Testing Nondeterministic and Probabilistic...

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Background Testing theory Nondeterminism Probabilities

Testing Nondeterministic and ProbabilisticProcesses

Matthew Hennessy

(joint work with Yuxin Deng, Rocco DeNicola, Rob van Glabbeek, Carroll Morgan, Chenyi Zhang)

BASICS, Shanghai October 09

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Outline

Background why bother ?

Testing theory

Testing nondeterministic processes

Testing Probabilistic and nondeterministic processes

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Outline


Testing theory



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Background

Goal: Specification and proof methodologies forprobabilistic concurrent systems

Nondeterminism + Probability – why necessary ?

◮ “Nondeterminism” intrinsic to specification development a la CSP

◮ underspecified components expressed using “nondeterminism”

COMP ⊓ OPTION ≤ COMP

underspecified more specified

◮ Analysis of concurrent systems requires “nondeterminism”

⊓ - internal choice of CSP

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Background









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Background









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Background









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Analysis of concurrent systems

Sys1:

Sys1 ⇐ (new s)(A | Sw)

A ⇐ up.U + s?down.D

Sw ⇐ s!stop

A Sws

up

down

Sys2:

Sys2 ⇐ (new s)(B | Sw)

B ⇐ s?(up.U + down.D) + s?down.D

Sw ⇐ s!stop

B Sws

up

down

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Analysis of concurent systems

In CSP theory:

Sys1 ≈ Sys2

semantically equivalent

Both equivalent to the nondeterministic

(up.U + down.D) ⊓ down.D

concurrency = nondeterminism + interleaving

probabilistic concurrency= probability + nondeterminism + interleaving

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In CSP theory:

Sys1 ≈ Sys2






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In CSP theory:

Sys1 ≈ Sys2






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In CSP theory:

Sys1 ≈ Sys2






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Outline


Testing theory



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Testing scenario

◮ a set of processes Proc

◮ a set of tests T

◮ a set of outcomes O

◮ Apply : T × Proc → P+(O) – the non-empty set of possible

results of applying a test to a process

Comparing sets of outcomes:

◮ O1 ⊑Ho O2 if for every o1 ∈ O1 there exists some o2 ∈ O2

such that o1 ≤ o2

◮ O1 ⊑Sm O2 if for every o2 ∈ O2 there exists some o1 ∈ O1

such that o1 ≤ o2

o1 ≤ o2 : means o2 is as least as good as o1

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Testing scenario

◮ a set of processes Proc

◮ a set of tests T

◮ a set of outcomes O

◮ Apply : T × Proc → P+(O) – the non-empty set of possible

results of applying a test to a process

Comparing sets of outcomes:

◮ O1 ⊑Ho O2 if for every o1 ∈ O1 there exists some o2 ∈ O2

such that o1 ≤ o2

◮ O1 ⊑Sm O2 if for every o2 ∈ O2 there exists some o1 ∈ O1

such that o1 ≤ o2

o1 ≤ o2 : means o2 is as least as good as o1

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Testing preorders

◮ P ⊑may Q if Apply(T , P) ⊑Ho Apply(T , Q) for every test T

◮ P ⊑must Q if Apply(T , P) ⊑Sm Apply(T , Q) for every test T

Standard testing:

Use as outcomes O = {⊤, ⊥} with ⊥ ≤ ⊤

Comparisions:

Possible outcome sets: {⊥} {⊥,⊤} {⊤}

◮ May: {⊥} <Ho {⊥,⊤} =Ho {⊤}

◮ Must: {⊥} =Sm {⊥,⊤} <Sm {⊤}

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Testing preorders



Standard testing:


Comparisions:


◮ May: {⊥} <Ho {⊥,⊤} =Ho {⊤}

◮ Must: {⊥} =Sm {⊥,⊤} <Sm {⊤}

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Testing preorders



Standard testing:


Comparisions:


◮ May: {⊥} <Ho {⊥,⊤} =Ho {⊤}

◮ Must: {⊥} =Sm {⊥,⊤} <Sm {⊤}

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Testing preorders



Probabilistic testing:

Use as O the unit interval [0, 1]Intuition: with 0 ≤ p ≤ q ≤ 1, passing a test with probability q

better than passing with probability p

Comparisions:

◮ May: O1 ⊑Ho O2 is every possibility p ∈ O1 can be improvedon by some q ∈ O2

◮ Must: O1 ⊑Sm O2 if every possibility q ∈ O2 is animprovement on some p ∈ O1

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Testing preorders



Probabilistic testing:

