CCS: Processes and Equivalences

1
CCS Processes and Equivalences
Reading Peled 8.1, 8.2, 8.5

• Mads Dam

2
Finite State Automata

• Coffee machine A1
• Coffee machine A2
• Are the two machines the same?

1kr
1kr
tea
coffee
1kr
1kr
1kr
tea
coffee
3
Little Language Refresher

• Automata recognize strings, e.g.
• 1kr tea 1kr 1kr coffee
• Language set of strings
• A1 language recognized by A1
• ? empty string
• ST concatenation of S and T
• S T union of S and T
• S Iteration of S
• S SS...S

• Ardens rule
• The equation
• X SX T
• has solution
• X ST
• If ? ? S the solution is unique

0 or more times
4
Now compute...

• For A1
• p1 1kr p2 ?
• p2 1kr p3 tea p1
• p3 coffee p1
• ?
• p2 1kr coffee p1 tea p1
• ? (etc, use Arden)
• p1 (1kr 1kr coffee 1kr tea)
• q1
• BUT p1, q1 should be different!
• SO need new theory to talk about behaviour
  instead of acceptance

A1
1kr
p2
p3
1kr
tea
p1
coffee
q3
1kr
A2
1kr
1kr
q4
q1
q2
tea
coffee
5
Process Algebra

• Calculus of concurrent processes
• Main issues
• How to specify concurrent processes in an
  abstract way?
• Which are the basic relations between concurrency
  and non-determinism?
• Which basic methods of construction ( operators)
  are needed?
• When do two processes behave differently?
• When do they behave the same?
• Rules of calculation
• Replacing equals for equals
• Substitutivity
• Specification and modelling issues

6
Process Equivalences

• Sameness of behaviour equivalence of states
• Many process equivalences have been proposed (cf.
  Peled 8.5)
• For instance q1 q2 iff
• q1 and q2 have the same paths, or
• q1 and q2 may always refuse the same
  interactions, or
• q1 and q2 pass the same tests, or
• q1 and q2 have identical branching structure
• CCS Focus on bisimulation equivalence

7
Bisimulation Equivalence

• Intuition q1 q2 iff q1 and q2 have same
  branching structure
• Idea Find relation which will relate two states
  with the same transition structure, and make sure
  the relation is preserved
• Example

q1
q2
a
a
a
b
c
b
b
c
c
8
Strong Bisimulation Equivalence

• Given Labelled transition system T (Q,?,R)
• Looking for a relation S ? Q ? Q on states
• S is a strong bisimulation relation if whenever
  q1 S q2 then
• q1 ?? q1 implies q2 ?? q2 for some q2 such
  that q1 S q2
• q2 ?? q2 implies q1 ?? q1 for some q1 such
  that q1 S q2
• q1 and q2 are strongly bisimilar iff q1 S q2 for
  some strong bisimulation relation S
• q1 ? q2 q1 and q2 are strongly bisimilar
• Peled uses bis for

9
Example
q1
p0
a
b
a
a
q0
p1
a
b
b
a
q2
p2
a
a
Does q0 p0 hold?
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Example
q0
p0
a
a
a
q1
p1
q2
c
c
b
b
p2
q3
q4
p3
Does q0 p0 hold?
11
Weak Transitions

• What to do about internal activity?
• ? Transition label for activity which is not
  externally visible
• q )? q iff q q0 ?? q1 ?? ... ?? qn q, n ? 0
• q )? q iff q )? q
• q )? q iff q )? q1 ?? q2 )? q (? ? ?)
• Beware that )? )? (non-standard notation)
• Observational equivalence, v.1.0 Bisimulation
  equivalence with ? in place of ?
• Let q1 ¼ q2 iff q1 q2 with )? in place of !?
• Cumbersome definition Too many transitions q )?
  q to check

12
Observational Equivalence

• Let S µ Q ? Q. The relation S is a weak
  bisimulation relation if whenever q1 S q2 then
• q1 ?? q1 implies q2 ?? q2 for some q2 such
  that q1 S q2
• q2 ?? q2 implies q1 ?? q1 for some q1 such
  that q1 S q2
• q1 and q2 are observationally equivalent, or
  weakly bisimulation equivalent, if q1 S q2 for
  some weak bisimulation relation S
• q1 ? q2 q1 and q2 are observationally
  equivalent/weakly bisimilar
• Exercise Show that ¼ ¼

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Examples
a
a
?
¼
a
a
?
¼
a
?
c
a
b
a
¼
b
a
?
c
?
c
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Examples
b
b
a
?
a
b
?
All three are inequivalent
a
?
15
Calculus of Communicating Systems - CCS

• Language for describing communicating transition
  systems
• Behaviours as algebraic terms
• Calculus Centered on observational equivalence
• Elegant mathematical treatment
• Emphasis on process structure and modularity
• Recent extensions to security and mobile systems
• CSP - Hoare Communicating Sequential Processes
  (85)
• ACP - Bergstra and Klop Algebra of Communicating
  Processes (85)
• CCS - Milner Communication and Concurrency (89)
• Pi-calculus Milner (99), Sangiorgi and Walker
  (01)
• SPI-calculus Abadi and Gordon (99)
• Many recent successor for security and mobility
  (more in 2G1517)

