CP5520  Advanced Databases and Applications Assignment, James Cook University, Australia
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1. Transaction Processing 
Consider schedules S_{1} and S_{2} below.
S_{1}: r_{2}(X), r_{2}(Z), r_{3}(Y), w_{2}(X), w_{2}(Z), w_{3}(Y), r_{1}(Z), r_{3}(X), w_{3}(X), r_{1}(X), r_{3}(Z), w_{1}(X), w_{3}(Z), w_{1}(Z)
S_{2}: r_{1}(X), r_{1}(Z), r_{2}(Y), w_{2}(Y), w_{1}(X), r_{3}(Y), w_{1}(Z), r_{2}(X), r_{3}(X), r_{2}(Z), w_{2}(Z), r_{3}(Z), w_{3}(Y), w_{3}(Z)
(a) Apply the basic timestamp ordering (BTO) algorithm to schedules S_{1} and S_{2}. Determine whether or not the algorithm allows the execution of the schedules, and discuss cascading rollback (if any).
Answer 
Schedule 1 
T_{1}

T_{2}

T_{3}


r2(X)



r2(Z)




r3(Y)


w2(X)



w2(Z)




w3(Y)

r1(Z)





r3(X)



w3(X)

r1(X)





r3(Z)

w1(X)





w3(Z)

w1(Z)



1) First of all, we will calculate the read and write time stamp of the data items X, Y, Z.
The time stamp of transactions T_{1}, T_{2}, T_{3} is 7, 1, 3 as the first operation of these transactions is issued on these timestamps.
To find Read time stamp of X, Y, Z
Read timestamp is the largest time stamp when a transaction successfully completes read operation.
R_TS(X)=7 as transaction T_{1} successfully completed read operation on it.
R_TS(Y)=3 as transaction T_{3} successfully completed read operation on it.
R_TS(Z)=7 as transaction T_{1} successfully completed read operation on it.
Write timestamp is the largest time stamp when a transaction successfully completes read operation.
W_TS(X)=7 as transaction T_{1} successfully completed write operation on it.
W_TS(Y)=3 as transaction T_{3} successfully completed write operation on it.
W_TS(Z)=7 as transaction T_{1} successfully completed write operation on it.
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2) Now, we will apply basic time stamp ordering protocol which says
if TS_(T) issues W_TS(A) then, it checks 2 conditions
i) If R_TS(A)>TS(T) or W_TS(A)>TS(T) then, abort and rollback T and reject write operation
ii) Otherwise, execute Write operation and update W_TS(A)=TS(T)
if TS_(T) issues R_TS(A) then, it checks 2 conditions
iii) If W_TS(A)>TS(T) then, abort and rollback T and reject write operation
iv) Otherwise, execute read operation and update R_TS(A)=Max(TS(T),current R_TS(A)).
a) Now, in this schedule transaction T_{2} issues read operation on data item X
W_TS(X)=7, TS(T2)=1 . Since, W_TS(X)> TS(T2) so, this read operation is aborted.
b) Now, transaction T_{2} issues read operation on data item Z
W_TS(Z)=7, TS(T2)=1 . Since, W_TS(X)> TS(T2) so, this read operation is aborted.
c) Now, transaction T_{3} issues read operation on data item Y
W_TS(Y)=3, TS(T3)=3 . Since, W_TS(X)=TS(T3) so, this read operation is executed.
R_TS(Y)=3
d) Now, transaction T_{2} issues write operation on data item X
W_TS(X)=7, TS(T2)=1 and. R_TS(X)=7, TS(T2)=1 Since, W_TS(X)>TS(T2) and also R_TS(X)>TS(T2) so, this write operation is aborted
e) Now, transaction T_{2} issues write operation on data item Z
W_TS(Z)=7, TS(T2)=1 and. R_TS(Z)=7, TS(T2)=1 Since, W_TS(Z)>TS(T2) and also R_TS(Z)>TS(T2) so, this write operation is aborted
f) Now, transaction T_{3} issues write operation on data item Y
W_TS(Y)=3, TS(T3)=1 and. R_TS(X)=3, TS(T3)=1 Since, W_TS(Y)>TS(T3) and also R_TS(Y)>TS(T3) so, this write operation is aborted
g) T_{1} issues read operation on data item Z
W_TS(Z)=7, TS(T1)=7.. Since, W_TS(Z)=TS(T1) so, this read operation is executed and R_TS(Z)=7.
h) Now, transaction T_{3} issues read operation on data item X
W_TS(X)=7, TS(T3)=3 . Since, W_TS(X)>TS(T3) so, this read operation is aborted.
i) Now, transaction T_{3} issues write operation on data item X
W_TS(X)=7, TS(T3)=3 and. R_TS(X)=7, TS(T3)=3 Since, W_TS(X)>TS(T3) and also R_TS(X)>TS(T3) so, this write operation is aborted
j) T_{1} issues read operation on data item X
W_TS(X)=7, TS(T1)=7.. Since, W_TS(X)=TS(T1) so, this read operation is executed and R_TS(X)=7.
k) T_{3} issues read operation on data item Z
W_TS(Z)=7, TS(T3)=3.. Since, W_TS(Z)>TS(T3) so, this read operation is aborted.
l) Now, transaction T_{1} issues write operation on data item X
W_TS(X)=7, TS(T1)=7 and. R_TS(X)=7, TS(T3)=7 Since, W_TS(X)=TS(T3) and also R_TS(X)=TS(T3) so, this write operation is executed and W_TS(X)=7
m) Now, transaction T_{3} issues write operation on data item Z
W_TS(Z)=7, TS(T3)=3 and. R_TS(Z)=7, TS(T3)=3 Since, W_TS(Z)>TS(T3) and also R_TS(Z)>TS(T3) so, this write operation is aborted
n) Now, transaction T_{1} issues write operation on data item Z
W_TS(Z)=7, TS(T1)=7 and. R_TS(Z)=7, TS(T3)=7 Since, W_TS(Z)=TS(T3) and also R_TS(Z)=TS(T3) so, this write operation is executed and W_TS(Z)=7
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Schedule 2 
T_{1}

