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3.13ACBC030109

# 3.13ACBC030109 - 1 3.13 ACBC Recall the reduced table for...

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1 3.13. ACBC. Recall the reduced table for AC from section 3.10. REDUCED AC 1. A . fA C . gA. INF. AL. ALF. FIN. NON. 2. A . fA C . gC. INF. AL. ALF. FIN. NON. 3. A . fA C . gB. INF. AL. ALF. FIN. NON. 4. B . fA C . gA. INF. AL. ALF. FIN. NON. 5. B . fA C . gC. INF. AL. ALF. FIN. NON. 6. B . fA C . gB. INF. AL. ALF. FIN. NON. Recall the reduced table for BC from section 3.8. REDUCED BC 1’. B . fB C . gB. INF. AL. ALF. FIN. NON. 2’. B . fB C . gC. INF. AL. ALF. FIN. NON. 3’. B . fB C . gA. INF. AL. ALF. FIN. NON. 4’. A . fB C . gB. INF. AL. ALF. FIN. NON. 5’. A . fB C . gC. INF. AL. ALF. FIN. NON. 6’. A . fB C . gA. INF. AL. ALF. FIN. NON. We can take advantage of symmetry through interchanging A with B as follows. Clearly (i,j’) and (j,i’) are equivalent, by interchanging A and B. So we can require that i j. Thus we have the following 21 ordered pairs to consider. We must determine the status of all attributes INF, AL, ALF, FIN, NON, for each pair. 1,1’. A . fA C . gA, B . fB C . gB. INF. AL. ALF. FIN. NON. 1,2’. A . fA C . gA, B . fB C . gC. ¬ INF. ¬ AL. ¬ ALF. FIN. NON. 1,3’. A . fA C . gA, B . fB C . gA. INF. AL. ALF. FIN. NON. 1,4’. A . fA C . gA, A . fB C . gB. INF. AL. ALF. FIN. NON. 1,5’. A . fA C . gA, A . fB C . gC. ¬ INF. ¬ AL. ¬ ALF. FIN. NON. 1,6’. A . fA C . gA, A . fB C . gA. INF. AL. ALF. FIN. NON. 2,2’. A . fA C . gC, B . fB C . gC. INF. AL. ALF. FIN. NON.

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2 2,3’. A . fA C . gC, B . fB C . gA. ¬ INF. ¬ AL. ¬ ALF. FIN. NON. 2,4’. A . fA C . gC, A . fB C . gB. ¬ INF. AL. ¬ ALF. FIN. NON. 2,5’. A . fA C . gC, A . fB C . gC. INF. AL. ALF. FIN. NON. 2,6’. A . fA C . gC, A . fB C . gA. ¬ INF. ¬ AL. ¬ ALF. FIN. NON. 3,3’. A . fA C . gB, B . fB C . gA. INF. AL. ALF. FIN. NON. 3,4’. A . fA C . gB, A . fB C . gB. INF. AL. ALF.
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3.13ACBC030109 - 1 3.13 ACBC Recall the reduced table for...

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