊓ τ2 = τ2 ⊓ τ1 B2b. τ1 ⊔ τ2 = τ2 ⊔ τ1 (commutativity)B3a. τ1 ⊓ (τ2 ⊔ τ3) = (τ1 ⊓ τ2) ⊔ (τ1 ⊓ τ3) B3b. τ1 ⊔ (τ2 ⊓ τ3) = (τ1 ⊔ τ2) ⊓ (τ1 ⊔ τ3) (distributivity)B4a. τ1 ⊓ (τ1 ⊔ τ2) = τ1 B4b. τ1 ⊔ (τ1 ⊓ τ2) = τ1 (absorption)B5a. τ ⊓ ● = τ B5b. τ ⊔ ○ = τ (identity)B6a. τ ⊓ τ = ○ B6b. τ ⊔ τ = ● (complement)

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Introduction

Tree Share Operators(cont.)

The partial join function ⊕:

τ1 ⊕ τ2 = τ3 def= τ1 ⊔ τ2 = τ3 ∧ τ1 ⊓ τ2 = ○

● ○ ○ ●⊕ ○ ● ○

Ei↦●1 ○2 ○3 ●4

⊕○1 ○2 ●3 ○4

⊕↦●1 ○2 ●3 ●4

Ri↦● ○ ●

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Introduction

τ1 ⊕ τ2 = τ3 def= τ1 ⊔ τ2 = τ3 ∧ τ1 ⊓ τ2 = ○

● ○ ○ ●⊕ ○ ● ○

Ei↦●1 ○2 ○3 ●4

⊕○1 ○2 ●3 ○4

⊕↦●1 ○2 ●3 ●4

Ri↦● ○ ●

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Introduction

τ1 ⊕ τ2 = τ3 def= τ1 ⊔ τ2 = τ3 ∧ τ1 ⊓ τ2 = ○

● ○ ○ ●⊕ ○ ● ○

Ei↦●1 ○2 ○3 ●4

⊕○1 ○2 ●3 ○4

⊕↦●1 ○2 ●3 ●4

Ri↦● ○ ●

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Introduction

Properties of ⊕

O = (T,⊕) for fractional permission in Separation Logic [Dockinset al. (2009)]:

J1. τ1 ⊕ τ2 = τ3 ⇒ τ1 ⊕ τ2 = τ ′3 ⇒ τ3 = τ ′3 (functionality)J2. τ1 ⊕ τ2 = τ2 ⊕ τ1 (commutativity)J3. τ1 ⊕ (τ2 ⊕ τ3) = (τ1 ⊕ τ2)⊕ τ3 (associativity)J4. τ1 ⊕ τ2 = τ3 ⇒ τ ′1 ⊕ τ2 = τ3 ⇒ τ1 = τ ′1 (cancellation)J5. ∃u. ∀τ. τ ⊕ u = τ (unit)J6. τ1 ⊕ τ1 = τ2 ⇒ τ1 = τ2 (disjointness)

J7. a⊕ b = z ∧ c ⊕ d = z ⇒ ∃ac, ad ,bc,bd .a b ac

ad bdbcc

ac ⊕ ad = a ∧ bc ⊕ bd = b ∧ ac ⊕ bc = c ∧ ad ⊕ bd = d (cross split)J8. τ ≠ ○⇒ ∃τ1, τ2. τ1 ≠ ○ ∧ τ2 ≠ ○ ∧ τ1 ⊕ τ2 = τ (infinite split)

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Introduction

The injection bowtie function & replaces ● with tree:

● ○ ○ ●& ○ ● =

○ ● ○ ○ ○ ●

Allow resources to be split uniformly:

τ1 ⋅ tree(`, τ2) def= tree(`, τ2 & τ1)

(τ1 ⊕ τ2) ⋅ tree(`, τ) ⇔ τ1 ⋅ tree(`, τ) ⋆ τ2 ⋅ tree(`, τ)τ1 ⋅ tree(`, τ2 & τ3) ⇔ (τ3 & τ1) ⋅ tree(`, τ2)

& can be hard to think about. Is this equation satisfiable?

v1 & v2 &○ ● ○ = ○ ● ○

& v2 & v1

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Introduction

● ○ ○ ●& ○ ● =

○ ● ○ ○ ○ ●

v1 & v2 &○ ● ○ = ○ ● ○

& v2 & v1

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Introduction

● ○ ○ ●& ○ ● =

○ ● ○ ○ ○ ●

v1 & v2 &○ ● ○ = ○ ● ○

& v2 & v1

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Introduction

● ○ ○ ●& ○ ● =

○ ● ○ ○ ○ ●

v1 & v2 &○ ● ○ = ○ ● ○

& v2 & v1

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Introduction

Properties of &

S = (&, ●) forms an Monoid with additional properties [Dockinset al. (2009)]:

M1. (τ1 & τ2) & τ3 = τ1 & (τ2 & τ3) (associativity)M2. τ & ● = ● & τ = τ (identity)M3. τ & ○ = ○ & τ = ○ (collapse point)M4. τ1 & (τ2 ◇ τ3) = (τ1 ◇ τ2) & (τ1 ◇ τ3), ◇ ∈ {⊓,⊔,⊕} (distributivity)M5. τ & τ1 = τ & τ2 ⇒ τ ≠ ○⇒ τ1 = τ2 (left cancellation)M6. τ1 & τ = τ2 & τ ⇒ τ ≠ ○⇒ τ1 = τ2 (right cancellation)

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Introduction

tree(x , τ)

(● ○ ⊕ ○ ●) ⋅ tree(x , τ)

● ○ ⋅ tree(x , τ) ○ ● ⋅ tree(x , τ)

(● ○ ⊕ ○ ●) ⋅ tree(x , τ)

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Decidability and Complexity results

Outline

1 Introduction

2 Decidability and Complexity resultsModel for Countable Atomless Boolean AlgebraFrom & to string concatenationTree Automatic Structures

3 Conclusion

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Model for Countable Atomless Boolean Algebra

Tree Shares as Countable Atomless Boolean Algebra

M = (⊔,⊓,2, ●, ○) is Countable Boolean Algebra because thedomain T is countable.

Atomless properties of M:Let τ1 ≠ τ2, we denote τ1 < τ2 iff τ1 ⊔ τ2 = τ2.M is atomless if for τ1 < τ3, there exists τ2 such thatτ1 < τ2 < τ3.Let τ1 =

○ ● ○ and τ3 = ● ○ then τ1 < τ3. We extend τ3

to the shape of τ1:

τ3Ei↦● ● ○

then replace one of the ● leaves of τ3 that is not in τ1 with

● ○:

● ● ○↦● ○ ● ○

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M = (⊔,⊓,2, ●, ○) is Countable Boolean Algebra because thedomain T is countable.Atomless properties of M:

Let τ1 ≠ τ2, we denote τ1 < τ2 iff τ1 ⊔ τ2 = τ2.

M is atomless if for τ1 < τ3, there exists τ2 such thatτ1 < τ2 < τ3.Let τ1 =

τ3Ei↦● ● ○

● ○:

● ● ○↦● ○ ● ○

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Let τ1 ≠ τ2, we denote τ1 < τ2 iff τ1 ⊔ τ2 = τ2.M is atomless if for τ1 < τ3, there exists τ2 such thatτ1 < τ2 < τ3.

Let τ1 =○ ● ○ and τ3 = ● ○ then τ1 < τ3. We extend τ3

τ3Ei↦● ● ○

● ○:

● ● ○↦● ○ ● ○

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Let τ1 ≠ τ2, we denote τ1 < τ2 iff τ1 ⊔ τ2 = τ2.M is atomless if for τ1 < τ3, there exists τ2 such thatτ1 < τ2 < τ3.Let τ1 =

τ3Ei↦● ● ○

● ○:

● ● ○↦● ○ ● ○

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Let τ1 ≠ τ2, we denote τ1 < τ2 iff τ1 ⊔ τ2 = τ2.M is atomless if for τ1 < τ3, there exists τ2 such thatτ1 < τ2 < τ3.Let τ1 =

τ3Ei↦● ● ○

● ○:

● ● ○↦● ○ ● ○

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Decidability of MThe first-order theory of M is decidable. The lower bound for itscomplexity is ⋃c<ω STA(∗,2cn,n) [Kozen (1980)].

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From & to string concatenation

Decidability of &

Decidability of S = (T,&)Let S = (T,&) then:

The existential theory of S is decidable in PSPACE.

The existential theory of S is NP-hard.

The general first-order theory over S is undecidable.

Decidability of S+ = (T/{○},&)Let S+ = (T/{○},&) then:

The existential theory of S+ is decidable in PSPACE.

The existential theory of S+ is NP-hard.

The general first-order theory over S+ is undecidable.

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Decidability of &

Decidability of S = (T,&)Let S = (T,&) then:

The existential theory of S is decidable in PSPACE.

The existential theory of S is NP-hard.

The general first-order theory over S is undecidable.

Decidability of S+ = (T/{○},&)Let S+ = (T/{○},&) then:

The existential theory of S+ is decidable in PSPACE.

The existential theory of S+ is NP-hard.

The general first-order theory over S+ is undecidable.

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Isomorphism between & and ⋅

To prove these results on S+ = (T/{○},&), we will construct anisomorphism between S+ equations and word equations.

In particular, we will transform & into string concatenation. Thetrick is that we must find an “alphabet” for S+ equations.