T_{2}

T_{3}

r1(X)



r1(Z)




r2(Y)



w2(Y)


w1(X)





r3(Y)

w1(Z)



Commit

r2(X)




r3(X)


r2(Z)



w2(Z)




r3(Z)



w3(Y)



w3(Z)

1) First of all, we will calculate the read and write time stamp of the data items X, Y, Z.
The time stamp of transactions T_{1}, T_{2}, T_{3} is 1, 3, 7 as the first operation of these transactions is issued on these timestamps.
To find Read time stamp of X, Y, Z
Read timestamp is the largest time stamp when a transaction successfully completes read operation.
R_TS(X)=7 as transaction T_{3} successfully completed read operation on it.
R_TS(Y)=7 as transaction T_{3} successfully completed read operation on it.
R_TS(Z)=7 as transaction T_{3} successfully completed read operation on it.
Write timestamp is the largest time stamp when a transaction successfully completes read operation.
W_TS(X)=1 as transaction T_{1} successfully completed write operation on it.
W_TS(Y)=3 as transaction T_{2} successfully completed write operation on it.
W_TS(Z)=7 as transaction T_{3} successfully completed write operation on it.
2) Now, we will apply basic time stamp ordering protocol which says
if TS_(T) issues W_TS(A) then, it checks 2 conditions
i) If R_TS(A)>TS(T) or W_TS(A)>TS(T) then, abort and rollback T and reject write operation
ii) Otherwise, execute Write operation and update W_TS(A)=TS(T)
if TS_(T) issues R_TS(A) then, it checks 2 conditions
i) If W_TS(A)>TS(T) then, abort and rollback T and reject write operation
ii) Otherwise, execute read operation and update R_TS(A)=Max(TS(T),current R_TS(A)).
a) Now, in this schedule transaction T_{1} issues read operation on data item X
W_TS(X)=1, TS(T1)=1 . Since, W_TS(X)= TS(T1) so, this read operation is executed and R_TS(X)=7.
b) Now, transaction T_{1} issues read operation on data item Z
W_TS(Z)=7, TS(T1)=1 . Since, W_TS(X)> TS(T1) so, this read operation is aborted.
c) Now, transaction T_{2} issues read operation on data item Y
W_TS(Y)=3, TS(T3)=3 . Since, W_TS(X)=TS(T3) so, this read operation is executed.
R_TS(Y)=3
d) Now, transaction T_{2} issues write operation on data item Y
R_TS(Y)=7, TS(T2)=3 Since, R_TS(X)>TS(T2) so, this write operation is aborted.
e) Now, transaction T_{1} issues write operation on data item X
R_TS(X)=7, TS(T1)=1 Since, R_TS(X)>TS(T1) so, this write operation is aborted.
f) Now, transaction T_{3} issues read operation on data item Y
W_TS(Y)=3, TS(T3)=3 . Since, W_TS(X)=TS(T3) so, this read operation is executed.
R_TS(Y)=3
g) T_{1} issues write operation on data item Z
W_TS(Z)=7, TS(T1)=1 Since, W_TS(Z)>TS(T1) so, this write operation is aborted.
h) Now, transaction T_{2} issues read operation on data item X
W_TS(X)=1, TS(T2)=3 . Since, W_TS(X)<TS(T3) so, this read operation is executed and R_TS(X)=7.