Use as O the unit interval [0, 1]Intuition: with 0 ≤ p ≤ q ≤ 1, passing a test with probability q

better than passing with probability p

Comparisions:

◮ May: O1 ⊑Ho O2 is every possibility p ∈ O1 can be improvedon by some q ∈ O2

◮ Must: O1 ⊑Sm O2 if every possibility q ∈ O2 is animprovement on some p ∈ O1

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Outline


Testing theory



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Nondeterministic processes

Intensional semantics:

A process is a state in an LTS

Labelled Transition Systems:

〈S , Actτ ,−→〉

◮ S - states

◮ −→ ⊆ S × Actτ × S

s1µ−→ s2: process s1 can perform action µ and continue as s2

s1τ−→ s2 special internal action

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Nondeterministic processes


A process is a state in an LTS

Labelled Transition Systems:

〈S , Actτ ,−→〉

◮ S - states

◮ −→ ⊆ S × Actτ × S

s1µ−→ s2: process s1 can perform action µ and continue as s2

s1τ−→ s2 special internal action

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Process calculi: Syntax for LTSs

Example process calculus CCS:

◮ 0 Do nothing

◮ µ.P Perform µ then act as P

◮ P | Q Run P and Q in parallel . . . communicating via complementary actions

◮ P + Q Nondeterministic choice between P and Q

◮ recursive definitions D ⇐ P

ActionsP µ−→ Q defined inductively

lots of other process calculi

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◮ 0 Do nothing







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◮ 0 Do nothing







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◮ 0 Do nothing







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Tests:Any process which may contain new report success action/state ω

T ⇐ a.ω + b.T + c . 0:

◮ requests an a action . . .

◮ after an arbitrary number of b actions . . .

◮ without doing any c action

Applying test T to process P :

◮ Run the combined process (T | P)

◮ Each execution succeeds or fails

◮ Each execution contributes ⊤ or ⊥ to Apply(T , P)

Nondeterministic (T | P) resolved to a set of deterministicexecutions

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T ⇐ a.ω + b.T + c . 0:









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T ⇐ a.ω + b.T + c . 0:









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Example

Test:T ⇐ a.ω + b.T + c . 0 Process:P ⇐ b.(a.Q + b.P)

Deterministic executions:

T | P τ−→bτ−→a ω | ⊤

T | P τ−→bτ−→b

τ−→a ω | ⊤


τ−→bτ−→a ω | ⊤

T | P . . . ⊤


τ−→b . . . . . .τ−→b . . . ⊥

Result:Apply(T , P) = {⊥, ⊤}

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Example



T | P τ−→bτ−→a ω | ⊤


τ−→a ω | ⊤


τ−→bτ−→a ω | ⊤

T | P . . . ⊤


τ−→b . . . . . .τ−→b . . . ⊥


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Example



T | P τ−→bτ−→a ω | ⊤


τ−→a ω | ⊤


τ−→bτ−→a ω | ⊤

T | P . . . ⊤


τ−→b . . . . . .τ−→b . . . ⊥


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Example



T | P τ−→bτ−→a ω | ⊤


τ−→a ω | ⊤


τ−→bτ−→a ω | ⊤

T | P . . . ⊤


τ−→b . . . . . .τ−→b . . . ⊥


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May-testing nondeterministic processes

Divergence not important:

P + τ.τ∞ ≃may P

Choice not important:

a

b c

≃may

a a

b c

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May-testing nondeterministic processes

Divergence not important:

P + τ.τ∞ ≃may P

Choice not important:

a

b c

≃may

a a

b c

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Must testing nondeterministic processes

Divergence catastrophic:

P + τ.τ∞ ≃must τ

∞

Internal choices not very important:

a

τ τ

τb

c

bc

≃must

a

τ τ

b c

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Must testing nondeterministic processes

Divergence catastrophic:

P + τ.τ∞ ≃must τ

∞

Internal choices not very important:

a

τ τ

τb

c

bc

≃must

a

τ τ

b c

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Characterising nondeterministic processes

Ingredients:

◮ Traces: a1a2 . . . an in Traces(P) whenever

P τ−→∗ a1−→ τ−→∗. . . . . .

τ−→∗ an−→ τ−→∗ P ′

◮ Divergences/convergences: P ⇓ whenever there is no infiniteexecution

P τ−→ τ−→ . . . . . .τ−→ . . . . . .

