Database Management Systems Paper Solution
Following questions have been asked in GATE CS exam.
1. Given the relations
employee (name, salary, deptno) and
department (deptno, deptname, address)
Which of the following queries cannot be expressed using the basic relational algebra
operations (U, -, x, π, σ, p)? (GATE CS 2000)
(a) Department address of every employee
(b) Employees whose name is the same as their department name
(c) The sum of all employees’ salaries
(d) All employees of a given department
Answer: (c)
Explanation:
The six basic operators of relational algebra are the selection(σ ), the projection(π), the Cartesian product (x)
(also called the cross product or cross join), the set union (U), the set difference (-), and the rename (p). These six
operators are fundamental in the sense that none of them can be omitted without losing expressive power. Many
other operators have been defined in terms of these six. Among the most important are set intersection, division,
and the natural join, but aggregation is not possible with these basic relational algebra operations. So, we cannot
run sum of all employees’ salaries with the six operations.
2. Given the following relation instance.
x y z
1 4 2
1 5 3
1 6 3
3 2 2
Which of the following functional dependencies are satisfied by the instance? (GATE CS 2000)
(a) XY -> Z and Z -> Y
(b) YZ -> X and Y -> Z
(c) YZ -> X and X -> Z
(d) XZ -> Y and Y -> X
Answer: (b)
Explanation:
A functional dependency (FD) is a constraint between two sets of attributes in a relation from a database. A FD X-
>Y require that the value of X uniquely determines the value of Y where X and Y are set of attributes. FD is a
generalization of the notion of a key.
Given that X, Y, and Z are sets of attributes in a relation R, one can derive several properties of functional
dependencies. Among the most important are Armstrong’s axioms, which are used in database normalization:
* Subset Property (Axiom of Reflexivity): If Y is a subset of X, then X ? Y
* Augmentation (Axiom of Augmentation): If X -> Y, then XZ -> YZ
* Transitivity (Axiom of Transitivity): If X -> Y and Y -> Z, then X -> Z
From these rules, we can derive these secondary rules:
* Union: If X -> Y and X -> Z, then X -> YZ
* Decomposition: If X -> YZ, then X -> Y and X -> Z
* Pseudotransitivity: If X -> Y and YZ -> W, then XZ -> W
In the above question, Y uniquely determines X and Z, for a given value of Y you can easily find out values of X
and Z.
So, Y -> X and Y -> Z hold for above schema.
From rule of augmentation we can say YZ->X. If we understand the notion of FD, we don’t need to apply axioms
to find out which option is true, just by looking at the schema and options we can say that (b) is true.
3. Given relations r(w, x) and s(y, z), the result of
select distinct w, x
from r, s
is guaranteed to be same as r, provided (GATE CS 2000)
(a) r has no duplicates and s is non-empty
(b) r and s have no duplicates
(c) s has no duplicates and r is non-empty
(d) r and s have the same number of tuples
Answer: (a)
Explanation:
The query selects all attributes of r. Since we have distinct in query, result can be equal to r only if r doesn’t have
duplicates.
If we do not give any attribute on which we want to join two tables, then the queries like above become
equivalent to Cartesian product. Cartisian product of two sets will be empty if any of the two sets is empty. So, s
should have atleast one record to get all rows of r.
4. In SQL, relations can contain null values, and comparisons with null values are treated as unknown.
Suppose all comparisons with a null value are treated as false. Which of the
following pairs is not equivalent? (GATE CS 2000)
(a) x = 5, not (not (x = 5)
(b) x = 5, x > 4 and x < 6, where x is an integer
(c) x < 5, not(x = 5)
(d) None of the above
Answer (c)
Explanation:
It doesn’t need much explanation. For all values smaller than 5, x < 5 will always be true but x = 5 will be false.
