SDS Foil no 1 Process Algebra Process Algebra – calculating with behaviours

SDSFoil no 1

Process Algebra

Process Algebra –calculating with behaviours

SDSFoil no 2

What

Languages for behaviour expressions:CCS, LOTOS, CSP, ACP, …with focus on concurrency and interactions

Semantics in terms of transition systems

Calculus with theorems and equivalences

Languages for behaviour expressions:CCS, LOTOS, CSP, ACP, …with focus on concurrency and interactions

Semantics in terms of transition systems

Calculus with theorems and equivalences

Language:

0,1, 2, 3, …

I, II, III, IV, ...

Language:

0,1, 2, 3, …

I, II, III, IV, ...

Semantics:

{}, {●} , {●●} , {●●●} , ...

0, succ(0), succ(succ(0)), ...

Semantics:

{}, {●} , {●●} , {●●●} , ...

0, succ(0), succ(succ(0)), ...

Calculus:

0+x = x

x+y = y+x

Calculus:

0+x = x

x+y = y+x

SDSFoil no 3

Why?

Reasoning about behaviours: understanding transforming and composing comparing deriving properties verifying and validatingUsing mathematics

Reasoning about behaviours: understanding transforming and composing comparing deriving properties verifying and validatingUsing mathematics

SDSFoil no 4

Reasoning about specification and design

•••

Specification

•••

Design

Verification

objects properties

Validation

Common representation

SDSFoil no 5

Rigid Synchronization Threes - RST

S2 = a; b + c; d S1=a; b; c S3 = a; (b + c ) S4 = a; b + a; c

a

b

c

a

b

c a

d

a

b

a

cb c

Behaviour S1 = action a followed by action b followed by action c

note the non-determinism here

SDSFoil no 6

Is S3 equivalent to S4?

S3 = a; (b + c ) S4 = a; b + a; c

a a

b

a

cb c

S3 ~ S4 ??

SDSFoil no 7

It depends ...

Traces: S3 and S4 represents the same traces

Traces: S3 and S4 represents the same traces

Observation: after a

• S3 will accept b and c

• S4 will accept b or c

Observation: after a

• S3 will accept b and c

• S4 will accept b or c

S3 = a; (b + c ) ~ S4 = a; b + a; c

a a

b

a

cb c

S3 = a; (b + c ) S4 = a; b + a; c

a a

b

a

cb c

Trace equivalence is less restrictive than observation equivalenceTrace equivalence is less restrictive than observation equivalence

1

SDSFoil no 8

Recursion

• Substitute ocurrence by definition

• loops represented by tail recursion

• Substitute ocurrence by definition

• loops represented by tail recursion

a

S1

S0 = a; S1 S1 = b + g; S3 S3 = d; S0

d

S0

S0 = a; (b + g; d; S0)

b

S3

g a

b g

d

S0

SDSFoil no 9

Algebra over RST

L- a set of actions

a, b, c L

Operations:

NIL: rst = NIL +: rst = rst + rst

prefix: rst = a; rst

Laws:

a + (b + c) = (a + b) + c

a + b = b + a

a + NIL = a

L- a set of actions

a, b, c L

Operations:

NIL: rst = NIL +: rst = rst + rst

prefix: rst = a; rst

Laws:

a + (b + c) = (a + b) + c

a + b = b + a

a + NIL = a

S T+ ST=

S

a

SDSFoil no 10

Parallel composition

B = b; DA = a; C

a

E = A | B

= a; (C | B) + b; (A | D)

C

b

D

a

C|B

b

A|D

A a Bb A a Bb

E

Note: a,b is visible outside E, but not connectedNote: a,b is visible outside E, but not connected

SDSFoil no 11

Interaction (or Synchronization)

• Complementing action pairs (a, a’) may interact

• Interaction is represented by - the unobservable action

• Interaction is synchronous or rigid - like pushing a button

• Complementing action pairs (a, a’) may interact

• Interaction is represented by - the unobservable action

• Interaction is synchronous or rigid - like pushing a button

B = a’; DA = a; C

a

E = A | B = a; (C | B) + a’; (A | D) + ;(C | D)

C

a’

D

a

C|B

a’

A|D

A a Ba’ Aa

E

C|D

Ba’

