
- DBMS Tutorial
- DBMS - Home
- DBMS - Overview
- DBMS - Architecture
- DBMS - Data Models
- DBMS - Data Schemas
- DBMS - Data Independence
- Entity Relationship Model
- DBMS - ER Model Basic Concepts
- DBMS - ER Diagram Representation
- DBMS - Generalization, Aggregation
- Relational Model
- DBMS - Codd's Rules
- DBMS - Relational Data Model
- DBMS - Relational Algebra
- DBMS - ER to Relational Model
- DBMS- SQL Overview
- Relational Database Design
- DBMS - Database Normalization
- DBMS - Database Joins
- Storage and File Structure
- DBMS - Storage System
- DBMS - File Structure
- Indexing and Hashing
- DBMS - Indexing
- DBMS - Hashing
- Transaction And Concurrency
- DBMS - Transaction
- DBMS - Concurrency Control
- DBMS - Deadlock
- Backup and Recovery
- DBMS - Data Backup
- DBMS - Data Recovery
- DBMS Useful Resources
- DBMS - Quick Guide
- DBMS - Useful Resources
- DBMS - Discussion
Find the canonical cover of FD {A->BC, B->AC, C->AB} in DBMS
Canonical cover is called minimal cover which is called the minimum set of FDs. A set of FD FC is called canonical cover of F if each FD in FC is a Simple FD, Left reduced FD and Non-redundant FD.
Simple FD − X->Y is a simple FD if Y is a single attribute.
Left reduced FD: X->Y is a left reduced FD if there are no extraneous attributes in X.{extraneous attributes: let XA->Y then A is a extraneous attribute if X_>Y}
Non-redundant FD − X->Y is a Non-redundant FD if it cannot be derived from F- {X->y}.
Problem
Find the canonical cover of FD {A->BC, B->AC, C->AB}.
Solution
Relational schema R(A,B,C) F: {A->BC, B->AC, C->AB}
Step 1 − Create a singleton right hand side
dependency A->BC will break into A->B, A->C.
F: { A->B A->C B->A B->C C->A C->B}
Step 2 − Remove extraneous attributes if any exists.
F:{ A->B A->C B->A B->C C->A C->B} NO extraneous attributes exists
Step 3 − Remove the redundant FD
F: { A->B A->C B->A B->C C->A C->B }
Remove B->A dependency and we can get A from B through B->C and C->A.
F= {A->B A->C B->C C->A C->B}
By removing C->B dependency we get B from C through C->A , A->B.
F={A->B B->C C->A A->C}
By removing A->C dependency we can determine C from A through A->B, B->C
Step 4 − The final canonical cover is as follows −
FC ={ A->B, B->C, C->A } [A]+ =BC [B]+=AC [C]+=AB.
- Related Articles
- What is the minimal set of functional dependencies or canonical cover of FD?
- Draw a line segment, say AB. Take any point C lying in between A and B. Measure the lengths of AB , BC , and AC. Is $AB = AC + CB $?
- Draw any line segment, say $\overline{AB}$. Take any point C lying in between. Measure the lengths of AB, BC, and AC. Is $AB=AC+CB$?
- $ABCD$ is a quadrilateral.Is $AB + BC + CD + DA > AC + BD?$
- If $a + b + c = 0$ and $a^2 + b^2 + c^2 = 16$, find the value of $ab + bc + ca$.
- If $a^2 + b^2 + c^2 = 16$ and $ab + bc + ca = 10$, find the value of $a + b + c$.
- If $a + b + c = 9$ and $ab + bc + ca = 23$, find the value of $a^2 + b^2 + c^2$.
- Choose the correct answer from the given four options:If \( \Delta \mathrm{ABC} \sim \Delta \mathrm{EDF} \) and \( \Delta \mathrm{ABC} \) is not similar to \( \Delta \mathrm{D} \mathrm{EF} \), then which of the following is not true?(A) \( \mathrm{BC} \cdot \mathrm{EF}=\mathrm{A} C \cdot \mathrm{FD} \)(B) \( \mathrm{AB}, \mathrm{EF}=\mathrm{AC} \cdot \mathrm{DE} \)(C) \( \mathrm{BC} \cdot \mathrm{DE}=\mathrm{AB} \cdot \mathrm{EF} \)(D) \( \mathrm{BC}, \mathrm{DE}=\mathrm{AB}, \mathrm{FD} \)
- If $a +b + c = 9$ and $ab + bc + ca = 26$, find the value of $a^3 + b^3 + c^3 - 3abc$.
- Draw any line segment, say AB. Take any point C lying in between A and B.Measure the length of AB, BC and AC.Is AB =AC+CB?
- $ABC$ is a triangle, right-angled at $C$. If $AB=25\ cm$ and $AC=7\ cm$, find $BC$.
- In an isosceles triangle $ABC$, if $AB\ =\ AC\ =\ 13\ cm$ and the altitude from $A$ on $BC$ is $5\ cm$, find $BC$."\n
- Construct a $∆ABC$ in which $BC = 3.6\ cm, AB + AC = 4.8\ cm$ and $\angle B = 60^o$.
- Construct a $∆ABC$ in which $AB + AC = 5.6\ cm, BC = 4.5\ cm$ and $\angle B = 45^o$.
- Construct a $\triangle ABC$ in which $BC = 3.4\ cm, AB - AC = 1.5\ cm$ and $\angle B = 45^o$.