i) Now, transaction T_{3} issues read operation on data item X
W_TS(X)=1, TS(T3)=7 Since, W_TS(X)<TS(T3) and so, this read operation is executed and R_TS(X)=7.
j) T_{2} issues read operation on data item Z
W_TS(Z)=7, TS(T2)=3.. Since, W_TS(Z)>TS(T3) so, this read operation is aborted.
k) Now, transaction T_{2} issues write operation on data item Z
W_TS(Z)=7, TS(T2)=7 and. R_TS(Z)=7, TS(T2)=7 Since, W_TS(X)=TS(T2) and also R_TS(X)=TS(T2) so, this write operation is executed and W_TS(Z)=7
l) Now, transaction T_{3} issues read operation on data item Z
W_TS(Z)=7, TS(T3)=7 Since, W_TS(Z)=TS(T3) so, this read operation is executed and R_TS(Z)=7
m) Now, transaction T_{3} issues write operation on data item Y
R_TS(Y)=&, TS(T3)=3 Since, R_TS(Y)>TS(T3) so, this write operation is aborted.
n) Now, transaction T_{3} issues write operation on data item Z
R_TS(Z)=7, TS(T3)=3 Since, R_TS(Z)>TS(T3) so, this write operation is aborted.
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(b) Testing the serializability of S_{1} and S_{2} by serialization graph technique to prove that the successful execution of a schedule under BTO will ensure the serializability of the schedule.
Answer  Serialization Graph
There is no cycle in the graph. Hence, it is serializable.
(c) Examine the recoverable characteristic of S_{1} and S_{2}. What schedule (S_{1} or S_{2}) can be executed under the strict timestamp ordering (STO) algorithm and write an equivalent strict schedule for it? We assume that a transaction will be be committed or aborted right after its last operation.
Answer  Recoverable schedules
Schedule 1
T_{1}

T_{2}

T_{3}


r2(X)



r2(Z)




r3(Y)


w2(X)



w2(Z)



Rollback

w3(Y)

r1(Z)





r3(X)



w3(X)

r1(X)





r3(Z)

w1(X)





w3(Z)

w1(Z)


Rollback

Rollback



Schedule 1 is recoverable with cascading rollback of all the transactions.
Schedule 2
T_{1}

T_{2}

T_{3}

r1(X)



r1(Z)




r2(Y)



w2(Y)


w1(X)





r3(Y)

w1(Z)



Rollback

r2(X)




r3(X)


r2(Z)



w2(Z)



Rollback

r3(Z)



w3(Y)



w3(Z)



Rollback

The schedule is recoverable after rollback of all the transactions.
Both the schedules can not be executed under strict time stamp ordering protocol because many operations of the transactions need to be aborted.
Schedule 1
So, this schedule can not be executed under strict two phase locking protocol as transactions t1 and t3 will keep waiting. Even, if the T2 release the lock after commit then, shrinking phase the schedule will start and no new lock can be taken.
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(d) Examine the twophase locking (2PL) technique on S_{1} and S_{2} with an assumption of unlock operations being done as late as possible. Are BTO schedules also 2PL schedules?
Answer  Strict 2PL for Schedule 1
T_{1}

