Process Algebra (2IF45) Assignments

Process Algebra (2IF45)

Assignments

Dr. Suzana Andova

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Groups (almost final)


Group 1. Stijn Fleuren, Jori Selen, John van Heur, Jordi Timmermans

Group 2. Nicky Gerritsen 0595838,Kevin van der Pol 0620300,

Group 3.

Group4. Luis Avila 0754588Pablo Puente 0761739

Group 5. Oerlemans, G.G. 0607213, Panhuijzen, I.W.F. 0570999, Poppelaars, J.J.G. 0518740Joost van Twist J. 0617198

Group 6. Johan Hendriks, 0588920Bas van der Oest, 0588728Roy van Doormaal, 0668932Peter Klerks, 0593253

Group 7. Tal Yosefa Milea (0757073), Sjoerd te Pas (0663210), Twan Vermeulen (0732529), Ahmed Ibrahim (0718605)

NOTE: Not all groups and students are listed here

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Details regarding the assignments


1. Assignments are not compulsory. You may chose to do the assignments, thus making it part of your final grade. Or you could chose to go only to the written exam. Advantages to do the assignments are:

• a part of the written exams will be covered by the assignments. • you will get deeper insight in the material• you will prepare for the (your “relaxed”) written part of the exam as a side (positive) effect• it may be that the assignments will be more extensive questions than those given at the exam, but you can work in a team and discuss it and learn more. Also you will have more time to think about a solution. • possibility to earn bonus points

2. The final grade will be calculated 40% from the assignment grade + 60% from the written exam.

3. The assignment will consists out of 3 smaller assignments, 1st related to SOS and axioms, 2nd process specification, and 3rd probabilistic/stochastic specification

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4. Way of working: Students within a group have to organize the work themselves. Each student has to be actively involved, from the beginning till the end.

5. Finalization of an assignment's part: For each assignment a strict deadline will be defined. A group delivers the solution, which they later defend. Defense will include all students from the group.

6. Assignments schedule: 1st: March 2nd April3rd May

They are obtained the first week of the corresponding month, to be delivered at the end of the month.

7. Assessment: The assignment grade will be based on: the quality of the delivered solution, the defense and peer-to-peer assessment of team members. Thus each student gets own grade.

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8. Questions?


Recursion

Dr. Suzana Andova

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LTSs Language Process terms


term

term1 term2

a

Deduction rules Terms built from constants, operators and variables

term

LTS

Set of Axioms (basic equalities)

Derivation gives more (derived) equalities

Language (signature)

Set of constants and operators

term

Process Terms (Specification)

term

8 Process Algebra (2IF45)

Bisimulation Term Equalities

term

term1 term2

a


term

LTSs. Equivalence relation

Set of Axioms (basic equalities) term1 = term2


Axiom ├ term3= term4



term


Process Terms (Specification)

term

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term

term1 term2

a






?coin.(!coffee.1 + !tea.0)

?coin.(!coffee.1 + !tea.0)

LTS

!tea !coffee

?coin

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term

term1 term2

a

Deduction rules





!coin.?coffee.1

Terms built from constants, operators and variables

LTS

!coin

User

?coffee

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term

term1 term2

a

Deduction rules

term

LTS





Process Terms


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term

term1 term2

a






?coin.(!coffee.x + !tea.0)

LTS

x

!tea !coffee

?coin

What is x?• 1• is !coffee.!coffee.1• !coffee.y

• ……

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term

term1 term2

a

Deduction rules

LTS





term?

!tea ?coin !coffee

“?coin.(!coffee +!tea).

?coin.(!coffee +!tea) .?coin.(!coffee +!tea)….”


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LTS

?coin.(!coffee.x + !tea.0)

x

!tea !coffee

?coin

What is X?• 1• is !coffee.!coffee.1• !coffee.y

• ……

How do we solve this?

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LTS

?

!tea ?coin !coffee

?coin.(!coffee +!tea).

?coin.(!coffee +!tea) .?coin.(!coffee +!tea)….

How do we solve this?Does the solution on the previous slide solves this question as well?


Recursive (Process) Variables, Recursive Equations, Recursive Specification

Example1.

?coin.(!coffee.X+ !tea.0)X = !coffee.!coffee.1

Example2.

Y = ?coin.(!coffee.Y +!tea.Y)

•Recursive specifications increase the specification power of equational theories. •By means of recursive specifications infinite processes can be specified.