SDSFoil no 12

Restriction (or hiding)

• The restriction operator \ serves to hide gates from the environment

• \a means hiding a and a’ from the environment

• The restriction operator \ serves to hide gates from the environment

• \a means hiding a and a’ from the environment

B = a’; DA = a; C

a

E = (A | B)\a = (C | D)\a

C

a’

D

A a Ba’ Aa

E

C|D\a

Ba’

(A | B)\a = A||B

SDSFoil no 13

Basic CCS

• - set of unmarked actions

• ’ - set of marked actions

• L = U ’ - set of visible actions (labels)

• Act = U - total set of actions (labels)

• E - a process expression

• a – an action in L

• – an action in Act

Operators:

E = NIL

E = E + E choice (sum) composition

E = a; E prefix (sequence) composition

E = E | E parallel composition

E = E \ a restriction - hiding a and a’ in st

E = E || E combined composition and restriction = S | T \A

where A is the set of labels that have complements in S and T

• - set of unmarked actions

• ’ - set of marked actions

• L = U ’ - set of visible actions (labels)

• Act = U - total set of actions (labels)

• E - a process expression

• a – an action in L

• – an action in Act

Operators:

E = NIL

E = E + E choice (sum) composition

E = a; E prefix (sequence) composition

E = E | E parallel composition

E = E \ a restriction - hiding a and a’ in st

E = E || E combined composition and restriction = S | T \A

where A is the set of labels that have complements in S and T

SDSFoil no 14

What is the structure defined by

• E = R | S | T

• E = R | S | T \ b

• E = R | S | T \ b\d\g

• E = R || S || T

Where:

???

• E = R | S | T

• E = R | S | T \ b

• E = R | S | T \ b\d\g

• E = R || S || T

Where:

???

bS

a

g

Tb’

g’

dR

b’

d’

SDSFoil no 15

The results

• E = R | S | T

• E = R | S | T \ b

• E = R | S | T \ b\d\g

• E = R || S || T

• E = R | S | T

• E = R | S | T \ b

• E = R | S | T \ b\d\g

• E = R || S || T

bS

a

g

Tb’

g’

dR

b’d’

E

bS

a

g

Tb’

g’

dR

b’d’

E

bS

a

g

Tb’

g’

dR

b’d’

E

a

SDSFoil no 16

Try to expand yourself

Given:

A = a; A1 + b; A2

B= a’; B1 + b’; B2

expand:

A|B =

and:

A|B\a,b =

Given:

A = b; A + ; a’; A

B = ; b’; B + a; B

expand:

A||B = (Models two-way synchronised method calls)

any problems? any lessons to be learned?

Given:

A = a; A1 + b; A2

B= a’; B1 + b’; B2

expand:

A|B =

and:

A|B\a,b =

Given:

A = b; A + ; a’; A

B = ; b’; B + a; B

expand:

A||B = (Models two-way synchronised method calls)

any problems? any lessons to be learned?

SDSFoil no 17

Semantics - Labeled Transition Systems

A labelled transition system (LTS) is a 4-tuple <S, L, T, s0 > where:

• S is a (countable) non-empty set of states;

• L is a (countable) set of observable actions;

• T S x (L U { }) x S is the transition relation;

• s0 S is the initial state

There are several notations:

• Synchronization Trees are (graphical) notation for LTS

• CCS contains textual notation for LTS

• LOTOS is a textual notation for LTS

A labelled transition system (LTS) is a 4-tuple <S, L, T, s0 > where:

• S is a (countable) non-empty set of states;

• L is a (countable) set of observable actions;

• T S x (L U { }) x S is the transition relation;

• s0 S is the initial state

There are several notations:

• Synchronization Trees are (graphical) notation for LTS

• CCS contains textual notation for LTS

• LOTOS is a textual notation for LTS

SDSFoil no 18

and this LTS:and this LTS:

Transitions and LTS

These expressions:S0 = a; S1

S1 = b + g; S3

S3 = d; S0

define these transitions:

These expressions:S0 = a; S1

S1 = b + g; S3

S3 = d; S0

define these transitions:

10 SS a

S0 = a; S1

S1 = b + g; S3

NIL S3 = d; S0

S0

a

gb

d

31 SS gNILS b1

03 SS d

SDSFoil no 19

Transition Rules

• Transition rules define the precise semantics of each combinator: prefix, +, |, \ , /,

