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S2 State Reduction and State Assignment

S2 State Reduction and State Assignment - Lets pick up the...

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Let’s pick up the process from the reduction of the state table: PS x 1 x 2 z 00 01 11 10 a b i c g 0 b b c f g 0 c h d d f 1 d h c e g 1 e b c i g 0 f f i i k 0 g j k g h 0 h e f c g 0 i i i i d 0 j b f c g 0 k a c e g 1 NS We have to make an implication table that compares all of the states pairwise. b c d e f g h i j k a b c d e f g h i j There is a state reduction technique called partitioning that is presented in your textbook. We are not studying it because it is, to some degree, iterative. However, there is one element of partitioning that we can use to help us with our implication table.
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Two equivalent states must have the same output for all input cases. So the first partition always separates the states into groups such that states in the same group have the same output for all input cases. With a single-output Moore machine, it is very easy to partition the states. z = 0: a, b, e, f, g, h, i, j z = 1: c, d, k It is impossible for any of the states in one of the groups to be equivalent to any of the states in the other group. We can put this information into the implication table right away, without having to do any state comparisons: b c d e f g h i j k a b c d e f g h i j This step eliminated about 40% of the comparisons before we compared a single pair of states. Furthermore, it gives us information we can use in the first pass of the implication table. Now, we perform the pairwise comparison. The table below shows the result of the first pass.
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b ci cf c d cd, de, fg e ci fi f bf, ci, gk bf, ci, fi, gk bf, ci gk g bj, cg, ki, gh bj, ck, fg, gh bj, ck, gi, gh fi, ik, gi, hk h be, fi be, cf be, cf, ci ef, fi, ci, gk ej, fk, gh, cg i bi, ci, dg bi, ci, fi, dg bi, ci, dg dk ij, ik, gi, dh ei, fi, ci, dg j fi cf cf, ci bf, fi, ci, gk bj, fk, cg, gh be bi, ci, dg, fi k ah, de, cd, fg ah a b c d e f g h i j From the implication table, we have found that a h j, b e, d k, and f i. The equivalence classes are {a, h, j}, {b, e}, {c}, {d, k}, {f, i}, and {g}. To represent the minimized state table, we must have one state from each of the equivalence classes. PS x 1 x 2 z 00 01 11 10 a b f c g 0 b b c f g 0 c a d d f 1 d a c b g 1 f f f f d 0 g a d g a 0 The reduced table has six states, and will require 3 bits of state. In the state assignment process, r = 3.
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This brings us to the state assignment. Our rationale is to make a state assignment based on maximizing the number of pairs of adjacent logic values. To do this, we will determine how many adjacencies result when two of the states in the reduced table are assigned adjacent state codes. We have 5 quantitative rules that we must apply. The first four rules must be applied prior to making an assignment, and the fifth rule must be applied after we have made an assignment based on the results from the first four rules. We will summarize the results of the first four rules in a new mileage chart. b
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S2 State Reduction and State Assignment - Lets pick up the...

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