U.G.C. NET Exam. 4 December, 2019 Paper II (COMPUTER SCIENCE & APPLICATIONS)

In given graph contains 9 nodes and 11 edges. The nodes are nothing but vertices. Here, In given diagram the number of rounds one the nodes. The nodes one 1, (2, 3), (4, 5), 6, 7, 8, 9, 10, 11
Count of total nodes in flow graph F are 9.

Cyclomatic complexity of a flow graph is the quantitative measure of the number of linearly independent paths in it.
for calculation of cyclomatic complexity, it is necessary to understand some parameters.
1-Region- In out case, there are 4 regions.
2-Edges- In out case, there are 9 nodes.
3-Nodes- In our case, there is 1 connected component.
5-Predicates- These are nodes that contain condition in our case, there are 3 predicates namely node 1, 2, 3, 6.
Formula- There are three ways of calculating the cyclamate complexity m,
1- m = number of regions
2- m = E - N + 2P
where
E - Number of Edges
N - Number of Nodes
P - Number of connected component
m = P + 1
3- where, P = number of predicate nodes
Calculation-
using formula 1, m = 4
using formula 2, m = 11–9+2(1) = 4
using formula 3, m = 3+1 = 4

REGION- region is an area that is bounded by edges and cannot be further subdivided. Region can be finite nature and can also be infinite nature planar graph can only have one infinite region. The graph in the data has 3 finite or closed regions.
Region 1- Bounded by edge 1-10-1
Region 2- Bounded by edges (2.3)-6-8-9-10-(4.5)-(2.3)
Region 3- Bounded by edges 6-7-9-8-6 The fourth region is infinite or open region Region 4-1- 11

Path 1: 1-11
Path 2: 1-2-3-4-5-10-1-11
Path 3: 1-(2, 3)-6-8-9-10-1-11
Path 4: 1-(2, 3)-6-7-9-10-1-11
Each of the above path introduces a new edge and Path 3 and Path 4 one longest independent paths with 8 Node.
Note that a path like 1-(2, 3)-4-5-10-1-2-3-6-8-9-10-1- 11 is not an independent path because it includes already specified edges and doesn't exclusively add new edges.

Predicate node is a node that contains a condition. This node has 2 or more outgoing edges from it.
In the given graph the nodes with a condition or with 2 or more outgoing edges one - 1, 2.3, 6. Hence, there are 3 predicates nodes.

A canonical cover is simplified and reduced version of the given set of functional dependencies minimal canonical cover is a cover for which removed of any single functional dependency destroys the covering property.
Data-
The functional dependencies over the universal relation R(A, B, C, D, E) are
A → BC = A → B, A → C
D → E
BC → D
A → D
Method- Following are the steps to find the minimal cover.
1- Write all FDs such that RHS of each FD has only one attribute.
A → B
A → C
D → E
BC → D
A →D
2- Remove transitive FDs and any other FDs where the RHS can be driven of LHS of the other FDs. In this case, A → D is a redundant FD. Hence the result becomes
A → B
A → C
D → E
BC → D
3- Finding the FDs with 2 attributes in LHS, we get the dependency BC → D. Try to eliminate the attributes using closure of both the attribute B and C. In this case, it is not possible to remove either B or C because both have different closures. Hence, minimal cover is
A → B
A → C
D → E
BC → D

A primary Rey is a special relational database table column (or combination of columns) designated to uniquely identify all table records.
Data- In the functional dependency set f of relation R, the key 'A' is not in the RHS of any FD.
Explanation:- It A is not in RHS of any FD, it is mandatory to includes it in the set of candidate key. Now, to make sure if 'A' alone is the primary key, we write its closure. The set of all those attributes which can be functionally determined from an attribute set is called as a closure of that attribute set.
A + = {A, B, C, D, E}
A derives all the attributes of relation R. Hence, it is the primary key of R.

Normal forms refer to type of functional dependencies.
INF- A relation is in 1 NF if it contains an atomic value.
2NF- A relation will be in 2NF if it is in 1NF and all non-key attributes are fully functional dependent on the primary key.
3NF- A relation will be in 3NF if it is in 2NF and no transition dependency exists. BCNF- A relation is in
BCNF if every functional dependency X → Y, X is the super key of the table.
Data
1- A is the primary key.
2- There is no partial dependency in R. Partial dependency occurs when a non-prime attribute is functionally dependent on part of a candidate key. Hence, R, is in 2NF Explanation- R is in 2NF and not in 3NF because D → E, BC → D neither have a super key in their LHS nor do they have a primary key in their RHS. If it is not in 3NF, it cannot be in BCNF.

Dependency A → D is derived from A → BC, BC → D. Hence it is transitive and redundant.