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DUBLIN CITY UNIVERSITY
```DUBLIN CITY UNIVERSITY
SEMESTER TWO EXAMINATIONS 2013
MODULE:
(Title & Code)
CA648/A/B Formal Programming
COURSE:
M.Sc. in Software Engineering (MSE)
M.Sc. in Computing (MCM)
Ph.D. Track (CAPT)
YEAR:
1,2,C
EXAMINERS:
(Including Telephone Nos.)
Dr. A. Grzenkowicz,
Dr. B. Lee,
Dr. G. Hamilton, Ext no. 5017.
TIME ALLOWED:
3 hours
INSTRUCTIONS:
All questions carry equal marks.
The use of programmable or text storing calculators is expressly forbidden.
Module Code: CA648/A/B
Semester Two Examinations 2013
PAGE 1 OF 12
QUESTION 1
[TOTAL MARKS: 20]
Consider the following partial correctness specification:
{N ≥ 0}
X := 0;
Y := 0;
WHILE X < N DO
BEGIN
X := X + 1;
Y := Y + (2 × X) − 1
END
{Y = N2 }
1(a)
[6 Marks]
Add appropriate annotations to this specification to allow it to be verified.
Solution:
This specification can be annotated as follows:
{N ≥ 0}
X := 0;
Y := 0; {X = 0 ∧ Y = 0 ∧ N ≥ 0}
WHILE X < N DO {Y = X2 ∧ X ≤ N}
BEGIN
X := X + 1;
Y := Y + (2 × X) − 1
END
{Y = N2 }
1(b)
[6 Marks]
List the verification conditions which would be generated for the annotated specification in 1(a).
Solution:
The following verification conditions will be generated from this specification:
1. N ≥ 0 ⇒ 0 = 0 ∧ 0 = 0 ∧ N ≥ 0
2. X = 0 ∧ Y = 0 ∧ N ≥ 0 ⇒ Y = X2 ∧ X ≤ N
3. Y = X2 ∧ X ≤ N ∧ ¬(X < N) ⇒ Y = N2
4. Y = X2 ∧ X ≤ N ∧ X < N ⇒ Y + (2 × (X + 1)) − 1 = (X + 1)2 ∧ (X + 1) ≤ N
1(c)
[8 Marks]
Verify this specification by showing that the verification conditions in 1(b) are
true.
Solution:
These verification conditions can be proved as follows:
Module Code: CA648/A/B
Semester Two Examinations 2013
PAGE 2 OF 12
1. N ≥ 0 ⇒ 0 = 0 ∧ 0 = 0 ∧ N ≥ 0
=
{ arithmetic, logical simplification }
True
2. X = 0 ∧ Y = 0 ∧ N ≥ 0 ⇒ Y = X2 ∧ X ≤ N
=
{ arithmetic }
X = 0 ∧ Y = 0 ∧ N ≥ 0 ⇒ 0 = 02 ∧ 0 ≤ N
=
{ arithmetic, logical simplification }
True
3. Y = X2 ∧ X ≤ N ∧ ¬(X < N) ⇒ Y = N2
=
{ arithmetic }
Y = X2 ∧ X = N ⇒ Y = N2
=
{ arithmetic, logical simplification }
True
4. Y = X2 ∧ X ≤ N ∧ X < N ⇒ Y + (2 × (X + 1)) − 1 = (X + 1)2 ∧ (X + 1) ≤ N
=
{ arithmetic }
Y = X2 ∧ X < N ⇒ Y + (2 × X) + 1 = X2 + (2 × X) + 1 ∧ (X + 1) ≤ N
=
{ logical simplification, arithmetic }
True
QUESTION 2
[TOTAL MARKS: 20]
Give a total correctness specification for the program in Question 1, and prove
the total correctness of the program.
