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DFA Minimization Algorithm for Formal Languages

Learn about the DFA minimization algorithm for constructing a smallest possible DFA to accept a given language in formal languages and automata theory.

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DFA Minimization Algorithm for Formal Languages

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  1. Fall 2008 The Chinese University of Hong Kong CSC 3130: Automata theory and formal languages DFA minimization algorithm 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. 0 1 1 1 1 q0 q1 q2 q3 0 0 0 Smaller DFA • Yes, we can do it with 4 states: • The state remembers the number of consecutive 1s at the end of the string (up to 3) • Can we do it with 3 states?

  4. Even smaller DFA? • Suppose we had a 3 state DFA M for L • We do not know what this M looks like… but 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? • What goes wrong if… • M ends up in same state on input 1 and input 111? • M ends up in same state on input e and input 11? M inputs: e, 1, 11, 111

  7. A smaller DFA • Suppose M ends up in the same state after reading inputs x = 1i and y = 1j, where 0 ≤ i < j ≤ 3 • Then after reading the continuationz = 13 – j • The state of yz should be accepting (yz = 111) • The state of xz should be rejecting (xz = e, 1, or 11) … but they are both in the same state! M inputs: 1i, 1j e, 1, 11, 111

  8. 0 1 1 1 1 q0 q1 q2 q3 0 0 0 No smaller DFA! • Conclusion • So, this DFA is minimal • In fact, it is the unique minimal DFA for L There is no DFA with 3 states for L

  9. DFA minimization • There is an algorithm to start with any DFA and reduce it to the smallest possible DFA • The algorithm attempts to identify classes of equivalent states • These are states that can be merged together without affecting the answer of the computation

  10. Examples of equivalent states 0 q2 0 q2 q0 q0 0, 1 0 0, 1 1 1 q1 1 0, 1 q3 q1 q3 0, 1 0, 1 q0, q1 equivalent q0, q1 equivalent q2, q3 also equivalent

  11. Equivalent and distinguishable states • Two states q, q’ are equivalent if • Here, d(q, w) is the state that the machine is in if it starts at q and reads the string w • q, q’ are distinguishable if they are not equivalent: ^ ^ For every string w, the states d(q, w) and d(q’, w) are either both accepting or both rejecting ^ For some string w, one of the states d(q, w),d(q’, w) is accepting and the other is rejecting

  12. 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 q3 distinguishable from q0, q1, q2 q3 is accepting, others are rejecting (q0, q2) and (q1, q2) distinguishablethey behave differently on input 0 q0, q1 equivalent

  13. DFA minimization algorithm • Find all pairs of distinguishable states as follows:  For any pair of states q, q’: If q is accepting and q’ is rejectingMark(q, q’) as distinguishable Repeat until nothing is marked: For any pair of states (q, q’): For every alphabet symbol a: If (d(q, a), d(q’, a)) are marked as distinguishableMark(q, q’) as distinguishable For any pair of states (q, q’): If (q, q’) is not marked as distinguishableMergeq and q’ into a single state  

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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

  20. Why does DFA minimization work? • We need to convince ourselves of three properties: • Consistency • The new DFA M’ is well-defined • Correctness • The new DFA M’ is equivalent to the original DFA • Minimality • The new DFA M’ is the smallestDFA possible for L M M’   

  21. Main claim • Proof outline: • First, we assume q, q’ are marked distinguishable, and show they must be distinguishable • Then, we assume q, q’ are not marked distinguishable, and show they must be equivalent Any two states q and q’ in M are distinguishable if and only if the algorithm marks them as distinguishable

  22. Proof of part 1 • First, suppose q, q’ are marked as distinguishable • Could it be that they are equivalent? • No, because recall how the algorithm works: w1 w2 wk-1 wk … q x x x x w1 w2 wk-1 wk … q’

  23. Proof idea for part 2 • Now suppose q, q’ are not marked as distinguishable • Could it be that they are distinguishable? Suppose so • Then for some string w, d(q, w) accepts, but d(q’, w) rejects • Working backwards, the algorithm will mark q and q’ as distinguishable at some point ^ ^

  24. Proof of part 2 w1 w1 wk-1 wk … q w1 w1 wk-1 wk … q’  For any pair of states q, q’: If q is accepting and q’ is rejectingMark(q, q’) as distinguishable

  25. Proof of part 2 w1 w2 wk-1 wk … q x x x x w1 w2 wk-1 wk … q’  Repeat until nothing is marked: For any pair of states (q, q’): For every alphabet symbol a: If (d(q, a), d(q’, a)) are marked as distinguishableMark(q, q’) as distinguishable

  26. Why does DFA minimization work? • We need to convince ourselves of three properties: • Consistency • The new DFA M’ is well-defined • Correctness • The new DFA M’ is equivalent to the original DFA • Minimality • The new DFA M’ is the smallestDFA possible for L M M’   

  27. Consistency of transitions • Why are the transitions between the merged states consistent? 0 0 q00 q0 1 A 0 1 q01 0 qe 1 0 q10 0 1 1 q1 0 1 B q11 C 1

  28. Proof of consistency B • Suppose there were two inconsistent transitions in M’ labeled a out of merged stateqA q5 a A q2 q3 q1 q7 a C q6

  29. Proof of consistency B • States q5 and q6 must be distinguishable because they were not merged together q5 a A q2 q3 q1 q7 a q6

  30. Proof of consistency B • Then, q3 and q7 must also be distinguishable, and this is impossible w q5 a A q2 q3 q1 q7 a C w q6

  31. Consistency of states • Why are the merged states either all accepting or all rejecting? 0 0 q00 q0 1 A 0 1 q01 0 qe 1 0 q10 0 1 1 q1 0 1 B q11 C 1

  32. Consistency of states • Because merged states are not distinguishable, so they are mutually equivalent 0 0 q00 q0 1 A 0 1 q01 0 qe 1 0 q10 0 1 1 q1 0 1 B q11 C 1

  33. Correctness • Each state of M’ corresponds to a class of mutually equivalent states in M 0 0 q00 q0 1 A 0 1 q01 0 qe 1 0 q10 0 1 1 q1 0 1 B q11 C 1

  34. Proof of correctness • Claim: After reading input w, • Proof: True in start state, and stays true after, because transitions are consistent • At the end, M accepts w in state qi if and only if M’ on input w lands in the state qA that represents qi • qA must be accepting by consistency of states M’ is in state qAif and only ifM is in one of the states represented by A

  35. Proof of minimality • Now suppose there is some smallerM’’ for L • By the pigeonhole principle, there is a state q’’ of M’’ such that All pairs of states in M’ are distinguishable M’ M’’ v q q’’ v, v’ v’ q’

  36. Proof of minimality • Since q and q’ are distinguishable, for some w… • But in M’’, d(q’’, w) cannot both accept and reject! • So, M’’cannot be smaller than M’ All pairs of states in M’ are distinguishable M’ M’’ v w q q’’ w ? v, v’ v’ q’ w

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