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Finite Automata CPSC 388 Ellen Walker Hiram College

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

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

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Parts of a DFA An alphabet ( ) A finite set of states (S) A deterministic set of transitions from one state to the next (T : S x S) One start state (s 0 S) One or more accepting states (A S)

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Deterministic Transition Table

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Testing some strings 1001reject 0101reject 10101accept 0reject 1accept 010101011accept

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Results

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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];

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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; }

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What Language is This?

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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

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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 -transitions (move w/o symbol)

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NFA: All strings containing “ab”

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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

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From NFA to DFA Each DFA state corresponds to a set of states in the NFA –DFA has max 2 n 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

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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

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

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Final NFA

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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)

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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!

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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

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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

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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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