T_{2}

T_{3}


Lock s(x)



r2(X)



Lock s(y)



r2(Z)

Lock s(Y)


Lock x(X)

r3(Y)


w2(X)



Lock x(Z)



w2(Z)



Unlock(x)



Unlock(Z)



Unlock(Y)




Lock x(Y)



w3(Y)

Lock S(Z)



r1(Z)





Lock S(X)



r3(X)



Lock x(X)



w3(X)

Wait





Lock s(Z)



r3(Z)



Lock x(Z)



w3(Z)



Unlock (X)



Unlock(Y)

Wait


Unlock(Z)

Lock s(X)



r1(X)



Lock x(X)



w1(X)



Lock x(Z)



w1(Z)



Unlock(X)



Unlock(Y)



Unlock(Z)



Strict 2PL for Schedule 2
T_{1}

T_{2}

T_{3}

Lock s(X)



r1(X)



Lock s(Z)



r1(Z)




Lock s(Y)



r2(Y)



Lock x(Y)



w2(Y)


Lock x(X)



w1(X)


Wait

Lock x(Z)


wait

w1(Z)

Wait

Wait

Unlock (X)

Wait

Wait

Unlock (Y)

Wait

Wait

Unlock (Z)

Wait

Wait


Wait

wait


Lock s(X)

wait


r2(X)

Wait


Lock s(Z)

wait


r2(Z)

Wait


Lock x(Z)

wait


w2(Z)

Wait


Unlock (X)

wait


Unlock (Y)

wait


Unlock (Z)

wait



Lock s(Y)



r3(Y)



Lock s(X)



r3(X)



Lock s(Z)



r3(Z)



Lock x(Y)



w3(Y)



Lock x(Z)



w3(Z)



Unlock (X)



Unlock (Y)



Unlock (Z)