◮ Failures/Acceptances: P accA whenever P ⇓ and

P τ−→∗ P ′ implies P ′ a=⇒ for some a in A

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Characterising nondeterministic processes

Ingredients:

◮ Traces: a1a2 . . . an in Traces(P) whenever

P τ−→∗ a1−→ τ−→∗. . . . . .

τ−→∗ an−→ τ−→∗ P ′

◮ Divergences/convergences: P ⇓ whenever there is no infiniteexecution

P τ−→ τ−→ . . . . . .τ−→ . . . . . .

◮ Failures/Acceptances: P accA whenever P ⇓ and

P τ−→∗ P ′ implies P ′ a=⇒ for some a in A

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Outline


Testing theory



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Probabilistic and nondeterministic processes


A process is a distribution in an pLTS

Probabilistic Labelled Transition Systems:

〈S , Actτ ,−→〉

◮ S - states

◮ −→ ⊆ S × Actτ ×D(S)

D(S): Mappings ∆ : S → [0, 1] with∑

s∈S ∆(s) = 1

s1µ−→ ∆: process s1

◮ can perform action µ

◮ with probability ∆(s2) it continues as process s2

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〈S , Actτ ,−→〉

◮ S - states

◮ −→ ⊆ S × Actτ ×D(S)


s∈S ∆(s) = 1




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〈S , Actτ ,−→〉

◮ S - states

◮ −→ ⊆ S × Actτ ×D(S)


s∈S ∆(s) = 1




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Example probabilistic processes

Pτ τ

12 1

2

τ

a

12 1

2

a

What is the probability of action a happening?

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Example probabilistic processes

Pτ τ

12 1

2

τ

a

12 1

2

a

What is the probability of action a happening?

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Probabilistic process calculi: Syntax for pLTSs

Example process calculus pCCS:

State terms S , T :

◮ 0

◮ µ.P

◮ S | T S + T

◮ recursive definitions

Process terms P, Q:

◮ S

◮ P p⊕ Q probabilistic choice between P and Q

Actionss µ−→ ∆ defined inductively

process terms are distributions over states

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State terms S , T :

◮ 0

◮ µ.P

◮ S | T S + T


Process terms P, Q:

◮ S




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State terms S , T :

◮ 0

◮ µ.P

◮ S | T S + T


Process terms P, Q:

◮ S




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State terms S , T :

◮ 0

◮ µ.P

◮ S | T S + T


Process terms P, Q:

◮ S




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State terms S , T :

◮ 0

◮ µ.P

◮ S | T S + T


Process terms P, Q:

◮ S




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Testing probabilistic processes

Tests:Any (prob. . . ) process which may contain report success

action/state ω

a.ω 14⊕ (b + c .ω):

◮ 25% of time requests an a action

◮ 75% requests a c action

◮ 75% requires that b is not possible in a must test


◮ Execute the combined process (T | P)

◮ Each execution contributes some probability p to Apply(T , P)

◮ Each execution is a deterministic resolution of (T | P)

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Testing probabilistic processes

Tests:Any (prob. . . ) process which may contain report success

action/state ω

a.ω 14⊕ (b + c .ω):

◮ 25% of time requests an a action

◮ 75% requests a c action

◮ 75% requires that b is not possible in a must test


◮ Execute the combined process (T | P)

◮ Each execution contributes some probability p to Apply(T , P)

◮ Each execution is a deterministic resolution of (T | P)

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Executing probabilistic nondeterministic processes (T | P)

◮ Choice points occur during an execution

◮ choices are made◮ statically◮ or dynamically

◮ choices are made◮ by schedulers◮ adversaries◮ policies

Executions:

◮ give deterministic behaviour - but may be probabilistic

◮ contribute a probability to Apply(T , P)

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Executions:



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Executions:



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Executions:



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Example of executions

T | P

s1 s2

tbd tgd

τ

34

14

τ

τ τ

ω

Static Policies:

pp1 : s1 7−→ s0

pp2 : s1 7−→ tbd

Possible results:

Using pp1:1

4+

3

4·1

4+ (

3

4)2 ·

1

4+ . . . + . . . = 1

Using pp2:1

4+

3

4· 0 =

1

4

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T | P

s1 s2

tbd tgd

τ

34

14

τ

τ τ

ω

Static Policies:

pp1 : s1 7−→ s0

pp2 : s1 7−→ tbd

Possible results:

