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Finite Automata

Finite Automata. CPSC 388 Ellen Walker Hiram College. Machine Model of Computation. Strings in the language are accepted Strings not in the language are rejected. Deterministic Finite Automaton. Accepts strings with an odd number of 1’s (e.g. 1011, 110111, 10, 01110). Parts of a DFA.

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Finite Automata

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  1. Finite Automata CPSC 388 Ellen Walker Hiram College

  2. Machine Model of Computation • Strings in the language are accepted • Strings not in the language are rejected

  3. Deterministic Finite Automaton Accepts strings with an odd number of 1’s (e.g. 1011, 110111, 10, 01110)

  4. Parts of a DFA • An alphabet (S) • A finite set of states (S) • A deterministic set of transitions from one state to the next (T : S x S S) • One start state (s0 S) • One or more accepting states (A  S)

  5. Deterministic Transition Table

  6. Testing some strings • 1001 reject • 0101 reject • 10101 accept • 0 reject • 1 accept • 010101011 accept

  7. Results

  8. Data for DFA in C++ Class State{ Public: string name; int index; // unique integer bool isFinal; //true if final state } State start; State next [MAXSTATES][MAXCHARS];

  9. Implementing DFA in C++ //Construct all the states, set isFinal //Create next array from transitions int machine(istream &sin){ //sin.get() reads a single character from sin for(State S=start; c=sin.get(); S=next[S.index][c]); //Action for S would go here... return S.isFinal; }

  10. What Language is This?

  11. Draw DFA for: • {aa, ab, bb} • (ab)* • (a+ab+aab)*b • All strings with 1 a and 2 b’s • All strings containing ab • All strings that don’t contain ab • All strings that don’t contain abc

  12. Deterministic vs. Non-Deterministic • Deterministic: exactly one new state for a given state,character pair • Non-deterministic: 0 or more new states for each state,character pair • Assume machine will always make the right choice! • Add e-transitions (move w/o symbol)

  13. NFA: All strings containing “ab”

  14. Are NFA’s More Powerful? • Yes if ... • There is at least one language that can be accepted by an NFA but no DFA • No if … • For every NFA, we can make a DFA that accepts the same language

  15. From NFA to DFA • Each DFA state corresponds to a set of states in the NFA • DFA has max 2n states if NFA has n states (still finite) • Transition from {s1,s2} on a is union of transitions from s1 on a and s2 on a

  16. Conversion Algorithm • Copy the transitions from the start state. • Choose one of the “new state” sets. Create a new row for that set, and construct its transitions (by unioning transitions from original table) • Repeat until every set in the table has a row

  17. Example: DFA for (a|b)*ab(a|b)* • Initial NFA table • Copy start state, create transition for {s1,s2}

  18. Final NFA

  19. Minimizing DFA • If 2 states have different isFinal, they are different • If 2 states have transitions on the same character to states that are known different, they are different • Otherwise, the states are the same and can be merged (repeat until all different)

  20. Minimization Example s1 != {s1,s3} s1 != {s1,s2,s3} (final) s1 != {s1,s2} (transition on b) {s1,s2} != {s1,s3} {s1,s2} != {s1,s2,s3} (final) We cannot say {s1,s3} != {s1,s2,s3}, so merge!

  21. Why minimize? • Implement machine with smaller table • Compare two machines • If both are minimized, if both compute the same language, then the machines will be identical except for state names

  22. DFA and Regular Expression • The set of languages accepted by DFA’s = the set of languages accepted by regular expressions • For every r.e. we can make a DFA that accepts the same language • For every DFA we can make a r.e. that accepts the same language

  23. For every r.e. there is a DFA... • If we can construct an NFA, that’s good enough • Prove by induction: • Base cases: , , a (element of ) -- build an NFA for each. • Step cases: r | s , rs , r* -- build resulting NFA from NFA’s from r and/or s

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