1. Answers to Midterm - Exam. Feb. 27, 2006
1.(20) Apply the quick sorting algorithm to the following
sequence:
1, 10, 9, 8, 5, 7, 6, 2, 4, 3,
and trace the sequence change step by step.
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2. The center element is 5. j i i = j = 5 main idea:
1st step:
2nd step:
3rd step:
4rd step:
from
center to 10 3 i j 9 4 8 2
i = j = 5
Smaller than 5
greater than 5
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3. 8 2 i = 3 j = 2 5th step: 1 4 5 7 6 9 9 10 3 6th step: 1 2 3 4 from
center to
5th step:
6th step:
from
center to 8 3 2
i = 3
j = 2
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4. center
7th step:
from to
i = 1
j = 1
8th step:
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5. (b) When the system crashes as shown in Fig. 1, how do we undo
2.(20) (a) Tell the difference between commit points and checking points. (10)
(b) When the system crashes as shown in Fig. 1, how do we undo
or redo the transactions? Here, we assume that all these transactions
are not interferenced and at a checking point all the made by the
transactions will be stored in the database. (10)
Time T1 T2 T3 T4 T5
Time of
checkpoint
failure
Fig. 1
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6. The DBMS force-writes all changes/updates made by a
2. (a) A transaction has committed when it has finished all this
work. At this point:
The DBMS force-writes all changes/updates made by a
transaction to the log
Then the DBMS force-writes a commit record for the
Transaction
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7. A DBMS will execute a checkpoint in order to simplify the
recovery process.
At a checkpoint any committed transactions will have their
database writes (updates/changes) physically written to the
database. (The changes made by unaccomplished transactions
may also be written to the database.)
This is a four-step process
Suspend transaction execution temporarily
The DBMS force-writes all database changes to the database
The DBMS writes a checkpoint record to the log and
force-writes the log to disk
Transaction execution is resumed
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8. 2. (b) T1 – redone T2 – no redo, no undo T3 – no redo, no undo
T5 – undone
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9. 3.(20)Given the relation schemas shown below, generate an optimal
query tree for the following query:
SELECT fname, lname
FROM EMPLOYEE, DEPARTMENT, PROJECT,
WORKS_ON
WHERE salary > and Pname = ‘Web Database’
and Pnumber = PNO and hours > 30
and dno = dnumber
and Dname = ‘CS’
and ssn = ESSN
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10. Dname, dnumber, mgrssn, mgrstartdate
DEPARTMENT
Dname, dnumber, mgrssn, mgrstartdate
EMPLOYEE
fname, minit, lname, ssn, bdate, address, sex, salary, superssn, dno
PROJECT
Pname, Pnumber, Plocation, Dnum
WORKS_ON
ESSN, PNO, HOURS
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11. fname, lname((salary >50000(Employee) Dname=’CC’(Department)
dno=Dnumber
((hours >30( Works_on)
Pname=’web database’(Project))
ssn=ESSN
PNO=Pnumber
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12. Question 3 fname, lname fname, lname PNO Pnumber
ssn=ESSN
Pname= ‘web database’
fname, lname ssn
ESSN, PNO
Project
Dnumber=dno
Dept.
Dname=’BC’
dnumber
fname, lname, ssn, dno
hours > 30
salary > 50000
Employee
Works_on
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13. Answers to Midterm - Exam. Feb. 27, 2006
4.(30) (i) Explain the difference among the recoverable, cascadeless
and strict schedules. (10)
  (ii) Given the following schedules, show which is recoverable,
non-recoverable, cascadeless or strict. (20)
(a) R1(X), W1(X), R2(Z), W2(Z), R1(Y), W1(Y), R2(Y), C1,
R2(X), W2(X), C2.
(b) R1(X), W1(X), R2(X), R1(Y), W2(X), C2, W1(Y), C1.
(c) R1(X), R2(X), W1(X), R2(Y), W2(Y), W2(X), C2, R1(Y),
W1(Y), C1.
(d) R1(X), W1(X), R2(Y), W2(Y), C1, R2(X), W2(X), C2.
(e) R1(X), R2(X), W1(X), R2(Y), W2(X), C2, A1.
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14. Recoverable: A schedule S is recoverable if no transaction T in S
commits until all transactions T’ that have written an item that T reads
have committed.
cascadeless: Every transaction in the schedule reads only items that
were written by committed transaction.
Strict: a transaction can neither read nor write an item X until the last
transaction that wrote X has committed or aborted.
(ii)
Recoverable
Non-recoverable
Cascadeless
Strict
cascadeless
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15. Answers to Midterm - Exam. Feb. 27, 2006
5.(5) (i)Consider the following transactions T1 and T2, which
execute concurrently. Is the result of this execution correct?
Explain your answer and try this for X = 10, Y = 20, N = 5
and M = 8. T1
Read(X)
X := X - N
Write(X)
Read(Y)
Y := Y + N
Write(Y)
Commit T2
Read(X)
X := X + M
Write(X)
Commit
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16. The result produced by the interleaved transaction is not correct.
Answer:
The result produced by the interleaved transaction is not correct.
It suffers the so-called lost-update problem.
After this interleaved execution, X becomes 5 and Y becomes 25.
The result of the serial execution is X = 13 and Y = 25.
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17. (ii) If the two transactions are executed interleavingly as follows and the
initial values for X and Y are 20 and 30, respectively, what are the values
of X and Y after the execution. Explain why the results are not correct. (5) T1
read_lock(Y)
read_item(Y)
unlock
Write_lock(X)
Read_item(X)
X :=X+Y
Write_item(X)
Unlock(X) T2
read_lock(X)
read_item(X)
unlock(X)
write_lock(Y)
read_item(Y)
Y:=X+Y
write_item(Y)
unlock(Y)
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18. The result produced by the interleaved transaction is not correct.
Answer:
The result produced by the interleaved transaction is not correct.
It is because both the transaction use the data before the transactions’ execution.
The changes made by the transactions are not accumulated. The result is not to
the serial execution.
After this interleaved execution, both X and Y are equal to 50.
Serial executions
T1 before T2: X = 50, Y = 80
T2 before T1: X = 70, Y = 50
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