5. Consider a schema R(A, B, C, D) and functional dependencies A -> B and C -> D. Then the decomposition
of R into R1 (A, B) and R2(C, D) is (GATE CS 2001)
a) dependency preserving and loss less join
b) loss less join but not dependency preserving
c) dependency preserving but not loss less join
d) not dependency preserving and not loss less join
Answer: (c)
Explanation:
Dependency Preserving Decomposition:
Decomposition of R into R1 and R2 is a dependency preserving decomposition if closure of functional
dependencies after decomposition is same as closure of of FDs before decomposition.
A simple way is to just check whether we can derive all the original FDs from the FDs present after
decomposition.
In the above question R(A, B, C, D) is decomposed into R1 (A, B) and R2(C, D) and there are only two FDs A -> B
and C -> D. So, the decomposition is dependency preserving
Lossless-Join Decomposition:
Decomposition of R into R1 and R2 is a lossless-join decomposition if at least one of the following functional
dependencies are in F+ (Closure of functional dependencies)
R1 ∩ R2 → R1
OR
R1 ∩ R2 → R2
In the above question R(A, B, C, D) is decomposed into R1 (A, B) and R2(C, D), and R1 ∩ R2 is empty. So, the
decomposition is not lossless.
Following Questions have been asked in GATE 2012 exam.
1) Which of the following statements are TRUE about an SQL query?
P: An SQL query can contain a HAVING clause even if it does not a GROUP BY clause
Q: An SQL query can contain a HAVING clause only if it has a GROUP BY clause
R: All attributes used in the GROUP BY clause must appear in the SELECT clause
S: Not all attributes used in the GROUP BY clause need to apper in the SELECT clause
(A) P and R
(B) P and S
(C) Q and R
(D) Q and S
Answer (B)
P is correct. HAVING clause can also be used with aggregate function. If we use a HAVING clause without a
GROUP BY clause, the HAVING condition applies to all rows that satisfy the search condition. In other words, all
rows that satisfy the search condition make up a single group. See this for more details.
S is correct. To verify S, try following queries in SQL.
CREATE TABLE temp
(
id INT,
name VARCHAR(100)
);
INSERT INTO temp VALUES (1, "abc");
INSERT INTO temp VALUES (2, "abc");
INSERT INTO temp VALUES (3, "bcd");
INSERT INTO temp VALUES (4, "cde");
SELECT Count(*)
FROM temp
GROUP BY name;
Output:
count(*)
--------
2
1
1
2) Given the basic ER and relational models, which of the following is INCORRECT?
(A) An attributes of an entity can have more that one value
(B) An attribute of an entity can be composite
(C) In a row of a relational table, an attribute can have more than one value
(D) In a row of a relational table, an attribute can have exactly one value or a NULL value
Answer (C)
The term ‘entity’ belongs to ER model and the term ‘relational table’ belongs to relational model.
A and B both are true. ER model supports both multivalued and composite attributes See this for more details.
(C) is false and (D) is true. In Relation model, an entry in relational table can can have exactly one value or a
NULL.
3) Suppose (A, B) and (C,D) are two relation schemas. Let r1 and r2 be the corresponding relation
instances. B is a foreign key that refers to C in r2. If data in r1 and r2 satisfy referential integrity
constraints, which of the following is ALWAYS TRUE?
Answer (A)
B is a foreign key in r1 that refers to C in r2. r1 and r2 satisfy referential integrity constraints. So every value that
exists in column B of r1 must also exist in column C of r2.
4) Which of the following is TRUE?
(A) Every relation in 2NF is also in BCNF
(B) A relation R is in 3NF if every non-prime attribute of R is fully functionally dependent on every key of R
(C) Every relation in BCNF is also in 3NF
(D) No relation can be in both BCNF and 3NF
Answer (C)
BCNF is a stronger version 3NF. So every relation in BCNF will also be in 3NF.
Please see GATE Corner for all previous year paper/solutions/explanations, syllabus, important dates, notes, etc.
