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Pair (2, 1) is marked on the first pass because 2 is an accepting state and 1 is not.

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Pair (2, 1) is marked on the first pass because 2 is an accepting state and 1 is not.

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  1. List all unordered pairs of distinct states (p, q). Make a sequence of passes through these pairs as follows. On the first pass, mark each pair (p, q) for which exactly one of the two states p and q is in A. On each subsequent pass, examine each unmarked pair (r, s); if there is a symbol σ ∈ such that δ(r, σ) = p, δ(s, σ) = q, and the pair (p, q) has already been marked, then mark (r, s). After a pass in which no new pairs are marked, stop. At that point, the marked pairs (p, q) are precisely those for which p q.

  2. 1 Pair (2, 1) is marked on the first pass because 2 is an accepting state and 1 is not.

  3. 1 1 Pair (6, 1) is marked on the first pass because 6 is an accepting state and 1 is not.

  4. 1 1 1 Pair (3, 2) is marked on the first pass because 2 is an accepting state and 3 is not.

  5. 1 1 1 1 Pair (4, 2) is marked on the first pass because 2 is an accepting state and 4 is not.

  6. 1 1 1 1 1 Pair (5, 2) is marked on the first pass because 2 is an accepting state and 5 is not.

  7. 1 1 1 1 1 1 Pair (7, 2) is marked on the first pass because 2 is an accepting state and 7 is not.

  8. 1 1 1 1 1 1 1 Pair (6, 3) is marked on the first pass because 6 is an accepting state and 3 is not.

  9. 1 1 1 1 1 1 1 1 Pair (6, 4) is marked on the first pass because 6 is an accepting state and 4 is not.

  10. 1 1 1 1 1 1 1 1 1 Pair (6, 5) is marked on the first pass because 6 is an accepting state and 5 is not.

  11. 1 1 1 1 1 1 1 1 1 1 Pair (7, 6) is marked on the first pass because 6 is an accepting state and 7 is not.

  12. a 1 1 2 1 1 a 1 1 1 1 1 1 Pair (4, 1) is marked on the second pass because δ(4, a) = 4 and δ(1, a) = 2. (4, 2) was marked in pass one.

  13. a 1 1 2 1 2 1 1 1 1 1 1 1 a Pair (4, 3) is marked on the second pass because δ(4, a) = 4 and δ(3, a) = 6. (6, 4) was marked in pass one.

  14. a 1 1 2 1 2 1 2 1 1 1 1 1 1 a Pair (5, 4) is marked on the second pass because δ(5, a) = 6 and δ(4, a) = 4. (6, 4) was marked in pass one.

  15. a 1 1 2 1 2 1 2 1 1 1 1 1 2 1 a Pair (7, 4) is marked on the second pass because δ(7, a) = 6 and δ(4, a) = 4. (6, 4) was marked in pass one.

  16. 1 1 2 1 2 1 2 1 1 1 1 1 2 1 Using the above information we can now determine the states in the minimum-state FA.

  17. 1 1 2 1 2 1 2 1 1 1 1 1 2 1 We will combine states 1, 3, 5, and 7.

  18. 1 1 2 1 2 1 2 1 1 1 1 1 2 1 We will also combine states 6 and 2.

  19. 1 1 2 2 1 1 2 2 1 2 1 1 1 1 1 2 1 State 4 will be a new state.

  20. Below is our minimum-state FA as outlined before: 1 1 2 1 2 1 2 a 1 1 1 1 1 2 1 4 b a a 1,3,5, 7 6,2 b b

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