Solution:
The total correctness specification is as follows:
[N ≥ 0]
X := 0;
Y := 0;
WHILE X < N DO
BEGIN
X := X + 1;
Y := Y + (2 × X) − 1
END
[Y = N2 ]
This specification can be annotated as follows:
[N ≥ 0]
X := 0;
Y := 0; {X = 0 ∧ Y = 0 ∧ N ≥ 0}
WHILE X < N DO {Y = X2 ∧ X ≤ N} [N − X]
BEGIN
X := X + 1;
Y := Y + (2 × X) − 1
END
[Y = N2 ]
The following verification conditions will be generated:
1. N ≥ 0 ⇒ 0 = 0 ∧ 0 = 0 ∧ N ≥ 0
Module Code: CA648/A/B
Semester Two Examinations 2013
PAGE 3 OF 12
2. X = 0 ∧ Y = 0 ∧ N ≥ 0 ⇒ Y = X2 ∧ X ≤ N
3. Y = X2 ∧ X ≤ N ∧ ¬(X < N) ⇒ Y = N2
4. Y = X2 ∧ X ≤ N ∧ X < N ⇒ (N − X) ≥ 0
5. Y = X2 ∧ X ≤ N ∧ X < N ∧ (N − X) = v ⇒ Y + (2 × X) − 1 = (X + 1)2 ∧ (X + 1) ≤ N ∧ (N − (X + 1)) < v
These verification conditions can be proved as follows:
1. N ≥ 0 ⇒ 0 = 0 ∧ 0 = 0 ∧ N ≥ 0
=
{ arithmetic, logical simplification }
True
2. X = 0 ∧ Y = 0 ∧ N ≥ 0 ⇒ Y = X2 ∧ X ≤ N
=
{ arithmetic }
X = 0 ∧ Y = 0 ∧ N ≥ 0 ⇒ 0 = 02 ∧ 0 ≤ N
=
{ arithmetic, logical simplification }
True
3. Y = X2 ∧ X ≤ N ∧ ¬(X < N) ⇒ Y = N2
=
{ arithmetic }
Y = X2 ∧ X = N ⇒ Y = N2
=
{ arithmetic, logical simplification }
True
4. Y = X2 ∧ X ≤ N ∧ X < N ⇒ (N − X) ≥ 0
=
{ arithmetic, logical simplification }
True
5. Y = X2 ∧ X ≤ N ∧ X < N ∧ (N − X) = v ⇒ Y + (2 × X) − 1 = (X + 1)2 ∧ (X + 1) ≤ N ∧ (N − (X + 1)) < v
=
{ arithmetic }
Y = X2 ∧ X < N ∧ (N − X) = v ⇒ Y + (2 × X) + 1 = X2 + (2 × X) + 1 ∧ (X + 1) ≤ N ∧ (N − X − 1) < v
=
{ logical simplification, arithmetic }
True
QUESTION 3
3(a)
[TOTAL MARKS: 20]
[4 Marks]
Describe what is meant by program refinement.
Solution:
Refinement is the process of constructing a program from a specification by following a number
of defined refinement laws which have been shown to be correct, thus ensuring that the resulting
program is correct by construction.
3(b)
[4 Marks]
Define the program refinement law for IF commands.
Solution:
[P, Q] ⊇ IF S THEN [P ∧ S, Q] ELSE [P ∧ ¬S, Q]
Module Code: CA648/A/B
Semester Two Examinations 2013
PAGE 4 OF 12
3(c)
[12 Marks]
Refine the following specification to a corresponding program:
[N ≥ 0, S =
N
X
2i ]
i=0
Solution:
[N ≥ 0, S =
N
X
2i ]
i=0
⊇
{ Block Law }
BEGIN
N
X
[N ≥ 0, S =
2i ]
i=0
EN D
⊇
{ Sequencing Law }
BEGIN
[N ≥ 0, N ≥ 0 ∧ X = 0];
N
X
[N ≥ 0 ∧ X = 0, S =
2i ]
i=0
EN D
⊇
{ Derived Assignment ` N ≥ 0 ⇒ N ≥ 0 ∧ 0 = 0 }
BEGIN
X := 0;
N
X
[N ≥ 0 ∧ X = 0, S =
2i ]
i=0
EN D
⊇
{ Sequencing Law }
BEGIN
X := 0;
[N ≥ 0 ∧ X = 0, N ≥ 0 ∧ X = 0 ∧ S = 1];
N
X
2i ]
[N ≥ 0 ∧ X = 0 ∧ S = 1, S =
i=0
EN D
⊇
{ Derived Assignment ` N ≥ 0 ∧ X = 0 ⇒ N ≥ 0 ∧ X = 0 ∧ 1 = 1 }
BEGIN
X := 0;
S := 1;
N
X
[N ≥ 0 ∧ X = 0 ∧ S = 1, S =
2i ]
i=0
EN D
⊇
{ Precondition Weakening ` N ≥ 0 ∧ X = 0 ∧ S = 1 ⇒ S =
X
X
2i ] ∧ X ≤ N }
i=0
BEGIN
X := 0;
S := 1;
X
N
X
X
[S =
2i ] ∧ X ≤ N, S =
2i ]
i=0
i=0
EN D
⊇
{ Postcondition Strengthening ` S =
X
X
i=0
Module Code: CA648/A/B
Semester Two Examinations 2013
2i ] ∧ X ≤ N ∧ ¬(X < N ) ⇒ S =
N
X
2i }
i=0
PAGE 5 OF 12
BEGIN
X := 0;
S := 1;
X
X
X
X
[S =
2i ] ∧ (N − X) ≥ 0, S =
2i ] ∧ X ≤ N ∧ ¬(X < N )]
i=0
i=0
EN D
⊇
{ While ` S =
