Application of TBDs

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Application of TBDs. Technical development Ordered TBDs Operations on ordered TBDs ( ,,) Reduced ordered TBDs. M odel checking == Manipulation of TBDs. Ordered TBDs. p 1. p 2. p 3. p n. p n+1. Ordered TBDs. p n+1. - p n+1. u. x. y. z. Example. A. - B. - B. - C. - PowerPoint PPT Presentation

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Application of TBDs

Technical developmentOrdered TBDsOperations on ordered TBDs (,,)Reduced ordered TBDs

Model checking == Manipulation of TBDs

p2 p3p1

Ordered TBDs

pn pn+1

Ordered TBDs

-pn+1pn+1 u

y zx

A

- B - C- B

- CA

D- D

- D

- A

- D

Example

A

- B - C- B

- CA

D- D

- D

- A

- D

Example

A

- D - C- B

- DC

D- D

- D

- D

- D

Example

D

Operations

Negation

Conjunction

Abstraction

s

s

t

sx

Negation

u

y zx

- u

y zx

Conjunction

pn+1

u

u

Conjunction

- pn+1

u

- pn+1

Conjunction

a

y zx

a

y’ z’x’

a

y zxy’ z’x’

Conjunction

- a

y zx

- a

y’ z’x’

- a

pn+1xz

x’z’

yz

y’z’

Conjunction

a

y zx

- a

y’ z’x’

- a

zxx’z’

yy’z’

Conjunction

a

y zx

b/-b

y’ z’x’

a

zx yb/-b

Conjunction

- a

y zx

b/-b

y’ z’x’

a

zx yb/-bb/-bb/-b

Abstraction

An abstraction of a TBD on a label u =Conjunction of a simplication on –u and a simplication on u

A

- B - C- B

- CA

D- D

- D

- A

- D

Simplification on a Label u/-u

Select all non-terminal nodes labeled with singed/unsigned uReplace the selected nodes with a simpler one according to given rules

Simplification for a node with label u

u

y zx xz

- u

yz

u

Simplification for a node with label -u

- u

y zx xz

- u

yz

u

Abstraction on u

Given a TBD.

(1) Make a simplification on –u and a simplification on u(2) Make a conjunction of the two simplifications

zu

Existential Abstraction on u

zu

Properties

s1

s2

t2

s1

s1 s2 t1 t2

s2

s1

t1

s2uu

Observation: comp(s)

spn

::

p1

pn+1

Quantified Boolean Formulas

Consider formulas with variables p1, p2, …, pn

pi

pi

pn+1 pn+1- pn+1

s st s

x

φ φΨ x. φ

φ is valid comp( ) holds s

u

y - pn+1x

x

Reduced Ordered TBDs

- pn+1 pn+1

- pn+1 y pn+1

x x y

x y y

y x y

x pn+1

y pn+1

Non-terminal

y pn+1

x>0

Not allowed

u

T’ - zT

T zT

- z

T

Reduction Rules for u

- z T- z - z

z Tz T

u

T z- z

Reduction Rules for uu

z T- z

- z zT z T- z

T’ TT T’ Tz

T’ T’T z T’T

- u

T’ - zT

T zT

z

- T

Reduction Rules for -u

- z T- z z

z Tz - T

- u

T z- z

Reduction Rules for -u- u

z T- z

- z zT z T- z

T’ TT T’ Tz

T’ T’T z T’T

- u

T zz

u

- T z- zz zT - z z- T

u

~y - z~x

~y z~x

- u u

- z

~x

- z

~y

Explanation on Some Rules (Semantics)

u

~y - z~x

- u u

- z - z

Explanation on Some Rules (1)

- y z- x

- y zx

y z- xy zx

- xx

- xx

- y

- yy

y

u

~y - z~x

- u u

- z - z

Explanation on Some Rules

- y z- x

y zx

- x

x

- y

y

u

~y - z~x - z

Explanation on Some Rules

- x z- x

x zx

- x

x

-u/u

u

T’ - zT

T zT

- z

T

Explanation on Some Rules

u

~y - z~x

- u u

- z - z

Explanation on Some Rules (2)

- y z- x

- y zx

y z- xy zx

- xx

- xx

- y

- yy

y

u

~y - z~x

- u u

- z - z

Explanation on Some Rules

- y z- x - x - y

y zx x y

u

~y - z~x

-u/u

- z

Explanation on Some Rules

- x z- x - x

x zx x

u

T’ - zT

T zT

- z

T

Explanation on Some Rules

- z T- z - z

z Tz T

Boolean Diagram Model Checking

m variables for representing states2m variables for representing transitions

Let n=2m

Construct a TBD for the formula representing the initial statesConstruct a TBD for the formula representing the transition relation

The rest follows from the CTL model checking techniques

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