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CSC 3130: Automata theory and formal languages

Fall 2009. The Chinese University of Hong Kong. CSC 3130: Automata theory and formal languages. DFA minimization. Andrej Bogdanov http://www.cse.cuhk.edu.hk/~andrejb/csc3130. Example. Construct a DFA over alphabet {0, 1} that accepts those strings that end in 111

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CSC 3130: Automata theory and formal languages

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  1. Fall 2009 The Chinese University of Hong Kong CSC 3130: Automata theory and formal languages DFA minimization Andrej Bogdanov http://www.cse.cuhk.edu.hk/~andrejb/csc3130

  2. Example • Construct a DFA over alphabet {0, 1} that accepts those strings that end in 111 • This is big, isn’t there a smaller DFA for this? 0 0 q000 1 0 q00 1 q001 q0 1 … 0 q01 … qe 1 q101 q10 0 … 1 q1 … 1 q11 1 q111 1

  3. Smaller DFA 0 1 1 1 1 q0 q1 q2 q3 0 0 0 Can we do it with 3 states?

  4. Even smaller DFA? • Suppose we had a 3 state DFA M for L… let’s imagine what happens when: • By the pigeonhole principle, on two of these inputs M ends in the same state M inputs: e, 1, 11, 111

  5. Pigeonhole principle • Here, balls are inputs, bins are states: Suppose you are tossing m balls into n bins, and m > n. Then two balls end up in the same bin. If you have a DFA with n states and you run it on m inputs, and m > n, then two inputs end up in same state.

  6. A smaller DFA • Suppose M ends up in the same state after reading inputs x = 1 and y = 11 • Then after reading one more1 • The state of x1 = 11 should be rejecting • The state of y1 = 111 should be accepting … but they are both the same state! 1, 11 M inputs: 11, 111 e, 1, 11, 111 “ends in 111”

  7. A smaller DFA • Suppose M ends up in the same state after reading inputs x = e and y = 1 • Then after reading 11 • The state of x1 = 11 should be rejecting • The state of y1 = 111 should be accepting … but they are both the same state! e, 1 M inputs: 11, 111 e, 1, 11, 111 “ends in 111”

  8. 0 1 1 1 1 q0 q1 q2 q3 0 0 0 No smaller DFA! • After looking at all possible pairs for x, y, x ≠ ywe conclude that • So, this DFA is minimal (e, 1) (e, 11) (e, 111) (1, 11) (1, 111) (11, 111) There is no DFA with 3 states for L

  9. 0 1 1 1 1 q0 q1 q2 q3 0 0 0 DFA minimization 0 0 q000 1 0 q00 1 q001 q0 1 … 0 q01 … qe 1 q101 q10 0 … 1 q1 … 1 q11 1 q111 1 We will show how to turn any DFA for L into the minimal DFA for L

  10. Minimal DFAs and distinguishable states • First, we have to understandminimal DFAs: reject accept 0 1 1 1 1 q0 q1 q2 q3 0 0 0 every pair of states is distinguishable minimal DFA

  11. Distinguishable states • Two states q and q’ are distinguishable if accept w1 w2 wk-1 wk … q reject w1 w2 wk-1 wk … q’ on the same continuation string w1w2...wk, one accepts, but the other rejects

  12. Examples of distinguishable states 1 0 1 1 0 1 q0 q1 q2 q3 0 0 (q0, q1) distinguishable by 01 (q0, q2) distinguishable by 1 (q0, q3) distinguishable by e DFA is minimal (q1, q2) distinguishable by 1 (q1, q3) distinguishable by e (q2, q3) distinguishable by e

  13. q2 q2 0 0 q0 0 0, 1 0, 1 q01 1 1 q1 1 q3 q3 0, 1 0, 1 Examples of distinguishable states (q0, q3) distinguishable by e (q1, q3) distinguishable by e indistinguishable pairs can be merged (q2, q3) distinguishable by e (q1, q2) distinguishable by 0 (q0, q2) distinguishable by 0 (q0, q1) indistinguishable

  14. Examples of distinguishable states 0 q2 q0 0, 1 0, 1 q23 q01 1 0, 1 0, 1 q3 q1 0, 1 (q0, q2) distinguishable by e (q1, q2) distinguishable by e (q0, q3) distinguishable by e (q1, q3) distinguishable by e (q0, q1) indistinguishable (q2, q3) indistinguishable

  15. Finding (in)distinguishable states If q is accepting and q’ is rejectingMark(q, q’) as distinguishable (x) Rule 1: x q q’ x q1 q1’ If (q1, q1’) are marked,Mark(q2, q2’) as distinguishable (x) a a Rule 2: x q2 q2’ Unmarked pairs are indistinguishable Merge them together Rule 3:

  16. Example of DFA minimization 0 q0 0 q00 q1 q0 1 0 1 q00 q01 0 qe 1 0 q01 q10 0 1 1 q10 q1 0 1 q11 q11 qe q0 q1 q00 q01 q10 1

  17. Example of DFA minimization 0 q0 0 q00 q1 q0 1 0 1 q00 q01 0 qe 1 0 q01 q10 0 1 1 q10 q1 0 1 q11 q11 x x x x x x qe q0 q1 q00 q01 q10 1  q11 is distinguishable from all other states

  18. Example of DFA minimization 0 q0 0 q00 q1 x x q0 1 0 1 q00 x q01 0 qe 1 0 q01 q10 0 1 1 x q10 q1 0 1 q11 q11 x x x x x x qe q0 q1 q00 q01 q10 1  q1 is distinguishable from qe, q0, q00, q10 On transition 1, they go to distinguishable states

  19. Example of DFA minimization 0 q0 0 q00 q1 x x q0 1 0 1 q00 x q01 0 qe 1 0 x x x q01 q10 0 1 1 x x q10 q1 0 1 q11 q11 x x x x x x qe q0 q1 q00 q01 q10 1  q01 is distinguishable from qe, q0, q00, q10 On transition 1, they go to distinguishable states

  20. Example of DFA minimization 0 q0 A 0 q00 q1 x x A q0 1 0 1 q00 A A x q01 0 qe 1 0 x x B x q01 q10 0 1 1 A A x A x q10 q1 0 1 B q11 q11 x x x x x x C qe q0 q1 q00 q01 q10 1  Merge states not marked distinguishableqe, q0, q00, q10 are equivalent → group Aq1, q01 are equivalent → group B q11 cannot be merged → group C

  21. Example of DFA minimization 0 q0 A 0 q00 q1 x x q0 1 A 0 1 q00 A A x q01 0 qe 1 0 x x B x q01 q10 0 1 1 A A x A x q10 q1 0 1 B q11 q11 x x x x x x C qe q0 q1 q00 q01 q10 1 0 1 1 minimized DFA: qA qB qC 1 0 0

  22. Food for thought • Why does method find all distinguishable pairs? w1 w2 wk-1 wk … q x x x x w1 w2 wk-1 wk … q’ Because we work backwards

  23. Food for thought • Why are there no inconsistencies when we merge? B q5 w a A q2 q3 q1 q7 a C w q6 Because we only merge indistinguishable states

  24. Food for thought • Why is there no smaller DFA? Suppose there is By the pigeonhole principle this must happen: M smaller DFA M’ v q q’’ v, v’ v’ q’

  25. Food for thought • Why is there no smaller DFA? But then M smaller DFA M’ v w q q’’ w ? v, v’ v’ q’ w Every pair of states is distinguishable q’’ cannot exist!

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