H�z`����0 !N����V����E�R�N�V��Y����'YA�3W-�_�ɗw>�P�������t7��V��r��� �w!dX[?��[���@e����7�w����!ߺ��˓).�t:\ӎ �7o�}-���oش|�ee��p=i�#^�W"9R����E 14 23 Before you go through this article, make sure that you have gone through the previous article on In this article, we will discuss how to find candidate keys of a given relation.A set of minimal attribute(s) that can identify each tuple uniquely in the given relation is called as a candidate key.These are candidate keys because each set consists of minimal attributes required to identify each student uniquely in the Student table.We can determine the candidate keys of a given relation using the following steps-Let R(A, B, C, D, E, F) be a relation scheme with the following functional dependencies-Here, the attributes which are not present on RHS of any functional dependency are A, C and F.If all essential attributes together can determine all remaining non-essential attributes, then-If all essential attributes together can not determine all remaining non-essential attributes, then-We will further understand how to find candidate keys with the help of following problems.The following practice problems are based on Case-01.Let R = (A, B, C, D, E, F) be a relation scheme with the following dependencies-Also, determine the total number of candidate keys and super keys.We will find candidate keys of the given relation in the following steps-= { C , E , F } ( Using C → F )= { A , C , E , F } ( Using E → A )= { A , C , D , E , F } ( Using EC → D )We conclude that CE can determine all the attributes of the given relation.So, CE is the only possible candidate key of the relation.There are total 6 attributes in the given relation of which-So, number of super keys possible = 2 x 2 x 2 x 2 = 16.Let R = (A, B, C, D, E) be a relation scheme with the following dependencies-Determine the total number of candidate keys and super keys.We will find candidate keys of the given relation in the following steps-= { A , B , C } ( Using AB → C )We conclude that AB can determine all the attributes of the given relation.There are total 5 attributes in the given relation of which-Consider the relation scheme R(E, F, G, H, I, J, K, L, M, N) and the set of functional dependencies-Also, determine the total number of candidate keys and super keys.We will find candidate keys of the given relation in the following steps-= { E , F , G , H } ( Using EF → G )= { E , F , G , H , I , J } ( Using F → IJ )= { E , F , G , H , I , J , K , L } ( Using EH → KL )= { E , F , G , H , I , J , K , L , M } ( Using K → M )= { E , F , G , H , I , J , K , L , M , N } ( Using L → N )We conclude that EFH can determine all the attributes of the given relation.So, EFH is the only possible candidate key of the relation.There are total 10 attributes in the given relation of which-So, number of super keys possible = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 128.Consider the relation scheme R(A, B, C, D, E, H) and the set of functional dependencies-We will find candidate keys of the given relation in the following steps-Watch video lectures by visiting our YouTube channel Given below are the examples of super keys since each set can uniquely identify each student in the Student table-All the attributes in a super key are definitely sufficient to identify each tuple uniquely in the given relation but all of them may not be necessary.A set of minimal attribute(s) that can identify each tuple uniquely in the given relation is called as a candidate key.Given below are the examples of candidate keys since each set consists of minimal attributes required to identify each student uniquely in the Student table-A primary key is a candidate key that the database designer selects while designing the database.Candidate key that the database designer implements is called as a primary key.Candidate keys that are left unimplemented or unused after implementing the primary key are called as alternate keys.Unimplemented candidate keys are called as alternate keys.Here, t_dept can take only those values which are present in dept_no in Department table since only those departments actually exist.Here, using partial key Emp_no, we can not identify a tuple uniquely but we can select a bunch of tuples from the table.A primary key comprising of multiple attributes and not just a single attribute is called as a composite key.Surrogate key is a key with the following properties-Mobile Number of students in a class where every student owns a mobile phone.Secondary key is required for the indexing purpose for better and faster searching.Watch video lectures by visiting our YouTube channel
/Root 15 0 R DBMS keys allow you to establish a relationship between and identify the relation between tables; Seven Types of DBMS keys are Super, Primary, Candidate, Alternate, Foreign, Compound, Composite, and Surrogate Key. �xu � �2� Constraints enforce limits to the data or type of data that can be inserted/updated/deleted from a table. In DBMS, there are following 5 different types of relational constraints-Tuple Uniqueness constraint specifies that all the tuples must be necessarily unique in any relation.This relation satisfies the tuple uniqueness constraint since here all the tuples are unique.This relation does not satisfy the tuple uniqueness constraint since here all the tuples are not unique.This relation does not satisfy the key constraint as here all the values of primary key are not unique.This relation does not satisfy the entity integrity constraint as here the primary key contains a NULL value.The following two important results emerges out due to referential integrity constraint-Consider the following two relations- ‘Student’ and ‘Department’.Here, relation ‘Student’ references the relation ‘Department’.To ensure the correctness of the database, it is important to handle the violation of referential integrity constraint properly.Watch video lectures by visiting our YouTube channel Liked this article? /Length 110 * Primary Key constraint * Foreign Key constraint * Unique Key constraint Many tables will have a primary key constraint and a table may only have one primary key constraint. endobj /FirstChar 32
0000002069 00000 n << %%EOF endobj x�c```f``Qb`e`PL`�g@ ~�+G�i��W3�8�:RZ�Ja�H@��z�X8/��_���~����;����XR8��fLa E���C*����xN��;��vB/j�:��CQ�C�*�(�ǔ7s!�c��? /Filter /FlateDecode By specifying NULL constraint, we can be sure that a particular column(s) cannot have NULL values.UNIQUE Constraint enforces a column or set of columns to have unique values. /T 77400
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