Please write comments if you find any of the answers/explanations incorrect, or you want to share more
information about the topics discussed above
1) Consider the following transactions with data items P and Q initialized to zero:
T1: read (P) ;
read (Q) ;
if P = 0 then Q : = Q + 1 ;
write (Q) ;
T2: read (Q) ;
read (P) ;
if Q = 0 then P : = P + 1 ;
write (P) ;
Any non-serial interleaving of T1 and T2 for concurrent execution leads to
(A) A serializable schedule
(B) A schedule that is not conflict serializable
(C) A conflict serializable schedule
(D) A schedule for which a precedence graph cannot be drawn
Answer (B)
Two or more actions are said to be in conflict if:
1) The actions belong to different transactions.
2) At least one of the actions is a write operation.
3) The actions access the same object (read or write).
The schedules S1 and S2 are said to be conflict-equivalent if the following conditions are satisfied:
1) Both schedules S1 and S2 involve the same set of transactions (including ordering of actions within each
transaction).
2) The order of each pair of conflicting actions in S1 and S2 are the same.
A schedule is said to be conflict-serializable when the schedule is conflict-equivalent to one or more serial schedules.
Source: Wiki Page for Schedule
In the given scenario, there are two possible serial schedules:
1) T1 followed by T2
2) T2 followed by T1.
In both of the serial schedules, one of the transactions reads the value written by other transaction as a first step.
Therefore, any non-serial interleaving of T1 and T2 will not be conflict serializable.
2) Consider the following relations A, B, C. How many tuples does the result of the following relational
algebra expression contain? Assume that the schema of A U B is the same as that of A.
Table A
Id Name Age
----------------
12 Arun 60
15 Shreya 24
99 Rohit 11
Table B
Id Name Age
----------------
15 Shreya 24
25 Hari 40
98 Rohit 20
99 Rohit 11
Table C
Id Phone Area
-----------------
10 2200 02
99 2100 01
(A) 7
(B) 4
(C) 5
(D) 9
Answer (A)
Result of AUB will be following table
Id Name Age
----------------
12 Arun 60
15 Shreya 24
99 Rohit 11
25 Hari 40
98 Rohit 20
The result of given relational algebra expression will be
Id Name Age Id Phone Area
---------------------------------
12 Arun 60 10 2200 02
15 Shreya 24 10 2200 02
99 Rohit 11 10 2200 02
25 Hari 40 10 2200 02
98 Rohit 20 10 2200 02
99 Rohit 11 99 2100 01
98 Rohit 20 99 2100 01
3) Consider the above tables A, B and C. How many tuples does the result of the following SQL query
contains?
SELECT A.id
FROM A
WHERE A.age > ALL (SELECT B.age
FROM B
WHERE B. name = "arun")
(A) 4
(B) 3
(C) 0
(D) 1
Answer (B)
The meaning of “ALL” is the A.Age should be greater than all the values returned by the subquery. There is no
entry with name “arun” in table B. So the subquery will return NULL. If a subquery returns NULL, then the
condition becomes true for all rows of A (See this for details). So all rows of table A are selected.
GATE-2015
1. Consider the following transaction involving two bank accounts x and y.
read (x) ; x := x – 50; write (x); read (y); y := y + 50; write (y)
The constraint that the sum of the accounts x and y should remain constant isthat of
(a) Atomicity
(b) Consistency
(c) Isolation
(d) Durability
Answer: option (b)
Explanation:
Consistency assures that after any transaction, the state of the database will remain valid. Database state refers
to all stored values within the database at a given instance. A valid database must satisfy all constraints specified
in the schema and over the database.
In the above transaction, 50 has been deducted from x, and y has been incremented by 50. For the database to be
valid after transaction, net amount x+y should remain constant before and after the transaction .This is ensured
by property consistency.