No, BTO are not 2PL
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2. Deductive Database
ANCESTOR(X, Y) : PARENT(X, Y)
ANCESTOR(X, Y) : PARENT(X, Z), ANCESTOR(Z, Y)
Notice that PARENT(X, Y) means that X, Y are humanbeings and Y is the (biological) parent of X; ANCESTOR(X, Y) means that Y is the ancestor of X.
Consider the following fact base: PARENT(john, steve), PARENT(john, olivia), PARENT(olivia, emma), PARENT(olivia, william).
(a) Construct a model theoretic interpretation of the above rules using the given facts.
Answer 
Rules
ANCESTOR(X, Y) : PARENT(X, Y)
ANCESTOR(X, Y) : PARENT(X, Z), ANCESTOR(Z, Y)
Known Facts
PARENT(john, steve) is true.
PARENT(john, olivia) is true.
PARENT(olivia, emma) is true.
PARENT(olivia, william) is true.
Parent (X,Y) is false for all other combinations of X,Y
Deduced facts
Ancestor(john, steve) is true.
Ancestor(john, olivia) is true.
Ancestor(olivia, emma) is true.
Ancestor(olivia, william) is true.
Ancestor(john, emma) is true.
Ancestor(john, william) is true.
Ancestor (X,Y) is false for all other combinations of X,Y
(b) State a new rule named as SIBLING(X, Y) and construct a proof theoretic interpretation of this rule to find all siblings.
Answer 
Rule:
SIBLING(X, Y):Parent (X,Z),Parent (Y,Z)
if Z is parent of X and also Z is parent Y then, X,Y will be sibling.
Known facts
PARENT(olivia, william) is true.
PARENT(olivia, emma) is true.
PARENT (tony, martin) is true.
PARENT (tony,olivia) is true.
PARENT (john, olivia) is true.
PARENT (john, steve) is true.
PARENT (jenny, tony) is true.
PARENT (david, tony) is true.
PARENT (michael, john) is true.
PARENT (dorothy, john) is true.
PARENT (sophia, jenny) is true.
PARENT (christopher, michael) is true.
Parent (X,Y) is not true for any other combination of X,Y.
Deduced Facts
Sibling (tony, john) is true.
Sibling((jenny, david) is true.
Sibling(Michael,Dorothy) is true.
Sibling(X,Y) is not true for any other combination of X,Y.
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(c) State a new rule named as DESCENDANT(X, Y) and construct a proof theoretic interpretation of this rule to find all descendants of Martin. Note DESCENDANT(X, Y) means Y is a descendant of X.
Answer 
DECENDENT(Z,X) : PARENT(Z, Y)
DECENDANT(Z,X) : PARENT(Z,Y), PARENT(Y,X)
DECENDANT(Z,X) states Z is decendant of X
Known facts
PARENT(olivia, william) is true.
PARENT(olivia, emma) is true.
PARENT (tony, martin) is true.
PARENT (tony,olivia) is true.
PARENT (john, olivia) is true.
PARENT (john, steve) is true.
PARENT (jenny, tony) is true.
PARENT (david, tony) is true.
PARENT (michael, john) is true.
PARENT (dorothy, john) is true.
PARENT (sophia, jenny) is true.
PARENT (christopher, michael) is true.
Deduced Facts
DECENADANT(William,Olivia) is true.
DECENADANT(emma,Olivia) is true.
DECENADANT(martin,tony) is true.
DECENADANT(Olivia,tony) is true.
DECENADANT(Olivia,john) is true.
DECENADANT(Steve,john) is true.
DECENADANT(tony,jenny) is true.
DECENADANT(tony,david) is true.
DECENADANT(john,michael) is true.
DECENADANT(john,dorothy) is true.
DECENADANT(jenny,sophia) is true.
DECENADANT(Michael,christopher) is true.
DECENADANT(emma,tony) is true.
DECENADANT(emma,john) is true.
DECENADANT(martin,jenny) is true.
DECENADANT(martin,david) is true.
DECENADANT(john,christopher) is true.
DECENDANT(X,Y) is not true for any other combination X,Y.
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(d) Given the following rules:
FIRST_COUSIN(X, Y) : PARENT(X, Z), PARENT(Y, T), SIBLING(Z, T)
COUSIN(X, Y) : FIRST_COUSINS(X, Y)
COUSIN(X, Y) : PARENT(X, Z), PARENT(Y, T), COUSIN(Z, T)
Note: Two people are first cousins if their parents are siblings. Cousins means any kind of cousins. Cousins can be second cousins who are the children of the two first cousins or third cousins who are the children of two second cousins etc.
1. Prove that FIRST_COUSIN(jenny, michael) is true.
2. Prove that COUSIN(sophia, christopher) is true.
Answer  FIRST_COUSIN(X, Y) : PARENT(X, Z), PARENT(Y, T), SIBLING(Z, T)
COUSIN(X, Y) : FIRST_COUSINS(X, Y)
COUSIN(X, Y) : PARENT(X, Z), PARENT(Y, T), COUSIN(Z, T)
Known facts
PARENT(olivia, william) is true.
PARENT(olivia, emma) is true.
PARENT (tony, martin) is true.
PARENT (tony,olivia) is true.
PARENT (john, olivia) is true.
PARENT (john, steve) is true.
PARENT (jenny, tony) is true.
PARENT (david, tony) is true.
PARENT (michael, john) is true.
PARENT (dorothy, john) is true.
PARENT (sophia, jenny) is true.
PARENT (christopher, michael) is true.
Sibling (tony, john) is true.
Sibling((jenny, david) is true.
Sibling(Michael,Dorothy) is true.
Sibling(X,Y) is not true for any other combination of X,Y.
1) Known Fact
PARENT (jenny, tony) is true.
PARENT (michael, john) is true.
Deduced Fact
From rule 1, FIRST_COUSIN(jenny, michael)is true.
2) Known fact
PARENT (christopher, michael) is true.
PARENT (sophia, jenny) is true.
Deduced Fact
From rule 1, FIRST_COUSIN(christopher, michael)is true.
Hence, From rule2 FIRST_COUSIN(christopher, michael) implies COUSIN(sophia, christopher)
Therefore, COUSIN(sophia, christopher) is true.
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