Using pp1:1

4+

3

4·1

4+ (

3

4)2 ·

1

4+ . . . + . . . = 1

Using pp2:1

4+

3

4· 0 =

1

4

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T | P

s1 s2

tbd tgd

τ

34

14

τ

τ τ

ω

Static Policies:

pp1 : s1 7−→ s0

pp2 : s1 7−→ tbd

Possible results:

Using pp1:1

4+

3

4·1

4+ (

3

4)2 ·

1

4+ . . . + . . . = 1

Using pp2:1

4+

3

4· 0 =

1

4

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More executionsT | P

s1 s2

tbd tgd

τ

34

14

τ

τ τ

ω

Arbitrary policies:combinations of

pp1 : s1 7−→ s0

pp2 : s1 7−→ tbd

Possible results:

Using pp1: 1

Using pp2:1

4

In general: p · 1 + (1 − p) ·1

4for some 0 ≤ p ≤ 1

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More executionsT | P

s1 s2

tbd tgd

τ

34

14

τ

τ τ

ω

Arbitrary policies:combinations of

pp1 : s1 7−→ s0

pp2 : s1 7−→ tbd

Possible results:

Using pp1: 1

Using pp2:1

4

In general: p · 1 + (1 − p) ·1

4for some 0 ≤ p ≤ 1

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Formalising executions I

From pLTSS to LTSs

∆ µ−→ Θ

◮ ∆ represents a cloud of possible process states

◮ each possible state must be able to perform µ

◮ all possible residuals combine to Θ

Examples:

◮ (a.b + a.c) 12⊕ a.d a−→ b 1

2⊕ d

◮ (a.b + a.c) 12⊕ a.d a−→ (b 1

2⊕ c) 1

2⊕ d

◮ (a.b + a.c) 12⊕ a.d a−→ (b p⊕ c) 1

2⊕ d

◮ (τ.a + τ.b) 12⊕ (τ.a + τ.c) τ−→ a 1

2⊕ (b 1

2⊕ c)

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From pLTSS to LTSs

∆ µ−→ Θ




Examples:

◮ (a.b + a.c) 12⊕ a.d a−→ b 1

2⊕ d

◮ (a.b + a.c) 12⊕ a.d a−→ (b 1

2⊕ c) 1

2⊕ d

◮ (a.b + a.c) 12⊕ a.d a−→ (b p⊕ c) 1

2⊕ d

◮ (τ.a + τ.b) 12⊕ (τ.a + τ.c) τ−→ a 1

2⊕ (b 1

2⊕ c)

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From pLTSS to LTSs

∆ µ−→ Θ




Examples:

◮ (a.b + a.c) 12⊕ a.d a−→ b 1

2⊕ d

◮ (a.b + a.c) 12⊕ a.d a−→ (b 1

2⊕ c) 1

2⊕ d

◮ (a.b + a.c) 12⊕ a.d a−→ (b p⊕ c) 1

2⊕ d

◮ (τ.a + τ.b) 12⊕ (τ.a + τ.c) τ−→ a 1

2⊕ (b 1

2⊕ c)

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From pLTSS to LTSs

∆ µ−→ Θ




Examples:

◮ (a.b + a.c) 12⊕ a.d a−→ b 1

2⊕ d

◮ (a.b + a.c) 12⊕ a.d a−→ (b 1

2⊕ c) 1

2⊕ d

◮ (a.b + a.c) 12⊕ a.d a−→ (b p⊕ c) 1

2⊕ d

◮ (τ.a + τ.b) 12⊕ (τ.a + τ.c) τ−→ a 1

2⊕ (b 1

2⊕ c)

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From pLTSS to LTSs

∆ µ−→ Θ




Examples:

◮ (a.b + a.c) 12⊕ a.d a−→ b 1

2⊕ d

◮ (a.b + a.c) 12⊕ a.d a−→ (b 1

2⊕ c) 1

2⊕ d

◮ (a.b + a.c) 12⊕ a.d a−→ (b p⊕ c) 1

2⊕ d

◮ (τ.a + τ.b) 12⊕ (τ.a + τ.c) τ−→ a 1

2⊕ (b 1

2⊕ c)

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From pLTSS to LTSs: formally

∆ µ−→ Θ

whenever

◮ ∆ =∑

i∈I pi · si , I a finite index set

◮ For each i ∈ I there is a distribution Θi s.t. siµ−→ Θi

◮ Θ =∑

i∈I pi · Θi

◮

∑

i∈I pi = 1

Note: in decomposition∑

i∈I pi · si states si are not necessarilyunique

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From pLTSS to LTSs: formally

∆ µ−→ Θ

whenever

◮ ∆ =∑

i∈I pi · si , I a finite index set

◮ For each i ∈ I there is a distribution Θi s.t. siµ−→ Θi

◮ Θ =∑

i∈I pi · Θi

◮

∑

i∈I pi = 1

Note: in decomposition∑

i∈I pi · si states si are not necessarilyunique

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Formalising executions IIExecuting (T | P) to Θ: (T | P) =⇒≻ Θ

(T | P) = ∆→

0 + ∆stop0

∆→

0τ−→ ∆→

1 + ∆stop1

. . . . . .

