CS375 2010 Fall Solution Key #3

CS375 2010 Fall Solution Key #3 - Solution Key #3 Page 1 of...

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CS375 2010 Fall Solution Key #3 1. Given F = {AB --> C, C --> B}, then F^+ = {A --> A, B --> B, C --> B, C --> C, C --> BC, AB --> A, AB --> B, AB --> C, AB --> AB, AB --> AC, AB --> BC, AB --> ABC, AC --> A, AC --> B, AC --> C, AC --> AB, AC --> AC, AC --> BC, AC --> ABC, BC --> B, BC --> C, BC --> BC, ABC --> A, ABC --> B, ABC --> C, ABC --> AB, ABC --> AC, ABC --> BC, ABC --> ABC} 2. Let G = {A --> C, AB --> C, C --> DI, CD --> I, EC --> AB, EI --> C}. Then NONREDUNDANT(G) sets F = {A --> C, AB --> C, C --> DI, CD --> I, EC --> AB, EI --> C}. Consider A --> C. F - {A --> C} does not logically imply A --> C since the closure of A under F - {A --> C} is A^+ = A and C is not a subset of A. Consider AB --> C. F - {AB --> C} logically implies AB --> C since the closure of AB under F - {AB --> C} is AB^+ = ABCDI and C is a subset of ABCDI. Thus, the FD AB --> C is removed and F = {A --> C, C --> DI, CD --> I, EC --> AB, EI --> C}. Consider C --> DI. F - { C --> DI } does not logically imply C --> DI since the closure of C under F - {C --> DI} is C^+ = C and DI is not a subset of C. Consider CD --> I. F - {CD --> I} logically implies CD --> I since the closure of CD under F - {CD --> I} is CD^+ = CDI and I is a subset of CDI. Thus, the FD CD --> I is removed and F = {A --> C, C --> DI, EC --> AB, EI --> C}. Consider EC --> AB. F - {EC --> AB} does not logically imply EC --> AB since the closure of EC under F - {EC --> AB} is EC^+ = ECDI and AB is not a subset of ECDI. Lastly, consider EI --> C.
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This note was uploaded on 11/15/2010 for the course CS 375 taught by Professor Dr.butz during the Fall '10 term at University of Regina.

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CS375 2010 Fall Solution Key #3 - Solution Key #3 Page 1 of...

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