A Super key is a combination of one or more than one attributes that can uniquely identify a row in the Relational Database
whereas
A Functionally dependency is used to show dependency of one attribute other other when one attribute can uniquely identify the other attribute
Like if we say A---> B then it means that A can uniquely determine B and Attribute B is functionally dependent on Attribute A
For Example
we have a Student Database
Stu_ID | Stu_name | Stu_Age |
1 | A | 20 |
2 | B | 25 |
3 | C | 16 |
4 | D | 16 |
Here Stu_ID attribute uniquely identifies each row student table So it the super key
And
if we know Stu_ID we can tell the student name associated with it. This is known as functional dependency
We can write it as Stu_ID->Stu_Name
ii)
When we are given set of Functional Dependencies then to find the candidate key . we need to take attribute closure of set of attributes and the set that covers all the attributes in the closure That is the Candidate Key
All those attributes which can be functionally determined from an attribute set is called as a closure of that attribute Denoted as { Attribute name } +.
For example we have A-->BC C---->A
{A} + ={ A,B,C} {B}+ ={ B} {C}+ ={ A ,C}
Now we have
? = {?,?,?,?,?,?,?,?,?,?,?,?,?}
? = {?→????,?→???,?→?,??→?,??→?}
***Important Note*** :-->If any Attribute is not in RHS of Any Functional Dependency then it must be include in Key
So we have A ,I .L that are not present in The RHS of any FD's
Lets see their Closure
{ AIL} += { A B C D E F G H I J K M }
Hence we can get all attributed in closure
So Candidate Key is AIL
This is how you can get the key from given set of FD's
I hope that you understood the Concept so DO Like the answer and have any doubt ask in comment . I will reply asap
Q2: Explain with example the difference between super keys and functional dependencies in relational databases. Show...
Q2: Explain with example the difference between super keys and functional dependencies in relational databases. Show how to find a key (super/candidate) for the following functional dependencies [10 points]: R = {A, B, C, D, E, F, G, H, I, J, K,L,M} F {A B C D E, E F G H I J,AI →K,AL →M} =
Hi,
This is relational databases question
I'm struggling with this homework question. Trying to better
understand what's being asked. Could you please explain your
answer. Many thanks
Consider the relation schema R-A, B, C, D) with an unknown set of functional dependencies 4.1 What is the maximum number of (candidate) keys that R may have? Design a set of functional depen- (2 dencies on R to achieve the maximum number of (candidate) keys. Justify your answers. Mark)
Write the complete proof.
Consider the relational schemas given below and the respective sets of functional dependencies valid in the schemas For each one of the relational schemas, determine the highest normal form, which is valid for a schema. Justify your answer If a schema is not in BCNF, then decompose it into a minimum number of schemas so that each one of them is in BCNF. Justify your answers. Justification must include the derivations of minimal keys from the...
Consider a relation R(A,B,C,D,E) with the following functional dependencies: 8. AB C BCD CDE DEA (a) Specify all candidate keys for R. (b) Which of the given functional dependencies are Boyce-Codd Normal Form (BCNF) violations'? (c) Give a decomposition of R into BCNF based on the given functional dependencies. (d) Give a different decomposition of R into BCNF based on the given functional dependencies. (e) Give a decomposition of R into 3NF based on the given functional dependencies.
Consider a...
Databases Consider the following functional dependencies that hold on R(ABCDEF): AB-> CD, E -> C, B -> EF a) Is R in 3NF? b) Explain your answer in detail and decompose the relation, as necessary, into a collection of relations that are in 3NF. c) Make sure you first find all the key(s) of R so that you will be able to tell whether an attribute is prime. *Please answer all parts of this question in detail for credit.
Databases question
Consider the following relation that stores information about
students living in dormitories at a college:
a. Using these assumptions and stating any others you need to
make, list all the non-trivial functional dependencies for this
relation
b. What are the candidate keys for this relation? Identify the
primary keys
c. Is this relation in third normal form? if not find a 3NF
lossless join decomposition of College that preserves
dependencies.
d.Is the relation or the resulting set of...
Language: SQL - Normalization and Functional
Dependencies
Part 4 Normalization and Functional Dependencies Consider the following relation R(A, B, C, D)and functional dependencies F that hold over this relation. F=D → C, A B,A-C Question 4.1 (3 Points) Determine all candidate keys of R Question 4.2 (4 Points) Compute the attribute cover of X-(C, B) according to F Question 43 (5 Points) Compute the canonical cover of F.Show each step of the generation according to the algorithm shown in class....
Consider the following schema and functional dependencies: SHIPPING (ShipName, ShipType, VoyageID, Cargo, Port, ArrivalDate) Key: ShipName, ArrivalDate FD1: ShipName > ShipType FD2: VoyageID > ShipName, Cargo FD3: ShipName, ArrivalDate > VoyageId, Port 1.Please list the final set of 3NF schema including all its keys. 2.Do any of the finalized 3NF schema have determinates that are not candidate keys? If yes, explain - which schema(s)? Why?
R = {Q1,Q2,Q3,Q4,Q5,Q6,Q7,Q8,Q9,QJ} Given these functional dependencies: {Q5,Q6} > {Q7} {Q5} > {Q8,Q9} {Q6} > {QJ} {QJ} > {Q1, Q2} {Q8} > {Q3, Q4} {Q3} > {Q7, Q1, Q2, Q5, Q6} a. Identify 2 primary keys to this table b. Assuming it is in 1NF, explain why it is not in 2NF.Make the required adjustments to convert table to 2NF.Show primary key of each table c. Assuming it is in 1NF, explain why your 2NF table is not in 3NF....
Table T1(A, B, C, D, E) functional dependencies: A -> B AC -> DE D -> C List candidate key(s), and how you found the key(s). Which functional dependency causes BCNF violation? Decompose T1 into BCNF tables. Please show all work here.