∆→

kτ−→ ∆→

(k+1)+ ∆stop(k+1)

. . . . . .

. . . . . .

Total: Θ =∑

∞

k=0 ∆stopk

◮ ∆stop: all states in ∆ which note: subdistributions

◮ are successful s ω−→

◮ or are stuck s 6 τ−→

◮ ∆→: all other states, which can proceed s τ−→

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(T | P) = ∆→

0 + ∆stop0

∆→

0τ−→ ∆→

1 + ∆stop1

. . . . . .

∆→

kτ−→ ∆→

(k+1)+ ∆stop(k+1)

. . . . . .

. . . . . .

Total: Θ =∑

∞

k=0 ∆stopk





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(T | P) = ∆→

0 + ∆stop0

∆→

0τ−→ ∆→

1 + ∆stop1

. . . . . .

∆→

kτ−→ ∆→

(k+1)+ ∆stop(k+1)

. . . . . .

. . . . . .

Total: Θ =∑

∞

k=0 ∆stopk





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(T | P) = ∆→

0 + ∆stop0

∆→

0τ−→ ∆→

1 + ∆stop1

. . . . . .

∆→

kτ−→ ∆→

(k+1)+ ∆stop(k+1)

. . . . . .

. . . . . .

Total: Θ =∑

∞

k=0 ∆stopk





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(T | P) = ∆→

0 + ∆stop0

∆→

0τ−→ ∆→

1 + ∆stop1

. . . . . .

∆→

kτ−→ ∆→

(k+1)+ ∆stop(k+1)

. . . . . .

. . . . . .

Total: Θ =∑

∞

k=0 ∆stopk





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Applying tests to processes: Apply(T , P)

◮ find all executions from (T | P): (T | P) =⇒≻ Θ

◮ calculate contribution of each Θ

Contribution of Θ:

◮ all states in Θ are successful s ω−→ or stuck s 6 τ−→

◮ V(Θ) =∑

{Θ(s) | s ω−→} weight of success

Apply(T , P) = { V(Θ) | (T | P) =⇒≻ Θ }

Problem: set of executions { Θ | (T | P) =⇒≻ Θ } difficultto calculate

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Contribution of Θ:


◮ V(Θ) =∑


Apply(T , P) = { V(Θ) | (T | P) =⇒≻ Θ }


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Contribution of Θ:


◮ V(Θ) =∑


Apply(T , P) = { V(Θ) | (T | P) =⇒≻ Θ }


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Contribution of Θ:


◮ V(Θ) =∑


Apply(T , P) = { V(Θ) | (T | P) =⇒≻ Θ }


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Alternative strategy

Recall:

◮ P ⊑pmay Q if Apply(T , P) ⊑Ho Apply(T , Q) for every test T

◮ P ⊑pmust Q if Apply(T , P) ⊑Sm Apply(T , Q) for every testT

Maybe:

◮ P ⊑pmay Q if sup(Apply(T , P)) ≤ sup(Apply(T , Q)) forevery test T

◮ P ⊑pmust Q if inf(Apply(T , P)) ⊑Sm inf(Apply(T , Q)) forevery test T

Strategy:

◮ calculate inf(Apply(T ,−)) and sup(Apply(T ,−)) directly

◮ do not calculate the entire set Apply(T ,−)31/37

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Recall:



Maybe:



Strategy:



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Recall:



Maybe:



Strategy:



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Example: Calculating the sup

T | P

s1 s2

tbd tgd

τ

34

14

τ

τ τ

ω

sup-equation:

rsup =3

4· s1 +

1

4· s2

s1 = max{rsup, tbd}

s2 = tgd

tbd = 0

tgd = 1

sup(Apply(T , P)) is least solution: rsup = 1

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T | P

s1 s2

tbd tgd

τ

34

14

τ

τ τ

ω

sup-equation:

rsup =3

4· s1 +

1

4· s2

s1 = max{rsup, tbd}

s2 = tgd

tbd = 0

tgd = 1


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T | P

s1 s2

tbd tgd

τ

34

14

τ

τ τ

ω

sup-equation:

rsup =3

4· s1 +

1

4· s2

s1 = max{rsup, tbd}

s2 = tgd

tbd = 0

tgd = 1


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Example: Calculating the inf