• Transition rules may be seen as rewrite rules

• You apply the rules systematically to find out what may happen

• Transition rules define the precise semantics of each combinator: prefix, +, |, \ , /,

• Transition rules may be seen as rewrite rules

• You apply the rules systematically to find out what may happen

α; E Eα

→

Act

E E’α

→

Com1 α

→

E|F E’|F

F F’α

→

Com2 α

→

E|F E|F’

E E’α

→

Com3

→

E|F E’|F’

F F’α’

→

Ej E’jα

→

Sumj

α

→

∑i in I Ej E’j (j in I)

E E’α

→

Resα

→

E\L E’\L

(α, α’ not in L)

SDSFoil no 20

Towards the expansion theorem

• only one transition at the time (interleaving semantics)

• include all possible transitions

• only one transition at the time (interleaving semantics)

• include all possible transitions

u = a’; u1

t | u = a; (t1 | u) + b; (t2 | u) + a’; (t | u1 ) + (t1 | u1)

a’ a

t1|u

a’

t2|u t1|u1

a b

t = a; t1 + b; t2

t1 t2 u1

b

t|u1

SDSFoil no 21

Expansion theorem

• t = i i ti

• u = k k uk

expansion without restrictions

• t | u = i i (ti | u) + k k (t | uk)

+ ik’ (ti | uk)

expansion with restrictions A = a1 \ a2 \ ...

• t | u \ A= i (i) i (ti | u) \ A

+k (k) k (t | uk) \A

+(ik’ ) (ti | uk) \A

general expansion

• t = (t 1 | t 2 | … | t m) \ A = … try it yourself

• t = i i ti

• u = k k uk

expansion without restrictions

• t | u = i i (ti | u) + k k (t | uk)

+ ik’ (ti | uk)

expansion with restrictions A = a1 \ a2 \ ...

• t | u \ A= i (i) i (ti | u) \ A

+k (k) k (t | uk) \A

+(ik’ ) (ti | uk) \A

general expansion

• t = (t 1 | t 2 | … | t m) \ A = … try it yourself

SDSFoil no 22

Why bother with expansion?

• Finding errors: deadlocks, livelocks, etc. (validating internal interfaces)

• Checking properties: liveness, recurrence, etc

• Finding the external behaviour of a composite system D

• Checking the behaviour of a composite system D against specifications S

• Comparing systems: is S and D equivalent? can S and D replace each other seen from E?

• Finding systems satisfying a specified behaviour

• Finding errors: deadlocks, livelocks, etc. (validating internal interfaces)

• Checking properties: liveness, recurrence, etc

• Finding the external behaviour of a composite system D

• Checking the behaviour of a composite system D against specifications S

• Comparing systems: is S and D equivalent? can S and D replace each other seen from E?

• Finding systems satisfying a specified behaviour

cD1

a

g

D2c’

g’

b’

Sa b’

DE

e a’

SDSFoil no 23

An example: specification and design of ATM

Specification:

ATM1 = gm; (m’; ATM1 + r’; ATM1)

Design:

ATMSYSTEM = ATM2 || BANK

where:

ATM2 = gm; ca’; (ok; m’; ATM2 + nok; r’; ATM2)

BANK = ca; (ok’; BANK + nok’; BANK)

Specification:

ATM1 = gm; (m’; ATM1 + r’; ATM1)

Design:


where:



ATM1gm

m’

r’

ATM2

gm

m’

r’BANK

ca

ok’nok’

ca’

oknok

ATMSYSTEM

gm

m’

r’

Is ATMSYSTEM equivalent to ATM1?

Is ATMSYSTEM compliant with ATM1?

can ATMSYSTEM replace ATM1?