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Isomorphism between & and ⋅

To prove these results on S+ = (T/{○},&), we will construct anisomorphism between S+ equations and word equations.

In particular, we will transform & into string concatenation. Thetrick is that we must find an “alphabet” for S+ equations.

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Review of Word Equations

Let A = {a,b, . . .} be the finite set of alphabet andV = {v1, v2, . . .} the set of variables.

A word w is a string in (A ∪ V)∗. A word equation E is a pairof words w1 = w2.

E has a solution if there exists a homomorphismf ∶ A ∪ V ↦ A∗ that maps each letter in A to itself.

For example, the equation v1v2ab = bav2v1 has a solution:

f (v1) = b, f (v2) = a

The satisfiability problem of word equation: checking whethera word equation E has a solution.

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f (v1) = b, f (v2) = a

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f (v1) = b, f (v2) = a

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f (v1) = b, f (v2) = a

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f (v1) = b, f (v2) = a

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Word Equation Results

Decidability and Complexity of Word Equation

The satisfiability problem of word equation is decidable. Thelower bound is NP-complete while the upper bound isPSPACE [Plandowski (1999)].

The satisfiability of a system of word equations with regularconstraints vi ∈ REGi can be reduced to the satisfiability of asingle word equation [Schulz (1990)].

The existential theory of string concatenation is decidablewith lower bound NP-complete and upper bound PSPACE.The first-order theory of string concatenation is undecidable(forklore).

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Tree factorization

Prime trees

A tree τ ∈ T/{●, ○} is prime iff τ = τ1 & τ2 then either τ1 = ● orτ2 = ●.

A tree share τ can be factorized into prime trees using &:

○○ ● ○ ● ○

= ○ ● &○ ● ○ ● ○

= ○ ● & ○ ● ● & ● ○

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Tree factorization

Prime trees

○○ ● ○ ● ○

= ○ ● &○ ● ○ ● ○

= ○ ● & ○ ● ● & ● ○

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Tree factorization

Prime trees

○○ ● ○ ● ○

= ○ ● &○ ● ○ ● ○

= ○ ● & ○ ● ● & ● ○

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Tree factorization

Prime trees

○○ ● ○ ● ○

= ○ ● &○ ● ○ ● ○

= ○ ● & ○ ● ● & ● ○

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Tree factorization

Prime trees

○○ ● ○ ● ○

= ○ ● &○ ● ○ ● ○

= ○ ● & ○ ● ● & ● ○

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Tree factorization(cont.)

Unique factorization

Let τ ∈ T/{○, ●} then there exists a unique sequence of prime treesτ1, . . . , τn such that:

τ = τ1 & . . . & τnFurthermore, the factorization problem is in PTIME.

Proof sketch: By induction on the structure of the tree.

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Infinite alphabet

Let Tp ⊂ T be the set of prime trees then Tp is countablyinfinite.

Tp is our alphabet for the word equation but we need toreduce it to finite alphabet.

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Infinite alphabet

Let Tp ⊂ T be the set of prime trees then Tp is countablyinfinite.

Tp is our alphabet for the word equation but we need toreduce it to finite alphabet.

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Infinite alphabet(cont.)

From infinity to finite

Let Σ be the set of word equations and inequations over infinitealphabet A then Σ has a solution iff it has a solution over somefinite alphabet B ⊂ A such that:

1 A(Σ) ⊂ B2 ∣B∣ = ∣A(Σ)∣ + n where n is the number of inequations in Σ.

The choice of the extra letters in B is not important.

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Example

v1 & v2 &○ ● ○ = ○ ● ○

& v2 & v1

v1 & v2 & ● ○ & ○ ● = ○ ● & ● ○ & v2 & v1

v1v2ab = bav2v1 solution: v1 = b, v2 = a

○ ● & ● ○ & ● ○ & ○ ● = ○ ● & ● ○ & ● ○ & ○ ●

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Example

v1 & v2 &○ ● ○ = ○ ● ○

& v2 & v1

v1 & v2 & ● ○ & ○ ● = ○ ● & ● ○ & v2 & v1

○ ● & ● ○ & ● ○ & ○ ● = ○ ● & ● ○ & ● ○ & ○ ●

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Example

v1 & v2 &○ ● ○ = ○ ● ○

& v2 & v1

v1 & v2 & ● ○ & ○ ● = ○ ● & ● ○ & v2 & v1

v1v2ab = bav2v1

solution: v1 = b, v2 = a

○ ● & ● ○ & ● ○ & ○ ● = ○ ● & ● ○ & ● ○ & ○ ●

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Example

v1 & v2 &○ ● ○ = ○ ● ○

& v2 & v1

v1 & v2 & ● ○ & ○ ● = ○ ● & ● ○ & v2 & v1

○ ● & ● ○ & ● ○ & ○ ● = ○ ● & ● ○ & ● ○ & ○ ●

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Example

v1 & v2 &○ ● ○ = ○ ● ○

& v2 & v1

v1 & v2 & ● ○ & ○ ● = ○ ● & ● ○ & v2 & v1

○ ● & ● ○ & ● ○ & ○ ● = ○ ● & ● ○ & ● ○ & ○ ●

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Find a decidable fragment for &

Since the first-order theory of S = (T,&) is undecidable, we wantto find a decidable fragment of & together with (⊔,⊓, 2).