T | P

s1 s2

tbd tgd

τ

34

14

τ

τ τ

ω

inf-equation:

rinf =3

4· s1 +

1

4· s2

s1 = min{rinf , tbd}

s2 = tgd

tbd = 0

tgd = 1

inf(Apply(T , P)) is least solution: rinf = 14

33/37

sfi



T | P

s1 s2

tbd tgd

τ

34

14

τ

τ τ

ω

inf-equation:

rinf =3

4· s1 +

1

4· s2


s2 = tgd

tbd = 0

tgd = 1


33/37

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T | P

s1 s2

tbd tgd

τ

34

14

τ

τ τ

ω

inf-equation:

rinf =3

4· s1 +

1

4· s2


s2 = tgd

tbd = 0

tgd = 1


33/37

sfi


Finitary pLTSs

Whenever

◮ set of states are finite

◮ set of actions are finite

In a finitary pLTS:

◮ execution sets {Θ | Apply(T , P) =⇒≻ Θ } are closed

◮ P ⊑may Q iff inf(Apply(T , P)) ≤ inf(Apply(T , Q)) for everytest T

◮ P ⊑must Q iff sup(Apply(T , P)) ⊑Sm sup(Apply(T , Q)) forevery test T

◮ inf(Apply(T ,−)) is least solution of inf-equation

◮ sup(Apply(T ,−)) is least solution of sup-equation

34/37

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Finitary pLTSs

Whenever



In a finitary pLTS:






34/37

sfi


Finitary pLTSs

Whenever



In a finitary pLTS:






34/37

sfi


Finitary pLTSs

Whenever



In a finitary pLTS:






34/37

sfi


Example

r1 = a.(τ.b + τ.c) r2 = a.b + a.c T = a.(b.ω 12⊕ c .ω)

r1

ω ω

τ

12

12

τ τ

τ

ττ

τ

r2

ω ω

τ τ

τ

12 1

2

τ

τ

12

12

τ

Apply(T ,r1)=

(

inf : 0

sup : 1Apply(T ,r2)=

(

inf : 12

sup : 12

So choice points do matter: r1 6≃pmay r2 r1 6≃pmust r2

35/37

sfi


Example

r1 = a.(τ.b + τ.c) r2 = a.b + a.c T = a.(b.ω 12⊕ c .ω)

r1

ω ω

τ

12

12

τ τ

τ

ττ

τ

r2

ω ω

τ τ

τ

12 1

2

τ

τ

12

12

τ

Apply(T ,r1)=

(

inf : 0

sup : 1Apply(T ,r2)=

(

inf : 12

sup : 12

So choice points do matter: r1 6≃pmay r2 r1 6≃pmust r2

35/37

sfi


Example

Pτ τ

12 1

2

τ

a

12 1

2

a

a.ω | P

ω ω

τ τ

12 1

2

τ

τ

12 1

2

τ

Apply(a.ω, P) =

{

inf : 12

sup : 1

P ≃pmay a. 0 P ⊑pmust a. 0 a. 0 6⊑pmust P

36/37

sfi


Example

Pτ τ

12 1

2

τ

a

12 1

2

a

a.ω | P

ω ω

τ τ

12 1

2

τ

τ

12 1

2

τ

Apply(a.ω, P) =

{

inf : 12

sup : 1


36/37

sfi


Example

Pτ τ

12 1

2

τ

a

12 1

2

a

a.ω | P

ω ω

τ τ

12 1

2

τ

τ

12 1

2

τ

Apply(a.ω, P) =

{

inf : 12

sup : 1


36/37

sfi


Example

Pτ τ

12 1

2

τ

a

12 1

2

a

a.ω | P

ω ω

τ τ

12 1

2

τ

τ

12 1

2

τ

Apply(a.ω, P) =

{

inf : 12

sup : 1


36/37

sfi


Coming up:

Reasoning techniques for probabilistic processes

Are these distinguishable by any test ?

P

d12

12

ab c

Q

d12

12

a b a c

37/37

sfi


Coming up:

Reasoning techniques for probabilistic processes

Are these distinguishable by any test ?

P

d12

12

ab c

Q

d12

12

a b a c

37/37

Date post:	22-Mar-2020
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