SDSFoil no 24

Expanding the ATMSYSTEM


where:



ATMSYSTEM = gm; (ca’; (ok; m’; ATM2 + nok; r’; ATM2))|| BANK = gm; ok; m’; ATM2 + nok; r’; ATM2)|| (ok’; BANK + nok’; BANK)

=gm; m’; ATM2 )||BANK) + r’; ATM2)||BANK)) =gm; (m’; (ATM2 ||BANK) + r’; (ATM2||BANK)))

Is ATMSYSTEM equivalent to ATM1 = gm; (m’; ATM1 + r’; ATM1)?


where:



ATMSYSTEM = gm; (ca’; (ok; m’; ATM2 + nok; r’; ATM2))|| BANK = gm; ok; m’; ATM2 + nok; r’; ATM2)|| (ok’; BANK + nok’; BANK)

=gm; m’; ATM2 )||BANK) + r’; ATM2)||BANK)) =gm; (m’; (ATM2 ||BANK) + r’; (ATM2||BANK)))

Is ATMSYSTEM equivalent to ATM1 = gm; (m’; ATM1 + r’; ATM1)?

SDSFoil no 25

Some behaviours are more equal than others

• Strong bisimulation: equally labelled transitions to states with equivalent behaviour (including transitions):

P ~ Q iff, for all α in Act:

(i) Whenever P P’ then, for some Q’ , Q Q’ and P’ ~ Q’

(ii) Whenever Q Q’ then, for some P’ , P P’ and P’ ~ Q’

• Weak bisimulation, observation equivalence ≈

P ≈ Q iff, for all α in Act:

(i) Whenever P P’ then, for some Q’ , Q Q’ and P’ ≈ Q’

(ii) Whenever Q Q’ then, for some P’ , P P’ and P’ ≈ Q’

(ignoring transitions)

• Trace equivalence: all traces are equal

• Strong bisimulation: equally labelled transitions to states with equivalent behaviour (including transitions):

P ~ Q iff, for all α in Act:

(i) Whenever P P’ then, for some Q’ , Q Q’ and P’ ~ Q’

(ii) Whenever Q Q’ then, for some P’ , P P’ and P’ ~ Q’

• Weak bisimulation, observation equivalence ≈

P ≈ Q iff, for all α in Act:

(i) Whenever P P’ then, for some Q’ , Q Q’ and P’ ≈ Q’

(ii) Whenever Q Q’ then, for some P’ , P P’ and P’ ≈ Q’

(ignoring transitions)

• Trace equivalence: all traces are equal

α

→

α

→

α

→

α

→

α

→

α^ α

→

α^

SDSFoil no 26

- Laws

• a; B ≈ a; B

• B + B ≈ B

• a; (B + C) + a;C ≈ a; (B + C)

• B ≈ B

but

• B ~ B

• a; B ≈ a; B

• B + B ≈ B

• a; (B + C) + a;C ≈ a; (B + C)

• B ≈ B

but

• B ~ B

SDSFoil no 27

Relabelling

B [b / a, c / b ] - replace a by b and b by c in B

• enables meaningful reuse of behaviour:

B [b / a, c / b ] - replace a by b and b by c in B

• enables meaningful reuse of behaviour:

Ba b’

ThreePlaceBuffer = (B | B [b/a, c/b] | B [c/a, d/b]) \b\c

= ?? expand it yourself!

B = a b’ B

a

B

b’

Ba b’

Bb c’

Bc d’

ThreePlaceBuffer

SDSFoil no 28

Data in CCS

• Buffer = in(x); Buffer’(x)

• Buffer’(x) = out(x)’; Buffer

• where x belong to some fixed value set V

• Buffer = in(x); Buffer’(x)

• Buffer’(x) = out(x)’; Buffer

• where x belong to some fixed value set V

SDSFoil no 29

Comparing with SDL

What is different?

• Syntax

• Components

• Communication/synchronisation

• Data

• Time

• Object-orientation

• Semantics

Mapping SDL on CCS - is it possible? How?

What is different?

• Syntax

• Components

• Communication/synchronisation

• Data

• Time

• Object-orientation

• Semantics

Mapping SDL on CCS - is it possible? How?

SDSFoil no 30

MSC and CCS

• From exam 1998: how to model this in CCS?• From exam 1998: how to model this in CCS?

a

b

ZYX

MSC A

– how to model each instance in CCS?

– how to model the medium, i.e. the rule that output must precede input?

– what is the resulting synchronisation tree when using the expansion theorem?

– what if the medium is an SDL fifo queue? Any deadlocks possible?

Documents

SDS Foil no 1 Process Algebra Process Algebra – calculating with behaviours