logging in or signing up Normalisation Example aSGuest782 Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 2079 Category: Others/ Misc License: All Rights Reserved Like it (1) Dislike it (1) Added: October 13, 2008 This Presentation is Public Favorites: 1 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript Normalisation Example : Normalisation Example CS2312 Normalisation Example : Normalisation Example BEER_DATABASE Additional Notes: Warehouses are shared by breweries. Each beer is unique to the brewer. Each brewery is based in a city. Minimal Sets of Functional Dependencies : Minimal Sets of Functional Dependencies A set of functional dependencies F is minimal if: 1. Every dependency F has a single determined attribute A 2. We cannot remove any dependency from F and still have a set of dependencies equivalent to F 3. We cannot replace and dependency X A in F with a dependency A X, where A X and still have a set of dependencies that is equivalent to F I.e. a canonical form with no redundancies (beer, brewery, strength, city, region, warehouse, quantity) beer brewery beer strength brewery city city region beer, warehouse, quantity Relational Synthesis Algorithm into 3NF: (beer, brewery, strength, city, region, {warehouse, quantity}) : Relational Synthesis Algorithm into 3NF: (beer, brewery, strength, city, region, {warehouse, quantity}) set D := { R } ; P. 426, P. 431 1. Find a minimal cover G for F 2. For each determinant X of a functional dependency that appears in G create a relation schema { X A1, X A2…X Am} in D where X A1, X A1, … X A1m are the only dependencies in G with X as the determinant; 3. Place any remaining (unplaced) attributes in a single relation to ensure attribute preservation property so we don’t lose anything. 4. If none of the relations contains a key of R, create one more relation that contains attributes that form a key for R. beer brewery (beer, brewery, strength) beer strength brewery city (brewery, city) city region (city, region) beer, warehouse, quantity (beer, warehouse, quantity) Step-wise normalisation: (beer, brewery, strength, city, region, {warehouse, quantity}) : Step-wise normalisation: (beer, brewery, strength, city, region, {warehouse, quantity}) beer brewery, strength partial dependency brewery city transitive dependency city region transitive dependency beer, warehouse, quantity repeating group 1NF remove repeating group (beer, brewery, strength, city, region, {warehouse, quantity}) (beer, warehouse, quantity) beer, warehouse, quantity (beer, brewery, strength, city, region) beer brewery, strength transitive dependency brewery city transitive dependency city region (beer, brewery, strength, city, region) : (beer, brewery, strength, city, region) beer brewery, strength brewery city transitive dependency city region transitive dependency 2NF no partial dependencies 3NF/BCNF no transitive dependencies (beer, brewery, strength, city, region) (city, region) city region (beer, brewery, strength, city) beer brewery, strength brewery city (brewery, city) brewery city (beer, brewery, strength) beer brewery, strength Take the most indirect transitive dependencies Using BNCF decomposition algorithm:(beer, brewery, strength, city, region, warehouse, quantity) : Using BNCF decomposition algorithm:(beer, brewery, strength, city, region, warehouse, quantity) beer brewery, strength partial dependency brewery city transitive dependency city region transitive dependency beer, warehouse, quantity Directly to BCNF take a violating dependency and form a relation from it. First choose a direct transitive dependency and its closure (beer, brewery, strength, city, region, warehouse, quantity) brewery city (brewery, city, region) brewery city city region transitive dependency (beer, brewery, strength, warehouse, quantity) beer brewery, strength partial dependency beer, warehouse, quantity Using BNCF decomposition algorithm:(beer, brewery, strength, city, region, warehouse, quantity) : Using BNCF decomposition algorithm:(beer, brewery, strength, city, region, warehouse, quantity) beer brewery, strength partial dependency brewery city transitive dependency city region transitive dependency beer, warehouse, quantity take a violating dependency and form a relation from it. First the partial dependency and its closure (beer, brewery, strength, city, region, warehouse, quantity) beer brewery, strength (beer, brewery, strength, city, region) beer brewery, strength brewery city transitive dependency city region transitive dependency normalise as before... 