A2 Flow Table Reduction

# Remember that none of the pairs of states above that

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Remember that none of the pairs of states above that failed the equivalence test can be compatible, either because they have different outputs, which is common to the definition of equivalence and compatibility, or because some link in the chain of compatibility eventually requires that a pair of states having different outputs be compatible, which is false for the same reason as stated previously. This means that we can incorporate the above information into our implication table for step 2:

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2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 We have the following compatibilities: 1-2, 2-4, 2-7, 3-5, 4-7, 6-8, 6-9, 6-10, 8-9, 8-10, 9-10. We can assemble the following maximally compatible sets of states: {1,2};{2,4,7};{3,5};{6,8,9,10}. All of these sets of states are “essential” to the cover, but it’s possible that we can eliminate state 2 from one of the sets that contains it. In fact, since all of the individual pairs of states are compatible by inspection, there are no possible closure violations for any compatibility class that we might choose. We can therefore investigate two choices for the reduced table, the one covering compatibility classes {1,2};{3,5};{4,7};{6,8,9,10}, and the one covering {1}; {2,4,7};{3,5};{6,8,9,10}. One or the other might be a better choice because it preserves don’t cares in the reduced flow table. Here is the reduced table based on the first choice of compatibility classes: 00 01 11 10 1,2 1 / 0 2 / 0 4 / - 3 / - 3,5 5 / 1 8 / - 6 / - 3 / 1 4,7 1 / - 2 / - 4 / 0 7 / 0 6,8,9,10 10 / 0 8 / 0 6 / 0 9 / 0 And here is the same table with new state names substituted for the old ones: 00 01 11 10 a a / 0 a / 0 c / - b / - b b / 1 d / - d / - b / 1 c a / - a / - c / 0 c / 0 d d / 0 d / 0 d / 0 d / 0 Notice how this table explicitly expresses the fact that once we get into state d , we can’t get out of it. This is
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