Derivation

Axioms, E ├ term_A= term_B,

where term_A and term_B may contain recursive variables from E



Recursive Equations X1 = …., X2 = ….., …. Recursive Specification E = {X1 = …., X2 = ….., …}

Recursive Equations and Rec. Specification in Equational Theory




Derivation

Axioms, E ├ term_A= term_B,

where term_A and term_B may contain recursive variables from E



Recursive Equations X1 = …., X2 = ….., …. Recursive Specification E = {X1 = …., X2 = ….., …}

Recursive Equations and Rec. Specification in Equational Theory

Example. E = { X = a.Y + c.0,Y = b.X}

BPA(A), E ├ X = a.Y +c.0 = a.(b.X) +c.0 = a.(b.(a.Y + c.))) + c.0


Generating LTSs for recursive specifications

Example. E = { X = a.Y + c.0,Y = b.X} with X being root variable.

X

a

Y 0

c b

Example. E2 = { X = a.(a.(X+1)) +1 } with X being root variable.

X

a

a.(X+1)

a a

X+1



Example. E = { X = X + a.0} with X being root variable.



Example. E = { X = X + a.0} with X being root variable.

What transitions can X execute? X can perform a transition iff X can perform a transition

If we substitute a.0 for X

0

a

0

a

0

a

If we substitute b.0 for X

0

b

0

b

0

a


Guarded recursive specification

What do we need take care?

That the recursive specification generates a unique LTS, in other words, it has a unique solution

How do we guarantee it?

We make sure that our specification is guarded.


Guarded recursive specification

What do we still miss?

Consider three specifications:E = {X = a.Y, Y = a.X} and F = {Z = a.a.Z} G = {Un = a.Un+1, n 0}

X

a

Y

a

Z

a

a.Z

a

U0 U1 U2

a a a…

Can we derive

BPA(A), E, F, G ├ X = Z or BPA(A), E, F, G ├ X = U0 or

BPA(A), E, F, G ├ Z = U0


Deriving equalities of recursive variables

Consider three specifications:E = {X = a.Y, Y = a.X} and F = {Z = a.a.Z} G = {Un = a.Un+1, n 0}

1. BPA(A), E, F, G ├ X = a.a.X which is exactly the form of Z 2. BPA(A), E, F, G ├ X = U0 cannot be derived directly. We introduce new variables X0, X1, … which are defined using X, as:

X0 = X, X1= a.X, X2 = a.a.X, …, Xn = an X, ….,

Now, we want to show that Xn can be rewritten in the form of Un for any n 0!BPA(A), E, F, G ├ X0 = X = a.a.X = a.X1

BPA(A), E, F, G ├ X1 = a.X = a.a.a.X = a.X2

and in general BPA(A), E, F, G ├ Xn = an X = an a.a.X = a.Xn+1

We conclude that Xn is in the same form as Un for any n 0

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Recursive principles for Deriving equalities of recursive variables

1. Every guarded recursive specification has a solution, this is called Restricted Recursive Definition Principle (RDP-)

2. Every guarded recursive specification has a unique solution, this is called Recursive Specification Principle (RSP)

1. BPA(A), E, F, G ├ X = a.a.X which is exactly the form of Z. Directly from the RSP and RDP- we can conclude

BPA(A), E, F, G, RDP-, RSP ├ X = Z 2. BPA(A), E, F, G ├ X = U0 cannot be derived directly. We introduce new

variables X0, X1, … which are defined using X, as:X0 = X, X1= a.X, X2 = a.a.X, …, Xn = an X, ….,

Now, we want to show that Xn can be rewritten in the form of Un for any n 0!BPA(A), E, F, G ├ X0 = X = a.a.X = a.X1

BPA(A), E, F, G ├ X1 = a.X = a.a.a.X = a.X2

and in general BPA(A), E, F, G ├ Xn = an X = an a.a.X = a.Xn+1

We conclude that Xn is in the same form as Un for any n 0. Directly from the RSP and RDP- we can conclude

BPA(A), E, F, G, RDP-, RSP ├ X0 = U0

and also BPA(A), E, F, G, RDP-, RSP ├ X = X0 = U0

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Specifying a Stack

Consider a finite set of data elements D = {d1, d2, …, dn} for some n natural number. Define a recursive specification that describes a stack with unlimited capacity. Elements from D can be added or removed from the stack from the top of the stack.

Steps: 1. First, define the set of atomic actions.

2. As a short hand notation you can use universal sum dD

3. Reason how many recursive variables you may need. 4. Specify the behaviour of the stack process.

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Specifying a Stack

Consider a finite set of data elements D = {d1, d2, …, dn} for some n natural number. Define a recursive specification that describes a stack with unlimited capacity. Elements from D can be added or removed from the stack from the top of the stack.

Steps: 1. First, define the set of atomic actions.

2. As a short hand notation you can use universal sum dD

3. Reason how many recursive variables you may need. 4. Specify the behaviour of the stack process.

Stack1 = S

S = dD push(d).Sd, for any d D

Sd = eD push(e).Sed + dD pop(d).S, for any d D and D*

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Process Algebra (2IF45) Assignments