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Tree Automatic Structures

Connection to Tree Automatic Structures

Let &τ be the unary function over trees such that

&τ(τ ′) = τ ′ & τ

Example:&○ ●

(● ● ○) = ● ● ○

& ○ ● =○ ● ○ ● ○

Tree automatic structure

Let T = (T,⊔,⊓, 2,&τ) then T is tree automatic, i.e., its domainand relations are recognized by tree automata. Consequently, thefirst-order theory of T is decidable [Blumensath (1999);Blumensath and Gradel (2004)].

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&τ(τ ′) = τ ′ & τExample:

&○ ●

(● ● ○) = ● ● ○

& ○ ● =○ ● ○ ● ○

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&τ(τ ′) = τ ′ & τExample:

&○ ●

(● ● ○) = ● ● ○

& ○ ● =○ ● ○ ● ○

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Conclusion

Outline

1 Introduction

2 Decidability and Complexity resultsModel for Countable Atomless Boolean AlgebraFrom & to string concatenationTree Automatic Structures

3 Conclusion

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Conclusion

Contributions

We show that M = (⊔,⊓,2, ●, ○) forms a Countably AtomlessBoolean Algebra.

We reduce & to string concatenation.

We show T = (T,⊔,⊓, 2,&τ) is tree-automatic.

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Conclusion

Future Work

Complexity of (T,⊓,⊔) (∃-theory and first-order theory).

Decidability of (T,⊓,⊔,&) (∃-theory).

Complexity of T = (T,⊔,⊓, 2,&τ) (∃-theory and first-ordertheory).

Extension of word equation to tree equation.

Thank you! ,

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Conclusion

Proof sketch:

Let f be a solution of Σ. For each inequation w1 ≠ w2 to hold,it suffices to have a single position where they differs.

Therefore, there is at most one letter ai /∈ A(Σ) in eachinequation that we need to preserve.

For other letters bi /∈ A(Σ), we simply replace them with aletter in A(Σ).As a result, the new solution satisfies the alphabet constraint.

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Conclusion

Proof sketch:

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Conclusion

Proof sketch:

For other letters bi /∈ A(Σ), we simply replace them with aletter in A(Σ).

As a result, the new solution satisfies the alphabet constraint.

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Conclusion

Proof sketch:

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References

A. Blumensath. Automatic Structures. PhD thesis, RWTHAachen, 1999.

A. Blumensath and E. Gradel. Finite presentations of infinitestructures: automata and interpretations. In Theory ofComputer Systems, pages 641–674, 2004.

John Boyland. Checking interference with fractional permissions.In Static Analysis, 10th International Symposium, SAS 2003,San Diego, CA, USA, June 11-13, 2003, Proceedings, pages55–72, 2003. doi: 10.1007/3-540-44898-5 4. URLhttp://dx.doi.org/10.1007/3-540-44898-5_4.

Robert Dockins, Aquinas Hobor, and Andrew W. Appel. A freshlook at separation algebras and share accounting. InProgramming Languages and Systems, 7th Asian Symposium,APLAS 2009, Seoul, Korea, December 14-16, 2009. Proceedings,pages 161–177, 2009. doi: 10.1007/978-3-642-10672-9 13.URL http://dx.doi.org/10.1007/978-3-642-10672-9_13.

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Conclusion

Dexter Kozen. Complexity of boolean algebras. In TheoreticalComputer Science 10, pages 221–247, 1980.

Matthew Parkinson. Local Reasoning for Java. PhD thesis,University of Cambridge, 2005.

Wojciech Plandowski. Satisfiability of word equations withconstants is in PSPACE. In 40th Annual Symposium onFoundations of Computer Science, FOCS ’99, 17-18 October,1999, New York, NY, USA, pages 495–500, 1999. doi:10.1109/SFFCS.1999.814622. URLhttp://dx.doi.org/10.1109/SFFCS.1999.814622.

Klaus U. Schulz. Makanin’s algorithm for word equations - twoimprovements and a generalization. In Word Equations andRelated Topics, First International Workshop, IWWERT ’90,Tubingen, Germany, October 1-3, 1990, Proceedings, pages85–150, 1990. doi: 10.1007/3-540-55124-7 4. URLhttp://dx.doi.org/10.1007/3-540-55124-7_4.

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