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CCS - Combinators

• The idea 7 elementary ways of producing or
  putting together labelled transition systems
• Pure CCS
• Turing complete can express any Turing
  computable function
• Value-passing CCS
• Additional operators for value passing
• Definable
• Convenient for applications

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Actions

• Names a,b,c,d,...
• Co-names a,b,c,d,...
• Sorry Overbar not good in texpoint!
• a a
• In CCS, names and co-names synchronize
• Labels l Names co-names
• ? 2 Actions ? Labels ?
• Define ? by
• l l, and
• ? ?

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Flow Graphs

• Often draw static process structures using simple
  diagrams like
• Only labels listed may be made available at
  external interface
• Composing flowgraphs
• The chb names are now internal

cha0
chb0
cha1
chb1
cha0
chb0
chb0
chc0
cha1
chb1
chb1
chc1
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CCS Combinators, II

• Nil 0 No transitions
• Prefix ?.P in.out.0 ?in out.0 ?out 0
• Definition A P Buffer in.out.Buffer
• Buffer ?in out.Buffer ?out Buffer

in
out
in
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CCS Combinators, Choice

• Choice P Q BadBuf in.(?.0
  out.BadBuf)
• BadBuf ?in ?.0 out.BadBuf
• ?? 0 or
• ?out BadBuf
• Obs No priorities between ?s, as or as
• CCS doesnt know which labels represent input,
  and which output
• May use ? notation

in
?
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Example Boolean Buffer

• 2-place Boolean Buffer
• Flow graph
• Buf2 Empty 2-place buffer
• Buf20 2-place buffer holding a 0
• Buf21 Do. holding a 1
• Buf2_00 Do. Holding 00
• ... etc. ...

• Buf2 in0.Buf20 in1.Buf21
• Buf20 out0.Buf2
• in0.Buf200 in1.Buf201
• Buf21 ...
• Buf200 out0.Buf20
• Buf201 out0.Buf21
• Buf210 ...
• Buf211 ...

in0
out0
in1
out1
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Example Scheduler

• ai start taski
• bi stop taski
• Requirements
• a1,...,an to occur cyclically
• ai/bi to occur alternately beginning with ai
• Any a_i/b_i to be schedulable at any time,
  provided 1 and 2 not violated

• Let X ? 1,...,n
• Schedi,X
• i to be scheduled
• X pending completion
• Scheduler Sched1,?
• Schedi,X
• ?j?Xbj.Schedi,X-j, if i ? X
• ?j?Xbj.Schedi,X-j
• ai.Schedi1,X?i, if i ? X

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Example Counter

• Basic example of infinite-state system
• Count Count0
• Count0 zero.Count0 inc.Count1
• Counti1 inc.Counti2 dec.Counti
• Can do stacks and queues equally easy try it!

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CCS Combinators, Composition

• Composition P Q Buf1 in.comm.Buf1
• Buf2 comm.out.Buf2
• Buf1 Buf2
• ?in comm.Buf1 Buf2
• ?? Buf1 out.Buf2
• ?out Buf1 Buf2
• But also, for instance
• Buf1 Buf2
• ?comm Buf1 out.Buf2
• ?out Buf1 Buf2

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Composition, Example

• Buf1 in.comm.Buf1
• Buf2 comm.out.Buf2
• Buf1 Buf2

comm.Buf1Buf2
out
comm
in
comm
?
Buf1Buf2
comm.Buf1out.Buf2
comm
in
out
comm
Buf1out.Buf2
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CCS Combinators, Restriction

• Restriction P L Buf1 in.comm.Buf1
• Buf2 comm.out.Buf2
• (Buf1 Buf2) comm
• ?in comm.Buf1 Buf2
• ?? Buf1 out.Buf2
• ?out Buf1 Buf2
• But not
• (Buf1 Buf2) comm
• ?comm Buf1 out.Buf2
• ?out Buf1 Buf2

27
CCS Combinators, Relabelling

• Relabelling Pf Buf in.out.Buf1
• Buf1 Bufcomm/out
• in.comm.Buf1
• Buf2 Bufcomm/in
• comm.out.Buf2
• Relabelling function f must preserve complements
• f(a) f(a)
• And ?
• f(?) ?
• Relabelling function often given by name
  substitution as above

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Example 2-way Buffers

• 1-place 2-way buffer
• Bufab a.b-.Bufab b.a-.Bufab
• Flow graph
• LTS

• Bufbc
• Bufabc/b,c-/b-,b-/a,b/a-
• (Obs Simultaneous substitution!)
• Sys (Bufab Bufbc)\b,b-
• Intention
• What went wrong?

a
b-
a
b-
b-
c
a-
b
a-
b
b
c-
b-.Bufab
b-
a
Bufab
b
a-.Bufab
a-