X
X
2i ] ∧ X ≤ N ∧ X < N ⇒ (N − X) ≥ 0 }
i=0
BEGIN
X := 0;
S := 1;
W HILE X < N DO
X
X
X
X
[S =
2i ] ∧ X ≤ N ∧ X < N ∧ (N − X) = v, S =
2i ] ∧ X ≤ N ∧ (N − X) < v]
i=0
i=0
EN D
⊇
{ Block Law }
BEGIN
X := 0;
S := 1;
W HILE X < N DO
BEGIN
X
X
X
X
2i ] ∧ X ≤ N ∧ (N − X) < v]
2i ] ∧ X ≤ N ∧ X < N ∧ (N − X) = v, S =
[S =
i=0
i=0
EN D
EN D
⊇
{ Sequencing Law }
BEGIN
X := 0;
S := 1;
W HILE X < N DO
BEGIN
X
X−1
X
X
[S =
2i ] ∧ X ≤ N ∧ X < N ∧ (N − X) = v, S =
2i ] ∧ X ≤ N ∧ (N − X) < v];
[S =
i=0
X−1
X
i=0
2i ] ∧ X ≤ N ∧ (N − X) < v, S =
i=0
2i ] ∧ X ≤ N ∧ (N − X) < v]
i=0
EN D
EN D
⊇
X
X
{ Derived Assignment ` S =
X
X
2i ] ∧ X ≤ N ∧ X < N ∧ (N − X) = v ⇒
i=0
(X+1)−1
S=
X
2i ] ∧ (X + 1) ≤ N ∧ (N − (X + 1)) < v }
i=0
BEGIN
X := 0;
S := 1;
W HILE X < N DO
BEGIN
X := X + 1;
X−1
X
X
X
i
[S =
2 ] ∧ X ≤ N ∧ (N − X) < v, S =
2i ] ∧ X ≤ N ∧ (N − X) < v]
i=0
i=0
EN D
EN D
Module Code: CA648/A/B
Semester Two Examinations 2013
PAGE 6 OF 12
⊇
{ Derived Assignment ` S =
X−1
X
2i ] ∧ X ≤ N ∧ (N − X) < v ⇒
i=0
S + 2X =
X
X
2i ] ∧ X ≤ N ∧ (N − X) < v }
i=0
BEGIN
X := 0;
S := 1;
W HILE X < N DO
BEGIN
X := X + 1;
S := S + 2X
EN D
EN D
QUESTION 4
[TOTAL MARKS: 20]
A library needs to keep track of customer borrowing information. Customers have
to join the library before they can use it, at which point they are given a unique
membership identifier. Members cannot borrow more than 5 books at any one
time, and cannot leave the library until they have returned all of their borrowed
books. The following events need to be handled:
join: allocate a previously unallocated membership identifier to a new member.
leave: remove the given membership identifier from the set of members; this event
is not enabled if the member has not returned all of their borrowed books.
borrow: allocate the given book to the given member; this event is not enabled
if the book is not available or the member has already borrowed their maximum allocation of 5 books.
return: return the given book from the given member to the library.
borrowed: give the set of books that are currently borrowed by the given member.
4(a)
[4 Marks]
Define the context for an Event-B specification of the library system.
Solution:
CONTEXT Library_ctx
SETS
BOOKS
MEMBERS
CONSTANTS
maxbooks
Module Code: CA648/A/B
Semester Two Examinations 2013
PAGE 7 OF 12
AXIOMS
axm1 : maxbooks ∈ N1
axm2 : maxbooks = 5
END
4(b)
[6 Marks]
Define the variables for an Event-B specification of the library system. Define a
suitable invariant for these variables, and show their initialisation, ensuring that
this initialisation satisfies the invariant.
Solution:
MACHINE Library
SEES Library_ctx
VARIABLES
members
loans
books
INVARIANTS
inv1 : members ⊆ MEMBERS
inv2 : loans ∈ BOOKS →
7 BORROWERS
inv3 : ∀x .x ∈ ran(loans) ⇒ card (loans −1 [{x }]) <= maxbooks
inv4 : books ⊆ BOOKS
EVENTS
Initialisation
begin
act1 : members := ∅
act2 : loans := ∅
act3 : books := ∅
end
4(c)
[10 Marks]
Specify the events for an Event-B specification of the library system, making use
of the definitions in 4(a) and 4(b).