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Normalisation Example aSGuest782 Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 2079 Category: Others/ Misc License: All Rights Reserved Like it (1) Dislike it (1) Added: October 13, 2008 This Presentation is Public Favorites: 1 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript Normalisation Example : Normalisation Example CS2312 Normalisation Example : Normalisation Example BEER_DATABASE Additional Notes: Warehouses are shared by breweries. Each beer is unique to the brewer. Each brewery is based in a city. Minimal Sets of Functional Dependencies : Minimal Sets of Functional Dependencies A set of functional dependencies F is minimal if: 1. Every dependency F has a single determined attribute A 2. We cannot remove any dependency from F and still have a set of dependencies equivalent to F 3. We cannot replace and dependency X A in F with a dependency A X, where A X and still have a set of dependencies that is equivalent to F I.e. a canonical form with no redundancies (beer, brewery, strength, city, region, warehouse, quantity) beer brewery beer strength brewery city city region beer, warehouse, quantity Relational Synthesis Algorithm into 3NF: (beer, brewery, strength, city, region, {warehouse, quantity}) : Relational Synthesis Algorithm into 3NF: (beer, brewery, strength, city, region, {warehouse, quantity}) set D := { R } ; P. 426, P. 431 1. Find a minimal cover G for F 2. For each determinant X of a functional dependency that appears in G create a relation schema { X A1, X A2…X Am} in D where X A1, X A1, … X A1m are the only dependencies in G with X as the determinant; 3. Place any remaining (unplaced) attributes in a single relation to ensure attribute preservation property so we don’t lose anything. 4. If none of the relations contains a key of R, create one more relation that contains attributes that form a key for R. beer brewery (beer, brewery, strength) beer strength brewery city (brewery, city) city region (city, region) beer, warehouse, quantity (beer, warehouse, quantity) Step-wise normalisation: (beer, brewery, strength, city, region, {warehouse, quantity}) : Step-wise normalisation: (beer, brewery, strength, city, region, {warehouse, quantity}) beer brewery, strength partial dependency brewery city transitive dependency city region transitive dependency beer, warehouse, quantity repeating group 1NF remove repeating group (beer, brewery, strength, city, region, {warehouse, quantity}) (beer, warehouse, quantity) beer, warehouse, quantity (beer, brewery, strength, city, region) beer brewery, strength transitive dependency brewery city transitive dependency city region (beer, brewery, strength, city, region) : (beer, brewery, strength, city, region) beer brewery, strength brewery city transitive dependency city region transitive dependency 2NF no partial dependencies 3NF/BCNF no transitive dependencies (beer, brewery, strength, city, region) (city, region) city region (beer, brewery, strength, city) beer brewery, strength brewery city (brewery, city) brewery city (beer, brewery, strength) beer brewery, strength Take the most indirect transitive dependencies Using BNCF decomposition algorithm:(beer, brewery, strength, city, region, warehouse, quantity) : Using BNCF decomposition algorithm:(beer, brewery, strength, city, region, warehouse, quantity) beer brewery, strength partial dependency brewery city transitive dependency city region transitive dependency beer, warehouse, quantity Directly to BCNF take a violating dependency and form a relation from it. First choose a direct transitive dependency and its closure (beer, brewery, strength, city, region, warehouse, quantity) brewery city (brewery, city, region) brewery city city region transitive dependency (beer, brewery, strength, warehouse, quantity) beer brewery, strength partial dependency beer, warehouse, quantity Using BNCF decomposition algorithm:(beer, brewery, strength, city, region, warehouse, quantity) : Using BNCF decomposition algorithm:(beer, brewery, strength, city, region, warehouse, quantity) beer brewery, strength partial dependency brewery city transitive dependency city region transitive dependency beer, warehouse, quantity take a violating dependency and form a relation from it. First the partial dependency and its closure (beer, brewery, strength, city, region, warehouse, quantity) beer brewery, strength (beer, brewery, strength, city, region) beer brewery, strength brewery city transitive dependency city region transitive dependency normalise as before... (beer, warehouse, quantity) beer, warehouse, quantity