GATE - 2015
2. With reference to the B+ tree index of order 1 shown below, the minimum number of nodes (including the root
node) that must be fetched in order to satisfy the following query: “Get all records with a search key greater than
or equal to 7 and less than 15” is ________.
Ans: 5
Explanation:
In a B+ tree, data pointers are stored only at the leaf nodes. The leaf nodes contain the value of the search field as
well as a pointer to the record/block that contains the record.In order to provide ordered access, the leaf nodes
have been linked.
In the above question, in order to fetch all records with key values greater than or equal to 7 and less than 15, we
first need to locate the leaf node that stores 7.To reach the node with key 7,we start from root with key value
9,follow left pointer to arrive at a node with key 5 (since search key 7 is less than 9). From node with key 5, we
follow the right pointer to reach the leaf node with key value 7. Now we move from 7 till 15 sequentially towards
right, thus traversing another 2 nodes.
Thus total number of nodes fetched in order to satisfy the query is 5.
GATE-2015
3. Consider two relations R1 (A, B) with the tuples (1, 5) (3,7)and R2(A, C) = (1, 7), (4, 9). Assume that R(A,B,C) is
the full natural outer join of R1 and R2. Consider the following tuples of the form (A,B,C)” a = (1, 5, null), b = (1,
null, 7), c = (3, null, 9), d = (4, 7, null), e = (1, 5, 7), f = (3, 7, null), g = (4, null, 9). Which one of the following
statements is correct?
(a) R contains a, b, e, f, g but not c, d
(b) R contains a, b, c, d, e, f, g
(c) R contains e, f, g but not a, b
(d) R contains e but not f, g
Ans: option (c)
Explanation:
R1 :
A
B
1
5
3
7
R2:
A
C
1
7
4
9
A full outer join of R1and R2 returns all rows of R1 and R2. Thus resulting relation
R(A,B,C) is as shown below :
A
B
C
1
5
7
3
7
null
4
null
9
Hence in the tuples correspond to e, f and g in the question.
GATE-2015
4. Consider a simple checkpointing protocol and the following set of operations in the log.
(start, T4); (write, T4, y, 2, 3); (start, T1); (commit, T4); (write, T1, z, 5, 7); (checkpoint);
(start, T2); (write, T2, x, 1, 9); (commit, T2); (start, T3); (write, T3, z, 7, 2);
If a crash happens now and the system tries to recover using both undo and redo operations, what are the
contents of the undo list and the redo list
(a) Undo: T3, T1; Redo: T2
(b) Undo: T3, T1; Redo: T2, T4
(c) Undo: none; Redo: T2, T4, T3; T1
(d) Undo: T3, T1, T4; Redo: T2
Answer: option (a)
Explanation:
The given set of operations can be rewritten as :
T1 T2 T3 T4
start
Write(y, 2, 3)
start
commit
write(z, 5, 7)
checkpoint checkpoint checkpoint checkpoint
start
write( x, 1, 9)
commit
start
write( z, 7, 2)
crash crash crash crash
Checkpoint is a mechanism where all the previous logs are removed from the system and stored permanently in
a storage disk. Checkpoint declares a point before which the DBMS was in consistent state, and all the
transactions were committed. During recovery the system reads the logs backwards from the end to the last
checkpoint.
(Reference :http://www.tutorialspoint.com/dbms/dbms_data_recovery.htm ) .
In the question, transactions T1 and T3 are uncommitted hence these transactions are undone during recovery.
T4 is committed before check pointing. Since T2 is committed after the checkpoint, it has to be redone.
5. A file is organized so that the ordering of data records is the same as or close to the ordering of data entries in
some index. Then that index is called
(a) Dense
(b) Sparse
(c) Clustered
(d) Unclustered
Ans: option (c)
Explanation:
With a clustered index the physical records (rows) are stored on the disk in the same order as that of index .
Hence every table can have exactly one clustered index.
All Other Gate Question Paper Year Wise with Answer Available in Soft Copy Format.
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