Solution:
Event join =
b
any
memberid
Module Code: CA648/A/B
Semester Two Examinations 2013
PAGE 8 OF 12
where
grd1 : memberid ∈ MEMBERS
grd2 : memberid ∈
/ members
then
act1 : members := members ∪ {memberid }
end
Event leave =
b
any
memberid
where
grd1 : memberid ∈ MEMBERS
grd2 : memberid ∈ members
grd3 : memberid ∈
/ ran(loans)
then
act1 : members := members \ {memberid }
end
Event borrow =
b
any
memberid
book
where
grd1
grd2
grd3
grd4
grd5
:
:
:
:
:
memberid ∈ MEMBERS
memberidid ∈ members
book ∈ BOOKS
book ∈
/ dom(loans)
card (loans −1 [{memberid }]) < maxbooks
then
act1 : loans := loans ∪ {book 7→ memberid }
end
Event return =
b
any
memberid
book
where
grd1
grd2
grd3
grd4
:
:
:
:
memberid ∈ MEMBERS
book ∈ BOOKS
memberid ∈ members
(book 7→ memberid ) ∈ loans
then
act1 : loans := {book } C
− loans
end
Module Code: CA648/A/B
Semester Two Examinations 2013
PAGE 9 OF 12
Event borrowed =
b
any
memberid
where
grd1 : memberid ∈ MEMBERS
grd2 : memberid ∈ members
then
act1 : books := loans −1 [{memberid }]
end
END
QUESTION 5
[TOTAL MARKS: 20]
5(a)
[7 Marks]
Write an Event-B specification for a program computing the product of all the
numbers in an array a : 1..n → N where n ≥ 1. The specification should have
a constant function which computes the desired result of the program and a result variable result. It should also have two abstract events Initialisation and
P roduct which give the appropriate precondition and postcondition respectively
for result.
Solution:
CONTEXT Product_ctx
CONSTANTS
n
a
product
AXIOMS
axm1 : n ∈ N1
axm2 : a ∈ 1 .. n → N
axm3 : product ∈ N → N
axm4 : (0 7→ 1 ) ∈ product
axm5 : ∀n, p.n 7→ p ∈ product ⇒ (n + 1 7→ p × a(n + 1 )) ∈ product
END
MACHINE Product
SEES Product_ctx
VARIABLES
result
INVARIANTS
Module Code: CA648/A/B
Semester Two Examinations 2013
PAGE 10 OF 12
inv1 : result ∈ N
EVENTS
Initialisation
begin
act1 : result := 1
end
Event Product =
b
begin
act1 : result := product(n)
end
END
5(b)
[8 Marks]
Give a refinement of the specification in 5(a) which adds variables index and productsofar, giving the value of the current index in the array, and the product of the
numbers in the array up to this index, respectively. Your refinement should define
a suitable variant and should also add two further events U pdate and P rogress.
The U pdate event should update the value of the productsofar variable. The convergent event P rogress should be used to ensure termination by decreasing the
variant. You should also refine the events Initialisation and P roduct to give
precise initial and final values for the result variable.
Solution:
MACHINE ProductR
REFINES Product
SEES Product_ctx
VARIABLES
result
index
productsofar
INVARIANTS
inv1 : index ∈ 0 .. n
inv2 : productsofar ∈ N
inv3 : productsofar = product(index )
VARIANT
n - index
EVENTS
Initialisation
Module Code: CA648/A/B
Semester Two Examinations 2013
PAGE 11 OF 12
begin
act1 : result := 1
act2 : index := 0
act3 : productsofar := 1
end
Event Product =
b
refines Product
when
grd1 : index ≥ n
then
act1 : result := productsofar
end
Event Progress =
b
Status convergent
when
grd1 : index < n
then
act1 : index := index + 1
end
Event Update =
b
when
grd1 : index < n
then
act1 : productsofar := productsofar × a(index )
end
END
5(c)
[5 Marks]
Give a program which computes the product of all the numbers in an array and
Solution:
result := 1;
index := 0;
productsofar := 1;
WHILE index < n DO
BEGIN
index := index + 1;
productsofar := productsofar × a(index)
END;
result := productsofar
Module Code: CA648/A/B
Semester Two Examinations 2013
PAGE 12 OF 12
```
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